MP Board Class 12th Maths Solutions Chapter 7 Integrals Ex 7.8

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 7 Integrals Ex 7.8 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 7 Integrals Ex 7.8

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MP Board Class 12th Maths Solutions Chapter 7 Integrals Ex 7.10

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 7 Integrals Ex 7.10 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 7 Integrals Ex 7.10

MP Board Class 12th Maths Solutions Chapter 7 Integrals Ex 7.10 1
MP Board Class 12th Maths Solutions Chapter 7 Integrals Ex 7.10 2
MP Board Class 12th Maths Solutions Chapter 7 Integrals Ex 7.10 3
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MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry

MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry

Electrochemistry NCERT Intext Exercises

Electrochemistry Class 12 NCERT Solutions help to score more marks.

Question 1.
How would you determine the standard electrode potential of the system Mg2+ | Mg ?
Answer:
The standard electrode potential of the system Mg2+/ Mg can be determined as explained below. Prepare Mg2+/ Mg system by dipping Mg rod in a solution of Mg2+ ions and connect it to standard hydrogen electrode.
Mg | Mg2+ || H+(aq) | H2(g); Pt
When magnesium is connected with SHE, oxidation takes place at the Mg electrode. Hence, the potential of the magnesium electrode is taken as – ve The emf of the cell, determined potentiometrically, is equal to the potential of the magnesium electrode because the potential of SHE is taken as zero.

Question 2.
Can you store copper sulphate solutions in a zinc pot?
Answer:
No, it is not possible. The E° values of the copper and zinc electrodes are as follows :
Zn2+(aq) + 2e– → Zn(s) ; E° = – 0·76 V
Cu2+(aq) + 2e– → Cu(s) ; E° = + 0·34 V
This shows that zinc is a stronger reducing agent than copper. It will lose electrons to Cu2+ ions and a redox reaction will immediately set in.
Zn(s) + Cu2+ (aq) → Zn2+(aq) + Cu(s)
Thus, copper sulphate solution cannot be stored in zinc pot.

Question 3.
Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
Answer:
Substances having higher E° values than Fe2+ ions can oxidise ferrous ions. Thus, Ag+, Hg2+, Br2, Cl2, F2 etc. can oxidise Fe2+ ions to Fe3+ ions.

Question 4.
Calculate the potential of the hydrogen electrode in contact with a solution whose pH is 10.
Solution:
For hydrogen electrode, H+ + e– → 1/2H2
Applying Nernst equation,
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 2

Question 5.
Calculate the emf of the cell in which the following reaction takes place :
Ni(s) + 2Ag+ (0.002 M) → Ni2+ (0.160 M) + 2Ag(s) Given that Ecell = 1.05 V.
Solution:
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 3

Question 6.
The cell in which the following reactions occurs :
2Fe3+(aq) + 2I(aq) → 2Fe2+(aq) + I2(s) has Ecell = 0-236 V at 298 K.
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Solution:
The two half reactions are :
2Fe3+ + 2e → 2Fe2+ and 2I → I2 + 2e
For the above reaction, n = 2
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 4

How to calculate emf of a cell PDF available in my website.

Question 7.
Why does the conductivity of a solution decrease with dilution?
Answer:
The number of ions per unit volume that carry the current in a solution decreases on dilution. Therefore the conductivity of a solution decreases with dilution.

Question 8.
Suggest a way to determine the \(\wedge_{m}^{0}\) value of water.
Answer:
Conductance of weak electrolytes can be determined by kohlraush’s law. Thus, molar conductance of water of infinite dilution can be determined by the molar conductances of NaOH, HCl and NaCl at infinite dilution.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 5

Question 9.
The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm2 mol-1 Calculate its degree of dissociation and dissociation constant Given \(\lambda_{\left(\mathbf{H}^{+}\right)}^{\circ}\) = 349.6 S cm2 mol-1 and \(\lambda_{\left(\mathrm{H} \mathrm{COO}^{-}\right)}^{\circ}\) = 54.6 S cm2 mol.
Solution:
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 6

Question 10.
If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire ?
Solution:
Q (coulomb) = 1 (ampere) × t (sec)
Q = 0.5 ampere × 2 × 60 × 60
= 3600 C
A flow of IF, i.e. 96500 C is equivalent to flow of 1 mole of electrons
i. e., = 6.023 × 1023 electrons
3600 C is equivalent to flow of electrons
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 7

Question 11.
Suggest a list of metals that are extracted electrolytically.
Answer:
Li, Na, Mg, Ba, Ca, Al

Question 12.
What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O-27 ? Consider the reaction :
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 8
Answer:
1 mole Cr2O-27 requires 6 moles electrons for reduction.
∴ Required charge = 6F
= 6 × 96500 coulomb.
= 579000 coulomb.

Question 13.
Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Answer:
Lead storage battery : It is a secondary cell i.e., a cell which is rechargeable because the products of cell reaction sticks to the electrode. It is also called as lead storage cells. It consists of six cells connected in series. Each cell consist of spongy lead anode and a grid of lead packed with lead dioxide (PbO2) acts as cathode. An aqueous solution of H2SO4 (38% by mass) acts as electrolyte. The reactions which takes place at electrodes can be represented as:
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 9
Concentration of H2SO4 decreases as sulphate ions are consumed to form PbSO4 during the working of the cell. As a result of this the density of solution also decreases.

Recharging the cell / battery: Lead storage battery can be recharged by connecting it to an external source of direct current. This reverses the flow of electron with the deposition of Pb on the anode and PbO2 on the cathode. That is, during recharging operation the cell behaves as electrolytic cell. Following reaction occurs during recharging.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 10

Question 14.
Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Answer:
Methanol (CH3OH), propane (C3H8)

Question 15.
Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Answer:
Formation of carbonic acid takes place on the surface of iron
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 11
In presence of H+ ion, oxidation of iron takes place
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 12
The electrons are used at other spot where reduction takes place.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 13
Overall reaction,
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 14

MP Board Solutions

Electrochemistry NCERT Textbook Exercises

Question 1.
Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.
Answer:
A metal with lesser standard potential (more reactive) can displace the other metal from solution of its salts.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 15

Question 2.
Given the standard electrode potentials, K+/K = -2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V, Mg2+/Mg = -2.37 V, Cr3+/Cr = -0.74V. Arrange these metals in their increasing order of reducing power.
Answer:
The lower the reduction potential, the higher is the reducing power. Hence, the reducing power of the given metals increases in the following order.
Ag < Hg < Cr < Mg < K.

Question 3.
Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+(aq)+ 2Ag(s) takes place. Further show :
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 16
(i) Anode i.e., Zn electrode is negatively charged.
(ii) Electrons are current carriers.

Question 4.
Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 17
Calculate the ∆rG°, and equilibrium constant of the reactions.
Solution:
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 18

Question 5.
Write the Nernst equation and emf of the following cells at 298 K :
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 19
Solution:
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 20
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 21
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 22

Question 6.
In the button cells widely used in watches and other devices the following reaction takes place:
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 23
Solution:
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 24

Question 7.
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Answer:
Conductivity: The conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section.

Molar conductivity: Molar conductivity of a solution at a dilution (V) is the conductance of all the ions produced from one mole of the electrolyte dissolved in V cm3 of the solution when the electrodes are one cm apart and the area of cross-section of the electrodes is so large that the whole of the solution is contained between them. It is usually represented by Λm.

Variation with concentration: The conductivity of a solution (Both for strong and weak electrolytes) decreases with a decrease in the concentration of the electrolyte i.e., on dilution. This is due to the decrease in the number of ions per unit volume of the solution on dilution. The molar conductivity of a solution increase in the decrease in the concentration of the electrolyte i.e., on dilution. This is due to the decrease in the number of ions per unit volume of the solution on dilution. The molar conductivity of a solution increases with decrease in the concentration of the electrolyte. This is because both the number of ions, as well as mobility of ions, increases with dilution. When concentration approaches zero, the molar conductivity is known as limiting molar conductivity.

Question 8.
The conductivity of 0.20 M solution of KCI at 298 K is 0.0248 Scm-1. Calculate its molar conductivity.
Solution:
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 25

Question 9.
The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500Ω What is the cell constant if the conductivity of 0.001M KCl solution at 298 K is 0.146 × 10-3 S cm-1.
Solution:
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 26

Question 10.
The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below :
Concentration/M 0.001 0.010 0.020 0.050 0.100 102 × kSm-1 1.237 11.85 23.15 55.53 106.74 for all concentrations and draw a plot between Λm and C1/2 Find the value calculate Λm and C1/2 Find the value calculate Λm of Λ0m.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 27
Λ0m = intercept on Λm axis = 124-0 S cm2 mol-1 (on extrapolation to zero concentration)
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 28

MP Board Solutions

Question 11.
The conductivity of 0.00241 M acetic acid is 7.896 × 10-5 S cm-1. Calculate its molar conductivity and if Λ0m for acetic acid is 390.5 S cm2 mol-1, what is its dissociation constant?
Solution:
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 29

Question 12.
How much charge is required for the following reductions :
(i) 1 mol of Al3+ to Al.
(ii) 1 mol of Cu2+ to Cu.
(iii) 1 mol of MnO4 to Mn2+.
Solution:
(i) The electrode reaction is :
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 30
(ii) The electrode reaction is :
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 31
(iii) The electrode reaction is :
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 32

Question 13.
How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten CaCl2.
(ii) 40.0 g of Al from molten Al2O3.
Solution:
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 33

Question 14.
How much electricity is required in coulomb for the oxidation of
1. 1 mole of H2O to O2
2. 1 mole of FeO to Fe2O3.
Answer:
1. 2H2O → 4H+ + O2 + 4e
2F of electricity is required for the oxidation of 1 mol of H2O

2. Fe2+ → Fe3+ + e
1F of electricity is required for the oxidation of 1 mole of FeO

Question 15.
A solution of Ni(NO3)2 is electrolyzed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
Answer:
According to Faraday’s first law:
W = ZIr [z = \(\frac { M }{ nF } \)]
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 34

Question 16.
Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3, and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
Solution:
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 35

Question 17.
Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible :
(i) Fe3+(aq) and I(aq)
(ii) Ag+(aq) and Cu(s)
(iii) Fe3+(aq) and Br(aq)
(iv) Ag(s) and Fe3+(aq)
(v) Br2(aq) and Fe2+(aq)
Answer:
A reaction is feasible if EMF of the cell is +ve.
Cathode : At which reduction occurs.
Anode : At which oxidation occurs.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 36
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 37

MP Board Solutions

Question 18.
Predict the products of electrolysis in each of the following:
(a) An aqueous solution of AgNO3 with silver electrodes.
(b) An aqueous solution of AgNO3 with platinum electrodes.
(c) A dilute solution of H2SO4 with platinum electrodes.
(d) An aqueous solution of CuCl2 with platinum electrodes.
Answer:
(i) At cathode:
The following reduction reactions compete to take place at the cathode
Ag+(aq) + e → Ag(s); Eθ = 0.80 V
H+ (aq) + e → \(\frac { 1 }{ 2 }\) H2 (g); Eθ = 0.00V
The reaction with a higher value of Eθ takes place of the cathode. Therefore, deposition of silver will take place at the cathode.

At anode:
The Ag anode is attacked by NO3 ions. Therefore, the silver electrode at the anode dissolves in the solution to from Ag+.

(ii) At cathode: Same as above
At anode: Anode is not attackable and hence OH ions have lower discharge potential than NO3 ions and OH ions react to give O2
OH → OH + e
4OH → 2H2O + O2 (g)
(iii) H2SO4 → 2H+ + SO2-4
HO2 ⇌ H+ + OH

At cathode:
2H++ 2e → H2
At anode: 4OH → 2H2O + O2 + 4e
i. e., H2 will be liberated at cathode and O2 at the anode.

(iv) CuCl2 → Cu2++2Cl
2nd PUC Chemistry Question Bank Chapter 3 Electrochemistry - 18
At Cathode: Cu2+ ions will be reduced in preference to H+ ions
Cu2+ + 2e → Cu
At anode: Cl’ ions will be oxidised in preference to OH ions.
2Cl → Cl2 + 2e
i.e., Cu will be deposited on the cathode and Cl2 will be liberated at the anode.

MP Board Solutions

Electrochemistry Other Important Questions and Answers

Electrochemistry Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
If specific conductance = K, R = Resistance, l = distance between the electrodes A = Cross-sectional area of a conductor Cm = mol L-1, Ceq = g.eq L-1 then specific resistance of electrolyte will be equal to :
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 38

Question 2.
K is equal to :
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 39

Question 3.
Molar conductivity Λm is equal to :
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 40

Question 4.
Cell constant is :
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 41

Question 5.
For electrolyte (v+ = v = 1) like NaCI:
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 42

Question 6.
Which among the following is not a conductor of electricity :
(a) NaCl(aq)
(b) NaCl(s)
(c) NaCl(mol)
(d) Ag.

Question 7.
By which of the following the resistance of the solution is multiplied to obtain cell constant:
(a) Specific conductance (K)
(b) Molar conductance (Λm)
(c) Equivalent conductance (Λeq)
(d) None of these.

Question 8.
Increase in equivalent conductance of the solution of an electrolyte by dilution is due to:
(a) Increase in ionic attraction
(b) Increase in molecular attraction
(c) Increase in association of electrolyte
(d) Increase in ionization of electrolyte.

Question 9.
If the specific conductance and observed conductance of an electrolyte is same then its cell constant will be :
(a) 1
(b) 0
(c) 10
(d) 1000

Question 10.
Unit of cell constant is :
(a) ohm-1 cm-1
(b) cm
(c) ohm cm
(d) cm-1.

Question 11.
Unit of specific conductance is :
(a) ohm-1
(b) ohm-1 cm-1
(c) ohm-2 cm-1 equivalent-1
(d) ohm-1 cm-2.

Question 12.
If the concentration of any solution is C gram equivalent/ litre and specific resistance is A, its equivalent conductance will be :
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 43

Question 13.
The coating of layer of zinc on iron to prevent it from corrosion is called :
(a) Galvanization
(b) Cathodic protection
(c) Electrolysis
(d) Photoelectrolysis.

Question 14.
A saturated solution of KNO3 is used for salt bridge because :
(a) Speed of K+ is more than NO3
(b) Speed of NO3 is more than K+
(c) Speed of both is nearly same
(d) Solubility of KNO3 is high in water.

Question 15.
Acts as dipolar in dry cells :
(a) NH4Cl
(b) Na2CO3
(c) pbSO4
(d) MnO2

Question 16.
Reduction is known as :
(a) Electronation
(b) De-electronation
(c) Protonation
(d) De-protonation.

Question 17.
What is the path of electric current in a Daniel cell when Zn and Cu electrodes are connected:
(a) From Cu to Zn inside the cell
(b) From Cu to Zn outside the cell
(c) From Zn to Cu inside the cell
(d) From Zn to Cu outside the cell.

Question 18.
In a cell containing Zn electrode and normal hydrogen electrode (NHE), Zn acts like:
(a) Anode
(b) Cathode
(c) Neither cathode nor anode
(d) Both anode and cathode.

Question 19.
If salt bridge is removed from half cells then voltage :
(a) Reduces and becomes zero
(b) Increases
(c) Immediately increases
(d) Does not change.

Question 20.
When lead storage battery is discharged :
(a) SO2 is released
(b) Pb is manufactured
(c) PbSO4 is used
(d) H2SO4 is used.

Question 21.
Process of Rusting of iron is :
(a) Oxidation
(b) Reduction
(c) Corrosion
(d) Polymerisation.

Question 22.
Value of standard potential of hydrogen electrode is :
(a) Positive
(b) Negative
(c) Zero
(d) No definite value.

Answers:
1. (b), 2. (c), 3. (d), 4. (b), 5. (a), 6 (b), 7. (a), 8. (d), 9. (a), 10. (d), 11. (b), 12. (a), 13. (a), 14. (c), 15. (d), 16. (a), 17. (d), 18. (a), 19. (a), 20. (d), 21. (c), 22. (c).

Question 2.
Fill in the blanks :

  1. Acetic acid is a ………………… electrolyte.
  2. Conductance of electrolyte ………………… with the increase in temperature.
  3. On increasing dilution the value of specific conductance of a solution …………………
  4. ………………… decreases with increase in size of ion.
  5. Unit of specific resistance is …………………
  6. Primary cells cannot be ………………… again.
  7. Device which converts chemical energy into electrical energy is known as ………………… cell.
  8. Amount of electric current which produces one gram equivalent of a substance is known as …………………
  9. In metallic conduction ………………… property remains unchanged.
  10. Reciprocal of resistance is known as …………………
  11. Conductance of 1 cm cube of a conductor is called …………………
  12. 1 Faraday is equal to ………………… coulomb.
  13. Rusting of iron is an example of …………………
  14. Potential of standard hydrogen is assumed to be …………………

Answers:

  1. Weak
  2. Increases
  3. Decreases
  4. Conductance
  5. ohm cm
  6. Charged
  7. Electrochemical
  8. Faraday
  9. Chemical property
  10. Conductance
  11. Specific conductance
  12. 96500 coulomb
  13. Corrosion process
  14. 0.0 volt
  15. Electrochemical cell.

Question 3.
Match the following :

I.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 44
Answers:

  1. (e)
  2. (c)
  3. (d)
  4. (b)
  5. (a).

II.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 45
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 46
Answer:

  1. (c)
  2. (d)
  3. (b)
  4. (a)
  5. (e)
  6. (f).

Question 4.
Answer in one word / sentence :

1. Give two examples of strong electrolyte.
2. Give two examples of weak electrolytes.
3. What is the effect of temperature on electrolytic conductivity ?
4. Write the formula of Kohlrausch’s law.
5. Write the formula of equivalent conductance.
6. Write the formula of molar conductance.
7. Device which converts electrical energy into chemical energy.
8. What is the potential of both the electrodes of the cell due to which electric current flows in the cell ?
9. What is the potential produced due to redox reaction between a metal electrode and its ions called ?
10. What is the unit of potential difference ?
11. Write the formula of cell constant.
12. Cell which can be recharged are known as.
13. State the unit of Equivalent conductance.
14. What is the chemical composition of rust ?
15. Write the relation between Electromotive force and Equilibrium constant of a cell.
16. What is the name of the reaction in which oxidation and reduction occur simultaneously ?
Answers:
1. Strong electrolyte : HCl, NaOH, NaCl
2. Weak electrolyte : CH3COOH, H2CO3,
3. Conductivity increases
4.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 47
5. Equivalent conductance
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 48
6.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 49
7. Electrolytic cell
8. Electromotive force
9. Electrode potential
10. Volt
11. Cell constant = \(\frac { l }{ A } \)
12. Secondary cell
13. Ohm cm2 gm eq-1
14. Fe2O3 . xH2O
15.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 50
16. Redox reaction.

MP Board Solutions

Electrochemistry Very Short Answer Type Questions

Question 1.
Write the definition of Electrochemical cell.
Answer:
System in which chemical energy is converted to electrical energy by oxidation reduction is known as electrochemical cell or voltaic cell.

Question 2.
What is an Electrolytic cell ?
Ans.
The container or system in which electrical energy is passed by which chemical reaction takes place thus, electrical energy is converted to chemical energy is known as electrolytic cell.

Question 3.
What is Electrode potential ?
Answer:
The potential difference developed between the electrodes and electrolyte of an Electrolytic cell is known as Electrode potential.

Question 4.
What is a strong electrolyte ? Write two examples.
Answer:
Electrolyte which completely dissociate in aqueous solution are known as strong electrolyte.
Example : NaCl, KCl, NH4Cl etc.

Question 5.
What is a weak Electrolyte ? Write two examples.
Answer:
Electrolyte which dissociate partially in aqueous solution are known as weak electrolyte.
Example : NH4OH, CH3COOH, HCN etc.

Question 6.
What is meant by standard electrode potential ?
Answer:
Standard electrode potential (E°) of a half cell is the potential difference when one electrode is dipped in molar solution of its ion at 298 K. If electrode is gaseous the pressure of gas must be one atmosphere.

In IUPAC system, reduction potential are known as standard electrode potential.

Question 7.
Write Ohm’s law.
Answer:
According to Ohm’s law “It states that potential difference across the conductor is directly proportional to the current (I) flowing through it” i.e.,
Mathematically, it can be written as :
I α V
V = IR (R = Resistance, unit = ohm, Ω)

Question 8.
What is cell constant ?
Answer:
For a conductivity cell, the ratio of distance between two electrodes (l) and area of cross-section of electrode (A) is called as cell constant.
∴ Cell constant = \(\frac { l }{ A } \) or x = \(\frac { l }{ A } \)
Unit of cell constant = cm-1.

Question 9.
What is galvanization ? Explain.
Answer:
Iron is coated with the layer of zinc to protect it from rusting. This process is known as galvanization. The galvanized iron articles keep their lustre due to the coating of invisible protective layer of basic zinc carbonate, ZnCO3.Zn(OH)2.

Question 10.
What is Electrochemical Equivalent ?
Answer:
Electrochemical equivalent of a substance is that mass of substance released or deposited on an electrode when a current of one ampere is passed for one second.

MP Board Solutions

Electrochemistry Short Answer Type Questions

Question 1.
What is salt bridge ? Write its two functions.
Answer:
‘U’ shaped tube filled with KCl or KNO3 in Agar-Agar solution or gelatin, is known as salt bridge. It connects the two half cell.
Functions : (i) It allows the flow of current by completing the circuit.
(ii) It maintains the electrical neutrality.

Question 2.
Derive relation between standard electromotive force and equilibrium constant
Answer:
Relation between standard electromotive force and equilibrium constant can be derived using van’t Hoff isochore. For any given reaction equilibrium constant IQ is equal to the ratio of rate constant of forward reaction and rate constant of backward reaction.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 51
Value of equilibrium constant Kc can be calculated using standard free energy change (∆G°) because
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 52

Question 3.
What do you understand by oxidation-reduction reactions ?
Answer:
Oxidation-Reduction reactions: Chemical reactions in which valency of elements changes are known as oxidation-reduction reactions. In this process both oxidation and reduction reactions occur simultaneously, in which one of the substance is oxidized and the other substance is reduced. Like
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 53
In this reaction FeCl3 is reduced to FeCl2 and SnCl2 is oxidized to SnCl4. In the reaction valency of Fe decreases and valency of Sn increases.

Question 4.
Write difference between Metallic conduction and Electrolytic conduction.
Answer:
Differences between Metallic conduction and Electrolytic conduction :

Metallic conduction:

  1. Metallic conduction takes place by movement of electrons.
  2. There is no chemical change.
  3. There is no transfer of matter.
  4. In metallic conduction conductivity decreases with increase in temperature.

Electrolytic conduction

  1. Electrolytic conduction takes place by movement of ions.
  2. Due to chemical change decomposition of electrolyte takes place.
  3. Transfer of matter takes place as ions.
  4. In electrolytic conduction conductivity increases with increase in temperature.

Question 5.
What are the difference between emf (Cell potential) and potential difference
Answer:
Difference between EMF and Potential difference :

EMF / Cell potential:

  1. It is the potential difference between the two terminals of the cell when no current is flowing in the circuit, i.e., in an open circuit.
  2. It is the maximum voltage which can be obtained from a cell.
  3. It can be measured by potentiaometrie method.
  4. Work performed by electromotive force is the maximum work done by a cell.
  5. It is responsible for continuous flow of current in electric circuit.

Potential difference:

  1. It is the difference of the electrodes potentials of the two electrodes when the cell is sending current through the circuit.
  2. It is the less than the maximum voltage as it is the difference of electrode potential.
  3. It can be measured by simple voltmeter also.
  4. Work performed by potential difference is less than the maximum work done by a cell.
  5. It is not responsible for the continuous flow of current in circuit.

Question 6.
What is specific conductance ? Give its unit.
Answer:
Specific conductivity: The reciprocal of resistivity is called specific conductiv¬ity. It is defined as the conductance between the opposite faces of one centimeter cube of a conductor. It is denoted by K (kappa).
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 54
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 55
The specific conductivity of a solution at a given dilution is the conductance of one cm cube of the solution. It is represented by K (kappa).
Note : The specific conductivity of a solu¬tion of electrolyte depends upon the dilution or molar concentration of the solution.

Question 7.
What is resistivity of any solution ?
Answer:
Resistivity : When current flow in the solution through two electrodes the resistance is proportional to length and inversely proportional to cross-sectional area A.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 56
The constant p (rho) is called resistivity or specific resistance.
Unit: If l is expressed in cm, A in cm2 and R in ohm, the unit of resistivity will be
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 57
or Resistivity of any solution is the resistance of 1 cm cube.

Question 8.
Differentiate between Electrochemical cell (Galvanic cell) and Electrolytic cell.
Answer:
Differences between Electrochemical and Electrolytic cells :

Electrochemical cell or Galvanic cell:

  1. It is a device to convert chemical energy into electrical energy.
  2. It consists of two electrodes in different compartments joined by a salt bridge.
  3. Redox reactions occurring in the cell are spontaneous.
  4. Free energy decreases with operation of cell, i.e., ∆G < 0.
  5. Useful work is obtained from the cell.
  6. Anode works as negative and cathode as positive electrodes.
  7. Electrons released by oxidation process at anode go into external circuit and pass to cathode.
  8. To set-up this cell, a salt bridge/porous pot is used.

Electrolytic cell:

  1. It is a device to convert electrical energy into chemical energy.
  2. Both the electrodes are in same solution.
  3. Redox reactions occurring in the cell are non-spontaneous.
  4. Free energy increases with operation of cell, i.e., ∆G > 0.
  5. Work is done on the system.
  6. Anode is positive and cathode is negative.
  7. Electrons enter into cathode electrode from external source and leave the cell at anode.
  8. No salt bridge is used in this cell.

Question 9.
What is equivalent conductance ?
Answer:
Equivalent conductance:
“Conductance of total ion produced by one gram equivalent of electrolyte in the solution is called equivalent conductance.” It is denoted by Λeq.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 58

Question 10.
What is molar conductance ?
Answer:
Molar Conductivity : The molar conductivity of a solution at definite concentration (or dilution) and temperature is the conductivity of that volume which contains one mole of the solute and is placed between two parallel electrodes 1 cm apart and having sufficient area to hold whole of the solution. It is denoted by Λm.
Mathematically,
Λm = K × V …(1)
Where V is the volume in ml in which one gram mole of substance is dissolved.
If M is molarity or m moles are dissolved in 1000 ml.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 59
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 60

Question 11.
Define cell constant Develop a relation between specific conductance and cell constant.
Answer:
Cell constant: In any conductive cell, the distance between two electrodes and surface area of electrode A are constant. The ratio of l and a is called cell constant i.e.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 61
Unit of cell constant is cm1 and it is generally expressed by x.
Relation between specific conductance and cell constant : For a conductor, the resistance R is directly proportional to length R and inversely proportional to area of cross – section of electrolyte.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 62

Question 12.
What are the factors which influence the electrical conductance of electrolytes ?
Answer:
Factor which influence electrical conductivity of electrolytes :

The main factor which influence the electrical conductivity are following:

1. Temperature: It influence following interactions.
(a) Interionic attractions: It depends upon the solute-solute interactions. Which is found between the ions of solute.
(b) Solvation of ions: It depends upon solute-solvent interactions. It is relation between ions of solute and solvent molecules.
(c) Viscosity of solvent: It depends upon solvent-solvent interactions. Solvent molecules are related with each other.
With increase in temperature all these three effects decrease and average kinetic energy of ions increases. Thus, with increase of temperature, resistance of solution decreases and hence conductance increases.

2. Nature of electrolyte : The conductance of solution depends upon the nature of electrolyte. On the basis of conductance measurement electrolytes are classified as strong electrolyte and weak electrolyte. Strong electrolytes have high value of conductance even at higher concentration also.

3. Dilution or concentration : It is main factor which influence electrical conductance. Effect of dilution or concentration can be studied indivisually in equivalent conductance, specific conductance and molar conductance. But for a general concept of electrical conductance of solution as the concentration is lowered or dilution increases, electrical conductance of whole solution increases.

Question 13.
On what factors does the various conductivities of an electrolytic solution depend ?
Answer:
Conductivities of electrolytic solution depend on the following factors :

  1. Dilution : On increasing dilution, value of specific conductance of a solution decreases, value of equivalent conductance and molar conductance increases.
  2. Nature of solvent: A solvent with high dielectric constant has high conductivity and with low dielectric constant has low conductivity.
  3. Number of ions present in solution: Conductivity of strong electrolytes is higher than the conductivity of weak electrolytes.
  4. Size of ion : In aqueous solution, small ions are heavily hydrated due to which their conductivity decreases.
  5. Effect of Temperature: With the increase in temperature conductivity increases.

Question 14.
With the increase in dilution how do specific conductance, Equivalent conductance and molar conductance change ?
Answer:
With the increase in dilution, specific conductance decreases. This is because by the increase in dilution number of ions present in 1 cm cube of solution decreases.
But, Equivalent conductance Λeq = K × V
and Molar conductance Λm = K × V
Equivalent conductance and molar conductance are the product of specific conductance and dilution. By the increase in dilution (or decrease in concentration) magnitude of K decreases but that of V increases.
Increase in magnitude of V is comparatively much more than the decrease in magnitude of K. Thus, by the combined effect of both, by the increase in dilution Λeq and Λm increases.

MP Board Solutions

Question 15.
What is an Electrolytic cell and how does it work ?
Answer:
Electrolytic cells : In these cells electric current is supplied through an external source, as a result of which chemical reactions take place which is called electrolysis like : Electrolysis of water, NaCl, Al2O3 etc. For example in Solvay trough cell electrode is immersed in sodium chloride solution and electric current is passed due to which NaCl electrolyses.

At mercury cathode sodium is released and at anode chlorine is released. Sodium forms amalgam with mercury and is taken out of the cell.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 63

Question 16.
What is meant by electromotive force of an electrochemical cell ?
Answer:
The difference in electrode potentials of the two electrodes of an electro- chemical cell is known as electromotive force or cell potential. It is expressed in volt.

Due to difference in potential electric current flows from an electrode of lower potential to an electrode of higher potential. EMF of the cell can be expressed in terms of reduction potential as :

Cell potential = Standard electrode potential – Standard electrode potential
of R.H.S. electrode of L.H.S electrode
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 64
EMF of a cell is measured by connecting the voltmeter between the two electrodes of a cell. EMF of a cell depend on the concentration of solutions of both half cells and nature of the two electrodes. For example, In Daniel cell, concentration of CuSO4 and ZnSO4 solutions in the two half cells is 1M and at 298 K EMF of the cell is 1.10 volt.

MP Board Solutions

Electrochemistry Long Answer Type Questions

Question 1.
What is standard hydrogen electrode ? How is it prepared ?
Answer:
Standard hydrogen electrode : This consists of gas at 1 atmospheric pressure bubbling over a platinum electrode immersed in 1 M HC1 at 25°C (298 K) as shown in figure. The platinum electrode is coated with platinum black to increase its surface. The hydrogen electrode thus con¬structed forms a half cell which on coupling with any other half cell begins to work on the principle of oxidation or reduction. Electrode depending upon the circumstances works both as anode or cathode.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 65

MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 66
Standard hydrogen electrode (SHE) is arbitrarily assigned a potential of zero.

Question 2.
Derive Nernst Equation for single electrode potential.
Answer:
Value of standard electrode potential given in electrochemical series is applicable only when the concentration of electrolyte is 1M and temperature is 298 K. But in electrochemical cells the concentration of electrolyte is not definite and electrode potential depends on concentration and temperature. In such condition single electrode potential can be expressed by Nernst equation.
For a reduction half reaction, Nernst equation can be expressed as follows :
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 67
Where E = Reduction electrode potential
E° = Standard electrode potential (Mn+ concentration 1M and at 298 K)
R = Gas constant = 8.31 JK-1 mol-1 = Temperature (in kelvin) = 298 K
n = Valency of metal ion, F = 1 Faraday (96,500 coulomb)
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 68
Equation (2) is Nernst equation for single electrode potential.

Question 3.
Write the Faraday’s laws of electrolysis.
Answer:
Faraday’s first law of electrolysis : The law states that, “The mass of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed.”
Thus, if W gm of the substance is deposited on passing Q coulomb of electricity, then
W α Q or W = ZQ
Where, Z is a constant of proportionality and is called electrochemical equivalent of the substance deposited. If a current of I ampere is passed for t second, then Q = I × t. So that,
W = Z × Q = Z × I × t
Thus, if Q = 1 coulomb, I = 1 ampere and t = 1 second, then W = Z. Hence, electrochemical equivalent of a substance may be defined as, “The mass of the substance deposited when a current of one ampere is passed for one second.”
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 69
As one faraday (96500 C) deposits one gram equivalent of the substance, hence electrochemical equivalent can be calculated from the equivalent mass.
Faraday’s second law of electrolysis : It states that, “When the same quantity of electricity is passed through solutions of different electrolytes connected in series, the weight of the substances produced at the electrodes are directly proportional to their equivalent mass.”

For example, for CuSO4 solution and AgNO3 solution connected in series, if the same quantity of electricity is passed, then
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 70

Question 4.
What is rusting of iron ? Describe Electrochemical theory of rusting.
Answer:
Corrosion: Process by which the layers of undesirable compounds are formed on the surface of a metal on its exposure to atmospheric condition are called corrosion. Rusting of iron is an example of corrosion, chemically it is Fe2O3xH2O.

Electrochemical theory of rusting :
Anode reaction : On one spot of iron sheet, oxidation takes place and this spot behaves as an anode.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 71
The electrons which are released at this spot travel through the metal and reach another spot on the metal which acts as cathode. These electrons cause the reduction of oxygen in the presence of hydrogen ions (H+). H+ ions are formed due to decomposition of carbonic acid formed by dissolution of CO2 in H2O.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 72
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 73

Question 5.
What is corrosion ? Write three factors affecting it and any three methods to prevent it.
Answer:
Corrosion : Process by which the layers of undesirable compounds are formed on the surface of a metal on its exposure to atmospheric conditions are called corrosion.

Factors affecting corrosion :

  1. Nature of metal: Reactive metal corrodes readily.
  2. Impurities in metal: Impure metal corrodes quickly to a greater extent.
  3. Environment: If environment around metal contains oxygens, carbon dioxide, moisture, salts and acidic gases like CO2, SO2, SO3 etc. then corrosion occurs quickly.

Methods to prevent corrosion :

1. Barrier protection : In this method, the surface of the metal is coated with paint, oil or grease. Due to this the surface of the metal remain unexposed to atmospheric conditions and hence corrosion is prevented. Surface of the metal can also be protected by :
(i) Coating metal surface with non-corroding metals like nickel and chromium is called electroplating.
(ii) Dipping iron article in molten metal like zinc. This process is called galvanization.

2. Sacrificial protection : The rusting of iron can be prevented by covering the iron with more electropositive metals like zinc. Zinc metal has more tendency to get oxidized as compared to iron. So, iron articles will not be harmed till the layer of zinc present on its surface hence zinc metal is called the sacrificial metal.

3. Antirust solution : The alkaline antirust solution are employed to prevent rusting, alkaline solution prevent the availability of H+ ions.

In this method, iron articles are dipped in alkaline sodium phosphate or chromate solution. Due to this an insoluble sticking film of iron phosphate is formed on the surface which prevents rusting.

MP Board Solutions

Question 6.
Describe dry cell with labelled diagram.
Answer:
Dry cell: It is a primary cell based on Leclanche cell invented by G. Leclanche in 1868. In a primary cell, the electrode reactions cannot be reversed by an external source of electrical energy. In this cell, the cell reaction takes place only once i.e., this cell is not rechargeable.

It is generally used in torches, transistors, radios, calculators, tape recorders, etc. It consists of a hollow zinc cylinder which is filled with a paste of NH4Cl and a little ZnCl2. This paste is made with the help of water. The zinc cylinder acts as anode while cathode is a graphite rod (Carbon). The carbon rod is surrounded by a black paste of MnO2 and carbon powder. The zinc cylinder has an outer insulation of cardboard case.

Dry cells are sealed with wax or other material to protect the moisture from evaporation. When the electrodes are connected, the cell operates.

The electrode reactions are complex. Metallic Zn is oxidized to Zn2+ and the electrons liberated are left on the container. The reactions which take place at electrodes can be represented as :
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 74

This reaction prevents polarization due to formation of ammonia. It also prevents the substantiaL increase of concentration of Zn2+ ions which would decrease the cell potential. This potential of dry cell is approximately 1.5 V.

Defect: Due to acidic nature of NH4Cl zinc container corrodes due to which holes develop through which the chemicals come out.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 75
Nowadays, the cells are made leakage resistant. In it KOH is used in place of NH4Cl by which zinc does not corrode.

Question 7.
What is Kohlrausch law ? Give its two applications.
Answer:
Kohlrausch in 1875 gave a generalisation known as Kohlrausch’s law, “At infinite dilution when the dissociation of the electrolyte is complete, each ion makes a definite contribution towards molar conductance of the electrolyte irrespective of the nature of the other ion with which it is associated.”
Or
“The value of molar conductance at infinite dilution is given by the sum of the contributions of ions (cation and anion).”
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 76
Where, λ+ and λ are ionic contributions or ionic conductances of cation and anion while v+ and v are the number of cations and anions in the formula unit of electrolyte.

Applications of Kohlrausch’s law :

(i) Calculation of molar conductance at infinite dilution for weak electrolytes :

Molar conductance or equivalent conductance of weak electrolytes cannot be obtained graphically by extrapolation method, since these are feebly ionized. Kohlrausch’s law enables indirect evaluation in such cases. For example, molar conductances of acetic acid can be obtained from the knowledge of molar conductances at infinite dilution of HCl, CH3COONa and NaCl which are strong electrolytes.

From Kohlrausch’s law, it is clear that
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 77
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 78

(ii) Determination of degree of dissociation :
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 79

Question 8.
Draw a labelled diagram of Daniel cell and explain cell reaction.
Or,
Draw a labelled diagram of electrochemical cell and write cell reaction.
Answer:
Electrochemical cell: In the redox reactions, the transfer of electrons between oxidizing and reducing agents occurs through wire and thus chemical energy changes into electrical energy. The device on which chemical energy changes into electrical energy is called electro chemical cell. These are also known as galvanic or voltaic cells. Working of these cells can be understood with the example of Daniel cell.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 80
Daniel cell : In this cell, Zn rod is dipped in ZnSO4 solution and Cu rod in copper sulphate solution. Both solutions are connected through KC1 salt bridge. When Zn and Cu electrodes are connected by wire and galvanometer, flow of electrons from Zn to Cu occurs. Zinc atoms change into Zn2+ and electrons reach at Cu electrode, where Cu2+ changes into Cu metal and this copper deposits on electrode.
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 81

MP Board Solutions

Electrochemistry Numerical Questions

Question 1.
On passing 5 ampere electric current for 30 minute through a container filled with AgNO3 10.07 gram silver is deposited, then determine the chemical equivalent of silver. If electrochemical equivalent of hydrogen is 0-00001036 then calculate the equivalent mass of Silver.
Solution:
Given : W = 10-07 gm, i = 5 ampere, t = 30 × 60 second
According to Faraday’s first law W = Zit
∴ 10.7 = Z × 30 × 60
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 82

Question 2.
What are weak electrolytes ? Give one example. Find out molar conductivity of LiBr aqueous solution infinite dilution when joint conductance of Li+ ion and Br ion are 38.7 Scm2 mol-1 and 78.40 Scm2 mol-1 respectively.
Solution:
Weak electrolytes : These are the substances which dissociate only to a small extent.
Examples: CH3COOH,NH4OH
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 83

Question 3.
What are strong electrolytes ? Find out the molar conductivity of aqueous solution of BaCl2 at infinite dilution when ionic conductance of Ba+2 ion and Cl ion are 127.30 Scm2 mol-1 and 76.34 Scm2 mol-1 respectively.
Solution:
Strong electrolytes : These are substances which dissociate almost completely into ions under all dilutions.
Examples: NaCl, HCl,CH3COONa
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 84

Question 4.
Calculate the molar conductance of Al2(SO4)3 if
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 85
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 86
Solution:
By the ionisation of Al2(SO4)3 solution
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 87
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 88

Question 5.
Molar conductivity of \(\frac { M }{ 30 } \) CH3COOH is 9.625 mho and molar conductance of CH3COOH at infinite dilution (Λm) is 385 mho. Calculate the percentage of dissociation of \(\frac { M }{ 30 } \) CH3COOH.
Solution:
Degree of dissociation
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 89

Question 6.
Calculate molar conductance of acetic acid at infinite dilution from following values:
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 90
Solution:
From Kohlrausch’s law,
MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry 91

MP Board Class 12th Chemistry Solutions

MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Alcohols, Phenols and Ethers NCERT Intext Exercises

Question 1.
Classify the following as primary, secondary and tertiary alcohols :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 1
Answer:
(i) 1°, (ii) 1°, (iii) 1°, (iv) 2°, (v) 2°, (vi) 3°.

Question 2.
Identify allylic alcohols in the above examples.
Answer:
Allylic alcohols are (ii) and (vi).

Question 3.
Name the following compounds according to IUPAC system :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 2
Answer:
(i) 3-chloromethyl-2-isopropyl pentan-1-ol.
(ii) 2,5-Dimethyl hexane-1,3-diol.
(iii) 3-Bromocyclohexan-l-ol.
(iv) Hex-l-en-3-ol.
(v) 2-Bromo-3 -methyl but-2-en- l-ol.

Question 4.
Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal ?
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 3
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 4

Question 5.
Write structures of the products of the following reactions:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 5
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 6
(ii) NaBH4 is weak reducing agent, it reduces aldehyde/ketones and not the esters.
Thus,
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 7

Question 6.
Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl -ZnCl2, (b) HBr and (c) SOCl2 : (i) Butan-l-ol , (ii) 2-Methylbutan-2-ol.
Answer:
(a) With ZnCl2-HCl (Lucas reagent) : Butan-l-ol (1° alcohol) does not react with Lucas reagent at room temperature. However, turbidity appears only after heating but- 2-methyl butan-2-ol (3° alcohol) reacts with Lucas reagent at room temperature immediately gives turbidity
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 8

(b) With HBr : Both the alcohols react with HBr to give corresponding alkyl bromides.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 9

(c) With SOCl2 : Both the alcohols react with SOCl2 to give corresponding alkyl chlorides.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 10
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 11

MP Board Solutions

Question 7.
Predict the major product of acid catalysed dehydration of:
(i) 1-methylcyclohexanol and (ii) butan-l-ol.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 12
According to Saytzeff rule, the highly substituted product is the major product.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 13

Question 8.
ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.
Answer:
(i) Phenoxide ion:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 14
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 15
Resonating structures of p-nitro phenoxide ion. In substituted phenols, the presence of electron withdrawing group (-R effect) such as -NO2 group, increases the acidic strength of phenol, ortho and para nitrophenoxide ions are more stable (because of additional resonance structures show in boxes) than phenoxide ion due to effective delocalisation of negative charge in phenoxide ion. As a result o-and p- nitrophenols are more acidic than phenols.

Question 9.
Write the equations involved in the following reactions :
(i) Reimer – Tiemann reaction
(ii) Kolbe’s reaction.
Answer:
(i) Reimer-Tiemann reaction : When phenol is treated with chloroform in presence of aqueous sodium hydroxide at 60°C, o-Hydroxy benzaldehyde (Salicylaldehyde) and p-Hydroxy benzaldehyde are formed. The ortho-isomer is the major product. This reaction is called Reimer-Tiemann reaction.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 16
If carbon tetrachloride is used in place of chloroform, salicylic acid is obtained as the main product.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 17

(ii) Kolbe- Schmidt reaction : When sodium salt of a phenol is heated with CO2 at 130°C. (403K) and 4-7 atm pressure, sodium salicylate is formed. This on acidification gives salicylic acid.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 18
Salicylic acid is the starting material for the manufacture of aspirin which is an important analgesic.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 19

Question 10.
Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-oI.
Answer:
Williamson’s synthesis is reaction of alkyl halide (1°) with sodium alkoxide to give ether by SN2 mechanism. Thus, alkyl halide should be derivde from ethanol and alkoxide ion from 3-methyl pentan-2-ol. The complete reaction is as follows :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 20

MP Board Solutions

Alkyne-reactions-cheat-sheet-summary-for-organic-chemistry-reactions.

Question 11.
Which of the following is an appropriate set of reactants for the preparation of l-methoxy-4-nitrobenzene and why ?
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 21
Answer:
Chemically both sets are equally probable. In set (A) the Br group is activated by the electron withdrawing effect of —NO2 group. Therefore, nucleophilic attack of CH3ONa followed by elimination of NaBr gives the desired ether.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 22
In set (B) nucleophilic attack of 4-nitrophenoxide ion on methylbromide gives the desired product.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 23

Illustration about Set line Chemical formula Calculator, Electrical panel and Light bulb with concept of idea.

Question 12.
Predict the products of the following reactions:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 24
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 25
When one of the groups in unsymmetrical ether is tertiary, then the halide formed is tertiary halide.

MP Board Solutions

Alcohols, Phenols and Ethers NCERT Textbook Exercises

Question 1.
Write IUPAC names of the following compounds :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 26
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 27
Answer:
(i) 2,2,4-Trimethyl pentan-3-ol
(ii) 5-Ethylheptane-2,4-diol
(iii) Butan-2,3-diol
(iv) Propane-1,2,3-triol
(v) 2-Methylphenol
(vi) 4-Methylphenol
(vii) 2,5-Dimethylphenol
(viii) 2,6-Dimethylphenol
(ix) l-Methoxy-2-methylpropane
(x) Ethoxybenzene
(xi) 1-Phenoxyheptane
(xii) 2-Ethoxybutane.

Question 2.
Write structures of the compounds whose IUPAC names are as follows :
(i) 2-Methylbutan-2-ol
(ii) l-Phenylpropan-2-ol
(iii) 3,5-Dimethylhexane -1, 3, 5-triol
(iv) 2,3 – Diethylphenol
(v) 1 – Ethoxypropane
(vi) 2-Ethoxy-3-methylpentane
(vii) Cyclohexylmethanol
(viii) 3-CycIohexylpentan-3-ol
(ix) Cyclopent-3-en-l-
(x) 4-Chloro-3-ethylbutan-l-ol.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 28
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 29

Question 3.
(i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
(ii) Classify the isomers of alcohols in question (i) as primary, secondary and tertiary alcohols.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 30
Isomers (ii), (vii) and (viii) contain chiral centers that can exhibit enantiomerism.

Question 4.
Explain, why propanol has a higher boiling point than hydrocarbon, butane?
Answer:
The molecules of butane are held together by weak van der Waals’ force of attraction while those of propanol are held together by stronger intermolecular hydrogen bonding. Therefore, the b.p. of propanol is much higher than that of butane.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 31

Question 5.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Hydrogen bonding between alcohol and water molecules is the reason.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 32

Question 6.
What is meant by the hydroboration-oxidation reaction? Illustrate it with an example.
Answer:
The addition of borane followed by oxidation is known as the hydroboration – oxidation reaction. For example, propan- l-ol is produced by the hydroboration – oxidation reaction of propene. In this reaction, propene reacts with deborane (BH3)2 to form trialkyl borane as an additional product. This additional product is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 33

Question 7.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.
Answer:
Three isomers are :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 34

MP Board Solutions

Question 8.
While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Answer:
Intramolecular H-bonding is present in o-hitro phenol and p-nitrophenol. In p- nitrophenol, the molecules are strongly associated due to the presence of intramolecular bonding. Hence, o-nitrophenol is steam volatile.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 35

Question 9.
Give the equations of reactions for the preparation of phenol from cumene.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 36

Question 10.
Write chemical reaction for the preparation of phenol from chlorobenzene.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 37

Question 11.
Write the mechanism of hydration of ethene to yield ethanol.
Answer:
Direct addition of water to ethene in presence of acid does not occur. Indirectly, ethene is first passed through cone. H2SO4 at room temperature to form ethyl hydrogen-sulfate, which is decomposed by water on heating to form alcohol.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 38
Mechanism : H2SO4 → H+ + OΘ SO2OH

Step-I: Protonation of alkene to form carbo-cation by the electrophilic attack of hydronium ion (H3O+).
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 39
Step-II: Nucleophilic attack by water on carbo-cation to yield protonated alcohol.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 40
Step-III: Deprotonation (loss of proton) to form an alcohol:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 41

Question 12.
You are given benzene, cone. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 42

Question 13.
Show how will synthesis:
(i) 1-Phenylethanol from a suitable alkene?
(ii) Cyclohexylmethanol using an alkyl halide by an SN2 reaction?
(iii) Pentan-1-ol using a suitable alkyl halide?
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 43
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 44

Question 14.
Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Answer:
The reactions showing acidic character of phenols are:
(i) Reaction with sodium : Phenol reacts with Na to give H2 gas.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 45
(ii) Reaction with NaOH : Phenol dissolves in NaOH to give sodium phenoxide and OH
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 46
Phenol is more acidic than ethanol. This is due to the reason that phenoxide ion left after the loss of a proton from phenol is stabilized by resonance (for structure refer to text acidic nature of phenol) while ethoxide ion (left after the loss of a proton from ethanol) is not.

Question 15.
Explain, why ortho nitrophenol is more acidic than ortho methoxy phenol?
Answer:
Due to the strong -R and -I effect of -NO2 group, the electron density in O-H bond decreases, and hence the loss of proton becomes easy.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 47
-R effect results in +ve charge on O-atom and hence facilitates the release of the proton. Moreover, o-nitrophenoxide formed after the loss of a proton is stabilized by resonance.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 48

o-nitrophenoxide ion is stabilized by resonance and hence o-nitrophenol is a stronger acid. On the other hand, due to the +R effect of the —OCH3 group the electron density in the O—H bond increases and this makes the loss of proton difficult.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 49
Furthermore, after the loss of proton o-methoxy- phenoxide ion left is destabilized by resonance.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 50
The two -ve charge repel each other and therefore destabilize the o-methoxyphenoxide ion. Thus, it is less acidic than o-nitrophenol.

Question 16.
Explain, how does the -OH group attached to a carbon of benzene ring activate it towards electrophilic substitution ?
Answer:
The – OH group is an electron-donating group. Thus, it increases the electron density in the benzene ring as shown in the given resonance structure of phenol. As a result, the benzene ring is activated towards electrophilic substitution.

Question 17.
Give equations of the following reactions:
(i) Oxidation of propan-l-ol with alkaline KMnO4 solution.
(ii) Bromine in CS2 with phenol.
(iii) Dilute HNO3 with phenol.
(iv) Treating phenol with chloroform in presence of aqueous NaOH.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 51
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 52

Question 18.
Explain the following with an example:
(i) Reimer-Tiemann reaction
(ii) Kolbe’s reaction
(iii) Williamson-ether synthesis
(iv) Unsymmetrical ether.
Answer:
(i) Reimer-Tiemann reaction: When phenol is treated with chloroform in presence of aqueous sodium hydroxide at 60°C, o-Hydroxy benzaldehyde (Salicylaldehyde) and p-Hydroxy benzaldehyde are formed. The ortho-isomer is the major product. This reaction is called the Reimer-Tiemann reaction.

MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 141
If carbon tetrachloride is used in place of chloroform, salicylic acid is obtained as the main product.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 142

(ii) Kolbe- Schmidt reaction: When sodium salt of phenol is heated with CO2 at 130°C. (403K) and 4-7 atm pressure, sodium salicylate is formed. This on acidification gives salicylic acid.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 143
Salicylic acid is the starting material for the manufacture of aspirin which is an important analgesic.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 144

(iii) Williamson-1 ether synthesis: This is the best method for the preparation of ethers because both symmetrical and unsymmetrical (Aliphatic as well as aromatic) ethers can be prepared. When haloalkane is treated with sodium alkoxide then ether is formed. It is an example of a nucleophilic substitution reaction in which halide ion (X-) of haloalkane is replaced by alkoxy or aroxy group.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 53
Example : Reaction of ethyl bromide with sodium ethoxide gives diethyl ether
C2H5Br + C2H5ONa → C2H5OC2H5 + NaBr
(iv) Unsymmetrical ether: An unsymmetrical ether is an ether where two groups on the two sides of an oxygen atom differ (i.e., have an unequal number of carbon atoms.
For example; ethyl methyl ether (CH3—O—CH2CH3)

Question 19.
Write the mechanism of acid dehydration of ethanol to yield ethene.
Answer:
The mechanism of acid dehydration of ethanol to yeild ethene involes the following three steps:

Step 1: Protonation of ethanol to form ethyl oxonium ion :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 54
Step 2: Formation of carbocation (rate determinning step):
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 55
Step 3: Elimination of a proton to form ethene:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 56
The acid consumed in step I is released in Step 3. After the formation of ethene, it is removed to shift the equilibrium in a forward direction.

MP Board Solutions

Question 20.
How are the following conversions carried out:
(i) Propene → Propan-2-ol.
(ii) Benzyl chloride → Benzyl alcohol.
(iii) Ethyl magnesium chloride → Propan-1-ol.
(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 57

Question 21.
Name the reagents used in the following reactions:

  1. Oxidation of primary alcohol to a carboxylic acid.
  2. Oxidation of a primary alcohol to aldehyde.
  3. Bromination of phenol to 2,4,6-tribromo-phenol.
  4. Benzyl alcohol to benzoic acid.
  5. Dehydration of propan-2-ol to propene.
  6. Butan-2-one to butan-2-ol.

Answer:

  1. Alkaline KMnO4
  2. PCC (Pyridinium chlorochromate)
  3. Aqueous Br2
  4. Na2Cr2O7/H2SO4
  5. 85% H3PO4
  6. NaBH4 or LiAlH4

Question 22.
Give a reason for the higher boiling point of ethanol in comparison to methoxymethane.
Answer:
Ethanol undergoes intermolecular H-boding due to the presence of the -OH group, resulting in the association of molecules. Extra energy is required to break these hydrogen bonds. On the other hand, methoxymethane does not undergo H-bonding. Hence, the bp of ethanol is higher than that of methoxymethane.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 58

Question 23.
Give IUPAC names of the following ethers:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 59
Answer:
(i) 1-Ethoxy-2-methylpropane
(ii) 2-Chloro-1-methoxyethane
(iii) 4-Nitro anisole
(iv) 1-Methoxypropane
(v) 1 -Ethoxy-4,4-dimethylcyclohexane
(vi) Ethoxybenzene.

Question 24.
Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
(i) 1-Propoxypropane
(ii) Ethoxybenzene
(iii) 2-Methoxy-2-methylpropane
(iv) 1-Methoxyethane
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 60

Question 25.
Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Answer:
Limitations of Williamson synthesis : (i) Better results are obtained, if the alkyl halide is primary. In case of secondary and tertiary alkyl halides, elimination completes over substitution. If a tertiary alkyl halide is used, an alkene is the only reaction product and no ether is formed. For example, the reaction of CH3ONa with (CH3)3C—Br gives exclusively 2-methylpropene.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 61

It is because alkoxides are not only nucleophile but strong bases as well. They react with alkyl halide leading to elimination reaction. Thus, in order to prepare methyl tertiary butyl ether, we must use methyl halide (primary) and sodium tertiary butoxide.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 62
(ii) Aryl halides and vinyl halides cannot be used as substrate for the preparation of aromatic aliphatic ether because aryl halide and vinyl halides are less reactive towards nucleophilic substitution reaction.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 63

Question 26.
How is 1-propoxypropane synthesized from propan-l-ol ? Write mechanism of this reaction.
Answer:
(a) It can be prepared by Williamson’s synthesis.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 64
(b) It can also be prepared by dehydration of propan-l-ol
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 65

MP Board Solutions

Question 27.
Preparation of ethers by acid dehydration of secondary or tertiary alcohol is not a suitable method. Give reason.
Answer:
The 1° alcohol gets protonated. Now it is attacked by another alcohol molecule. The mechanism is SN2 (Nucleophilic attack)
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 66
The 2° and 3° alcohols also get protonated. But another molecule of alcohol cannot attack it due to steric hindrance. The protonated alcohol loses a molecule of H2O to form a stable 2° or 3° carbo-cation. The carbocation prefers to lose a proton to form alkene.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 67
Similarly, the 3° alcohol (CH3)3COH forms isobutane.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 68

Question 28.
Write the equation of the reaction with hydrogen iodide :
(i) 1-Propoxypropane
(ii) Methoxybenzene and
(iii) Benzyiethylether.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 69

Question 29.
Explain the fact that in aryl alkyl ethers:
(i) The aikoxy group activates the benzene ring towards electrophilic substitution and
(ii) It directs the incoming substituents to ortho and para positions in benzene ring.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 70
In aryl alkyl ethers, due to the +R effect of the aikoxy group, the electron density in the benzene ring increases as shown in the following resonance structure.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 71
Thus, benzene is activated towards electrophilic substitution by the aikoxy group.

(ii) It can also be observed from the resonance structure that the electron density increases more at the ortho para positions than at the meta position. As a result, the incoming substituents are directed to the ortho and para positions in the benzene ring.

Question 30.
Write the mechanism of the reaction of HI with methoxymethane.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 72
Protonated ether undergoes SN2 attack by 1- ion and gives a mixture of methyl iodide and methyl alcohol. However, if HI is taken in excess, the methyl’alcohol formed in (ii) is also converted into methyliodide by the following mechanism.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 73

Question 31.
Write equations of the following reactions:
(i) Friedel-Craft’s reaction – alkylation of anisole.
(ii) Nitration of anisole.
(iii) Bromination of anisole in ethanoic acid medium.
(iv) Friedel-Craft’s acetylation of anisole.
Answer:
(i) Friedel-Crafts reaction – alkylation of anisole : Anisole undergoes Friedel- Crafts reaction i.e., the alkyl and aryl group is introduced at ortho and para-positions by reaction with alkyl halide and aryl halide in the presence of anhydrous aluminium chloride (Lewis acid) as catalyst.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 74

(ii) Nitration of anisole : Anisole reacts a mixture of concentrated sulphuric and nitric acids to yield a mixture of ortho and para nitroanisole.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 75

(iii) Bromination of anisole in ethanoic acid medium : Phenetole or anisole undergoes bromination with bromine in ethanoic acid even in absence of iron (III) bromide catalyst. It is due to the activation of benzene ring by the ethoxy group, para-isomer is obtained in 90% yield.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 76

(iv) Friedel-Craft’s acetylation of anisole :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 77

Question 32.
Show how would you synthesize the following alcohols from appropriate alkenes ?
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 78
Answer:
(i) Alcohols on dehydration give alkene. Dehydrate the given alcohol. The alkene formed will give the desired alcohol on adding a molecule of H2O. The addition of H2O molecule takes place according to MarkownikofFs rule. In case dehydration of the given alcohol gives two alkenes, then see which alkene will give the desired alcohol.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 79
Both the alkene give the desired alcohol on adding a molecule of H2O.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 80
(ii) Two alkenes are formed. They will give the desired alcohol on adding a H2O molecule.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 81

Addition of H2O molecule to pent-1-ene gives the desired alcohol. Remember that -OH group comes to that carbon atom of the double bond which contains less number of H-atoms. In case of pent-2-one, both the carbon atoms of double bond have one H-atom. Therefore, -OH group may come to either of the carbon atom. This alkene will give pentan-2-ol as well as pentan-3-ol.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 82
Addition of H2O molecule to 2-methylcy-clohexyl but-2-ene will give the desired alcohol -OH group comes to that carbon atom of double bond which contains less number of H- atom.

Question 33.
When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 83
Give a mechanism for this reaction.
(Hint : The secondary carbocation formed in step-II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)
Answer:
Alcohol gets protonated and then a H2O molecule is given out to give a carbocation.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 84
Now, the 2° carbocation rearrange to form more stable 3° carbocation, one H—atom migrates from the adjacent carbon atom to C+. It is called 1,2-shift. Then addition of Br takes place.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 85

MP Board Solutions

Alcohols, Phenols and Ethers Other Important Questions and Answers

Alcohols, Phenols and Ethers Objective Type Questions

Choose the correct answer:

Question 1.
Sodium dissolves easily in alcohol because :
(a) Alcohol has higher density than water
(b) Alcohol is lighter than water
(c) Alcohol is neutral
(d) Alcohol is amphoteric.
Answer:
(d) Alcohol is amphoteric.

Question 2.
Most acidic among the four compound is :
(a) Phenol
(b) o-nitrophenol
(c) m – nitrophenol
(d) p – nitrophenol.
Answer:
(c) m – nitrophenol

Question 3.
The reaction for the formation of salicylaldehyde from phenol is:
(a) Rosenmund reaction
(b) Friedel Crafts reaction
(c) Reimer-Tiemann reaction
(d) Wurtz’s reaction.
Answer:
(c) Reimer-Tiemann reaction

Question 4.
Most effective reagent which converts propanol-2 to propanone:
(a) LiAlH4
(b) Cu/300°C
(c) CO2
(d) K2Cr2O7.
Answer:
(b) Cu/300°C

Question 5.
Carbolic acid is :
(a) Phenol
(b) Phenyl benzoate
(c) Phenylacetate
(d) Methyl salicylate.
Answer:
(a) Phenol

Question 6.
Which compound is known as oil of wintergreen:
(a) Phenyl benzoate
(b) Phenyl salicylate
(c) Phenylacetate
(d) Salol.
Answer:
(d) Salol.

Question 7.
The reaction of Lucas reagent is fastest with:
(a) (CH3)3 C-OH
(b) (CH3)2 CHOH
(c) CH3-(CH2)2 OH
(d) CH3CH2OH.
Answer:
(a) (CH3)3 C-OH

Question 8.
At low temperature phenol reacts with Br2 in CS2 to give
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 86
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 87

Question 9.
Which compound is obtained on passing vapours of ethanol on hot Al2O3:
(a) Ethyl ether
(b) Acetone
(c) Acetaldehyde
(d) Ethane.
Answer:
(a) Ethyl ether

Question 10.
Which of the compound is Aspirin :
(a) Acetylsalicylic acid
(b) Salicylic acid
(c) Acetamide
(d) Salicylamide.
Answer:
(a) Acetylsalicylic acid

Question 11.
The following compound reacts with phthalic acid to give an acid-base indicator:
(a) Chlorobenzene
(b) Phenol
(c) Alcohol
(d) Ether.
Answer:
(b) Phenol

Question 12.
Bakelite is formed when phenol is condensed with:
(a) HCHO
(b) CH3CHO
(c) C6H5CHO
(d) CH3COCH3.
Answer:
(a) HCHO

Question 13.
Used as an anaesthetic:
(a) CH3OH
(b) C2H5OH
(c) CH3—CHO
(d) (C2H5)2 O.
Answer:
(d) (C2H5)2 O.

Question 14.
Lucas reagent is:
(a) Conc. HCl
(b) Conc. H2SO4
(c) Anhydrous ZnCl2
(d) Conc. HCl and anhydrous ZnCl2.
Answer:
(d) Conc. HCl and anhydrous ZnCl2.

Question 15.
Ether and alcohol can be distinguished by the following:
(a) Reaction with Na
(b) Reaction with PCl5
(c) Reaction with 2, 4 dinitrophenyl hydrazine
(d) None of these.
Answer:
(a) Reaction with Na

Question 16.
Which is used for poisoning the alcohol:
(a) Methyl alcohol
(b) Ethyl alcohol
(c) Glycerine
(d) All the above.
Answer:
(a) Methyl alcohol

Question 17.
Gives Libermann’s nitroso test:
(a) C6H5OH
(b) CH3-OH
(c) C2H5—OH
(d) CH3-O-CH3
Answer:
(a) C6H5OH

Question 18.
Which is identified by Lucas reagent:
(a) Phenol
(b) Ether
(c) Aldehyde
(d) Alcohol.
Answer:
(d) Alcohol.

Question 19.
Alcohols are soluble in water. Its main reason is:
(a) O—H bond
(b) Hydrogen bond
(c) Covalent bond
(d) Electrovalent bond.
Answer:
(b) Hydrogen bond

Question 20.
Which is formed on heating ethyl alcohol with bleaching powder:
(a) Diethyl ether
(b) Phenol
(c) Chlorobenzene
(d) Chloroform.
Answer:
(d) Chloroform.

Question 2.
Fill in the blanks :

  1. The general formula of ether is ……………….
  2. The main product of Kolbe-Schmidt reaction of phenol is ……………….
  3. By the hydrogenation of phenol ………………. is formed.
  4. Alcohol reacts with I2 and a base to give a yellow precipitate of ……………….
  5. Phenol on being heated with Zn powder forms ……………….
  6. On heating formaldehyde with ………………. bakelite is formed.
  7. Diethyl ether is used as an ……………….
  8. On heating RX with NaOR, ROR is formed. Name of this reaction is ……………….
  9. Alcohol is ………………. whereas phenol is of ………………. nature.
  10. On heating alcohol with cone. H2SO4 for 160 – 170°C ………………. is formed.
  11. By the dehydration of ethyl alcohol ………………. and ………………. are obtained.
  12. Rectified spirit is a mixture of ………………. % alcohol and ………………. water.

Answer:

  1. CnH2n+1OCnH2n+1
  2. Salicylic acid
  3. Cyclohexanol
  4. Iodoform (CHI3),
  5. Benzene
  6. Phenol
  7. Anesthetic
  8. Williamson synthesis
  9. Neutral, acidic
  10. Alkene
  11. Ethylene, diethyl ether
  12. 95-5%, 4-5%.

Question 3.
Match the following :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 88
Answer:

  1. (f)
  2. (d)
  3. (e)
  4. (g)
  5. (b)
  6. (a)
  7. (j)
  8. (c)
  9. (h)
  10. (i).

Question 4.
Answer in one word/sentence:

  1. Diethyl ether does not reacts with Na. Why?
  2. Fire caused due to ether cannot be extinguished by water. Why?
  3. Which is formed on burning ether?
  4. Write name of the enzyme which converts maltose to glucose.
  5. Sulphuric ether is known as.
  6. Reaction of ether with HI is used for the detection of what?
  7. Phenol reacts with Br2 in presence of CS2 to form.
  8. In Victor Meyer method, 1° alcohol gives which colour with base.
  9. Phenol reacts with phthalic anhydride in presence of H2SO4 to form.
  10. Which gas is obtained during fermentation?
  11. Name the primary alcohol which gives the iodoform test.
  12. Name the reaction in which phenol reacts with chloroform and sodium hydroxide to form salicylaldehyde.

Answer:

  1. Absence of acidic H-atom
  2. Lighter than water and insoluble
  3. CO2 and H2O
  4. Maltase
  5. Diethyl ether
  6. Alkoxy (Ziesel)
  7. o-and p-bromophenol,
  8. Red
  9. Phenolphthalein
  10. CO2
  11. C2H5—OH
  12. Reimer-Tiemann reaction.

MP Board Solutions

Alcohols, Phenols and Ethers Short Answer Type Questions

Question 1.
Why ether is less soluble in a saturated solution of NaCl? Explained.
Answer:
Ether is a weak polar compound and a saturated solution of NaCl decreases the polarity of the ether. Na+ and Cl ions attract H2O molecules more strongly than ether molecules, thus the solubility of ether decreases in the saturated solution of NaCl.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 89

Question 2.
Your teacher gives you two bottles, one with ethanol and the other with methanol without labelling.
(i) Write a suitable test to distinguish them.
(ii) By using the same test, can you distinguish between propan-1-ol and ethanol? Write the chemical equation.
Answer:
(i) The liquid which gives iodoform test is ethanol. Methanol does not give iodoform test. Ethanol reacts with alkaline solution of Iodine (I2/NaOH) to form yellow precipitate of iodoform.
CH3CH2OH + 3I2 + 4NaOH → CHI3 + HCOONa + 3H2O + 3NaI
CH3OH + 3I2 + 4NaOH -» No reaction

(ii) Yes. Ethanol gives iodoform test while propan -1 – ol does not give iodoform test.
CH3 – CH2CH2OH + I2 + NaOH → No reaction

Question 3.
Give equations for the preparation of ethyl alcohol by starch and write name of enzymes.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 90
Enzymes : (1) Diastase, (2) Maltase, (3) Zymase.

Question 4.
What is Lucas reagent ? How are primary, secondary and tertiary alcohol identified by it ? Explain.
Answer:
Mixture of anhydrous ZnCl2 and cone. HCl is known as Lucas reagent.

  1. Tertiary Alcohol : On adding Lucas reagent in alcohol at normal temperature, immediately white oily precipitate of Alkyl chlorides is formed, then it is tertiary alcohol.
  2. Secondary Alcohol: If on adding Lucas reagent in alcohol, at normal temperature, a white oily precipitate of alkyl chloride is obtained after 5 minutes, then it is secondary alcohol.
  3. Primary Alcohol: Primary alcohol does not show any reaction with Lucas reagent at normal temperature.

Question 5.
Why the b.p. of alcohol are higher than ethers and alkene ?
Or,
C2H5OH and CH3OCH3 both have same molecular formula (C2H6O) but the b.p. of alcohol is 78.4°C and b.p. of ether is -240°C. Explain the reason.
Answer:
In case of C2H5OH there is strong intermolecular hydrogen bonding between the molecules of alcohol. So alcohols (C2H5OH) required much energy to evaporate than ether molecules. In other words, we can say that the C2H5OH molecules are in associated form due to H-bonding so the b.p. of C2H5OH is very higher than ether and alkene.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 91

Question 6.
What do you understand by Methylated spirit or denaturing alcohol ?
Answer:
Methylated Spirit : Ordinary rectified spirit is known as industrial alcohol.

Methylated spirit is 90% ethanol to which nauseating materials like methyl alcohol, pyridine or mineral naphtha have been added. These materials are added so that the ethanol will not be used for beverage purposes. This process is called denaturing. It is of two types :

It is used for the preparation of spirit, varnish etc. By it misuse of ethanol as a beverage is controlled.

Question 7.
Explain the manufacture of CH3OH by water gas.
Answer:
From water gas : Steam is passed over red hot coke when water gas is formed.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 92
Water gas is mixed with half its volume of hydrogen, compressed to about 200 atm and passed over a catalyst which is a mixture of oxides of copper, zinc and chromium at 300°C.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 93

Question 8.
Give the chemical equation of the following conversion:
(i) Diethyl ether from ethanol
(ii) Ethanol from diethyl ether
(iii) Ethyl acetate from ethanol
(iv) Ethanol from glucose.
Answer:
(i) Diethyl ether from Ethanol:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 94
(ii) Ethanol from Diethyl ether :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 95
(iii) Ethyl acetate from Ethanol :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 96
(iv) Ethanol from Glucose :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 97

Question 9.
Differentiate between Phenol and Alcohol and write Libermann’s reaction related to phenol.
Answer:
Differences between Phenol and Alcohol:

Phenol:

  1. Physical properties: Characteristic phenolic odour, sparingly soluble in water.
  2. It is acidic and dissolves in bases to form salt.
  3. On oxidation, hybrid coloured product is formed.
  4. Produce characteristic colour with Ferric chloride.
  5. It does not react with halogen acid.
  6. With PCl5, mainly form triaryl phosphate.

Alcohol:

  1. Pleasant odour, fairly soluble in water.
  2. It is neutral and do not reacts with bases.
  3. It can easily oxidize to Aldehydes and ketones.
  4. It does not reacts with ferric chloride.
  5. Forms Alkyl halide.
  6. Alkyl chloride are formed.

Libermann’s Reaction: On adding few drops of concentrated sulphuric acid and little sodium nitrite in phenol first dark blue colour is produced on adding water colour becomes red and on adding an alkali red colour again changes to blue colour.

Question 10.
Pure phenol is a colourless solid but why it is converted into pink after some time?
Or,
What change in colour is observed in phenol in presence of oxygen? Explain with reaction.
Answer:
In the presence of air pure phenol oxidises into quinone.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 98
This quinone again combines with two molecules of phenol by H-bond and gives pink phylloquinone.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 99

Question 11.
Name the enzymes present in yeast which can convert sucrose into glucose and fructose and then to ethyl alcohol.
Answer:
Invertase can convert sucrose to glucose and fructose Zymase can convert glucose and fructose to ethanol and carbon dioxide.

Question 12.
Boiling point of higher than the corresponding alkane. Why ?
Answer:
Boiling point of alcohols is much higher than hydrocarbons of nearly similar molecular mass due to intermolecular hydrogen bond. Alcohols molecules associate kilo calories mole-1. Thus, extra energy is required for the separation of these molecules, which lead to an increase in boiling point. Hydrocarbons do not form hydrogen bond, thus their boiling point is comparatively less.

Question 13.
Ethyl alcohol and phenol both contain — OH group. What is the reason that phenol is acidic and alcohol has alkaline effect?
Or,
Ethyl alcohol and phenol both contain — OH group. What is the reason that phenol is acidic and alcohol is neutral in nature?
Answer:
Explanation of acidic nature of phenol: One possible explanation why phenols are stronger acids as compared to alcohols is that phenols exist as a resonance hybrid.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 100
Due to resonance, the oxygen atom gets a positive charge and attracts the electron pair of the O—H bond and thus facilitates the release of a proton. The phenoxide ion formed after the release of a proton is also stabilized by resonance.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 101
In alcohols, no resonance is possible hence the hydrogen atom is more firmly linked to the oxygen.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 102

MP Board Solutions

Question 14.
Write Victor Meyer method to distinguish primary, secondary and tertiary alcohol.
Answer:
Victor Meyer’s method:

  • The given alcohol is converted into an iodide by concentrated HI or red phosphorus and iodine.
  • The iodide is treated with silver nitrite to form nitroalkane.
  • Nitroalkane is finally treated with nitrous acid (NaNO2 + H2SO4) and made alkaline with KOH.

If a blood red colour is obtained, the original alcohol is primary.
If a blue colour is obtained, the alcohol is secondary.
If no colour is produced, the alcohol is tertiary.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 103

Question 15.
What is Williamson’s continuous etherification process? Is it a continuous process? Explain. Give labeled diagram.
Or,
Describe the laboratory method of preparation of diethyl ether. How ether thus obtained is purified?
Answer:
Laboratory Method for the Preparation of Diethyl Ether (Sulphuric Ether):
Diethyl ether is prepared in the laboratory and industry by the Williamson continuous etherification process, i.e., by heating ethanol (in excess) with concentrated sulphuric acid.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 104
Sulphuric acid is regenerated in the reaction hence, it appears as if only a small amount of acid may convert excess alcohol into the ether. So, this method is called Williamson’s continuous etherification process but actually, we cannot get ether continuously.

This is due to the following two reasons:

  1. Water formed in the reaction dilutes the acid and its reactivity decreases.
  2. A part of sulphuric acid is reduced by alcohol into sulphur dioxide.

Method: Ethanol and H2SO4 (2:1) are taken in a flask and heated on sand bath at ‘ 140°C. Ethanol is added at the same rate at which ether distilled over and is collected in a receiver cooled in ice-cold water.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 105
Purification: Ether contain ethanol, water and sulphuric acid as impurities. It is washed with NaOH to remove sulphuric acid and then agitated with 50% solution of calcium chloride to remove the alcohol. It is then washed with water, dried over anhydrous calcium chloride and redistilled.

Question 16.
Write a note on fermentation.
Answer:
Fermentation: In this method, molasses or starch are used as raw materials. This is the old method and is used at present also. Fermentation (Latin-fermentare means to boil) proceeds with fast evolution of CO2 producing much foam giving the appearance as if the solution is boiling. “It is a process of decomposition of complex large molecules into simple and small molecules slowly and by the enzyme.” Enzymes are of many kinds and are present in yeast which is a living and complex substance containing several types of bacteria which are called enzymes. It is a good ferment. The enzymes present in yeast are zymase, maltase, invertase, etc.

When yeast is mixed in glucose solution and kept at proper conditions, ethyl alcohol is formed as a result of fermentation.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 106

Favourable conditions for Fermentation:

  1. Favourable temperature: It is between 25-35°C.
  2. Other substances: Some inorganic salts like ammonium sulphate or ammonium nitrate function as food for fermentation.
  3. Concentration: Solution should be dilute (Concentration 8-10%).
  4. Air: The process occur in presence of air.

Question 17.
(i) How can we obtain phenol from benzene diazonium chloride?
(ii) What is the reaction of diethyl ether with HI acid?
Answer:
(i) Phenols are prepared by hydrolysis of diazonium salts by water, dil. acids etc.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 107
(ii) The reaction of diethyl ether with cone. HI acid, on heating gives one molecule of ethyl iodide and one molecule of ethyl alcohol.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 108

Question 18.
Give two reactions that show the acidic nature of phenol.
Answer:
(a) The reactions showing acidic character of phenol are:
(i) Reaction with sodium: Phenol reacts with sodium to give H2 gas.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 109
(ii) Reaction with NaOH: Phenol dissolve in NaOH to give sodium phenoxide.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 110

(b) Comparison of Acidity of Phenol with alcohol.
C2H5OH + NaOH → No reaction
But phenol reacts with NaOH and exhibits its strong acidic nature.
Ionization of Ethanol and Phenol is as follows :
On the other hand ethoxide ion and ethanol do not represent resonance thus negative charge is on oxygen of ethoxide ion whereas charge displacement occur in phenoxide ion.
pKa value of ethanol is 15.9 where of phenol is 10.0. Thus, phenol is many more times’ more acidic than ethanol.

MP Board Solutions

Alcohols, Phenols and Ethers Long Answer Type Questions

Question 1.
Explain the mechanism of dehydration of alcohol.
Answer:
Dehydration of alcohol:
(i) When ethyl alcohol is heated in excess of cone. H2SO4 molecule of water is eliminated and alkene is formed.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 111
Mechanism : (i) Protonation of alcohol by H2SO4
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 112
(ii) Removal of water
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 113

(iii) Elimination of β-hydrogen in the form of proton by base (bisulphate ion)
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 114
Stability of the carbocation (I) determines the case of dehydration and order of stability of carbocation is :
CH3 < C2H5 < Isopropyl < Tertiary butyl

Question 2.
How is ethyl alcohol obtained by molasses? Explain in brief.
Or
What are molasses? How is alcohol obtained by fermentation? Explain. Tell favourable conditions of fermentation. Draw labelled diagram of coffee still.
Answer:
From molasses: Molasses is the syrupy solution of sugar left after the separa¬tion of cane sugar or beet sugar crystals from the concentrated juice.

The different steps of the manufacture processes are:

(a) Dilution: The molasses is diluted with water so that a concentration of 8-10 percent sugar is obtained in the solution. This is acidified with dilute sulphuric acid to retard other bacterial growth. A solution of ammonium salts is also added which acts as food for the ferment.

(b) Alcoholic fermentation: The dilute solution obtained above [From step (a)] is taken in big fermentation tanks and some yeast is added. The mixture is kept for a few days and the temperature is maintained at about 30°C. The fermentation reaction starts and the enzyme Invertase (From Yeast) converts sucrose into glucose and fructose which are then converted into ethanol by Zymase (From Yeast).
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 115
The fermentation is completed in about 3 days. The carbon dioxide is collected as a by-product.

(c) Distillation: The fermented liquor is technically called wash or wort which contains about 9-10 percent ethanol. It is then distilled in a continuous still called Coffey’s still. It consists of two tall fractionating columns which are called analyser and the rectifier. It works on the counter-current principle and the steam and washes travel in opposite directions through the still.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 116
The steam goes upwards in the analyser and takes away the alcohol vapours from the down coming dilute alcohol. The mixture leaves the analyser from the top and enters the rectifier at the base. Here it heats the wash flowing through the pipes on its way to the analyser. Most of the steam condenses and the alcohol vapours condense in the condenser. The distillate contains 90% alcohol.

(d) Rectification: Wash is rectified by fractional distillation.

Question 3.
How can you change the following:
(i) Methanol to ethanol
(ii) Ethanol to methanol
Answer:
(i) Methanol to ethanol:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 117
(ii) Ethanol to methanol:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 118

Question 4.
Differentiate primary, secondary and tertiary alcohol by oxidation and dehydrogenations method.
Answer:
1. Oxidation: The oxidizing agents generally used for oxidation of alcohols are acid dichromate, acid or alkaline KMnO4, and dilute HNO3.

(i) A primary alcohol is easily oxidized to an aldehyde and then to an acid both containing the same number of carbon atoms as the original alcohol.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 119

(ii) A secondary alcohol on oxidation gives a ketone with the same number of carbon atoms as the original alcohol, ketones are oxidized with difficulty but prolonged action of oxidizing agents produce carboxylic acids containing fewer carbon atoms than the original alcohol.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 120

(iii) A tertiary alcohol is resistant to oxidation in neutral or alkaline solutions but is readily oxidized by an acid oxidizing agent giving a mixture of ketone and acid each having lesser number of carbon atoms than the original alcohol.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 121

2. Dehydrogenation (Action of hot reduced copper at 300°C): Different types of alcohols give different products when their vapours are passed over Cu gauze at 300°C.
Primary alcohols lose hydrogen and yield an aldehyde.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 122
Secondary alcohols lose hydrogen and yield a ketone.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 123
Tertiary alcohols are not dehydrogenated but lose a water molecule to give alkenes.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 124

Question 5.
How can you obtain the following compounds from phenol :
(i) 2, 4, 6-Tribromophenol
(ii) Picric acid
(iii) Aniline
(iv) Benzene
(v) Phenolphthalein
(vi) p-cresol, o-cresol.
Answer:
(i) Phenol to Tribromophenol: Phenols readily react with halogens to give polyhalogen substituted compounds. Phenol gives white precipitate of 2, 4, 6-tribromo- phenol with bromine water.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 125
(ii) Phenol to Picric acid : Nitration : On nitration, phenols give a variety of products depending upon the conditions.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 126
Nitration of phenol with conc. HNO3 in presence of conc. H2SO4 gives, 2,4,6-Trinitro-phenol (Picric acid).
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 127
(iii) Phenol to Aniline:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 128
(iv) Phenol to Benzene:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 129
(v) Phenol to Phenolphthalene: Phenol condenses with phthalic anhydride in presence of conc. H2SO4 give phenolphthalein which is an indicator for acid-base titrations and is used as a laxative in medicine.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 130
(vi) Phenol to ortho and para cresol:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 131

Question 6.
Give the chemical equation of the following conversion:
(i) Diethyl ether from ethanol
(ii) Ethanol from diethyl ether
(iii) Ethyl acetate from ethanol
(iv) Ethanol from glucose.
Answer:
(i) Diethyl ether from Ethanol:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 132
(ii) Ethanol from Diethyl ether:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 133
(iii) Ethyl acetate from Ethanol:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 134
(iv) Ethanol from Glucose :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 135

Question 7.
Describe the manufacture method of methanol by the destructive distillation of wood.
Answer:
Manufacture of methanol: By the destructive distillation of wood: Dried wood is heated (350°C) in closed retorts for about 3 hours. Different products obtained are given ahead:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 136
Recovery of methyl alcohol from pyroligneous acid : Pyroligneous acid contains methyl alcohol (2-4%), acetone (0-5 to 1%), acetic acid (about 10%) and water.

The pyroligneous acid is taken in a copper vessel and distilled. The vapours are passed through hot milk of lime which retains acetic acid as non-volatile calcium acetate. Methyl alcohol and acetone vapours pass over and are condensed. This mixture is subjected to fractional distillation to separate methyl alcohol (b.p. 65°C) and acetone (b.p. 56°C). Methyl alcohol obtained by fractional distillation is about 95% pure. To purify further methyl alcohol is treated with anhydrous calcium chloride when it forms solid derivative CaCl2.4CH3OH. This solid derivative is separated and methyl alcohol is recovered by distillation.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 137

Question 8.
Give equations for three methods of preparation of phenol.
Answer:
Methods of preparation of phenol:
(i) By the hydrolysis of Benzene diazonium salts: Benzene diazonium salt is formed by aromatic primary amine (aniline) with nitrous acid at 0-5°C. On boiling aqueous solution of this salt phenol is formed.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 138
(ii) By alkaline fusion of sodium benzene sulphonate : On fusing sodium benzene sulphonate with NaOH, sodium phenoxide is formed which on acidification forms phenol.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 139
(iii) Rasching method : On heating benzene with mixture of HCl and air to 230°C in the presence of Cu catalyst chlorobenzene is formed which on hydrolysis form phenol.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 140

MP Board Class 12th Chemistry Solutions

MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules

MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules

Biomolecules NCERT Intext Exercises

Question 1.
Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain.
Answer:
Glucose and sucrose have – OH groups while cyclohexane and benzene do not. Thus glucose and sucrose can form hydrogen bonds with water and are soluble in water. While cyclohexane and benzene do not form hydrogen bonds with water and thus insoluble in water.

Question 2.
What are the expected products of hydrolysis of lactose?
Answer:
Galactose and glucose are the products of hydrolysis of lactose.
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 1

Question 3.
How do you explain the absence of aldehyde group in the penta acetate of D-glucose?
Answer:
The pentacetate of glucose does not react with hydroxylamine indicating the absence of the free – CHO group.MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 2

Question 4.
The melting points and solubility in water of amino acids are generally higher than that of the corresponding haloacids. Explain.
Answer:
The amino acids exist as Zwitter ionsMP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 3Due to this dipolar salt like character they have strong dipole – dipole attractions or electrostatic attractions. There fore, their melting points are higher than haloacids which do not have salt like character. Further, due to salt like character, they interact strongly with H2O. As a result, solubility in water of amino acids is higher than that of the corresponding haloacids which do not have salt like character.

Question 5.
Where does the water present in the egg go after boiling the egg ?
Answer:
About three-fourth of an egg is plain water in which the albumin protein and fats are suspended. At room temperature, the protein strands are tightly folded in a complex three-dimensional shape. Because the individual proteins are able to move freely, both the egg white and yolk remain liquid. But when egg is boiled, the protein strands straighten out.

The protein ends, normally protected within the folds, become exposed and the open ends join together in bridge-like bonds forming a giant structure. These new bonds prevent the molecules to move thereby making the egg material solid and the water seems to be disappeared.

Question 6.
Why cannot vitamin C be stored in our body ?
Answer:
Vitamin C is soluble in water. It cannot be stored in our body because it is easily excreted in the urine.

Question 7.
What products would be formed when a nucleotide from DNA containing thiamine is hydrolysed?
Answer:
D – 2 – deoxyribose (sugar), thymine (nitrogenous base) and phosphoric acid are formed when a nucleotide from DNA containing thymine is hydrolysed.

Question 8.
When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Answer:
When RNA is hydrolysed, there is no relationship between the quantities of four bases, i.e., cytosine (C), guanine (G), adenine (A) and uracil (U) are obtained, therefore, the base-pairing principle, i.e., (A) pairs with (T) and (C) pairs with (G) is not followed (as in DNA). This suggest that unlike DNA (a double-stranded structure), the RNA has a single-stranded molecule.

Biomolecules Ncert Textbook Exercises

Question 1.
What are monosaccharides?
Answer:
Monosaccharides are carbohydrates which cannot be hydrolysed further to give simpler units of polyhydroxy aldehydes or ketones.

Question 2.
What are reducing sugars?
Answer:
Carbohydrates which reduce Fehling’s solution to red p.pt of Cu2O or Tollen’s reagent to shining metallic Ag are called reducing sugars. All monosaccharides (both aldoses and ketoses) and disaccharides except sucrose are reducing sugars.

Question 3.
Write two main functions of carbohydrates in plants.
Answer:

  • Cellulose is a predominant constituent of cell wall of plant cells.
  • Starch is the main food storage polysaccharide of plants.

Starch is stored in seeds and acts as the reserve food material for the small plant unless it is capable of making its own food by photosynthesis.

Question 4.
Classify the following into monosaccharides and disaccharides: Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Answer:
Monosaccharide: Ribose, 2-deoxyribose, galactose and fructose.
Disaccharide: Maltose and lactose.

MP Board Solutions

Question 5.
What do you understand by the term glycosidic linkage?
Answer:
The oxygen (ethereal) linkage through which two monosaccharide units are joined together by the loss of a water molecule to form a molecule of a disaccharide is called glycosidic linkage. For example, sucrose (a disaccharide) is composed of C1 of α – glucose and C2 of β – fructose through the glycosidic linkage.
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 3

Question 6.
What is glycogen? How is it different from starch?
Answer:
Glycogen-like starch is a condensation polymer of α – D glucose. Just as glucose is stored in plants in the form of starch it is stored in the form of glycogen in human beings. It is present in liver, muscles and brain. When body needs glucose, enzymes breakdown glycogen to provide glucose. It is also present in yeast and fungi.

The main difference between glycogen and starch is that starch is made up of amylose and amylopectin whereas glycogen consists of a structure almost same as amylopectin with large number of branching, (amylose has a linear structure and amylopectin has a branched structure with branching after every 25-30 molecules). Glycogen has branching after every 10-15 molecules.

Question 7.
What are the hydrolysis products of:
1. Sucrose
2. Lactose
Answer:
1. Glucose and fructose:
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 5

2. D-Glucose and D-Galactose:
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 6

Question 8.
What is the basic structural difference between starch and cellulose ?
Answer:
Starch consists of two components :
amylose and amylopectin. Amyiose is a long linear polymer of 200-1000 α-D-(-) glucose units held by C1 – C4 glycosidic linkage. It is soluble in water. Amylopectin is a branched chain polymer of α-D-(+) glucose linkage whereas branching occurs by C1– C6 glycosidic linkage. It is insoluble in water. On the other hand cellulose is a straight chain polysaccharide composed only of β – D- (+) glucose units which are formed by glycosidic linkage between C1 of one glucose unit and C4 of next glucose unit.

Question 9.
What happens when D – glucose is treated with the following reagents:
(i) HI
(ii) Bromine water
(iii) HNO3?
Answer:
(i) When glucose is treated with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in straight chain.

MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 33

(ii) On heating glucose with bromine water, it gets oxidized to six carbon carboxylic acid, gluconic acid.
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 8

(iii) Glucose on treatment with nitric acid gives a dicarboxylic acid, saccharic acid.
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 9

Question 10.
Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
Answer:
Following reactions could not be explained by open chain structure of D – glucose:

  1. The penta-acetate of glucose does not react with hydroxylamine indicating the absence of free – CHO group.
  2. Despite having the aldehyde group
    • Glucose does not form the hydrogen sulphide addition product with NaHSO3.
    • Glucose does not give Schifts test.

Question 11.
What are essential and non-essential amino acids ? Give two examples of each type.
Answer:
(a) Essential amino acids:
The amino acids which our body cannot make and must be obtained through diet.
Examples:
Valine, Isoleucine, Arginine, Lysine, Threonine etc.

(b) Non-essential amino acids:
These are the amino acids which our body can make.
Examples:
Glycine, Alanine, Glutamic acid, Aspartic acid, Glutamine, Serine etc.

MP Board Solutions

Question 12.
Define the following as related to proteins :
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation.
Answer:
(i) Peptide linkage:
A peptide bond is an amide linkage formed between – COOH group of one a-amino acid and NH2 group of the other a-amino acid by loss of a molecule of water. It units two amino acids unit in a peptide bond (molecule).
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 10

(ii) Proteins are the polymers of a-amino acids. These polymers (also known as polypeptide) consist of amino acids linked with each other in a specific sequence. This sequence of amino acids is known as the primary structure of proteins. Any change in this sequence of amino acids (i.e., primary structure) creates a different proteins.

(iii) A process that changes the physical and biological properties of proteins without affecting the chemical composition of proteins is called denaturation. The denaturation is caused by certain physical and chemical treatment such as changes in pH, temperature, presence of some salts or certain chemical agents.

Question 13.
What are the common types of secondary structure of proteins ?
Answer:
The conformation of polypeptide chain assumed as a result of hydrogen bonding is called secondary structure of proteins. The two types of secondary structures are a-helix and β – pleated sheet structure. (For detail refer your NCERT text-book.)

Question 14.
What type of bonding helps in stabilizing the a-helix structure of proteins ?
Answer:
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 32
(Refer your NCERT text-book for details and structure.)

Question 15.
Differentiate between globular and fibrous proteins.
Answer:
Difference between globular and fibrous protein:
Globular protein:

  • They have foilded ball like structure.
  • They have three-dimensional structure
  • They are soluble in water and aq. solution of salt and base.
  • These proteins are inactive towards temperature and pH value.

Fibrous protein:

  • These molecule have long threads like structure.
  • They have sheet like structure.
  • These are insoluble in water.
  • Fibrous protein are active towards temperature and pH value.

Question 16.
How do you explain the amphoteric behaviour of amino acids ?
Answer:
Due to dipolar or Zwitter’s ion structure, amino acids are amphoteric in nature. The acidic character of the amino acids is due to the – NH3 group and the basic character is due to the – COOH group as shown below:
Acidic character:
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 11

Basic character:
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 12

Question 17.
What are enzymes ?
Answer:
The enzymes are globular proteins which act as bio-catalyst. They are very specific and efficient in their action. Almost all the reactions that occur in the body are catalyzed by enzymes.

Question 18.
What is the effect of denaturation on the structure of proteins ?
Answer:
During denaturation, 2° and 3° structures of proteins are destroyed but 1° structure remains intact. As a result of denaturation, the globular proteins (soluble in H2O) are converted into fibrous proteins (insoluble in H2O)) and their biological activity is lost. The coagulation of egg white on boiling is a common example of denaturation.

MP Board Solutions

Question 19.
How are vitamins classified ? Name the vitamin responsible for the coagulation of blood.
Answer:
On the basis of solubility in water or fat, the vitamins are generally classified into two types:

  • Water soluble vitamins:
    These include vitamin B-complex. (B1 B2, B3, B4, B6, B12 and nicotinic acid etc.) and vitamin C etc.
  • Fat soluble vitamins:
    These include vitamins A, D, E and K. Liver cells are rich in fat soluble vitamins. Vitamin K is responsible for coagulation of blood.

Question 20.
Why are vitamin-A and vitamin-C essential to us ? Give their important sources.
Answer:
Vitamin – A:
It is essential for us because its deficiency causes Night blindness and Xerophthalmia (Hardening of cornea of eye).

Sources:
Carrots, milk, butter, fish liver oil, egg yolk, green and yellow vegetables.

Vitamin – C:
It is essential for us because its deficiency causes Scurvy (bleeding gums), Tooth decay’s and Pyorrhoea etc.

Sources:
Lemon, orange (citrous fruits), amla, tomatoes, potatoes and green leafy vegetables.

Question 21.
What are nucleic acids ? Mention their two important functions.
Answer:
Nucleic acids are long chain polymers of nucleotides. They are also called poly-nucleotides. Nucleic acids are mainly of two types, the deoxyribonucleic acid (DNA) and ribonucleic acid (RNA).

Functions:
(i) DNA is responsible for transmission of hereditary effects from one generation to another. This is due to unique property of replication, during cell division and two identical DNA strands are transferred to the daughter cells.

(ii) DNA and RNA are responsible for synthesis of all proteins needed for the growth and maintenance of our body. Actually, the proteins are synthesized by various RNA molecules (m-RNA and t-RNA etc.) in the cell but the message for the synthesis of a particular protein is present in DNA.

Question 22.
What is the difference between a nucleoside and a nucleotide ?
Answer:
A nucleoside contains only basic component of nucleic acids namely a pentose sugar and a nitrogenous base. A nucleotide contains all the three basic components of nucleic acids namely a phosphoric acid group, a pentose sugar and a nitrogenous base.
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 15
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 16

Question 23.
The two strands in DNA are not identical but are complementary. Explain.
Answer:
The two strands in DNA molecule are held together by hydrogen bonds beetween purine base of one strand and pyrimidine base of the other and vice-versa. Because of different sizes and geometries of the bases, the only possible pairing in DNA are G (guanine) and C (cytosine) through three H-bonds i.e., (C = G) and between A (adenine) and T (thiamine) through two H-bonds i.e., (A=T) (for figure refer your NCERT text book). Due to this base pairing principle, the sequence of bases in one strand automatically fixes the sequence of bases in the other strand. Thus, the two strands are complimentary and are not-identical.

MP Board Solutions

Question 24.
Write the important structural and functional differences between DNA and RNA.
Answer:
Differences between DNA and RNA:
DNA:

  • Occurs mainly in the nucleus of the cell.
  • It contains the sugar deoxyribose.
  •  Does not contain nitrogenous base, uracil.
  • It has a double strand helix
  • It is responsible for the transmission of heredity character.
  • Alkaline hydrolysis is quite slow.
  • Ratio A/T = 1 and G/C = 1.

RNA:

  • Occurs in the cytoplasm of the cell.
  • It contains the sugar ribose.
  • Does not contain nitrogenous base thymine.
  • It has double as well as single strand helix.
  • It helps in protein biosynthesis.
  • Alkaline hydrolysis takes place readily.
  • Such ratio is not present.

Question 25.
What are the different types of RNA found in the cell ?
Answer:
There are three types of RNAs:

  • Ribosomal RNA (r-RNA)
  • Messenger RNA (m-RNA)
  • Transfer RNA (t-RNA).

MP Board Solutions

Biomolecules Other Important Exercises

Biomolecules Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Which protein transports oxygen in blood flow:
(a) Haemoglobin
(b) Insulin
(c) Albumin
(d) Myoglobin.
Answer:
(d) Myoglobin.

Question 2.
Beri-beri diseases is caused due to deficiency of which vitamin:
(a) Vitamin – A
(b) Vitamin – C
(c) Vitamin – B2
(d) Vitamin – D
Answer:
(c) Vitamin – B2

Question 3.
Enzyme which enhances the conversion of glucose to ethanol is:
(a) Zymase
(b) Invertase
(c) Maltase
(d) Diastase.
Answer:
(a) Zymase

Question 4.
In human body carbohydrate is stored:
(a) In the form of glucose
(b) In the form of glycogen
(c) In the form of starch
(d) In the form of fructose.
Answer:
(b) In the form of glycogen

Question 5.
Change in optical rotation of a freshly prepared solution of sugar after some time is called:
(a) Optical activity
(b) Inversion
(c) Specific rotation
(d) Mutation.
Answer:
(b) Inversion

Question 6.
Formula of most familiar disachharide is:
(a) C10H18O9
(b) C10H20O10
(c)C18H22O11
(d) C12H22O11.
Answer:
(d) C12H22O11.

MP Board Solutions

Question 7.
The following statement is false in relation to Ribose:
(a) It is a polyhydroxy compound
(b) It is a aldehydic sugar
(c) It contain 6 carbon atoms
(d) It has optical rotation.
Answer:
(c) It contain 6 carbon atoms

Question 8.
In the formation of carbohydrate are required:
(a) 2 carbon
(b) 3 carbon
(c) 4 carbon
(d) 6 carbon.
Answer:
(d) 6 carbon.

Question 9.
Haemoglobin is:
(a) Enzyme
(b) Globular protein
(c) Vitamin
(d) Carbohydrate.
Answer:
(d) Carbohydrate.

Question 10.
Which carbohydrate is an essential constituent of plant cells:
(a) Cellulose
(b) Starch
(c) Sucrose
(d) Vitamin.
Answer:
(b) Starch

Question 11.
How many subunits are present in haemoglobin:
(a) 2
(b) 3
(c) 4
(d) 5.
Answer:
(b) 3

Question 12.
Starch is a polymer of:
(a) Glucose
(b) Sucrose
(c) Both (a) and (b)
(d) None of these.
Answer:
(a) Glucose

MP Board Solutions

Question 13.
Which sugar is present in maximum amount in human blood:
(a) d-fructose
(b) d-glucose
(c) Sucrose
(d) Lactose.
Answer:
(b) d-glucose

Question 14.
Element present in Vitamin B12 is:
(a) Pb
(b) Zn
(c) Fe
(d) CO.
Answer:
(d) CO.

Question 15.
Amount of glucose in blood is determined by:
(a) Tollen’s reagent
(b) Benedict solution
(c) Alkaline iodine solution
(d) Bromine water.
Answer:
(b) Benedict solution

Question 16.
Deficiency of which Vitamin causes rickets:
(a) Vitamin C
(b) Vitamin B
(c) Vitamin A
(d) Vitamin D.
Answer:
(d) Vitamin D.

Question 17.
In metabolic processes which of the following provide maximum energy :
(a) Protein
(b) Vitamin
(c) Thiamine
(d) Pyrimidine.
Answer:
(d) Pyrimidine.

Question 18.
Vitamin B1 is:
(a) Riboflavin
(b) Cobaltamine
(c) Thiamine
(d) Pyrimidine.
Answer:
(a) Riboflavin

MP Board Solutions

Question 19.
Deficiency of Vitamin C leads to:
(a) Scurvy
(b) Rickets
(c) Pyorrhoea
(d) Anaemia.
Answer:
(a) Scurvy

Question 20.
Most effective energy reservoir in a living cells is:
(a) A.M.P.
(b) A.T.P.
(c) A.D.P.
(d) U.D.P.
Answer:
(b) A.T.P.

Question 21.
Disaccharide present in milk is:
(a) Sucrose
(b) Lactose
(c) Maltose
(d) Cellulose.
Answer:
(b) Lactose

Question 22.
Which is not glyceroid:
(a) Fat
(b) Oil
(c) Phospholipid
(d) Soap.
Answer:
(d) Soap.

Question 23.
Which is not found in R.N.A.:
(a) Thymine
(b) Uracil
(c) Adenine
(d) Guanine.
Answer:
(a) Thymine

Question 24.
Enzymes are:
(a) Nitrogen containing complex compound
(b) Carbohydrate
(c) Co-ordination compounds
(d) Metallic compound.
Answer:
(a) Nitrogen containing complex compound

Question 25.
Chemical name of vitamin C is:
(a) Cyano cobaltamine
(b) Ascorbic acid
(c) Tocopherol
(d) Biotin.
Answer:
(b) Ascorbic acid

MP Board Solutions

Question 2.
Fill in the blanks:

  1. By the oxidation of glucose …………… molecules of ATP are produced.
  2. The breaking of complex molecules in organisms is known as ……………
  3. In hyperglycaemia the amount of …………… in blood increases.
  4. Deficiency of …………… leads to eye disease.
  5. Deficiency of iodine leads to …………… disease.
  6. Blood …………… the temperature of the entire body.
  7. …………… hormone balances the amount of sugar in blood.
  8. …………… is responsible for the clotting of blood.
  9. Denaturation does not affect …………… structure of protein.
  10. Protein is a polymer of ……………
  11. …………… is the basic unit of protein.
  12. …………… is not present in D.N.A.
  13. Haemoglobin is a …………… compound of iron.
  14. Oil and fats obtained from organisms are known as ……………

Answers:

  1. 38
  2. Catabolism
  3. Sugar
  4. Vitamin A
  5. Goitre
  6. Balance
  7. Insulin
  8. Vitamin K
  9. Primary
  10. Amino acids
  11. Amino acid
  12. Uracil
  13. Complex
  14. Lipid.

MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 19

Question 3.
Match the following:
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 34
Answer:

  1. (f) Vitamin A
  2. (c) Vitamin D
  3. (b) Vitamin E
  4. (e) Vitamin B
  5. (d) Vitamin B12
  6. (a) Biotin
  7. (g) Lactose.

Question 4.
Answer in one word / sentence:

  1. Write the chemical name of Vitamin C.
  2. Tell the source of Vitamin K.
  3. Is responsible for clotting of blood ?
  4. Which bond links amino acids together ?
  5. How many amino acids are synthesized by human body ?
  6. Cellulose is a linear polymer of which glucose ?
  7. In RNA molecule which pyrimidine is present in place of Thymine ?
  8. Lactose on hydrolysis gives.
  9. Glucose contains Pyranose ring whereas Fructose contain.
  10. In polysaccharides, monosaccharide units are linked to each other by which bond ?
  11. Which protein helpful for clotting of blood known as ?
  12. Write one example of Monosaccharide Carbohydrate.
  13. What is the name of Disaccharides sugar present in milk ?

Answers:

  1. Ascorbic acid
  2. Green leafy vegetables
  3. Vitamin K (Phylloquinone)
  4. Peptide bond
  5. Ten
  6. 2-glucose
  7. Uracil
  8. Glucose and Lactose
  9. Furanose ring
  10. Glycosidic
  11. Fibrinogen
  12. Glucose or Fructose
  13. Lactose.

MP Board Solutions

Biomolecules Short Answer Type Questions

Question 1.
What is the effect of deficiency of proteins on our body ?
Answer:
Deficiency of protein causes following diseases:

  • Anaemia:
    The deficiency of haemoglobin protein in the body creates lack of enough blood which causes paleness in body.
  • Kwashiorkar:
    This disease is mainly found in children, in this body of patient swells up and deforms.

Question 2.
Define carbohydrates.
Answer:
Substances which are polyhydroxy aldehydes or polyhydroxy ketones or which give these compounds on hydrolysis are called carbohydrates.

Question 3.
Sucrose on enzymatic hydrolysis gives one mole of glucose and one mole of fructose.

  1. Which enzyme catalyses this hydrolysis?
  2. Write down the monosaccharides present in maltose and lactose.
  3. Which is the sweetest sugar, glucose, or fructose?

Answer:

  1. Invertase
  2. Maltose contains two units of glucose while lactose is glucose + galactose.
  3. Fructose

Question 4.
Write the diseases caused by the following Vitamins:
(a) Vitamin-A, (b) Vitamin-B, (c) Vitamin-D, (d) Vitamin-E.
Or,
Write the functions of the following vitamins:
(a) Vitamin-A, (b) Vitamin-D, (c) Vitamin-E, (d) Vitamin-K.
Answer:
Diseases caused by the above Vitamins:
(a) Vitamin-A: Night blindness
(b) Vitamin-B: Beri-beri, loss of appetite and vigour
(c) Vitamin-D: Rickets, osteomalacia
(d) Vitamin-E: Loss of reproductive ability.

Functions caused by the above Vitamins:
(a) Vitamin-A: For vision and growth develops resistance against diseases.
(b) Vitamin-D: For bones, control of metabolism of calcium and phosphorus.
(c) Vitamin-E: Virility in man and reproduction.
(d) Vitamin-K: Coagulation of blood.

MP Board Solutions

Question 5.
Write a note on the structure of DNA.
Answer:
The three-dimensional structure of DNA was elucidated by James Watson and Francis Crick. They proposed that the DNA chain consists of two strands of polynucleotides coiled around each other in the form of the double helix.

The nucleic acid backbone consists of alternating sugar-phosphate residues. One of the four bases is attached to each of the sugar residues in the backbone. The bases are held together by hydrogen bonding between them. The hydrogen bonding is very specific as the structure of bases is such that they permit only one mode of pairing with other bases.

The pairing is A – T, C – G. i.e., adenine pairs only with thymine, cytosine pairs only with guanine. Two strands of DNA are complementary to each other in the sense that the sequence of bases in one strand automatically decides that of the other.

Question 6.
What are enzymes? Give applications.
Answer:
Enzymes are complicated organic molecules which are produced by living cells and are also called biological catalysts. They are complex nitrogenous molecules, unlike inorganic catalysts, they take part in the reaction and get destroyed.
Example:
Enzyme invertase catalyse the hydrolysis of sucrose.
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 17

Factors affecting the activity of catalyst:

  • Temperature
  • pH
  • Concentration of enzyme
  • Concentration of substance
  • Concentration of the product formed.

Functions of enzymes:
Enzymes play a very important role in many biological reactions. The names of some important enzymes and the biochemical reactions catalysed by them are given below:

1. Role of enzymes in the process of digestion:
Enzymes play an important role in digestion resulting in the breakdown of macromolecules into simple molecules.

2. Industrial applications:
Enzymes play an important role in many industrial processes such as the manufacture of ethanol from sugar or starch, the manufacture of acetic acid, the manufacture of cheese, leather, tanning, etc.

Question 7.
What are monosaccharides? Explain with examples.
Answer:
These are the simplest carbohydrates which cannot be hydrolysed into small molecules. Their general formula is CnH2nOn (there are some exceptions also). Here n = 2 to 10. These are sweet in taste and soluble in water. These contain either one – CHO group or one > C = O group.

Aldopentose – Ribose, Zylose etc. (C5H10O5)
Aldohexose – Glucose, Galactose, Mannose etc. (C6H12O6)
Ketohexose – Fructose, Sorbose etc. (C6H12O6).

MP Board Solutions

Protein Molecular Weight Calculator makes it easy for you to determine the theoretical yield value of the chemical reaction in fraction of seconds.

Question 8.
What are proteins?
Answer:
The word protein is derived from the greek word protious (Protious = to take the first) i.e. first or very important. Proteins are high molecular mass nitrogen containing complex organic compounds found in the protoplasm of all animal and plants.
Chemically protein is the condensation polymer of a-amino acid.

Question 9.
What are carbohydrates? Which unit of carbohydrate provide energy to human body?
Answer:
Carbohydrates are compound of carbon, hydrogen and oxygen. General formula of carbohydrate is (CxH2O)y where x and y are integers. These compounds have ratio of hydrogen and oxygen 2: 1 like water (H2O). Therefore the name of these compounds is given carbohydrates. Examples of carbohydrates are glucose (C6H12O6), fructose (C6H12O6), sucrose (C12H22O11) etc.

Glucose is the unit of carbohydrates which is responsible to provide energy. Glucose decomposes slowly with the help of oxydase enzyme present in the body into CO2 and water. Energy is released in this process.
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 18

Question 10.
Fibrous protein present in hair is elastic while in silk it is not elastic. Explain.
Answer:
Protein present in hair has α – helical structure. If it is stretched, the hydrogen bonder broken, thereby increasing the length of the helix. On releasing the tension, the hydrogen bonus are formed again, giving back the original shape. So proteins having α – helix structure are elastic. Proteins in silk have β – pleated structure and they are not elastic because stretching leads to pulling of the peptide (covalent) bonds.

Question 11.
Explain the formation of amide and peptide bonds in the structure of proteins.
Answer:
(1) Amide bond:
Protein is a complicated organic compound which is formed by the combination of different amino acids. When the – CONH – link resulting from the reaction between a compound containing a carboxylic group (- COOH) and that with an amine group (- NH2) it forms an amide bond.
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 19

(2) Peptide linkage:
When the – CONH – like resulting from the reaction between two molecules of a-amino acids, is called peptide bond. Such a bond involves only the alpha-amino and a-carboxylic group. Thus, two molecules of a-amino acid on reaction yield a peptide bond.
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 20

Question 12.
Define polysaccharides. Give examples also.
Answer:
Polysaccharides are natural isomers which have a molecular weight of up to many lacks, General formula of polysaccharides is (C6H10O5)where the value of n is from 12 to thousands. These are complex material which is formed by condensations of monosaccharides. These compounds contain glycosidic bonds.
Example: Starch, cellulose, etc.

MP Board Solutions

Question 13.
What is invert sugar?
Answer:
Cane – sugar is dextrorotatory [D or +] which gives an equimolar mixture of monosaccharides. This mixture is laevorotatory [D or-]. Hence the mixture of glucose and fructose obtained as a result of hydrolysis of cane-sugar is called invert sugar.
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 22

Question 14.
What are disaccharides? Write the general formula of disaccharides.
Answer:
Disaccharides are sugar which is formed by the combination of two monosaccharides by the removal of one molecule of water. Both monosaccharides are of hexose type in which one unit is glucose. Thus, disaccharides are of aldose-aldose or aldose-ketose type. The general formula of disaccharides is C12H22O11
Example: Sucrose, maltose, lactose, etc.

Question 15.
Give the pyranose structure of sucrose and maltose.
Answer:
In sucrose two monosaccharides (glucose and fructose) are combined through a glycosidic bond which is present between C1 of α-glucose and C2 of β-fructose.
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 23
Whereas maltose molecule is obtained by the condensation of two glucose molecules.
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 24

Question 16.
Clarify the structure of protein.
Answer:
Molecule of protein is made up of amino acids. Basically protein molecules are linear polymers of amino acids. Their entire structure can be expressed in four steps:

  1. Primary structure:
    It provides knowledge of mutual combination of various amino acids in a polypeptide chain.
  2. Secondary structure:
    It provides knowledge of conformations of peptide chains of protein.
  3. Tertiary structure:
    It explains how protein molecule coils to obtain a specific shape.
  4. Quarternary structure:
    By this, arrangement of two polypeptide chains with respect to each other is known.

Question 17.
What do you understand by nucleoside and nucleotide ?
Answer:
Nucleoside: When a purine or pyrimidine base joins with a pentose sugar molecule, then it is known as nucleoside.
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 25

Question 18.
What are carbohydrates ? Explain mono, di and polysaccharides with example.
Answer:
Carbohydrates are polyhydric aldehydes or polyhydric ketones or substances which on hydrolysis produce such compounds. The carbohydrates are classified into three major classes on the basis of hydrolysis.

1. Monosaccharides:
These are the simples carbohydrates. They cannot be hydrolysed further to give more simple carbohydrates. Monosaccharides have the general formula (CH2O)n These are crystalline solids, soluble in water and are sweet in taste. Example: Glucose, Fructose.

2. Disaccharides:
Disaccharides are sugar which are formed by combination of two monosaccharides by removal of one molecule of water. Both monosaccharides are of hexose type – in which one unit is glucose. General formula of disaccharides is C12H22O11.
Example: Sucrose, Maltose etc.

3. Polysaccharides:
General formula of polysaccharides is (C6H10O5)n where value of n is from 12 to thousands. These are complex material which are formed by condensation of monosaccharides. Example: Starch, Cellulose etc.

MP Board Solutions

Question 19.
What happens when protein is denatured ?
Answer:
Denaturation of Protein:
Disruption of tertiary structure of protein is called denaturation. These reactions take place by heating in presence of acids or highly concentrated salts or heavy metals. Denaturation does not affect the primary structure of protein. Denaturation takes place in the rearrangement of secondary and tertiary structures. As a result of this, protein losses its biological actvity.

During denaturation the protein molecule uncoils from an ordered and specific conformation into a more disordered conformation. Denaturation takes place when proteins are heated or treated with acids, bases, alcohols, KI, urea, acetone etc. or when exposed to UV or X-ray radiations.

Question 20.
Write functions of vitamin B and name two diseases caused due to its deficiency.
Answer:
Vitamin B include a group of vitamins which are as follows:
‘Vitamin-B1 (Thiamine), Vitamin – B2 (Riboflavin), Vitamin B3 (Pentothenic acid), Vitamin-B6 (Pyridoxine), Vitamin B12 (Cyano cobalamin)

Vitamin B1 (Thiamine)- Source: husk of rice, green vegetables, egg.
Function: Maintenance of activity of nervous system.

Deficiency disease:
(i) Beri-Beri (swelling in hands and legs).
(ii) Gastric (disorder in digestion).

Vitamin B2 (Riboflavin)- Source: egg yolk, milk etc.
Function: Growth of body.
Deficiency disease: Dermatitis, weakness in eye sight.

Question 21.
What are vitamins ? Name the vitamins whose deficiency lead to following diseases:

  • Haemorrhage (No clotting of blood)
  • Night blindness
  • Anaemia
  • Ricket
  • Pyrohea
  • Loss of reproductive ability
  • Convulsions.

Answer:
Vitamins are complex organic molecules which act as essential nutrients for body. Though these vitamins are not formed in our body but deficiency of these vitamins causes various diseases in our body. The vitamins are classified according to solubility.

  • Fat soluble vitamins: Vitamin A, D, E and K.
  • Water soluble vitamins: Vitamin C and vitamin B complex.

Name the vitamins deficiency lead to following diseases :

  • Haemorrhage (No clotting of blood) : Vitamin K (Phylloquinone)
  • Night blindness: Vitamin A (Retinol)
  • Anaemia: Vitamin B12 (Cyanocobalamin)
  • Ricket: Vitamin D (Calciferol)
  • Pyorrhoea: Vitamin C (Ascorbic acid)
  • Loss of reproductive ability: Vitamin E (a -Tocopherol)
  • Convulsions: Vitamin B6 (Pyridoxine).

MP Board Solutions

Question 22.
Write about the following:

  1. Testosterone
  2. Thyroxine
  3. Insulin
  4. Cortisone.

Answer:

  1. Testosterone:
    Testosterone hormone is secreted by the gland testes and its function is to control the reproductive organs in males.
  2. Thyroxine:
    This hormone is secreted by the thyroid gland. Its function is to control metabolic activity and growth.
  3. Insulin:
    Insulin hormone is secreted by pancreas and its function is to control the amount of glucose in the blood.
  4. Cortisone:
    Cortisone is secreted by adrenal cortex and its function is to control the metabolism of protein and water.

Question 23.
Write the name of two-two diseases caused due to deficiency of retinol, thiamine, ascorbic acid and riboflavin.
Answer:
Retinol: Night blindness, Xerophthalmia
Thiamine : Beri-Beri, growth of body stops
Ascorbic acid: Scurvy, Tooth loss
Riboflavin: Weak eye sight, dermititis.

Question 24.
Write two-two source of vitamin A and C. Write one-one disease caused due to its deficiency.
Answer:

  • Vitamin Source Disease
  • Vitamin-A Milk, butter, cheese Night blindness
  • Vitamin-C Lemon, Tomato Scurvy.

Question 25.
Write the functions of vitamin A, D, E and K.
Answer:

  • Vitamin-A : It participate in the formation of visible pigment named Rhodopsin and Iodopsin.
  • Vitamin-D : It is useful in the formation of bones.
  • Vitamin-E : It prevents the breaking of RBCs.
  • Vitamin-K : It is helpful in the clotting of blood.

Question 26.
Write functions and sources of the following bio-molecules/elements :

  1. Protein
  2. Carbohydrates
  3. Fat
  4. Calcium.

Answer:

  1. Protein:
    Formation of new tissues, and their repairing with the body
    Source: Milk, egg, meat, cheese, fish.
  2. Carbohydrates: Carbohydrates provides energy to the body.
    Source: Rice, potato, fruits, cane sugar etc.
  3. Fats: They provides energy to the body.
    Source: Ghee, oils, milk, egg. etc.
  4. Calcium: Increase the bones and teeth.
    Source: Green leafy vegetables, milk.

MP Board Solutions

Question 27.
Write two Differences between a – Amino acid and Protein.
Answer:
Differences between a – Amino acid and Protein:
α – Amino acid:

  • They are simple compounds having amino and Carboxylic acid group.
  • On combining amino acid gives dipeptide,polypeptide and then protein e.g. glucose, lysine etc.

Protein:

  • Proteins are complex nitrogenous compounds.
  • Protein on hydrolysis gives amino acid. e.g. Haemoglobin, casein etc.

Question 28.
Explain in brief:
(a) Name of two enzymes and their function.
(b) Name of two water soluble enzymes and disease caused due to their deficiency.
Answer:
(a) Name of enzyme and function:
1. Amylase (ptyalin): It converts starch into glucose:
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 26

2. Pepsin: It converts protein into amino acid:
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 27

(b) Water soluble vitamin:
1. Vitamin B1: Thiamine
Deficiency disease: Beri-Beri

2. Vitamin C: Ascorbic acid
Deficiency disease: Scurvy Other examples are vitamin B2, B6, B12 and K.

Biomolecules Long Answer Type Questions

Question 1.
Write a note on Nucleic acid.
Answer:
Nucleic acid:
It is found in nucleus of the cell. It has large amount of phosphorus. Nucleic acid are poly nucleotides which is formed by the combination of various nucle otide units. Each nucleotide is formed by three chemical components:

  1. Phosphate group
  2. Pentose ribose sugar or De – oxyribose
  3. Heterocyclic base: Like derivative of pyrimidine (Thiamine, uracil, cytosine) and derivatives of purine (Adinine and Guanine).

Nucleic acid are of two types:
(A) DNA: De-oxy ribonucleic acid.
(B) RNA: Ribonucleic acid.

Components of DNA:

  • De-oxyribose sugar molecule
  • Phosphoric acid molecule
  • Nitrogenous base:

It is of two types:

  1. Pyrimidine base: It includes cytosine (C) and Thymine (T)
  2. Purine base: It includes Adenine (A) and Guanine (G)

Components of RNA:
RNA contains Ribose and nitrogen base like Adenine (A), Guanine (G), Uracil (U) and Cytosine (C).

MP Board Solutions

Question 2.
What are carbohydrates ? Write its classification and four main functions.
Answer:
Definition:
Optically active polyhydroxy aldehydes or ketones or substances which yield these on hydrolysis are known as carbohydrates.
Example: Glucose, starch, cellulose, sucrose etc.

Classification of Carbohydrate:
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 28

Functions of carbohydrates:

  • It is the main structural component of cell.
  • It acts as a bio-fuel and provides energy to organisms for doing work.

MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 29

  • Carbohydrate is stored in the liver as glycogen, which hydrolysis to provide the energy required.
  • Cellulose is found in grass and plants which provide energy to animals grazing grass because these animals possess specific enzymes which hydrolyses cellulose to glucose.

Question 3.
Give the diseases caused by ascorbic acid, thiamine retinol and nicotinic acid.
Or
Give the source and diseases caused by Vitamin A, B, C and D.
Answer:
Name of Vitamins, its functions and deficiency diseases are given in the ahead chart: Chemical names of Vitamins, their sources, functions and diseases due to deficiency
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 30

Question 4.
Give differences between monosaccharide, disaccharide and polysaccharide.
Answer:
Difference between monosaccharide, disaccharide and polysaccharide:
MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules - 31

MP Board Class 12th Chemistry Solutions

MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 7 Integrals Miscellaneous Exercise Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 1
MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 2
MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 3
MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 4
MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 5

MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 6
MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 7
MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 8
MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 9
MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 10

MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 11
MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 12
MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 13
MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 14
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MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 31
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MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 33
MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 34
MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 35
MP Board Class 12th Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 36

MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.3

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MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.3

MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.3 1
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MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.3 5
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MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.3 11
MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.3 12

MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.6

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MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.6

MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.6 1
MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.6 2
MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.6 3
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MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.6 6
MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.6 7
MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.6 8
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MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.6 11
MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.6 12
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MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.6 26
MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.6 27
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MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.6 29
MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.6 30

MP Board Class 12th Business Studies Important Questions Chapter 5 संगठन

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MP Board Class 12th Business Studies Important Questions Chapter 5 संगठन

संगठन Important Questions

संगठन वस्तुनिष्ठ प्रश्न

प्रश्न 1.
सही विकल्प चुनकर लिखिए –

प्रश्न 1.
निम्नलिखित में कौन-सा अंतरण का तत्व नहीं है –
(a) उत्तरदेयता
(b) अधिकार
(c) उत्तरदायित्व
(d) अनौपचारिक संगठन
उत्तर:
(d) अनौपचारिक संगठन

प्रश्न 2.
कार्य करते हुए अंतःक्रिया में अचानक बना सामाजिक संबंध तंत्र कहलाता है –
(a) औपचारिक संगठन
(b) अनौपचारिक संगठन
(c) विकेन्द्रीकरण
(d) अंतरण।
उत्तर:
(b) अनौपचारिक संगठन

प्रश्न 3.
संगठन को देखा नहीं जा सकता –
(a) प्रबंध के एक कार्य के रूप में
(b) संबंधों के एक ढाँचे के रूप में
(c) एक प्रक्रिया के रूप में
(d) उद्यम वृत्ति के रूप में।
उत्तर:
(d) उद्यम वृत्ति के रूप में।

प्रश्न 4.
निम्नलिखित में से कौन-सा संगठन प्रक्रिया का एक चरण नहीं है –
(a) क्रियाओं का विभाजन
(b) कार्य सौंपना
(c) कार्यों एवं विभागों का निर्माण
(d) नेतृत्व प्रदान करना।
उत्तर:
(d) नेतृत्व प्रदान करना।

प्रश्न 5.
निम्नलिखित में से कौन संदेशवाहन के अधिकारिक मार्ग का प्रयोग नहीं करता –
(a) औपचारिक संगठन
(b) अनौपचारिक संगठन
(c) कार्यात्मक संगठन
(d) प्रभागीय संगठन
उत्तर:
(b) अनौपचारिक संगठन

प्रश्न 6.
किस संगठन में क्रियाओं को उत्पादों के आधार पर समूहों में बाँटा जाता है –
(a) विकेन्द्रित संगठन
(b) प्रभागीय संगठन
(c) कार्यात्मक संगठन
(d) केन्द्रित संगठन।
उत्तर:
(b) प्रभागीय संगठन

प्रश्न 7.
एक लंबा ढाँचा होता है –
(a) प्रबंध की सिकुड़ी हुई शृंखला
(b) प्रबंध की फैली हुई श्रृंखला
(c) प्रबंध की कोई भी श्रृंखला
(d) इनमें से कोई नहीं।
उत्तर:
(a) प्रबंध की सिकुड़ी हुई शृंखला

प्रश्न 8.
उत्पादन रेखा पर आधारित सामूहिक क्रिया अंग है –
(a) अंतरित संगठन का
(b) प्रभागीय संगठन का
(c) कार्यात्मक संगठन का
(d) स्वायत्त शासित संगठन का।
उत्तर:
(b) प्रभागीय संगठन का

प्रश्न 9.
केन्द्रीयकरण से तात्पर्य है –
(a) निर्णय लेने के अधिकारों को सुरक्षित रखना
(b) निर्णय लेने के अधिकारों का बिखराव करना
(c) प्रभागों का लाभ केन्द्र बनाना
(d) नये केन्द्रों अथवा शाखाओं को खोलना।
उत्तर:
(a) निर्णय लेने के अधिकारों को सुरक्षित रखना

प्रश्न10.
प्रबंध के विस्तार से तात्पर्य है –
(a) प्रबंधकों की संख्या में वृद्धि
(b) एक प्रबंधक की नियुक्ति के समय की सीमा जिसके लिये उसे नियुक्ति दी गई है
(c) एक उच्चाधिकारी के अंतर्गत कार्य करने वाले अधीनस्थों की गणना
(d) शीर्ष प्रबंध के सदस्यों की गणना।
उत्तर:
(c) एक उच्चाधिकारी के अंतर्गत कार्य करने वाले अधीनस्थों की गणना

MP Board Solutions

प्रश्न 2.
रिक्त स्थानों की पूर्ति कीजिए

  1. अधिकार अंतरण या भारार्पण का आशय …………. सौंपना है।
  2. विकेन्द्रीकरण में …………. के वितरण पर बल दिया जाता है।
  3. विकेन्द्रीकरण अधिकारियों के कार्यभार में …………. करता है।
  4. कार्यात्मक संगठन संरचना की अवधारणा …………. के मस्तिष्क की उपज है।
  5. बैंक …………. संरचना संगठन का उदाहरण है।
  6. संगठन संरचना से आशय संस्था में …………. संबंधों की व्याख्या करने से है।
  7. संस्था को उसके द्वारा उत्पादित वस्तुओं के आधार पर विभक्त करना ……….. संगठन कहलाता है।
  8.  …………. संगठन में अधिकार तथा कर्तव्यों की स्पष्ट व्याख्या की जाती है।

उत्तर:

  1. अधिकार
  2. सत्ता
  3. कमी
  4. टेलर
  5. भौगोलिक
  6. अधिकार कर्तव्य
  7.  संभागीय
  8. औपचारिक।

प्रश्न 3.
एक शब्द या वाक्य में उत्तर दीजिए –

  1. उच्च अधिकारियों द्वारा अधीनस्थों में अपने कार्यों-अधिकारों का वितरण, सीमित रूप से किया जाना क्या कहलाता है ?
  2. संस्था में शीघ्र निर्णयन को प्रोत्साहित करने हेतु क्या अपनाया जाना चाहिए?
  3. ऐसी संस्था जो विभिन्न प्रकार के उत्पाद बनाती है उसे कौन-सा संगठन अपनाना चाहिए ?
  4. किस स्थिति में अधिकारों का हस्तांतरण होता है परन्तु उत्तरदायित्वों का नहीं?
  5. संगठन का वह स्वरूप जो कर्मचारियों के मध्य पारस्परिक संबंध के कारण स्वतः ही विकसित होता है, उसे क्या कहते हैं ?
  6. उस संगठन का नाम बताइए जो नियमों एवं कार्य विधियों पर आधारित है।
  7. संगठनात्मक ढाँचे में किस प्रकार के संबंध को दिखाया जाता है ?
  8. प्रबंध का संगठन का कार्य प्रबंध के किस कार्य के बाद आता है ?
  9. विभागीय संगठन प्रक्रिया का कौन-सा चरण है ?
  10. अनौपचारिक संगठन में सदस्यों में किस प्रकार का संबंध होता है ?

उत्तर:

  1. भारार्पण
  2. विकेंद्रीयकरण
  3. संभागीय संगठन
  4. भारार्पण
  5. अनौपचारिक संगठन
  6. औपचारिक संगठन
  7. अधिकारी व अधीनस्थ संबंध
  8. नियोजन के बाद
  9. दूसरा चरण
  10. मधुर संबंध।

प्रश्न 4.
सत्य या असत्य बताइये

  1.  संगठन की स्थापना निम्न स्तर के प्रबन्ध द्वारा होती है।
  2. संगठन भ्रष्टाचार को जन्म देता है।
  3. अधिकार का भारार्पण दिया जा सकता है।
  4. जवाबदेही का भारार्पण नहीं किया जा सकता है।
  5. संगठन का प्रबन्ध में वही महत्व है जो मानव शरीर में हड्डियों के ढाँचे का होता है।

उत्तर:

  1. असत्य
  2. असत्य
  3. सत्य
  4. सत्य
  5. सत्य।

प्रश्न 5.
सही जोड़ी बनाइये –
MP Board Class 12th Business Studies Important Questions Chapter 5 संगठन - 1
उत्तर:

  1. (e)
  2. (a)
  3. (b)
  4. (c)
  5. (d)

संगठन लघु उत्तरीय प्रश्न

प्रश्न 1.
संगठन के उद्देश्य बतलाइये।
उत्तर:
संगठन के उद्देश्य –

1. उत्पादन में मितव्ययिता – संगठन का प्रमुख उद्देश्य उत्पादन में मितव्ययिता लाना है अर्थात् न्यूनतम लागत पर अधिकतम उत्पादन करना प्रत्येक उपक्रम का उद्देश्य होता है।

2. समय तथा श्रम में बचत – संगठन में श्रेष्ठ मशीनें व यंत्र तथा श्रेष्ठ प्रणाली अपनाई जाती है जिससे कार्य के समय व श्रम में काफी बचत की जाती है। संगठन द्वारा बड़े पैमाने में उत्पादन के लाभ भी लिए जा सकते हैं।

3. श्रम तथा पूँजी में मधुर संबंध – श्रमिकों तथा प्रबंध के बीच मधुर संबंध स्थापित करना भी संगठन का एक उद्देश्य है, इस हेतु कुशल संगठनकर्ता की नियुक्ति कर श्रम व पूँजी के हितों की रक्षा की जाती है।

4. सेवा भावना – वर्तमान सामाजिक चेतना एवं जन जागरण के कारण प्रत्येक व्यवसायी का उद्देश्य “प्रथम सेवा फिर लाभ” हो गया है वैसे भी प्रत्येक व्यवसायी को समाज सेवा कर अपने उत्तरदायित्व का निर्वहन करना चाहिए।

MP Board Solutions

प्रश्न 2.
संगठन के कोई चार सिद्धान्तों का उल्लेख कीजिए।
उत्तर:
संगठन के सिद्धान्त –

1. विशिष्टीकरण का सिद्धान्त प्रत्येक सदस्य की क्रियाएँ विशेष कार्य को पूरा करने तक ही सीमित होनी चाहिए, अर्थात् एक व्यक्ति को एक कार्य सौंपा जाये तो इससे कुशलता में वृद्धि होगी।

2. एकरूपता का सिद्धान्त – प्रत्येक पद से संबंधित अधिकारों तथा उत्तरदायित्वों में एकरूपता होनी चाहिए, परन्तु एक अधिकारी के अधिकार दूसरे से टकराने नहीं चाहिए, इससे संस्था का अनुशासन ढीला होगा और कार्य कुशलतापूर्वक सम्पन्न नहीं हो सकेगा।

3. अपवाद का सिद्धान्त – इस सिद्धान्त का प्रतिपादन प्रो. टेलर वैज्ञानिक प्रबंध के जन्मदाता ने किया था। इस सिद्धान्त के अनुसार दिन-प्रतिदिन के कार्यों को करने के लिये अधीनस्थों को अधिकार दे दिये जाने चाहिए तथा अपवादपूर्ण एवं महत्वपूर्ण मामलों पर निर्णय करने के कार्य को उच्च अधिकारियों पर छोड़ देना चाहिये।

4. सरलता का सिद्धान्त – संगठन का ढाँचा सरल होना चाहिये ताकि प्रत्येक कार्य के निष्पादन में कमसे-कम समय एवं व्यय लगे। सरलता के अभाव में संदेशों के आदान-प्रदान में भी कठिनाई सामने आती हैं तथा कार्य शीघ्रता व सरलता से नहीं हो पाएगा।

प्रश्न 3.
कार्यात्मक संगठन संरचना के गुण/लाभ समझाइये।
उत्तर:
कार्यात्मक संगठन ढाँचे के गुण / लाभ निम्नलिखित हैं

  1. कार्यात्मक संगठन में व्यावसायिक क्रियाओं का विभाजन तर्क संगत होता है।
  2. प्रत्येक विभाग के प्रबन्धक अपने विभाग के कार्य के लिए उत्तरदायी होते हैं।
  3. कार्यात्मक संगठन द्वारा सर्वोच्च प्रबन्ध का उपक्रम पर नियंत्रण रखना आसान हो जाता है।
  4. इस ढाँचे का आधार श्रम विभाजन है अर्थात् प्रबन्धकों को कार्य उनकी योग्यता एवं सूची के अनुसार दिये जाते हैं।

प्रश्न 4.
संगठन के कोई चार लाभ लिखिए।
उत्तर:
संगठन वह तंत्र है, जिसकी सहायता से प्रबंध व्यवसाय का संचालन, समन्वय तथा नियंत्रण करता है। यह प्रबंध की आधारशिला है। संगठन का महत्व निम्नलिखित बातों से स्पष्ट हो जाता है

1. उपक्रम के विकास में सहायक-श्रेष्ठ संगठन के माध्यम से उपक्रम का विकास तेज गति से होने लगता है, बड़े पैमाने के उत्पादन की सफलता के पीछे श्रेष्ठ संगठन का हाथ होता है।

2. समन्वय स्थापित होना-संगठन की सहायता से विभिन्न विभागों, उपविभागों, विभिन्न व्यक्तियों एवं क्रियाओं में उचित समन्वय एवं सामंजस्य स्थापित होता है। जिससे प्रशासन द्वारा निर्धारित नीति का पूर्ण परिपालन संभव होता है।

3. भ्रष्टाचार पर नियंत्रक-स्वस्थ एवं कुशल संगठन भ्रष्टाचार पर प्रभावी नियंत्रण करने में भी सफल रहता है, जिससे कर्मचारियों की दक्षता बढ़ती है और उनका मनोबल तथा उत्साह बढ़ता रहता है।

4. मनोबल में वृद्धि-कुशल संगठन से कर्मचारियों के मनोबल में वृद्धि होती है। प्रत्येक व्यक्ति अपने अधिकार, दायित्व तथा कर्तव्य से सुपरिचित रहता है तथा वह उपक्रम की नीतियों व उद्देश्यों को भली-भाँति समझ जाता है।

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प्रश्न 5.
श्रेष्ठ संगठन से प्रशासकीय तथा प्रबंधकीय क्षमता में वृद्धि होती है। स्पष्ट कीजिए।
उत्तर:
प्रशासन द्वारा उपक्रम की नीतियों एवं लक्ष्यों का निर्धारण किया जाता है व इनका क्रियान्वयन प्रबंध द्वारा संगठन के सहयोग से किया जाता है। श्रेष्ठ संगठन में सहयोग, समन्वय, कार्यनिष्ठा व अनुशासन की भावना भरी होती है, जिससे संगठन के मानवीय प्रयासों को एक निश्चित दिशा देकर अधिक सार्थक व प्रभावी बनाया जा सकता है। उपक्रम की क्रियाओं एवं उद्देश्यों को सरलता व शीघ्रता से समय पर पूर्ण कराया जा सकता है। इस प्रकार श्रेष्ठ संगठन से प्रशासकीय व प्रबंधकीय क्षमता में वृद्धि होती है।

प्रश्न 6.
रेखा एवं कर्मचारी संगठन क्या है ? इसकी तीन विशेषताएँ लिखिए।
उत्तर:
रेखा एवं कर्मचारी संगठन में काम का विभाजन स्वतंत्र विभागों में किया जाता है तथा उत्तरदायित्व का विभाजन भी लम्बवत् होता है किन्तु कार्यदक्षता तथा सहकारिता को प्रोत्साहन देने के लिए प्रत्येक विभाग में विशेषज्ञ नियुक्त किये जाते हैं, जो परामर्श का कार्य करते हैं।
रेखा व कर्मचारी संगठन की विशेषताएँ

  1. इस संगठन में अधिकार व उत्तरदायित्व ऊपर से नीचे की ओर प्रवाहित होता है।
  2. इस संगठन में योग्य कर्मचारियों को उन्नति के अच्छे अवसर प्राप्त होते हैं।
  3. इसमें सोचने एवं परामर्श देने व करने के कार्य दोनों पृथक्-पृथक् होते हैं।

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प्रश्न 7. रेखा संगठन के दोष लिखिए।
उत्तर:
रेखा संगठन के दोष :

  1. इस संगठन में पक्षपात की आशंका बनी रहती है। इसमें सर्वोच्च अधिकारी अपने अधीनस्थों की नियुक्ति, पदोन्नति आदि में पक्षपात कर सकता है।
  2. आधुनिक उद्योग इतने जटिल व मिश्रित हैं कि एक ही व्यक्ति सारे कार्य दक्षतापूर्वक सम्पन्न नहीं कर सकता है,
  3. विशिष्टीकरण इसमें संभव नहीं है।
  4. दक्ष अधिकारी के बिना विभिन्न विभागों में समन्वय स्थापित करना कठिन हो जाता है।
  5. अनुशासन पर अधिक ध्यान देने के कारण इस प्रणाली में तानाशाही की बुराइयाँ आ जाती हैं।
  6. उच्चाधिकारी के गलत निर्णय लेने पर उपक्रम विफल हो सकता है।

प्रश्न 8.
रेखा संगठन के लाभों को समझाइये।
उत्तर:
रेखा संगठन के प्रमुख लाभ निम्नलिखित हैं

  1. इस संगठन में प्रत्येक सदस्य को अपने उत्तरदायित्व व कर्तव्यों की स्पष्ट जानकारी रहती है।
  2. इस संगठन में निर्णय प्रायः एक व्यक्ति द्वारा लिए जाते हैं, अतः निर्णय सरलतापूर्वक व शीघ्र होते हैं।
  3. विभिन्न विभागों को एक ही व्यक्ति के निर्देशन में कार्य करना होता है, अतः उनमें समन्वय आसानी से किया जा सकता है।
  4. इस संगठन प्रणाली में आवश्यकतानुसार समायोजन भी किया जा सकता है अर्थात् यह लोचपूर्ण है।

प्रश्न 9.
कार्यात्मक संगठन प्रभागीय संगठन से किस प्रकार असमानता रखता है ?
उत्तर:
संगठनात्मक ढाँचा (Organisational structure)- संगठनात्मक ढाँचा संगठन में विभिन्न पदों के बीच अधिकार और उत्तरदायित्व संबंध प्रदर्शित करता है और साथ ही स्पष्ट करता है कि कौन किसको रिपोर्ट करेगा। हर्ले (Hurley) के अनुसार, “संगठन ढाँचे, एक संस्था में विभिन्न पदों एवं उन पदों पर काम करने वाले व्यक्तियों के मध्य संबंधों के स्वरूप होते हैं ।” संगठन ढाँचे को प्रायः संगठन चार्ट पर प्रदर्शित किया जाता है। संगठन ढाँचे के दो रूप हैं-

  1. कार्यात्मक ढाँचा तथा
  2.  प्रभागीय ढाँचा।

प्रश्न 10.
औपचारिक संगठन तथा अनौपचारिक संगठन किस प्रकार आपस में संबंधित हैं?
उत्तर:
अनौपचारिक संगठन और औपचारिक संगठन में संबंध (Relation between formal organization and informal organization)- अनौपचारिक संगठन औपचारिक संगठन का एक अंग होता है। औपचारिक संगठन के अंदर सदस्य अपने कार्यों को एक-दूसरे के सहयोग से पूरा करते हैं। वे एक-दूसरे से बातचीत करते हैं। इससे उनमें मैत्रीपूर्ण संबंध बन जाते हैं। मित्रता के आधार पर सामाजिक समूह का एक संजाल (Network) बन जाता है।

इस संजाल को अनौपचारिक संगठन कहते हैं। इस प्रकार औपचारिक संगठन अनौपचारिक संगठनों को जन्म देते हैं। अनौपचारिक संगठन के अंदर काम करने वाले सदस्यों के बीच सामाजिक संबंधों की एक व्यवस्था होती है। औपचारिक ढाँचे के अंदर ही अनौपचारिक संगठन की उत्पत्ति होती है।

औपचारिक संगठन में अनौपचारिक संगठन की उत्पत्ति के कई आधार होते हैं- जैसे समान रूचि, भाषा, संस्कृति आदि। ये संगठन पूर्व-नियोजित नहीं होते। ये लोगों की आवश्यकताओं और संस्था के वातावरण के अनुसार स्वतः ही बन जाते हैं।

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प्रश्न 11.
केंद्रीयकरण और विकेंद्रीयकरण की अवधारणा अधिकार अंतरण की अवधारणा से संबंधित है। कैसे?
उत्तर:
अधिकारों के केंद्रीयकरण से अभिप्राय उच्च प्रबंध स्तरों पर अधिकारों के संकेद्रण (Centralization) से है। यहाँ अधिकांश निर्णय उच्च स्तर के प्रबंधकों के द्वारा लिए जाते हैं। अधीनस्थ प्रबंधको के द्वारा निर्देशों के अनुसार करते हैं। इसके विपरीत अधिकार विकेंद्रीयकरण से अभिप्राय केवल केंद्रीय बिंदुओं पर ही प्रयोग किये जा सकने वाले अधिकारों को छोड़कर शेष सभी अधिकारों को व्यवस्थित रूप से निम्न स्तरों को सौंपने से है।

केंद्रीयकरण और विकेंद्रीकरण की दोनों अवधारणाएँ अधिकार अंतरण की अवधारणा से संबंधित हैं। जब अधिकार अधीनस्थों को हस्तांतरण नहीं हो जाते और सारे अधिकार उच्च प्रबंध पर संकेद्रित हैं, तब केंद्रीयकरण कहलायेगा और जब अधिकारों का अंतरण अधीनस्थों को किया जाता है तब विकेंद्रीयकरण की स्थिति उत्पन्न होती है।

प्रश्न 12.
अनौपचारिक संगठन औपचारिक संगठन की किस प्रकार सहायता करता है ? स्पष्ट कीजिए।
उत्तर:
अनौपचारिक संगठन का जन्म औपचारिक संगठन से होता है, जब व्यक्ति अधिकारिक तौर पर बतलाई गई भूमिकाओं से परे आपस में मेल-मिलाप से कार्य करते हैं। जब कर्मचारी सगंठन की ओर नहीं धकेला जा सकता बल्कि वे मैत्रीपूर्ण व सहयोगपूर्ण विचारों से एक ग्रुप बनाने की ओर झुकते हैं। यह उनके आपसी हितों की अनुरूपता को प्रकट करता है।

ऐसे समूहों का प्रयोग संगठन की उन्नति तथा सहयोग के लिए किया जा सकता है। ऐसे समूह उपयोगी तथा कर्मचारियों एवं उच्चाधिकारियों के मध्य झगड़े सुलझाने का कार्य अनौपचारिक संप्रेषण के द्वारा भली-भाँति किया जा सकता है। प्रबंधकों को औपचारिक तथा अनौपचारिक दोनों प्रकार के संगठनों का युक्ति पूर्ण उपयोग करना चाहिए ताकि संगठन का कार्य सुगमतापूर्वक चल सके। औपचारिक संगठन को यदि भली-भाँति नियंत्रित किया जाए तो औपचारिक संगठन के द्वारा बनाये गये उद्देश्यों की प्राप्ति में अनौपचारिक संगठन अत्यंत उपयोगी है।

प्रश्न 13.
केंद्रीयकरण एवं विकेंद्रीयकरण में अंतर्भेद कीजिए।
उत्तर:
बहुत से संगठनों में सभी निर्णयों को लेने में शीर्ष प्रबंध की मुख्य भूमिका होती है जबकि अन्य संगठनों में यह अधिकार प्रबंध के निम्नतम स्तर को भी दिया जाता है। जिन उपक्रमों में निर्णय लेने का अधिकार केवल शीर्ष स्तरीय प्रबंध को ही होता है वे केंद्रीकृत संगठन कहलाते हैं। जबकि उन संगठनों में जहाँ इस प्रकार के निर्णयों को लेने में निम्न स्तर तक के प्रबंध को भागीदार बनाया जाता है, विकेंद्रीकृत संगठन कहते हैं।

विकेंद्रीयकरण से तात्पर्य उस विधि से है जिसमें निर्णय लेने का उत्तरदायित्व सोपानिक क्रम में विभिन्न स्तरों में विभाजित किया जाता है। सरल शब्दों में विकेंद्रीयकरण का अर्थ संगठन के प्रत्येक स्तर पर अधिकार अंतरण करना होता है। निर्णय लेने का अधिकार निम्नतम स्तर तक के प्रबंध को दिया जाता है जहाँ पर वास्तविक रूप में कार्य होना है। दूसरे शब्दों में निर्णय लेने का अधिकार आदेश की श्रृंखला में नीचे तक दिया जाता है।

केंद्रीयकरण (Centralization) – जिस संगठन में निर्णय लेने का अधिकार केवल उच्चस्तरीय प्रबंधन को ही होता है तो वह संगठन केंद्रीकृत कहलाता है। कोई भी संस्था कभी भी न तो पूर्णरूपेण केंद्रीकृत हो सकती है और न विकेंद्रीकृत। जब कोई संस्था आकार तथा जटिलताओं की ओर अग्रसर होती है तो यह देखा गया है · कि वे संस्थाएँ निर्णयों में विकेंद्रीयकरण को अपनाती हैं। यह इसलिए होता है क्योंकि बड़ी-बड़ी संस्थाओं में जहाँ कर्मचारियों को प्रत्यक्ष तथा अतिनिकट से कार्य संचालन में आल्पित किया जाता है उनका ज्ञान तथा अनुभव उन उच्चस्तरीय प्रबंधकों से कहीं अधिक होता है जो संस्थान से अप्रत्यक्ष रूप से जुड़े हुए होते हैं।

श्रेष्ठ नियंत्रण (Best control)- विकेंद्रीयकरण से प्रत्येक स्तर पर कार्य निष्पादन के मूल्यांकन का अवसर मिलता है जिससे प्रत्येक विभाग व्यक्तिगत रूप से उसके परिणामों के लिए जवाबदेह बनाया जा सकता है। संगठन के उद्देश्यों की उपलब्धि किस सीमा तक हुई या समस्त उद्देश्यों को प्राप्त करने में प्रत्येक विभाग कितना सफल हो सका इसका भी निर्धारण किया जा सकता है। सभी स्तरों से प्रतिपुष्टि द्वारा भिन्नताओं का विश्लेषण करने तथा सुधारने में सहायता मिलती है। विकेंद्रीयकरण में निष्पादन की जवाबदेही एक बड़ी चुनौती है। इस चुनौती का सामना करने के लिए संतुलन अंक कार्ड तथा प्रबंध सूचना विधि।

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प्रश्न 14.
विकेंद्रीयकरण के महत्व लिखिए।
उत्तर:
एक संगठन निम्न कारणों से विकेंद्रीयकरण होना पसंद करता है

1. उच्च अधिकारियों की अत्यधिक कार्यभार से मुक्ति (More capacity utilization)विकेंद्रीयकरण के अतंर्गत दैनिक प्रबंधकीय कार्यों को अधीनस्थों को सौंप दिया जाता है। इसके फलस्वरूप उच्च प्रबंधकों के पास पर्याप्त समय बचता है जिसका प्रयोग वे नियोजन, समन्वय, नीति निर्धारण, नियंत्रण आदि में कर सकते हैं।

2. विभिन्नीकरण में सुविधा (Ease to expansion)- इस बात से इंकार नहीं किया जाता कि एक व्यक्ति का नियंत्रण सर्वश्रेष्ठ होता है लेकिन इसकी भी एक सीमा होती है। सीमा का अभिप्राय व्यवसाय के आकार से है अर्थात् जब तक व्यवसाय का आकार छोटा है उच्च स्तर पर सभी अधिकारियों को केंद्रित करके व्यवसाय को कुशलतापूर्वक चलाया जा सकता है लेकिन जब उसमें उत्पादन की जाने वाली वस्तुओं की मात्रा अधिक हो जाती है तब केंद्रीय नियंत्रण से काम नहीं चल सकता क्योंकि अकेला व्यक्ति सभी वस्तुओं की समस्याओं की ओर पर्याप्त ध्यान नहीं दे सकता।

3. प्रबंधकीय विकास (Managerial development)- विकेंद्रीयकरण का अभिप्राय है निम्नतम स्तर के प्रबंधकों को भी अपने कार्यों के संबंध में निर्णय लेने के अधिकार होना। इस प्रकार निर्णय लेने के अवसर प्राप्त होने से सभी स्तरों के प्रबंधकों के ज्ञान एवं अनुभव में वृद्धि होती है और इस प्रकार कहा जा सकता है कि यह व्यवस्था प्रशिक्षण का काम करती है।

4. कर्मचारियों के मनोबल में वृद्धि (Increase in employees morale)- विकेंद्रीयकरण के कारण प्रबंध में कर्मचारियों की भागीदारी बढ़ती है। इससे संस्था में उनकी पहचान बनती है। जब संस्था में किसी व्यक्ति की पहचान बने अथवा उसका महत्व बढ़े तो उसके मनोबल में वृद्धि होना स्वाभाविक है। मनोबल में वृद्धि होने से वे अपनी इकाई की सफलता के लिए बड़े से बड़ा त्याग करने से भी नहीं घबराते।

प्रश्न 15.
औपचारिक तथा अनौपचारिक संगठन में अंतर स्पष्ट कीजिए।
उत्तर:
औपचारिक संगठन एवं अनौपचारिक संगठन में अन्तर:
औपचारिक संगठन

  1. इसका निर्माण किसी योजना को पूरा करने के लिए विचार-विमर्श करके किया जाता है। होता है।
  2. इसे किसी तकनीकी उद्देश्य को पूरा करने के उद्देश्य से बनाया जाता है।
  3. इसका आकार बड़ा हो सकता है।
  4. इसमें सत्ता का प्रवाह ऊपर से नीचे की ओर होता है।

अनौपचारिक संगठन

  1. इसका निर्माण स्वतः ही सामाजिक संबंधों द्वारा
  2. इसका निर्माण सामाजिक संतोष प्राप्त करने
  3. यह प्रायः छोटे आकार का होता है।
  4. इसमें सत्ता का प्रवाह नीचे से ऊपर की ओर होता है।

प्रश्न 16.
औपचारिक संगठन से क्या आशय है ? इसकी कोई चार विशेषताएँ बताइए।
उत्तर:
औपचारिक संगठन – इसका आशय एक ऐसे संगठन से है जिसमें प्रत्येक स्तर के प्रबन्धकों के अधिकारों, कर्तव्यों तथा दायित्वों की स्पष्ट सीमा निर्धारित होती है। इस संगठन को संगठनात्मक लक्ष्यों की प्राप्ति हेतु प्रबंध द्वारा बनाया जाता है।

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प्रश्न 17.
अनौपचारिक संगठन किसे कहते हैं ? विशेषताओं सहित बताइए।
उत्तर:
अनौपचारिक संगठन से अभिप्राय ऐसे संगठन से है जिसकी स्थापना जानबूझकर नहीं की जाती अपितु, अनायास ही सामान्य हितों, संबंध, रुचियों तथा धर्म के कारण हो जाती है।
अर्ल. पी. स्ट्रांग के अनुसार, “अनौपचारिक संगठन एक ऐसी सामाजिक संरचना है जिसका निर्माण व्यक्तिगत आवश्यकताओं की पूर्ति के लिए किया जाता है।”
अनौपचारिक संगठन की विशेषताएँ- इसकी प्रमुख विशेषताएँ निम्नलिखित हैं.

1. निर्माण स्वतः-इसका निर्माण जानबूझकर नहीं किया जाता बल्कि व्यक्तियों के आपसी संबंधों तथा रुचियों के आधार पर स्वयं हो जाता है।

2. व्यक्तिगत संगठन-व्यक्तिगत संगठन से अभिप्राय है कि इसमें व्यक्तियों की भावनाओं को ध्यान में रखा जाता है, उन पर किसी भी बात को थोपा नहीं जाता।

3. अधिकार-इसमें अधिकार व्यक्ति से जुड़े रहते हैं तथा उनका प्रवाह नीचे से ऊपर की ओर या समतल रूप में चलता है।

4. स्थायित्व का अभाव-इसमें जब तक व्यक्ति एक समूह में है तभी तक वह संगठन है जब व्यक्ति अलग होता है तो संगठन समाप्त हो जाता है। इसमें स्थायित्व का अभाव रहता है।

प्रश्न 18.
अधिकार अन्तरण या भारार्पण से क्या तात्पर्य है ? इसके प्रमुख तत्वों ( प्रक्रिया) को समझाइए।
उत्तर:
अधिकार अंतरण से आशय, अधीनस्थों को निश्चित सीमा के अन्तर्गत कार्य करने का केवल अधिकार देना है। अधिकार अंतरण प्रक्रिया का एक महत्वपूर्ण अंग है। इसकी आवश्यकता उस समय उत्पन्न होती है जब एक प्रबंधक के पास कार्यभार अधिक होने के कारण वह सभी कार्यों का अधीनस्थों में विभाजन करता है। अतः उसे अधिकार अंतरण का सहारा लेना पड़ता है। जब कुछ कार्यों का निष्पादन करने के लिए दूसरे व्यक्तियों को अधिकृत किया जाता है तो इसे अधिकार अंतरण कहते हैं।
विभिन्न विद्वानों ने अधिकार अंतरण को निम्न प्रकार से परिभाषित किया है

  1. एफ. जी. मूरे, “अधिकार अंतरण से अभिप्राय दूसरे लोगों को कार्य सौंपना और उसे करने के लिए अधिकार देना है।”
  2. मैस्कॉन, “अधिकार अंतरण किसी व्यक्ति को कार्य तथा सत्ता सौंपना है, जो उनके लिए दायित्व ग्रहण करता है।”

अधिकार अंतरण के तत्व-अधिकार अंतरण के पाँच तत्व हैं

1. कार्यभार सौंपना – अधिकार अंतरण प्रक्रिया का पहला कदम कार्यभार सौंपा जाना है क्योंकि कोई भी अधिकारी इतना सक्षम नहीं होता कि वह अपना सारा कार्य स्वयं पूरा कर ले इसलिए अपने कार्य का सफलतापूर्वक निष्पादन करने के लिए वह अपने कार्य का विभाजन करता है, विभाजन के समय अधीनस्थों कीयोग्यता एवं कुशलता का ध्यान में रखा जाना अत्यंत आवश्यक होता है।

2. अधिकार प्रदान करना-कार्य का सफलतापूर्वक निष्पादन करने हेतु अधिकार सौंपे जाते हैं क्योंकि . जब तक अधीनस्थों को अधिकार प्रदान नहीं किए जाएँगे तब तक कार्यभार सौंपना अर्थहीन होता है। अतः कार्य को पूरा करने के लिए अधिकार सौंपे जाने चाहिए।

3. प्रत्यायोजन की स्वीकृति-जिस व्यक्ति को कार्य दिया गया है वह उस कार्य को स्वीकार या अस्वीकार कर सकता है। यदि अधिकार अंतरण स्वीकार नहीं किया जाता तो प्रत्यायोजक प्रबन्धक किसी अन्य अधीनस्थ व्यक्ति को कार्य सौंपने की कार्यवाही करेगा। इसलिए प्रत्यायोजन की स्वीकृति अत्यंत आवश्यक होती है।

4. जवाबदेही निर्धारित करना-जवाबदेही, अधीनस्थों को सही ढंग से कार्य करने के लिए जिम्मेदार ठहराता है, प्रत्येक अधीनस्थ केवल उस अधिकारी के समक्ष ही जवाबदेह होता है जिससे उसे कार्य करने के अधिकार प्राप्त होते हैं।

MP Board Solutions

प्रश्न 19.
विकेन्द्रीयकरण से क्या तात्पर्य है ? विकेन्द्रीयकरण के कौन-कौन से लाभ होते हैं ?
उत्तर:
विकेन्द्रीयकरण से तात्पर्य है, निर्णय लेने के अधिकार को संगठन के निम्न स्तर पर पहुँचाना है। विकेन्द्रीयकरण, अधिकार अंतरण का ही विस्तृत रूप है। जब किसी उच्च अधिकारी के द्वारा अपने अधीनस्थों को अपेक्षाकृत अधिक भाग में अधिकारों का भारार्पण किया जाता है, तो यह विकेन्द्रीयकरण कहलाता है। इसके अंतर्गत केवल ऐसे अधिकार जो उच्च अधिकारियों के लिए सुरक्षित रखना जरुरी है, को छोड़कर शेष सभी अधिकार अधीनस्थों को स्थाई रूप से सौंप दिए जाते हैं।
विकेन्द्रीयकरण के लाभ-विकेन्द्रीकरण के लाभों को निम्न बातों से स्पष्ट किया जा सकता है

1. उच्च अधिकारियों का कार्य भार कम होना-विकेन्द्रीयकरण की सहायता से उच्च अधिकारियों के कार्यभार में बहुत कमी आ जाती है, वे अपना पूरा ध्यान महत्वपूर्ण कार्यों में लगा सकते हैं उन्हें छोटे-छोटे कार्यों में उलझना नहीं पड़ता है, जिस कारण व्यावसायिक उपक्रम उच्च अधिकारियों की योग्यता, कुशलता तथा विवेक का अधिकारिक लाभ उठा सकते हैं।

2. विविधीकरण की सुविधा-विकेन्द्रीयकरण में विविधीकरण की पर्याप्त सुविधा होती है क्योंकि अलग-अलग क्रियाओं के लिए अलग-अलग अध्यक्षों की नियुक्ति की जा सकती है।

3. अनौपचारिक संबंधों का विकास-विकेन्द्रीयकरण के द्वारा अनौपचारिक संबंधों का विकास होता है।

4. अभिप्रेरण-विकेन्द्रीयकरण के द्वारा कर्मचारियों को स्वतंत्र रूप से निर्णय लेने का अधिकार होता है जिससे वह बेहतर परिणाम लाने के लिए अभिप्रेरित होते हैं।।

प्रश्न 20.
कार्यात्मक संगठन तथा प्रभागीय संगठन में अन्तर स्पष्ट कीजिए।
उत्तर:
कार्यात्मक संगठन तथा प्रभागीय संगठन में अन्तर –

प्रश्न 21.
औपचारिक संगठन के लाभ तथा दोषों का वर्णन कीजिए।
उत्तर:
औपचारिक संगठन के लाभ (Advantages of Formal Organisation)

1. व्यवस्थित कार्यवाही (Systematic Working)-इस ढाँचे का परिणाम एक संगठन की व्यवस्थित और सरल कार्यवाही होता है।

2. संगठनात्मक उद्देश्यों की प्राप्ति (Achievement of Organisational Objectives)-इस ढाँचे को संगठनात्मक उद्देश्यों को प्राप्त करने के लिए स्थापित किया जाता है।

3. कार्य का दोहराव नहीं (No Overlapping of Work)-औपचारिक संगठनात्मक ढाँचे में विभिन्न विभागों और कर्मचारियों के बीच कार्य व्यवस्थित ढंग से विभाजित होता है। इसलिए कार्य के दोहराव का कोई अवसर नहीं होता है।

4. समन्वय (Coordination)-इस ढाँचे का परिणाम विभिन्न विभागों की क्रियाओं को समन्वित करना होता है।

औपचारिक संगठन के दोष (Disadvantages of Formal Organization)

1. कार्य में देरी (Delay in Action) – सोपान श्रृंखला और आदेश श्रृंखला का अनुसरण करते समय कार्यों में देरी हो जाती है।

2. कर्मचारियों की सामाजिक आवश्यकताओं की अवहेलना करता है (Ignores Social Needs of Employees)-यह ढाँचा कर्मचारियों की सामाजिक और मनोवैज्ञानिक आवश्यकता को महत्व नहीं देता ‘जिससे कर्मचारियों के अभिप्रेरण में कमी हो सकती है।

3. केवल कार्य पर बल (Emphasis on Work Only) – यह ढाँचा केवल कार्य को महत्व देता है इसमें मानवीय संबंधों, सृजनात्मकता, प्रतिभाओं इत्यादि को महत्व नहीं दिया जाता है।

प्रश्न 22.
विकेंद्रीयकरण एवं भारार्पण में अंतर लिखिए।
उत्तर:
विकेंद्रीयकरण एवं भारार्पण में अंतर
MP Board Class 12th Business Studies Important Questions Chapter 5 संगठन - 4

प्रश्न 23.
अधिकार अंतरण तथा विकेन्द्रीयकरण में अंतर स्पष्ट कीजिए।
उत्तर:
अधिकार अंतरण तथा विकेन्द्रीयकरण में अंतर

अधिकार अंतरण

  1. उच्च अधिकारी के द्वारा सत्ता का अधीनस्थों को हस्तांतरण, अधिकार अंतरण कहलाता है। करण कहलाता है।
  2. इसमें अंतिम उत्तरदायित्व अधिकार सौंपने उत्तरदायित्व कावाले का ही होता है अर्थात् इसमें उत्तर भी हस्तांतरण हो जाता है।दायित्व का हस्तांतरण नहीं होता।
  3. यह सभी संस्थाओं के लिए आवश्यक होता है क्योंकि अधीनस्थों से काम लेने के लिए उन्हें अधिकार देना पड़ता है।
  4. अधिकार अंतरण, विकेन्द्रीयकरण पर नहीं होता।

विकेन्द्रीयकरण

  1. पूरे संगठन में सत्ता का फैलाव करना विकेन्द्रीय
  2. इसमें अधिकार के साथ-ही-साथ
  3. विकेन्द्रीयकरण प्रत्येक संस्था के लिए आवश्यक नहीं है।
  4. विकेन्द्रीयकरण, अधिकार अंतरण के बिना संभव आधारित नहीं होता है।

प्रश्न 24.
अधिकार एवं उत्तरदायित्व में अन्तर स्पष्ट कीजिए।
उत्तर:
अधिकार एवं उत्तरदायित्व में अन्तर –

अधिकार

  1. अधिकार का अर्थ है, निर्णय लेने की शक्ति।
  2. अधिकार वह शक्ति है जो अधीनस्थों के कर्तव्य को पूरा करने से है।
  3. अधिकार का प्रत्यायोजन किया जा सकता है।
  4. अधिकार, संस्था में औपचारिक पद के कारण उत्पन्न होता है।

उत्तरदायित्व

  1. उत्तरदायित्व का अर्थ है, सौंपे गये कार्य को सही
  2. उत्तरदायित्व का आशय किसी व्यक्ति द्वारा अपने व्यवहार को प्रभावित करती है।
  3. उत्तरदायित्व दूसरों को सौंपे जा सकते हैं।
  4. उत्तरदायित्व, उच्चाधिकारी अधीनस्थ सम्बन्ध से उत्पन्न होता है।

प्रश्न 25.
प्रभावी भारार्पण की अनिवार्य अपेक्षाओं को समझाइए।
उत्तर:
प्रभावी भारार्पण की अनिवार्य अपेक्षाएँ – अधिकार सौंपने की व्यवस्था को प्रभावी बनाने के लिए निम्नांकित निर्देशों का पालन करना चाहिए

1. अधिकार और दायित्वों की स्पष्ट सीमा – अधीनस्थों को जो कार्य सौंपे जाएँ तो उन्हें यह पूरी तरह मालूम होना चाहिए कि उनके दायित्व तथा अधिकारों का क्षेत्र क्या है तथा उनकी सीमा कहाँ तक है।

2. अधिकार और दायित्वों की स्पष्ट व्याख्या – अधीनस्थों को स्पष्ट रूप से यह ज्ञान होना चाहिए कि उन्हें क्या करना है तथा सौंपे गए कार्य के लिए कितने अधिकार दिये गये हैं तथा उनसे किस प्रकार के कार्य की आशा की जाती है।

3. उचित नियन्त्रण – अधिकार सौंपने की व्यवस्था के पश्चात् भी वरिष्ठ अधिकारी का उत्तरदायित्व समाप्त नहीं होता। इसलिए उच्च अधिकारियों को अपने अधीनस्थों के कार्यों पर उचित नियंत्रण रखना चाहिए।

4. योग्य व्यक्ति का चयन – अधिकार हमेशा योग्य व्यक्ति को ही सौंपने चाहिए तभी सौंपा गया कार्य सही ढंग से पूर्ण होता है।

प्रश्न 26.
संगठन के उद्देश्य लिखिए।
उत्तर:
किसी भी व्यावसायिक एवं औद्योगिक इकाई में संगठन के निम्नलिखित उद्देश्य होते हैं…

1. उत्पादन में मितव्ययिता (Economy in production) संगठन का प्रमुख उद्देश्य, उत्पादन में मितव्ययिता लाना है, अर्थात् न्यूनतम लागत पर अधिकतम उत्पादन करना प्रत्येक उपक्रम का उद्देश्य होता है, यहाँ यह ध्यान रखना है कि न्यूनतम लागत के साथ-साथ वस्तु की गुणवत्ता (Quality) व मात्रा (Quantity) में गिरावट नहीं आनी चाहिये।

2. समय व श्रम में बचत (Saving in time and labour)-संगठन में श्रेष्ठ मशीनें व यंत्र तथा श्रेष्ठ प्रणाली अपनाई जाती है जिससे कार्य के समय व श्रम में काफी बचत हो जाती है, संगठन द्वारा बड़े पैमाने में उत्पादन के लाभ भी लिये जा सकते हैं।

3. श्रम व पूँजी में मधुर सम्बन्ध (Cordial relations between labour and capital)- श्रमिकों व प्रबन्ध के बीच मधुर सम्बन्ध स्थापित करना भी संगठन का एक उद्देश्य है, इस हेतु कुशल संगठनकर्ता की नियुक्ति कर श्रम व पूँजी के हितों की रक्षा की जाती है।

4. सेवा भावना (Spirit of service) वर्तमान सामाजिक चेतना एवं जन-जागरण के कारण प्रत्येक व्यवसायी का उद्देश्य ‘प्रथम सेवा फिर लाभ’ हो गया है वैसे भी प्रत्येक व्यवसायी को समाज सेवा कर अपने उत्तरदायित्व का निर्वहन करना चाहिये। इसलिये वर्तमान में प्रत्येक संगठन चाहे वह आर्थिक हो या अनार्थिक सभी का उद्देश्य सेवा करना होता है।

MP Board Solutions

प्रश्न 27.
संगठन प्रक्रिया का संक्षेप में विवेचन कीजिये।
उत्तर:
संगठन के निर्माण के प्रमुख चरण

1. उद्देश्यों की स्थापना-संगठन के साथ उद्देश्यों का होना अति आवश्यक है। संगठन के उद्देश्यों में उन कार्यों की स्पष्ट व्याख्या होनी चाहिए जिनके कारण संगठन की संरचना की जाती है।

2. क्रियाओं का निर्धारण-संगठन का दूसरा कदम क्रियाओं का निर्धारण एवं उनका विभाजन करना होता है। कार्यों को उपकार्यों में विभाजित कर प्रत्येक कार्य की स्पष्ट व्याख्या करनी चाहिए।

3. क्रियाओं का वर्गीकरण-तृतीय चरण में क्रियाओं का वर्गीकरण किया जाता है। इस हेतु विभिन्न क्रियाओं को अलग-अलग समूह में बाँट दिया जाता है। जैसे-क्रय, विक्रय, विज्ञापन, निरीक्षण, उत्पादन, मजदूरी, वेतन आदि।

4. कार्य का विभाजन-इसके अन्तर्गत सही कार्य के लिए सही व्यक्ति के सिद्धान्त के अनुसार कार्य विभाजन किया जाता है। साथ ही कर्मचारियों की जिम्मेदारी भी निश्चित की जाती है।

MP Board Solutions

प्रश्न 28.
विकेन्द्रीयकरण के उद्देश्य लिखिए।
उत्तर:
विकेन्द्रीयकरण के निम्न उद्देश्य होते हैं

1.शीर्ष स्तर पर कार्यभार में कमी-सामान्यतः निर्णय उच्च प्रबन्धक या शीर्ष स्तर पर लिये जाते हैं। संगठन का विस्तार होने पर शीर्ष स्तर का कार्य और बढ़ जाता है अतः इस कार्यभार को कम करने के लिये विकेन्द्रीयकरण किया जाता है। इससे शीर्ष अधिकारियों का कार्यभार कम होगा और वे अति महत्वपूर्ण कार्यों के लिये अच्छा निर्णय ले सकते हैं।

2. प्रजातान्त्रिक व्यवस्था के लाभ प्राप्त करने के लिये-विकेन्द्रीयकरण से अधिकारों का फैलाव छोटे-छोटे स्तर तक हो जाता है इससे प्रजातान्त्रिक व्यवस्था के समस्त लाभ सभी वर्ग को प्राप्त होते हैं।

3. कार्यों के शीघ्र निष्पादन के लिये-उच्चाधिकारियों के अधिक कार्यों को कम कर दिया जाये तो वे अपने कार्यों को ठीक ढंग से व शीघ्र उसका निपटारा कर सकते हैं इसी के साथ प्रतिस्पर्धा में आगे बढ़ने के लिये प्रत्येक स्तर पर अधिकारों का होना आवश्यक है इससे कार्य शीघ्र पूर्ण किये जा सकते हैं।

4.सर्वांगीण विकास-विकेन्द्रीयकरण से शीर्ष स्तर एवं निम्न स्तर के समस्त अधिकारियों को अधिकार प्राप्त होने से उन्हें अपने विकास का पूर्ण अवसर प्राप्त होता है। इसमें पूर्ण कार्यक्षमता से कार्य निष्पादित किये जा सकते हैं। अतः संगठन के सर्वांगीण विकास के लिये भी विकेन्द्रीयकरण आवश्यक है।

प्रश्न 29.
अनौपचारिक संगठन के विशेषताओं का उल्लेख कीजिए।
उत्तर:
अनौपचारिक संगठन में सामान्यतः निम्न विशेषतायें होती हैं

  1. इन संगठनों को बनाया नहीं जाता अपितु विशेष परिस्थितियों के कारण ये बन जाते हैं।
  2. इन संगठनों के समान उद्देश्य होते हैं।
  3. ये अस्थायी होते हैं कभी भी संगठन समाप्त हो जाते हैं।
  4. इनका अपना कोई चार्ट, विधान या मैन्यूवल नहीं होता अपितु सामान्य परम्परा ही इनके कानून होते हैं।
  5. ये संगठन व्यक्तिगत सम्बन्धों पर आधारित होते हैं।
  6. ये वैध एवं अवैध दोनों हो सकते हैं ।
  7. इसमें अधिकार एवं सत्ता को ऊपर से नीचे समर्पित किया जाता है।

प्रश्न 30.
संगठन की विशेषताएँ लिखिए।
उत्तर:

1. दो या दो से अधिक व्यक्ति का होना-एक व्यक्ति अपने आपको संगठन नहीं कह सकता। अतः संगठन हेतु कम से कम दो व्यक्तियों का होना अति आवश्यक है। अधिकतम संख्या पर कोई प्रतिबन्ध नहीं है।

2. निश्चित उद्देश्य का होना-संगठन में निश्चित उद्देश्य का होना अति आवश्यक है, बिना उद्देश्य के व्यक्तियों का समूह, संगठन नहीं हो सकता। जैसे कि किसी स्थान पर एक हजार लोग बिना उद्देश्य के एकत्र हैं तब वह संगठन न होकर मात्र भीड़ (Crowd) होगी।

3. लक्ष्य का पूर्व निर्धारित होना- संगठन के लिये यह आवश्यक है कि लक्ष्य या उद्देश्य पूर्व निर्धारित हो, लक्ष्य की प्राप्ति के लिये ही संगठन बनाया जाता है। अत: संगठन के पूर्व लक्ष्य (Targets) निर्धारित होना चाहिये।

4. व्यक्तियों का समूह होना-पशु-पक्षी या अन्य प्राणियों का समूह संगठन नहीं हो सकता। संगठन केवल व्यक्तियों का ही हो सकता है। यद्यपि फर्म या कम्पनियों के भी विधान द्वारा व्यक्ति (Person) कहा गया है, किन्तु ये सब कृत्रिम (Artificial) व्यक्ति हैं इसलिये 4 या 14 कम्पनियों का एक साथ कार्य करने को समूह (Group) कहेंगे संगठन नहीं।

MP Board Solutions

प्रश्न 31.
संगठन के किन्हीं चार सिद्धांतों को संक्षिप्त रूप में समझाइए।
उत्तर:
संगठन के सिद्धांत (Principles of Organization)

1.सोपानिक सिद्धांत (Principle of scalar chain)- कौन व्यक्ति किस प्रबंधक की अधीनस्थता में काम करेगा और किससे आदेश लेगा इस बात का स्पष्ट उल्लेख होना चाहिए ताकि कोई भी व्यक्ति इस आदेश श्रृंखला का उल्लंघन न करे।

2.अपवाद का सिद्धांत (Principle of exception)- प्रत्येक अधीनस्थ को इतने अधिकार दिये जायें कि वह दैनिक कार्यों को स्वयं निपटा सके। केवल असाधारण मामले ही अधीक्षक के पास आने चाहिए। इस प्रकार प्रबंधक का समय अधिक महत्वपूर्ण मामलों के लिए सुरक्षित रहता है।

3. लोच का सिद्धांत (Principle of flexibility)- संगठन में पर्याप्त लोच होनी चाहिए ताकि इसमें परिस्थितियों व समय के अनुसार आसानी से परिवर्तन किये जा सकें।

4. अधिकार एवं दायित्व (Rights and responsibility)- संगठन के प्रत्येक सदस्य को उसके अधिकार व दायित्व साथ-साथ दिये जाने चाहिए। अधिकार व दायित्व में तालमेल के बिना कोई भी व्यक्ति कुशलतापूर्वक कार्य नहीं कर पायेगा। अधीनस्थों को पर्याप्त अधिकार सौंपे जाने चाहिए।

प्रश्न 32.
रेखा एवं रेखा तथा कर्मचारी संगठन में अंतर लिखिए।
उत्तर:
रेखा एवं रेखा तथा कर्मचारी संगठन में अन्तर
(Difference between Line and Line and Staff Organization)
MP Board Class 12th Business Studies Important Questions Chapter 5 संगठन - 5

प्रश्न 33.
अधिकार, उत्तरदायित्व तथा उत्तरदेयता/जवाबदेही में तुलना करें।
अथवा
अधिकार प्रत्यायोजन के तत्वों में तुलना करें।
उत्तर:
अधिकार, उत्तरदायित्व एवं उत्तरदेयता में अंतर
MP Board Class 12th Business Studies Important Questions Chapter 5 संगठन - 2

MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise

MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 1
MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 2
MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 3
MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 4
MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 5
MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 6

MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 7
MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 8
MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 9
MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 10
MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 11