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MP Board Class 11th Chemistry Important Questions Chapter 4 Chemical Bonding and Molecular Structure

Chemical Bonding and Molecular Structure Important Questions

Chemical Bonding And Molecular Structure Very Short Answer Type Questions

Question 1.
What type of bond is present generally in same atoms?
Answer:
Covalent bond.

Question 2.
Which hybridization is present in ammonia (NH₃)?
Answer:
sp³.

Question 3.
What is the reason for high boiling point of water?
Answer:
Presence of H – bond between molecules of water.

Question 4.
What is the dipole moment of C0₂?
Answer:
Zero.

Question 5.
What type of bonds are directional?
Answer:
Covalent bonds.

Question 6.
What is the bond angle in water?
Answer:
104°5′.

Question 7.
What is the structure of [Ni(CN)₄]^2-?
Answer:
Square planner.

Question 8.
Which bond is present in s – s overlapping?
Answer:
cr (Sigma) bond.

Question 9.
What is the structure of diamond?
Answer:
Crystal lattice (Tetrahedral).

Question 10.
Which bond is present in sidewise overlapping of p – p orbitals?
Answer:
n – bond.

Question 11.
Which type of hybridization found in PCl₅?
Answer:
sjy’d hybridization.

Question 12.
What is full form of LCAO?
Answer:
Linear Combination of Atomic Orbitals.

Question 13.
What is the structure of NH₃?
Answer:
Trigonal bipyramidal.

Question 14.
What is dipole moment (µ) of a linear covalent molecule?
Answer:
Zero.

Question 15.
What is the unit of electron gain enthalpy?
Answer:
eV per atom or kJ per mole.

Question 16.
What is the formula of bond order?
Answer:
Bond order = \(\frac{1}{2}\) (N_b – N_a).

Question 17.
What is the symbol of superoxide and peroxide?
Answer:
Superoxide – O^–₂ and peroxide – O^2-₂.

Question 18.
What do you mean by bond order?
Answer:
The number of electrons present between two atoms of molecules or ions.

Question 19.
More polarizing power and more polarisability increases which property of molecule?
Answer:
Covalent property.

Question 20.
What is determined by Born – Haber cycle?
Answer:
Lattice energy.

Question 21.
What is full form of VSEPR?
Answer:
Valence Shell Electron Pair Repulsion.

Chemical Bonding And Molecular Structure Short Answer Type Questions – I

Question 1.
What is the electronic theory of covalency? Write its main postulates?
Answer:
The electronic theory of valency and its main postulates are as follows:

1. The covalency of any element depends upon the no.of electrons present in its valence shell.
2. All the elements have the tendency to acquire the nobel gas configuration.
3. The electrons present in valence shell are called valence electrons and when the electron comes out than the vacant space is called kernel.
4. If an element is unstable, than it works for its stability for this it gives and takes the electron. On this basis the bonds are of three types:
   1. Ionic bond
   2. Covalent bond and
   3. Co – ordinate bond.

Question 2.
What do you mean by lone pair of electron?
Answer:
The electron pair which present in the valence orbital of the element and does not take part bond formation is called lone pair electron.
Example: In the H₂O atom the lone pairs of electrons on an atom:

Bond pairs = 2, Lone pair electrons = 2.

Question 3.
What do you mean by dipole moment?
Answer:
Dipole moment is defined as, the product ofthe magnitude of charge on any one of the atoms and distance between them. It is represented by Greek letter µ(mu).
Mathematically, dipole moment is expressed as µ = e × d
Where, e is charge on any one of the atoms and d is distance between the atoms.
As e is of the order of 10^-10 esu while d is of the order of 10^-8 cm µ is of the order 10¹⁸ esu cm and this unit of p is known as Debye (D). Thus,
1D = 1 × 10^-18esu cm

Question 4.
Is He₂ molecule is possible? Clearify it.
Answer:
He₂ molecule is not possible.
2He → 1s²
Two electrons are present in Is orbital of He atom. It is complete and stable orbital and so it cannot accept an extra electron. The bond order is zero. So the formation of He₂ molecule is impossible.

Question 5.
What is the total number of σ (sigma) and π (pi) bond are present in following molecules:

1. C₂H₂
2. C₂H₄

Answer:

1. C₂H₂
2. C₂H₄

Question 6.
What do you mean by hydrogen bond?
Answer:
Hydrogen bond is defined as the electrostatic force of attraction which exists between the covalently bonded hydrogen atom of one molecule and the electronegative atom of the other molecule. Hydrogen bond is of two types:

1. Intermolecular hydrogen bonding
2. Intra – molecular hydrogen bonding.

Question 7.
At normal temperature H20 is in liquid state but H2S is in gaseous state. Why?
Answer:
Hydrogen bonding affects the physical state of the molecule. For example, H₂0 and H₂S are the hydrides of group 16 elements, but H₂O is liquid whereas H₂S is gas at room temperature. Actually, H₂O fulfils the (MPBoardSolutions.com) condition of hydrogen bonding therefore, in H₂O hydrogen bonding is found and they get associated, grow in molecular size and exist in liquid state. The H₂S molecule, the magnitude of hydrogen bonding is negligible, therefore, molecules remains separated from each other and acquires gaseous state at room temperature.

Question 8.
Why, viscosity of glycerol is more than ethanol?
Answer:
In ethanol molecule, one hydroxyl gap is present whereas in every glycerol, three OH – groups are present which form hydrogen bond. Therefore, in glycerol hydrogen bond formation is more in comparison to ethanol. That is why, the viscosity of glycerol is more than ethanol.

Question 9.
Why σ – bond is stronger than π – bond?
Answer:
The strength of any bond depends upon the overlapping limit, σ – bonds are formed by axial overlapping whereas σ – bonds are formed by sidewise overlapping. So, the extent of overlapping in σ – bond is more than π – bond. So σ – bond is more stronger than π – bond.

Question 10.
Why HF molecule is more polar than HI?
Answer:
The electronegativity of F is more than I. So, the displacement of electron in covalency in HF is more than HI, resultant the deviation of charges in HF is greater than HI. So, HF is more polar than HI.

Question 11.
C – Cl bond is polar but CCl₄ is non – polar, why? Give the reason?
Answer:
In C – Cl bond the electronegativity of Cl is more than C, due to this the electrons in covalent bond shifts towards Cl, due to which partial +ve charge appears on C and partial -ve charge develops once and this bond become polar in nature. Whereas the structure of CCl₄ is symmetric, due to which the dipole moment of C – Cl bond cancelled each other. So, CCl₄ molecule is non – polar.

Question 12.
What do you mean by resonance?
Answer:
Some compounds cannot be represented by a single definite structure, rather more than one structure is required by none of them is able to explain all the known properties of the compound alone. Thus, the various structures written for a compound to explain the known properties of it completely are called resonating structure. This phenomenon is called resonance.

Question 13.
What are the conditions for resonance?
Answer:
The conditions are as follows:

1. The heat of formation of each resonating structure should be same.
2. The arrangement of atoms in each formula should be same.
3. The number of unpaired electron in each structures should be same.

Question 14.
What is resonance energy ?
Answer:
The difference between the actual energy of the resonance hybrid and the most stable one of the resonating structures is called resonance energy.

Question 15.
Determine the bond order in N₂?
Answer:
N₂ (14 electrons): The electronic configuration
σ1s², σ^*1s², σ2s², σ^*2s², (π2p_x² = π2p_y²), σ2p^z2
Bond order = \(\frac{1}{2}\) (N_b – N_a) = \(\frac{1}{2}\) × (10 – 4) = 3.

Question 16.
On the basis of molecular orbital theory explain that the Be₂ atom is not formed?
Answer:
₄Be: Electronic configuration = 1s² 2s²
Be₂ (molecule) (4 + 4 = 8e^–)
Electronic configuration = σ1s², σ^*1s², σ2s², σ^*2s²
Bond order = \(\frac{1}{2}\) (N_b – N_a) = \(\frac{1}{2}\) × (4 – 4) = 0.
So, the Be₂ molecule does not formed.

Question 17.
Write the definition of hybridization?
Answer:
The process of intermixing of atomic orbitals of nearly equal energy and proper symmetry giving rise to equal number of new orbitals of same energy is called hybridization and the orbitals so formed hybridized orbital.

Question 18.
What do you understand by lattice energy and solvation energy?
Answer:
Lattice energy:
Once the gaseous ions are formed, the ions of opposite charges come close together and pack up three – dimensionally in a definite geometric pattern to form ionic crystal (Crystal lattice). Since, the packing of ions of opposite charges takes place as a result of attractive force between them, the process is accompanied (MPBoardSolutions.com) with the release of energy referred to as lattice enthalpy. Lattice enthalpy may be defined as; the amount of energy released when one mole of ionic solid is formed by the close packing of its constituents. It is denoted by ∆_LH and negative in nature.

Solvation energy:
The energy released when an ion get soluble in water is called solvation energy.

Question 19.
Whose boiling points are high electrovalent compounds or covalent compounds, why?
Answer:
High boiling points: The boiling points of ionic solids are very high. This is due to strong electrostatic force of attraction between the oppositely charged ions. To change the physical state of ionic compounds, high temperature is required.

Question 20.
Why BaS0₄ is insoluble in water?
Answer:
The solubility of any ionic compound depends upon the lattice energy and solvation energy. If the lattice energy of any compound is more than solvation energy, than ionic compound is insoluble in water. The solvation energy of BaS04 is less than lattice energy. So it is insoluble in water.

Question 21.
If Be – H bond is polar, the dipole moment of Be – H₂ is zero. Why?
Answer:
BeH₂ is linear. The bond moment present in opposite direction cancelled each other. That is why Be the dipole moment of BeH₂ is zero.

Chemical Bonding And Molecular Structure Short Answer Type Questions – II

Question 1.
Why ice is lighter than water? Explain?
Or, Density of ice is less than water. Why?
Answer:
Density of ice is less than water:
In ice each oxygen atom is tetrahedrally surrounded by four hydrogen atoms in which two hydrogen atoms are linked to oxgyen atom by covalent bond and other two hydrogen atoms are linked by hydrogen bond.

The molecules of H₂O are not packed closely. This H gives rise to open cage like structure for ice having a larger volume for the given mass of water. Thus, density of ice is less than water. Ice is actually hydrogen bonded crystal. (MPBoardSolutions.com) Three dimensional structure of protein and nucleic acids like biologically important substances is due to hydrogen bond. Energy of hydrogen bond is between 3.5 kJmol^-1 and 8 kJmol^-1. Thus, hydrogen bond is stronger than van der Waals force and weaker than covalent bond.

Question 2.
Write the rules of hybridization?
Answer:
Conditions for hybridization: Following are the conditions for hybridization:

1. The orbitals of one and same atom participate in hybridization. Only the orbitals and not electrons get hybridized.
2. The energy difference between the hybridizing orbitals should be small.
3. Promotion of electron is not essential prior to hybridization.
4. It is not necessary that only the half-filled orbitals may participate in hybridiza¬tion. In some cases, even the filled orbitals may participate in hybridization.

Question 3.
Explain covalent bond with example?
Answer:
Lewis – Langmuir theory:
Lewis and Langmuir suggested that, atoms may combine by sharing of electrons in their outermost shell to complete their respective octet. The shared electrons becomes the property of both the atoms. This types of linkage is known as covalent linkage or covalent bond. Thus,

1. The force which binds atoms of same or different elements by mutual sharing of electrons is called a covalent bond.

2. This type of bond is formed between two similar non-metalic elements (A, A) or (B, B) or dissimilar atoms (A and B).
Example:
Chlorine molecule:
Both the chlorine atoms (Z = 17) contain 7 electrons in their valence shells and short in one electron each. They share one electron pair in which an electron is contributed by both as shown below:

Question 4.
Differentiate between atom and ion?
Answer:
Differences between Atom and Ion:
Atom:

1. Atoms are electroneutral.
2. Not present in free state.
3. Takes part in chemical reaction.
4. No. of e^– and proton is same.

Ion:

1. Ions are charged.
2. Ions present in free state.
3. Not take part in chemical reaction.
4. No. of e^– is more than proton.

Question 5.
Differentiate between Sigma (σ) and Pi (π) bond?
Answer:
Differences between Sigma (σ) and Pi (π) bond:
Sigma (σ) bond:

1. This bond is formed by end to end or head on overlapping of orbitals along the inter nuclear axis.
2. This is formed by overlapping of s – s, s – p or p – p orbitals.
3. Overlapping is large, hence it is strong bond.
4. Free rotation about σ – bond is possible.
5. Electron cloud is symmetrical about inter nuclear axis.

Pi (π) bond:

1. This bond is formed by the sidewise overlapping of orbitals.
2. This is formed by overlapping of p – p orbitals only.
3. Overlapping is small, hence it is weak bond.
4. Free rotation about a π – bond is not possible.
5. Electron cloud of π – bond is unsymmetrical.

Question 6.
Write difference between s – and p – orbitals?
Answer:
Differences between s – and p – orbitals:
s – orbital:

1. They are oval – symmetrical.
2. Non – directional
3. 1 = 0 and m = 0

p – orbitals:

1. They are dumbelled shape and lines symmetry of axis.
2. Directional
3. l = 1 and m = -1, 0, +1

Question 7.
Explain inter molecular and intramolecular hydrogen bonds with example?
Answer:
Types of hydrogen bond:
1. Intermolecular hydrogen bonding:
When these atoms (hydrogen and electronegative atom) are of different molecules, it is called intermolecular hydrogen bonding as in H₂O, HF, C₂H₅OH etc.

Hydrogen bond is represented by dotted lines. Many molecules of HF associates and form (HF)_n. On the same way molecules of water and alcohols are linked with hydrogen bonds.

2. Intramolecular hydrogen bonding:
If these atoms (hydrogen and electronegative atoms) are present in same molecule, this type of hydrogen bonding is called intramolecular hydrogen bonding.
e.g., o – nitrophenol

Question 8.
Among NH₃ and NF₃ whose dipole moment is high. Why?
Answer:
Both these molecules have pyramidal shape with one lone pair of electron on nitrogen atom. As fluorine is more electronegative than hydrogen therefore N – F bond should be more polar than N – H bonds. Consequently, the resultant dipole moment of NF₃ should be much larger than that of NH₃. However, the dipole moment of NH₃ (µ = 1.47D) is larger than that of NF₃ (µ = 0.24D).

The anomalous behaviour can be explained due to the presence of lone pair on nitrogen. In case of NH₃, the orbital dipole due to lone pair of electron and the bond moments of three N – H bonds are in same direction. Therefore, it adds on the resultant dipole moment of the N – H bonds. On the other hand in case of NF₃, the orbital dipole moment is in the opposite direction to resultant dipole moment of three N – F bonds. Thus, the lone pair moment cancels the resultant N – F bond moments as shown in figure. Consequently, the dipole moment of NF₃ is low.

Question 9.
Is according to following equation, is the hybridization changes in B and N:
BF₃ + NH₃ → F₃B.NH₃.
Answer:
In BF₃ three bonded pair and zero lone pair electrons are present. Due to this B is sp² hybridized and in NH₃ three bonded pair and one lone pair of electron is present. So, N is sp₃ hybridized. After reaction the hybridization of B becomes sp₃ but the hybridization of N remains same as N gives its lone pair to B atom.

Question 10.
Explain the change in hybridization in A1 atom in following reaction:
AlCl₃ + cl^– → AlCl^–₄
Answer:
The electronic configuration of Al is:
At ground state = ₁₃Al = ls² 2s² 2p⁶ 3s² 3p_x¹
At excited state = 1s², 2s², 2p⁶ 3s² 3p_x¹ _y¹
In the formation of AlCl₃, Al is sp² hybridized and its geometry is trigonal bipyramidal. Whereas in the formation of AlCl₄^–, due to inclusion of 3p_z orbital. Al is sp³ hybridized and its geometry is tetrahedral.

Question 11.
What are the postulates of orbital overlap concept of covalent bond?
Answer:
According to this concept:
1. The covalent bond is formed due to partial overlap of the two half-filled atomic orbitals of the valence shells of the combining atoms. Partial overlap means that a part of the electron cloud of each of the two half – filled domic orbitals becomes common. As a result the probability of finding electrons in the region of overlap is much more at the other places. This reduces the intemuclear repulsion and hence decreases the energy.

2. The orbitals undergoing overlap must have electrons with opposite spins.

3. Greater the extent of overlapping, stronger is the bond formed.

4. Larger the size of the orbitals, less effective is the overlapping and thus weaker is the bond formed.

Question 12.
Explain that the geometry of PCl₅ is trigonal bipyramidal and that of IF₅ is pyramidal?
Answer:
PCl₅: P is central metal atom.

As P in PCl₅ in sp³d hybridized so its geometry is square pyramidal.
IF₅: Central Metal atom is I (Z = 53)

In IF₇ , I is sp³d²

Question 13.
What are σ bond and π bond? Explain with example?
Answer:
Hybridization is defined as, “The process of intermixing of atomic orbitals of nearly equal energy and proper symmetry giving rise to equal number of new orbitals of same energy is called hybridization and the orbitals so formed hybridized orbitals.”

Sigma bond:
The bond formed by overlapping of two orbitals along their axis is called a sigma (σ) bond. The line joining the two nuclei of the combining atoms is called the intemuclear axis or bond axis.
Example: This type of overlapping takes place between s – s orbital, s – p orbital and pz – pz orbitals.

Pi – bond:
The bond formed by the lateral overlapping of two p – orbitals (p_x – p_x) (p_y – p_y) is called π – bond.

It is important to note that overlapping at both the lobes of the p – orbital occurs in Pi – bond whereas in case of sigma bond the overlap occurs in a single region.

Question 14.
Give the reason for the difference in properties of two allotropes diamond and graphite of carbon?
Answer:
Diamond and graphite are two allotropes of carbon. But due to difference in C arrangement their properties are different. In diamond the C atom is sp3 hybridized. Every C atom attached with four cations form tetrahedral geometry. So it forms a lattice structure and so hard and have high melting point.

In graphite every C atom is sp² hybridized, i.e. each C is surrounded by three cations and fourth valency of C is unstable. In graphite different layers are present which are joined together with weak vander Waals’ forces. That is why graphite is soft and due to presence of free electron it conducts electricity.

Question 15.
Explain the hybridized structure of acetylene by diagram?
Answer:
Formation of ethyne or acetylene (HC = CH): In the formation of acetylene molecule, each carbon atom undergoes ip – hybridization leaving two 2p – orbitals in the original unhybridized state. The two sp – hybrid orbitals of carbon atom are linear and are directed at an angle of 180°. Whereas the two unhybridized p – orbitals remain perpendicular to ip – hybrid orbital and also perpendicular to each other.

In the formation of acetylene, ip – hybrid orbital of One C – atom overlap with ip – hybrid orbital of another C – atom along the intemuclear axis forming a σ – bond. The second sp – hybrid orbital of each C – atom overlaps with the half – filled 1s – orbital of H – atom again along intemuclear axis thus forming a-bonds. (MPBoardSolutions.com) Each of the two unhybridized orbitals of both the carbon atoms overlap. Sidewise to form two π – bonds. Thus, all the carbon and hydrogen atoms are linear and there is electron cloud above and below, in the front and at the back of the C – C axis. In other words, there is electron cloud all around the intemuclear axis thus giving a cylindrical shape as represented in fig.

Question 16.
With tetrahedral geometry CH₄ molecule have a possible geometry of square plannar. In which the H atoms are present at the four corners of the square. Explain that the CH₄ molecule is not have square plannar geometry?
Answer:
The electronic configuration of C is:
In ground state ₆C = 1s² 2s² 2p_x¹2p_y¹
In excited state ₆C = 1s² 2s¹2p_x¹ 2p_y¹2p_z¹
sp³ hybridisation.
In CH₄ molecule carbon is sp³ hybridized. So its geometry is tetrahedral. For square plannar geometry dsp² hybridization is necessary. But due to absence of d – orbital in C atom. This geometry is impossible. With this according to VSEPR concept the bonded electrons in C atom is present at four comers of tetrahedron. The bond angle in tetrahedron is 109°28′ and in square plannar 90°. So in case of tetrahedral geometry the repulsion of electrons is less than in square plannar geometry.

Question 17.
On the basis of hybridization explain that the structure of BeCl₂ is linear?
Answer:
Formation of BeCl₂:
In the compound (BeF₂, BeCl₂, etc.), beryllium shows a covalency of two. In order to explain the formation of two equivalent bonds with beryllium its 25 – electron from the ground state (₄Be, Is² 2s²) is excited to 2p – orbital (1s²2s¹ 2p¹]) 2s and 2p – orbitals get mixed up to two equivalent sp – hybrid orbitals which make an angle of 180° with each other and oriented linearly. Each sp – orbitals overlap with half – filled p – orbital of chlorine (1s² 2s² 2p⁶ 3s² 3p_x² 3p_y² 3p_z¹) atoms to form two sigma bonds. Thus, the shape of BeCl₂ is linear.

Question 18.
Explain the formation of N₂ molecule on the basis of orbital theory as overlapping?
Answer:
π – bond formation of N₂ molecules:
Nitrogen molecule has a triple – bond consiting of one σ and two π – bonds (\(N\overset { \pi }{ \underset { \pi }{ \equiv N } } \)). Nitorgem atom has three half – filled p – orbitals.
₇N : 1s² 2s² 2p_x¹2p_y¹ 2p_z¹

When 2p_x orbital of each nitrogen atom overlaps co – axially, a σ – bond is formed. The 2p_y and 2p_z orbitals of one N atom overlap N atom to overlap laterally to form two π – bonds.

Chemical Bonding And Molecular Structure Long Answer Type Questions – I

Question 1.
Write the postulates of Valence bond theory? Write its limitation also?
Answer:
Valence bond theory was given by Heitler and London which is modified by Pauling and Slater. The postulates of the theory are:

1. The covalent bonds are formed by the partial overlapping of atomic orbitals (half filled).
2. In the orbitals taking part in overlapping electrons with opposite spin are present.
3. Strength of bond depends upon the extent of overlapping.
4. Strong directional bonds are formed between the orbitals of same stability, same energy and same symmetry.

Limitations of Valence Bond Theory:

1. According to this theory, no unpaired e^– is present in O² molecules. So the nature of O² is paramagnetic but O² is diamagnetic.
2. Not explain about the formation of coordinate bonds.
3. This theory is failed to explain the formation of H₂⁺ molecules.
4. Doesn’t give any information about resonance.

Question 2.
Write the condition for the formation of molecular orbitals by linear combination of atomic orbitals?
Answer:
Conditions for the combination of Atomic Orbitals:
Molecular orbital is formed by the linear combination of atomic orbitals. There are certain conditions for the effective linear combination of atomic orbitals. These conditions are:

1. The combining atomic orbitals should have same or nearly same energy: This means that in the formation of a homonuclear diatomic molecule Is atomic orbital of one atom will undergo linear combination with 1s atomic orbital of the other atom, but not with the 2s atomic orbital because the energy of the 2s orbital is appreciably higher than that of 1s atomic orbital. Similarly, because of the energy difference between 2s and 2p atomic orbitals, they will also not combine to form molecular orbitals.

2. There should be maximum overlap of atomic orbitals:
Greater the overlap, greater will be the charge density between the nuclei of a molecular orbital. This condition is often referred to as the principle of maximum overlap.

3. The atomic orbitals should have the same symmetry about the molecular axis:
This condition is known as symmetry condition for the combination of atomic orbitals. Taking the Z – axis as the molecular axis, the following pairs of atomic orbitals will not combine to form any molecular orbital, because of their different symmetries.

1. s – p_x pair
2. s – p_y pair
3. p_x – p_y pair and
4. ↔p_y – p_z pair.

This means that, s – s, px- px, py -py and Pz~Pz combinations are allowed because combining atomic orbitals have the same symmetry. A ,pz orbital, however, is able to combine with an s-orbital since, they have the same symmetry.

Question 3.
Differentiate between Ionic compounds and Covalent compounds?
Answer:
Differences between Ionic and Covalent compounds:

Question 4.
On what factors the formation of ionic bond depends?
Answer:
Factors influencing Ionic Bond formation:
The formation of ionic bond (Electrostatic force of attraction) depends upon the following factors:
1. Ionisation enthalpy:
One of the combining atoms (metal) must have low ionisa-tion enthalpy. So that, cation formation becomes easy. For example, alkali and alkaline earth metals of periodic table has tendency to form positive ion because they have comparatively low ionisation energy.

2. Electron gain enthalpy:
The electrons released in the formation of cation are to be accepted by the other atom taking part in the ionic bond formation. The electron accepting tendencies of an atom depends upon the electron gain enthalpy. (MPBoardSolutions.com) It may be defined as: Energy released when an isolated gaseous atom takes up an electron to form an anion. Greater the negative electron gain enthalpy, easier will be the formation of anion or negative ion. The halogen present in group 17 have the maximum tendency to form anions as they have very high negative electron gain enthalpy.

3. Lattice energy:
The amount of energy released when one mole of ionic solid is formed by the close packing of its constituents. It is denoted by ∆_LH and negative in nature.
A_(g)⁺ + B_(g)^– → A⁺B_(s)^– + Lattice enthalpy (∆_LH)
Thus, greater the magnitude of -ve lattice energy, more will be the stability of the ionic bond or ionic compound.

Question 5.
Explain the main points of Molecular orbital theory?
Answer:
The main points of Molecular orbital theory are:
1. In a molecule, electrons are present in new orbitals called molecular orbitals. These molecular orbitals are characterised by a set of quantum numbers just like atomic orbitals.

2. Molecular orbitals are formed by combination of atomic orbitals of equal energies (in case of homonuclear molecules) or of comparable energies (in case of heteronuclear molecules).

3. The number of molecular orbitals formed is equal to the number of atomic orbitals undergoing combination.

4. Two molecular orbitals are formed by combination of two atomic orbitals one of these two molecular orbitals has a lower energy and the other has a higher energy than either of the combining atomic orbitals. The molecular orbital with lower energy is called bonding molecular orbital and the other is called antibonding molecular orbital.

Question 6.
What is resonance? Explain with example?
Answer:
When properties of a molecule are not explained by one structure and two or more than two structures are assigned to express its characteristics, it is said that molecule is resonance hybrid of these structures and this property is known as resonance. Different resonating structures are exhibited by using sign (↔) in between these structures.
Example: Carbon dioxide (CO₂) is represented by following three structures:

Chemical Bonding And Molecular Structure Long Answer Type Questions – II

Question 1.
Show the molecular orbital energy levels of N₂ by diagram?
Answer:
N₂ molecule: Each nitrogen atom contains seven electron. Thus total 14 electrons are filled in seven orbitals of increasing energy.

Molecular orbital structure of N₂ molecule will be as follows:

Where KK represents closed shell structure (σ _1s)²(\(({ \sigma _{ 1s } })^{ 2 }\))(\(({ \sigma ^{ * }_{ 1s } })^{ 2 }\)
This structure shows that it contains 10 bonding and 4 antibonding electrons.
Bond order = \(\frac{1}{2}\) [N_b – N_a] = \(\frac{1}{2}\) [10 – 4] = 3.
Value of bond order is more. Hence value of bond energy should also be high. Experimental value of bond energy is 945 kJ mol^-1 which proves the presence of paired electrons in nitrogen molecule. Thus, it is a diamagnetic molecule.

How to Calculate Bond Order. Calculations Step by Step.

Question 2.
Give the important applications of dipole moment?
Answer:
Applications of Dipole moment:
1. Comparison of relative polarity:
It is possible by comparing the value of dipole moment e.g., HF (1.98 D) is more polar than HCl (1.03D).

2. Predicting the nature of molecules:
Molecules with specific dipole moments are polar in nature while those with zero value are non – polar. Thus, BeF₂ (µ – 0D) is non – polar while H₂O (µ = 1.84D) is polar.

3. Calculation of percentage ionic character:
% ionic character = \(\frac{Observed dipole moment}{Caluculated dipole moment}\) × 100 (100% ionic character)
or % I.C. = \(\frac { \mu _{ obs } }{ \mu _{ cal } } \) × 100
For example, the observed dipole moment of HCl molecule is 1.03D. For 100% ionic character i.e., complete transfer of electron charge on H⁺ and Cl^– ions would be equal to one unit (4.8 × 10^-10e.s.u.)× (1.275 × 10^-8 each. The bond length of H – Cl bond is 1.275 × 10^-8cm. Therefore, dipole moment for complete electron transfer
µ = q × d = (4.8 × 10^-10 e.s.u) × (1.275 × 10^-8cm)
= 6.12 × 10^-18 e.s.u cm = 6.12D
Observed dipole moment, µ(obs) = 1.03D
% ionic character = \(\frac { 1.03 }{ 6.12 } \) × 100 = 16.83 %

4. Dipole moment of symmetric molecules is zero, but they have two or more than two polar bonds. It is applied for the measurement of symmetry.

5. Distinction between ortho, meta and para isomers of aromatic compounds:
In general, the dipole moments follow the order: ortho > meta > para e.g., In dichlorobenzene, the dipole moments of o, m and p isomers are: 2.54 D, 1.48 D and 0 respectively.

Question 3.
What is valence shell electron pair repulsion theory? Write its limitations? Or, Explain valence shell electron pair repulsion theory with example?
Answer:
VSEPR theory:
On the basis of this theory, “When central atom is surrounded by only bonded electron pairs, in that case the geometry of the molecule will be ordinary, but if the central atom is surrounded by bonded electron pairs as well as lone pairs or non – bonded pairs, then the geometry of die molecule will become abnormal.” This (MPBoardSolutions.com) means that the repulsion between the non – bonded or lone pair of electrons and bonded electron pairs become greater than that of repulsion between only bonded electron pairs. Repulsion

Between the electron pairs is in the following order:
lone pair – lone pair > lone pair – bonded pair > bonded pair – bonded pair

Shapes of some molecules accroding to VSEPR theory:
1. Shape of CH₄:
In methane, central atom carbon is surrounded by four electron pairs and four C – H bonds. This molecule has tetrahedral geometry. Shared electrons are at the comers of tetrahedron for maximum separation. Bond angle is 109°28′.

2. Shape of H_2O:
In water molecule central oxygen atom is surrounded by four electron pairs, two of which are lone pair of electrons. Thus, Ip – lp and lp – bp repulsion exist. Due to this, bond angle reduces to 104.5° in place of 109°28′ and shape becomes V shaped.

3. Shape of NH₃:
In ammonia, central N atom is surrounded by four electron pairs, so shape of molecule is tetrahedral. According to VSEPR theory, the four groups around the central atom of ammonia should be tetrahedrally arranged at bond angle of 109°28’.

But, the measured bond angle is 107°. This is explained on the basis of repulsive effect of the lone pair of electrons on bonding electrons. In ammonia molecule there is a lone pair of electrons on the N atom. Thus, the shape of NH₃ molecule is distorted and it looks like pyramidal and it is polar in nature.

Example:

Limitations of VSEPR theory:
VSEPR theory no doubt, theoretically gives the shapes of simple molecules but could not explain them and also has limited application. To overcome these limitations, two important theories based on quantum mechanical principles are commonly used. These are:

1. Valence bond theory (VBT) and
2. Molecular orbital theory (MOt).

Question 4.
Structure of two molecules are given:

1. Among these which contain intermolecular hydrogen bond and intramolecular hydrogen bond?
2. The melting point of any compound depends upon hydrogen bonding also, on the basis of this explain which one have high M.P.
3. The solubility of any compound depends upon its tendency to formed H – bonding with water among these who will form H – bond with water easily.

Answer:
1. Compound (a) forms intramolecular H – bond. When H – bond is present between the atoms of same molecule than it is called intramolecular H – bond.
In ortho nitrophenol [(a)] H – bond is present between two O – atoms.

In compound (b) inter molecular H – bond forms.
In p – nitrophenol [(b)] between – N0₂ and – OH vacant space is present. So among H – atom of one molecule and O – atom of another molecule, H – bond is formed.

2. The melting point of compound (b) is high because many molecules are forming H – bond.

3. Due to intramolecular H – bonding, compound (a) will not form H – bond with water so it is less soluble in water. Whereas compound (b) forms H – bond easily with water so soluble in water.

Question 5.
Draw molecular orbital diagram for 0₂ molecule?
Answer:
Oxygen molecule, (0₂): Each oxygen atom has eight electrons. When two oxygen atom combine, molecular orbitals are formed. These molecular orbitals have following configuration:

Its molecular orbital diagram is:

From the above configuration we have,
N_b = 10, N_a = 6
∴ Bond order = \(\frac{1}{2}\) [N_b – N_a] = \(\frac{1}{2}\) [10 – 6] = 2
Hence, there is a double bond in oxygen molecule. Due to the presence of two unpaired electrons it is paramagnetic.

MP Board Class 11 Chemistry Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 5 States of Matter

States of Matter Important Questions

States of Matter Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
At the time of opening the bottle of ammonia, it can be recognized from a distance because:
(a) It is very reactive
(b) It diffuses very fast
(c) It possess a pungent smell
(d) It is lighter than air
Answer:
(b) It diffuses very fast

Charles’ Law Calculator uses formula that relates the initial and final volume and temperature values of an ideal gas at constant pressure.

Question 2.
Who established the relationship between density and rate of diffusion of a gas:
(a) Boyle
(b) Charles
(c) Graham
(d) Avogadro
Answer:
(c) Graham

Question 3.
The value of in Calorie (approx):
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 4.
Value of gas constant R is:
(a) 8314 × 10⁷ ergs/degree/mol
(b) 83.14 × 10⁶ ergs/degree/mol
(c) 83.14 × 10⁵ ergs/degree/mol
(d) 8.314 × 10⁷ ergs/degree/mol
Answer:
(d) 8.314 × 10⁷ ergs/degree/mol

Question 5.
Absolute temperature is:
(a) 0° C
(b) – 100° C
(c) – 273° C
(d) – 373°C
Answer:
(c) – 273° C

Question 6.
To get general gas equation which two laws are combined:
(a) Charle’s law and Dalton’s law
(b) Graham’s law and Dalton’s law
(c) Boyle’s law and Charle’s law
(d) Avogadro’s law and Dalton’s law
Answer:
(c) Boyle’s law and Charle’s law

Question 7.
Which is correct in the following:
(a) r.m.s. velocity = 0.9212 × average velocity
(b) average velocity = 0.9212 × r.m.s. velocity
(c) r.m.s. velocity = 0.9013 × average velocity
(d) average velocity = 0.9013 × r.m.s. velocity
Answer:
(b) average velocity = 0.9212 × r.m.s. velocity

Question 8.
At constant volume the pressure of monoatomic gas depends on:
(a) Thickness of the wall of the vessel
(b) Absolute temperature
(c) Atomic number of the element
(d) Number of valence electron.
Answer:
(b) Absolute temperature

Question 9.
Behaviour of real gases near to that of ideal gases if:
(a) Temperature is low
(b) Pressure is high
(c) Low pressure and high temperature
(d) Gas is monoatomic
Answer:
(c) Low pressure and high temperature

Question 10.
At mountains of high altitude, water boils at lower temperature because of:
(a) Low atmospheric pressure
(b) High atmospheric pressure
(c) Hydrogen bonding in water is more strong at height
(d) Water vapour is lighter than liquid
Answer:
(a) Low atmospheric pressure

Question 11.
If molecular masses of two gases A and B are 16 and 64 respectively ratio of rates of diffusion A and B will be:
(a) 1:4
(b) 4:1
(c) 2:1
(d) 1:2
Answer:
(c) 2:1

Question 12.
Gases deviate from ideal behavior at high pressure because:
(a) At pressure number of colloision of molecule increases
(b) Attraction between molecules increases at high pressure
(c) Size of molecule decreases at high pressure.
(d) Molecule become steady at high pressure.
Answer:
(b) Attraction between molecules increases at high pressure

Question 13.
Distance on which magnitude of energy is minimum is called:
(a) Atomic radius
(b) Lattice radius
(c) Critical distance
(d) Molecular distance
Answer:
(a) Atomic radius

Question 14.
Diffusion rate of methane is twice than that of gas x, molecular mass of gas x will be:
(a) 64
(b) 32
(c) 40
(d) 8.0
Answer:
(a) 64

Question 15.
Part of van der Waals’ equation which illustrate the inter molecule force of real gases:
(a) (V – b)
(b) RT
(c) [P+\(\frac{a}{VL}\)]
(d) (RT)^-1
Answer:
(c) [P+\(\frac{a}{VL}\)]

Question 16.
Temperature which is same in both Celsius scale and Fahrenheit scale:
(a) 0°C
(b) 32°F
(c) – 40°C
(d) 40°C
Answer:
(c) – 40°C

Question 2.
Fill in the blanks:

1. With increase in temperature viscocity ………………………..
2. S.I. unit of surface tension is ………………………….
3. ………………………… is more heavy among dry air and moist air
4. AT N.T.P., volume of gas equal to Avogrado’s numbers ………………………….
5. Unit of vander Waals’ constant ‘b’ is ………………………..
6. …………………………. has minimum value among average velocity, root mean square velocity and most probable velocity
7. Process of diffusion of air from the puncture of an autommobile tube is known as ……………………….
8. …………………………… is obtained if a graph is plotted between volume and absolute temperature
9. Apparatus used for measuring gas pressure is …………………………
10. S.I. unit of viscosity is ………………………….
11. Absolute temperature of ideal gas is ………………………… to the average kinetic theory of the molecule.
12. A liquid which is permanently super cooled is called ………………………..
13. According to kinetic theory, the average kinetic energy of a gas is directly proportional to its …………………………. temperature.
14. The kinetic energy of one mole of a gas is equal to …………………………..
15. Root mean square velocity is …………………………..

Answer:

1. Decreases
2. Nm^-1
3. Dry air
4. 22.41
5. Litre/mol
6. Most probable velocity
7. Effiission
8. Straight line
9. Monometer
10. Nm^-2S
11. Proportional
12. Glass
13. Absolute
14. \(\frac{3}{2}\) RT
15. \(\sqrt { \frac { 3PV }{ M } } \) or \(\sqrt { \frac { 3RT }{ M } } \)

Question 3.
Answer in one word/sentence:

1. Vaporization of ethanol is faster as compared to water?
2. The resistance produced in the flow of a liquid is called?
3. Unit of surface tension is?
4. Molecular mass of two gases A and B are 36 and 64 respectively. What will be the ratio of diffusion of the two gases?
5. Write S.I. unit of pressure?
6. If rate of diffusion of oxygen is r, then tell the rate of diffusion of hydrogen?
7. Compressibility factor of ideal gases is?
8. Vander Waals’ equation is?
9. Adiabatic expression of ideal gas is?
10. Number of electron in outer most orbit of crypton is?

Answer:

1. Value of molar vaporization enthalpy of water is high
2. Viscosity
3. Dyne per cm
4. 4:3
5. pascal
6. r₁ = 4r [\(\frac { r_{ 1 } }{ r_{ 2 } } \) = \(\sqrt { \frac { m_{ 2 } }{ m_{ 1 } } } \)]
7. 1.0
8. [P + \(\frac { an^{ 2 } }{ V^{ 2 } } \)] [V – nb] = nRT
9. q = 0
10. 8 electron

Question 4.
Match the following:
[I]

Answer:

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

[II]

Answer:

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

States of Matter Very Short Answer Type Questions

Question 1.
The resistance produced in the flow of a liquid is called?
Answer:
Viscosity.

Question 2.
Unit of surface tension is?
Answer:
Dyne per cm.

Question 3.
Write S.I. unit of pressure?
Answer:
Pascal.

Question 4.
Name the scientist who developed relation between density and rate of diffusion of gas?
Answer:
Graham’s.

Question 5.
What is the Value of gas constant in S.I. unit?
Answer:
8.314 JK^-1 mol^-1

Question 6.
1 Pascal is equal to?
Answer:
1 Nm².

Question 7.
Who established the relationship between density and rate of diffusion of gas?
Answer:
Graham’s.

Question 8.
Absolute temperature is?
Answer:
~273°C.

Question 9.
The value of R in calorie (approx)?
Answer:
2.

Question 10.
Behaviour of real gases is near to that of ideal gases of?
Answer:
Low pressure and high temperature.

Question 11.
At mountains or high altitude, water boils at lower temperature because of?
Answer:
Low atmospheric pressure.

Question 12.
To get general gas equation. Which two laws are combined?
Answer:
Boyle’s law and Charle’s law.

Question 13.
At constant volume the pressure of gas depends upon?
Answer:
On absolute temperature.

Question 14.
Part of van der Waals’ equation which illustrate the inter – molecular forces of real gases?
Answer:
P + \(\frac { a }{ V^{ 2 } } \).

Question 15.
Temperature which is same in both Celsius scale and fahrenheit scale?
Answer:
– 40°C.

Question 16.
What is the formula for kinetic theory of gases?
Answer:
PV = \(\frac{1}{3}\) mnv².

Question 17.
The average kinetic energy of gases is proportional to?
Answer:
Absolute temperature.

Question 18.
What is the kinetic energy of 1 mole of gas?
Answer:
\(\frac{3}{2}\) RT.

Question 19.
What is the formula of Root mean square velocity?
Answer:
\(\sqrt { \frac { 3PV }{ M } } \) or \(\sqrt { \frac { 3RT }{ M } } \).

Question 20.
Which type of Crystals are diamond and ice?
Answer:
Diamond – Covalent crystal
Ice – Ionic crystal.

Question 21.
What is Critical temperature?
Answer:
The temperature above which the gas cannot be liquefied.

Question 22.
Poise is the unit of which basic property?
Answer:
Viscosity (1 poise = dynes/cm²s).

Question 23.
What is the unit of a (volume correction), b (pressure correction) in van der Waals’ equation?
Answer:
a (Volume correction) = atm L² mol^-2
b (Pressure correction) = L mol^-1

States of Matter Short Answer Type Question – I

Question 1.
Explain Anisotropic and Isotropic?
Answer:
Anisotropic:
The crystalline solid exhibits different physical properties in the , three direction. In this way, crystalline solids are called anisotropic.

Isotropic:
The amorphous solid exhibits same physical properties in all directions. Due to this property they are called isotropic.

Question 2.
Define the term absolute zero? Write its value in centigrade scale?
Answer:
The lowest possible temperature at which all the gases are supposed to occupy zero volume is called absolute zero. The actual value of absolute zero is – 273.15°C. It is , related to temperature in centigrade scale by this relation.
t°C = t + 273 K.

Question 3.
What is unit cell?
Answer:
Smallest unit is a crystal which is formed by systematic arrangements of constituent particles as atom, molecule or ions, is called unit cell. The unit cell generates the whole lattice translation.

Question 4.
What is crystal lattice?
Answer:
Geometry or shape of any crystal in which unit cells are arranged systematically and three dimensionally is called crystal lattice.

Question 5.
How does volume of balloon used for weather study, change with height?
Answer:
At height, atmospheric pressure decreases. The volume of gas inside the balloon increases with decrease in pressure. A stage comes when due to decrease in atmospheric pressure in larger extent, volume increases and balloon bursts.

Question 6.
In winter season, a layer of ice is formed in the lake but the fishes and other organisms present in the lake remain alive. Why?
Answer:
The maximum density of water is at 4°C but below 4°C temperature the density decreases. When the temperature of lake decreases then the water present on the surface become denser and goes downward. This occurs upto the level when the temperature rises to t 4°C. (MPBoardSolutions.com) If the temperature of the upper layer is less than 4°C, the water remains on the upper surface and converts into ice slowly and the water below the surface remains as such due to high density and remains as liquid. That is why, the fishes and micro – organisms remains alive.

Question 7.
Why is the density of hot gas is less in comparison to cold gas?
Answer:
According to Charle’s law, volume of any gas of definite mass is directly proportional to absolute temperature. On increasing the temperature volume of gas also increases, but increase in volume results decrease in density.

Question 8.
Why one feel sluggish, breathlessness and headache at high altitude?
Answer:
At high altitude, the pressure is less and the corresponding volume of air is more, Thus, air becomes less dense at high altitude and the oxygen in air becomes insufficient for normal breathing. This causes what is known as altitude sickness.

Question 9.
What is Critical temperature?
Answer:
The temperature to which gas must be cooled before it can be liquefied by com-pression is known as critical temperature and is represented by Tc.
Example: Critical temperature of CO₂ gas is 31.1°C or 304.1K.

Question 10.
What is critical pressure and critical volume?
Answer:
Critical pressure:
The minimum pressure required to liquefy the gas at its critical temperature is known as critical pressure and denoted by P_c.
Example: Critical pressure of CO₂ is 72.8 atm.
Critical Volume: The volume occupied by 1 mole of gas at the critical temperature and critical pressure is known as critical volume and denoted by V_c.
Example: Critical volume of C0₂ gas is 94 cm³/mol.

Question 11.
Why are tyres of automobile inflated to lesser pressure in summer than in winter?
Answer:
As the automobiles move the temp, of tyre increases due to friction against the road. Consequently the air inside the tyre expand thereby the pressure exerted by air against the wall of tyre also increases. (MPBoardSolutions.com) In summer, there is increase in temperature hence increase in pressure is much more. This may leads bursting at tyre. In order to check bursting of tyre, the tyre are inflated with looser amount of air than in winter.

Question 12.
What would be the S.I. unit for the quantity PV²T²/n?
Answer:
\(\frac { PV^{ 2 }T^{ 2 } }{ n } \) = \(\frac { (Nm^{ -2 })(m^{ 3 })^{ 2 }(K)^{ 2 } }{ mol } \) = Nm⁴K²mol^-1.

Question 13.
Explain on the basis of Charle’s law that minimum possible temperature is – 273°C
Answer:
According to Charle’s law:
V_t = V₀ [1 + \(\frac { t }{ 273 } \)]
At t = – 273°C V_t = V₀ [ 1 – \(\frac { 273}{ 273 } \)] = 0
Therefore at – 273°C, the volume of gas becomes 0 and below this temperature the volume becomes – ve which is meaningless.

Question 14.
Why are liquid drops spherical?
Answer:
Small drops are spherical in shape:
Surface tension tries to decrease the surface area of a given liquid for a given volume. Therefore, drops of liquid are spherical because for a given volume sphere has minimum volume.

Question 15.
What is Root Mean Square velocity and Average velocity?
Answer:
1. Root Mean Square velocity:
It is defined as root of, mean of, square of, velocity of large no. of molecules of same gas. It is denoted by V.
V = \(\sqrt { \frac { v_{ 1 }^{ 2 }+v_{ 2 }^{ 2 }+v_{ 3 }^{ 2 }…..v_{ n }^{ 2 } }{ n } } \)
2. Average velocity:
It is defined as average of, velocity of all the molecules present in gas. It is denoted by V_a.
V_a = \(\frac { v_{ 1 }+v_{ 2 }+v_{ 3 }…..v_{ n } }{ n } \)

Question 16.
Explain the difference between Evaporation and Boiling?
Answer:
Differences between Evaporation and Boiling:
Evaporation:

1. Evaporation decreases spontaneously and occur at all temperatures.
2. Evaporation is a process of the liquid surface.
3. Evaporation is a slow process.

Boiling:

1. Boling takes place only when the vapour pressure of this liquid becomes equal to atmospheric pressure.
2. Boling is a process of the entire liquid and occurs in the form of bubbles inside the liquid.
3. Boling is a fast process.

Question 17.
What do you mean by compressibility factor of gases?
Answer:
The ratio of observed volume and caluculated volume of a gas at a given temperature and pressure is known as compressibility factor. It is denoted by Z.
Thus,

or Z = \(\frac { PV }{ nRT } \)
For ideal gases, PV = nRrt
∴For ideal gas, Z = 1.

Question 18.
What is the effect of pressure on melting of ice?
Answer:
By increasing pressure there occur tremendous increase in kinetic energy of molecules, due to this at low temperature, the molecules move freely, hence on increasing pressure the ice below its melting point converted into liquid.

Question 19.
Mountaineers carry oxygen cylinders with them at the lance of climbing mountains. Why?
Answer:
Atmospheric pressure is relatively low at heights. Quantity of oxygen is low in mountain and climbers feel difficulties in breathing. Therefore, they carry oxygen cylinders along with them.

Question 20.
Define viscosity of liquid. Explain the effect of temperature on viscosity?
Answer:
Resistance in flow of any liquid is called viscosity. Such resistance is produced due to internal friction of different layers of liquid. (MPBoardSolutions.com) When temperature is increased, the cohesive force, which opposes liquid flow, decreases and molecular velocity increases. Due to this, viscosity decreases.

Question 21.
On same temperature when ether and water pour on different hands, then ether seems to be more colder than water. Why?
Answer:
In ether the intermolecular attractive forces between the molecules is less in comparison to water, so ether evaporates more quickly than water and the energy required for evaporation is absorbed from hand, that is why ether seems to be colder.

Question 22.
What is surface tension? Write its S.L unit?
Answer:
It is an important property of a liquid which is related with interatomic attraction force. The molecules present inside the liquid is attracted equally by molecules present in all direction. (MPBoardSolutions.com)
But molecule present on the surface of liquid is attracted by molecules at bottom and in sides, as a result the molecules at surface are pulled downward and nature of surface is to lessen the area. Due to compactness, the surface of liquid behaves as a stretched membrane. This effect is called surface tension.

“Surface tension is a measure of work which is necessary to increase the unit cross-section of liquid.”
Its S.I. unit is Joule/metre² or Newton metre.

Question 23.
The compressibility factor Z of a gas is as follows:

1. What will be the value of Z for an ideal gas?
2. What will be the effect on Z above Boyle’s temperature for real gas?

Answer:

1. For ideal gas, compressibility factor Z = 1.
2. Above Boyle’s temperature, real gases show positive deviation. So, Z > 1.

Question 24.
What is Ideal gas? Write its characteristics?
Answer:
Ideal Gas:
The gases which obey Ideal gas equation under all conditions of temperature and pressure is called ideal gas.
Characteristics:

1. At constant temperature, product of pressure and volume of ideal gas are always constant. Therefore, horizontal line should be obtained in a graph. If graph is plotted between PV and P.
2. If an ideal gas is called at constant pressure, then its volume requestly decreased and become at – 273°C.
3. There is no force of attraction between gas molecules.
4. The compressibility factor of ideal gas is equal to one.

Question 25.
What is Real gas? What are its properties?
Answer:
Gas which does not obey Boyle’s law, Charle’s law and Ideal gas equation strictly is called Real gas.
The gases which does not follow ideal gas equation behaviour under all condtions of temperature and pressure called real gas.
Properties:

1. They do not follow gas law at low temperature and high pressure.
2. At – 273°C their volume is not zero because most of the gases converted into liquid on cooling.
3. The attractive force between gas molecule is negligible.
4. The compressibility factor of real gas is not equal to zero.

States of Matter Short Answer Type Questions – II

Question 1.
State and explain Boyle’s Law?
Answer:
According to this law: “At constant temperature, the volume of a known amount of gas is inversely proportional to the pressure.”
P ∝ \(\frac{1}{V}\) (at constant temperature)
⇒P = Constant × \(\frac{1}{V}\)
⇒ PV = Constant.
Thus, “at constant temperature, product of volume of a given mass of gas and pressure remain constant.”
At initial condition, P₁V₁ = K …………….. (1)
At final condition, P₂V₂ = K ………………. (2)
From eqn. (1) and eqn. (2).
P₁V₁ = P₂V₂

Question 2.
What is the nature of gas constant R?
Answer:
We know that,
PV = nRT
R = \(\frac{PV}{nT}\)

Question 3.
Explain concept of absolute zero from Charle’s law?
Answer:
Charle’s law: According to this law “At a constant pressure the volume of certain mass of a gas increases or decreases by of its previous volume for every 1°C changes in temperature (increases or decreases).”
Suppose V₀ is the volume of certain mass of gas at 0°C then,
Volume of the gas at 1°C temperature = V₀ [1 + \(\frac{1}{273}\)]
Volume of the gas at t°C temperature = V₀ [1 + \(\frac{t}{273}\)]
Volume of the gas at – 1°C temperature = V₀ [1 – \(\frac{t}{273}\)]
Volume of gas at – 273°C = V₀ [1 – \(\frac{273}{273}\)] = 0
Thus decrease of temperature results in the decrease in volume of the gas and ultimately the volume should become zero at – 273°C. (MPBoardSolutions.com) This lowest possible temperature at which all the gases are suppossed to occupy zero volume is called absolute zero.

Question 4.
What is Gay Lussac’s law?
Answer:
Gay Lussac law:
According to this law, “At constant volume the pressure of a given mass of gas is directly proportional to its absolute temperature.”
P ∝ T (Mass and volume are constant)
⇒ P = K × T
⇒ \(\frac{P}{T}\) = K
Suppose at initial condition,
\(\frac { P_{ 1 } }{ T_{ 1 } } \) = K
At final condition,
\(\frac { P_{ 2 } }{ T_{ 2 } } \) = K
From eqn. (1) and (2),
\(\frac { P_{ 1 } }{ T_{ 1 } } \) = \(\frac { P_{ 2 } }{ T_{ 2 } } \)

Question 5.
State and explain Avogadro’s law?
Answer:
According to this law, “Equal volume of all gases under identical conditions of temperature and pressure contain equal number of molecules”.
V ∝N (at constant temperature and pressure) …………. (1)
At constant temperature and pressure number of moles of a gas n is directly proportional to number of molecules N.
Hence, N ∝n
⇒\(\frac{V}{n}\) = Constant
Suppose at initial condition volume of gas is V₁ and no. of mole of gas is n₁ hence
\(\frac { V_{ 1 } }{ n_{ 1 } } \) = Constant …………… (2)
Similarly at final condition no. of moles and volume of gas is n₂ and V₂, hence
\(\frac { V_{ 2 } }{ n_{ 2 } } \) = Constant …………… (3)
From eqns. (2) and (3)
\(\frac { V_{ 1 } }{ n_{ 1 } } \) = \(\frac { V_{ 2 } }{ n_{ 2 } } \)

Question 6.
Write the applications of Graham’s law of diffusion?
Answer:
1. To determine the density and molecular weight of a gas:
If the time of diffusion and density of a gas is known and the time of diffusion of other gas is known, then the density and molecular weight of other gas can be calculated.

2. Marsh gas indicator:
The persons working in the mines get aware by the leakage of the poisonous gases by this indicator.

3. In separation of gases:
The gases can be separated easily due to difference in the rate of diffusion of gases.

4. Smell:
Bad smell and poisonous gases get separated due to diffusion in air.

Question 7.
Derive Charle’s law on the basis of Kinetic gas theory?
Answer:
According to Kinetic gas theory, kinetic energy of gases is directly proportional to absolute temperature.
K.E ∝T
\(\frac{1}{2}\) mnv² ∝ T
⇒\(\frac{1}{2}\) mnv² = KT
⇒\(\frac{3}{2}\) × \(\frac{1}{3}\) mnv² = KT
⇒\(\frac{1}{3}\) mnv² = \(\frac{2}{3}\) KT
⇒PV = \(\frac{2}{3}\) KT,
⇒V = \(\frac{2}{3}\) \(\frac{K}{P}\).T
At constant pressure \(\frac{2}{3}\) \(\frac{K}{P}\) = constant
V = constant ∝ T
V ∝ T

Question 8.
How are rates of diffusion of different gases compared?
Answer:
Let two gases are A and B, equal volume V of both gases diffuse in times t_A and t_B respectively.
Rate of diffusion of gas A,
r _A = \(\frac { V }{ t_{ A } } \)
Rate of diffusion of gas B,
r _B = \(\frac { V }{ t_{ B } } \)
∴\(\frac { r_{ A } }{ r_{ B } } \) = \(\sqrt { \frac { d_{ B } }{ d_{ A } } } \)
From equation (3) and equation (4),
\(\frac { t_{ B } }{ t_{ A } } \) = \(\sqrt { \frac { d_{ B } }{ d_{ A } } } \).

Question 9.
The ratio between the rate of diffusion of an unknown gas (x) s&d CO₂ Is 40:45 Find out the molecular mass of unknown gas (x).
Solution:
According to Graham’s law of diffusion,
\(\frac { r_{ 1 } }{ r_{ 2 } } \) = \(\sqrt { \frac { M_{ 2 } }{ M_{ 1 } } } \)
Given, r₁:r₂ = 40:45 M₂(CO₂) = 44, M₁ = ?
⇒\(\frac{40}{45}\) = \(\sqrt { \frac { 44 }{ M_{ 1 } } } \)
⇒ \(\frac { (40)^{ 2 } }{ (45)^{ 2 } } \) = \(\sqrt { \frac { 44 }{ M_{ 1 } } } \)
M₁ = \(\frac { 44\times 45\times 45 }{ 40\times 40 } \) = 55.68.

Question 10.
If relative density of chlorine is 36, diffusion of 25 volume of hydrogen takes 40 sec. under the condition how much time will be taken for the diffusion of 30 volume of chlorine?
Solution:
Hydrogen d₁ = 1, r₁ = \(\frac{25}{40}\), Chlorine d₂ = 36, r₂ = \(\frac{30}{t}\)

States of Matter Long Answer Type Questions

Question 1.
State and explain Graham’s law of diffusion?
Answer:
Graham’s Law of Diffusion:
The rate of diffusion of gas under similar condition of temperature and pressure is inversely proportional to the square roots of their density.”
Thus, Rate of diffusion ∝\(\frac { 1 }{ \sqrt { density } } \)
⇒r ∝\(\frac { 1 }{ \sqrt { d } } \)
If the rate of diffusion of two gases are r₁ and r₂ and their density are d₁ and d₂ respectively.
r₁ = K \(\frac { 1 }{ \sqrt { d_{ 1 } } } \)
r₁ = K \(\frac { 1 }{ \sqrt { d_{ 2 } } } \)
\(\frac { r_{ 1 } }{ r_{ 2 } } \) = \(\sqrt { \frac { d_{ 2 } }{ d_{ 1 } } } \)
∵ M. Mass = 2 × Vapour density = \(\frac { Molecular\quad mass }{ 2 } \) = Vapour density
\(\frac { r_{ 1 } }{ r_{ 2 } } \) = \(\sqrt { \frac { d_{ 2 } }{ d_{ 1 } } } \) = \(\sqrt { \frac { M_{ 2 } }{ M_{ 1 } } } \).

Question 2.
Write difference between Real and Ideal gas?
Answer:
Differences between Ideal gas and Real gas:
Ideal gas:

1. Ideal gas obeys die equation, PV = RT at all temperature and pressure.
2. There is no ideal gas, they are hypothetical.
3. Total volume of gas molecules is supposed to be negligible in comparison to total volume.
4. There is no attraction force between gas molecules.

Real gas:

1. Real gas obeys PV = RT only at low pressure and high temperature.
2. All existing gases are real gases and show deviation from ideal gas behaviour, positive or negative.
3. Volume of gas molecules are not negligible in comparison to total volume.
4. Gas molecules attract each other.Therefore, total pressure is less than ideal gas.

Question 3.
What are the main differences between Crystalline and Amorphous solids?
Answer:
Differences between Crystalline and Amorphous solids:
Crystalline:

1. In it the constituent particles are arranged in regular manner.
2. Melting point of crystalline solid is fixed.
3. They are an anisotropic i.e. some of their physical properties are different in different directions.
4. They are rigid and their shape is not distorted by mild distorting tone.
5. They have a definite heat of fusion.
6. They are true solid in real meaning.

Amorphous solids:

1. In it the constituent particles are arranged in irregular manner.
2. Melting point of amorphous solid is not fixed.
3. They are Isotropic i.e. their physical properties are same in all directions.
4. They are not very rigid, they can be distorted easily.
5. They do not have definite heat of fusion.
6. They are super cooled liquid.

Question 4.
Using state equation clarify that the density of a gas at given temperature is proportional to pressure of gas?
Answer:
PV = nRT
PV = \(\frac{m}{M}\) RT,
Or P = \(\frac{mRT}{VM}\),

Or P = \(\frac{dRT}{M}\)
Or d = \(\frac{PM}{RT}\)
if T = known constant
∴d ∝P

Question 5.
Derive Ideal gas equation on the basis of kinetic gas equation?
Answer:
According to Kinetic gas theory, “ The average kinetic energy of gas molecules is directly proportional to absolute temperature.”
Average Kinetic energy = \(\frac{1}{2}\) mnv²
\(\frac{1}{2}\) mnv² ∝ T
⇒\(\frac{1}{2}\) mnv² = KT
⇒\(\frac{3}{2}\) × \(\frac{1}{3}\) mnv² = KT
⇒\(\frac{1}{3}\) mnv² = \(\frac{2}{3}\) KT,
⇒\(\frac{PV}{T}\) = R
⇒ PV = RT.

Question 6.
Explanation of Dalton’s law on the basis of Kinetic gas theory?
Answer:
If the volume of container V litre and no. of moles of gas A is n₁ ar.d mass of each particles in m₁ R.M.S. velocity is V₁ then,
P_A = \(\frac{1}{3}\) \(\frac { m_{ 1 }n_{ 1 }v_{ 1 }^{ 2 } }{ V } \)
Kinetic gas equation PV = \(\frac{1}{3}\) mnv²
For B gas number of moles = n₂
Mass = m₂
R.M.S. Velocity = v₂
P_B = \(\frac{1}{3}\) \(\frac { m_{ 2 }n_{ 2 }v_{ 2 }^{ 2 } }{ V } \)
If both the gases are kept in same container total pressure of mixture
P = \(\frac{1}{3}\) \(\frac { m_{ 1 }n_{ 1 }v_{ 1 }^{ 2 } }{ V } \) + \(\frac{1}{3}\) \(\frac { m_{ 2 }n_{ 2 }v_{ 2 }^{ 2 } }{ V } \)
⇒ P = P_A + P_B
so in general, for more than two gases
P = P_A + P_B + P_C + …………………..
It is Daltons law of partial pressure.

MP Board Class 11 Chemistry Important Questions

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 3 Matrices Ex 3.3 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 3 Matrices Ex 3.3

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 4 Determinants Ex 4.1 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 4 Determinants Ex 4.1

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 3 Matrices Ex 3.1 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 3 Matrices Ex 3.1

MP Board Class 7th Science Solutions Chapter 8 Winds, Storms and Cyclones

Winds, Storms and Cyclones Intext Questions

Question 1.
I wonder why the winds shown in the figure are not in the exact north – south direction?

Answer:
The winds would have flown in the north – south direction from north to south or from south to north. A change in direction from however, caused by the rotation of the earth.

Question 2.
I want to know what these winds do for us?
Answer:
The winds from the oceans carry water and bring rain. It is a part of the water cycle.

Activities

Activity – 1
Blow the balloons:
Take two balloons of approximately equal size. Put a little water into the balloons. Blow up both the balloons and tie each one to a string. Hang the balloons 8 – 10 cm apart on a cycle spoke or a stick. Blow in the space between the balloons.

Question 1.
What did you expect? What happens?
Answer:
We expected that balloons would move apart. But the balloons come closer.

Activity – 2
Can you blow and lift?
Hold a strip of paper, 20 cm long and 3 cm wide, between your thumb and forefinger as shown in the paper. Paheli Thinks that the strip will be lifted up. Boojho thinks that the strip will bend down.

Question 1.
What do you think Will happen to the paper?
Answer:
Paper strip will be lifted up.

Question 2.
Were the observations along the lines you thought?
Answer:
Yes.

Question 3.
Do you get the feeling that the increased wind speed is accompanied by a reduced air pressure?
Answer:
Yes.

Activity – 3
Take two paper bags or empty paper cups of the same size. Hang the two bags in the inverted position on the two ends of a metal or wooden stick. Tie a piece of thread in the middle of the stick. Hold the stick by the thread (See Fig.) as in a balance. Put a burning candle below one of the bags as shown in the figure. Observe what happens.

Question 1.
Why is the balance of the bags disturbed?
Answer:
The bag below which the candle is lighted, is pushed up by the rising hot air from above the candle flame.

Question 2.
Does this activity indicate that warm air rises up?
Answer:
Yes.

Question 3.
Does the disturbance of the balance suggest that the warm air is lighter than the cold air?
Answer:
Yes.

350 Degrees F to C, 176.667 °C. The Celsius (ºC) degree or Fahrenheit (ºF) degree temperature scales are used to convert temperature.

Winds, Storms and Cyclones Text Book Exercises

Question 1.
Fill the missing word in the blank spaces in the following statements:

1. Wind is ……………. air.
2. Winds are generated due to ……………. heating on the earth.
3. Near the earth’s surface ……………. air rises up whereas air comes down.
4. Air moves from a region of ……………. pressure to a region of pressure.

Answer:

1. Moving
2. Uneven
3. Warm, cooler
4. High, low.

Question 2.
Suggest two methods to find out wind direction at a given place?
Answer:

1. By wind direction indicator.
2. By watching the direction of movement of a paper released in air.

Question 3.
State two experiences that made you think that air exerts pressure (other than those given in the text).
Answer:

1. Compressed air is used in the brake system for stopping trains.
2. Blowing air in a balloon makes it expand.

Question 4.
You want to buy a house. Would you like to buy a house having windows but no ventilators ? Explain your answer.
Answer:
No, a house which has no ventilators is not a healthy house to live in. Basically ventilators provide a path for warm air to go out of the rooms.

Question 5.
Explain why holes are made in hanging banners and hoardings?
Answer:
We know that air exerts pressure, so that due to this pressure banners and hoardings flutter when the wind is blowing. The holes are made in the banners and hoardings as wind pass through that holes and they does not become loose and fall down.

Question 6.
How will you help your neighbours in case cyclone approaches your village/town?
Answer:
I will help by following ways:

1. By warning everyone about the coming danger.
2. Searching for shelter.
3. Moving people fast to safe places.
4. Managing first aid facility.

Question 7.
What planning is required in advance to deal with the situation created by a cyclone?
Answer:
The following planning is required in advance to deal with the situation created by a cyclone:

1. Listening carefully to warnings being transmitted on TV and radio.
2. Setting up cyclone warning system,
3. Moving to cyclone shelter.
4. Storing food in water – proof bags.
5. Keeping an emergency kit ready.

Question 8.
Which one of the following place is unlikely to be affected by a cyclone.

1. Chennai
2. Mangaluru (Mangalore)
3. Amritsar
4. Puri.

Answer:
3. Amritsar.

Question 9.
Which of the statements given below is correct?

1. In winter the winds flow from the land to the ocean.
2. In summer the winds flow from the land towards the ocean.
3. A cyclone is formed by a very high – pressure system with very high – speed winds revolving around it.
4. The coastline of India is not vulnerable to cyclones.

Answer:
1. In winter the winds flow from the land to the ocean.

Extended Learning – Activities and Projects

Question 1.
You can perform the Activity 8.5 (of textbook) in the chapter slight differently at home. Use two plastic bottles of the same size. Stretch one balloon on the neck of each bottle. Keep one bottle in the sun and the other in the shade. Record your observations. Compare these observations and the result with those of Activity 8.5 of text book
Answer:
Do yourself.

Question 2.
You can make your own anemometer?
Answer:
Collect the following items:
4 small paper cups (used ice cream cups), 2 strips of cardboard (20 cm long and 2 cm wide), gum, stapler, a sketch pen and a sharpened pencil with eraser at one end. Take a scale draw crosses on the cardboard strips as shown in the Fig. (a). This will give you the centres of the strips.

Fix the strips at the centre, putting one over the other so that they make a plus (+) sign. Now fix the cups at the ends of the strips. Colour the outer surface of one cup with a marker or a sketch pen. All the 4 cups should face in the same direction.

Push a pin through the centre of the strips and attach the strips and the cups to the eraser of. the pencil. Check that the strips rotate freely when you blow on the cups. Your anemometer is ready. Counting the number of rotations per minute will give you an estimate of the speed of the wind.

To observe the changes in the wind speed, use it at different places and different times of the day. If you do not have a pericil with attached eraser you can use the tip of a ball pen. The only condition is that the strips should rotate freely. Remember that this anemometer will indicate only speed changes. It will not give you the actual wind speed.

Question 3.
Collect articles and photographs from newspapers and magazines about storms and cyclones. Make a story on the basis of what you learnt in this chapter and the matter collected by you?
Answer:
Do with the help of your subject teacher.

Question 4.
Suppose you are a member of a committee, which is responsible for creating development plan of a coastal state. Prepare a short speech indicating the measures to be taken to reduce the suffering of the people caused by cyclones?
Answer:
Do with the help of your subject teacher.

Question 5.
Interview eyewitness to collect the actual experience of people affected by a cyclone?
Answer:
Do with the help of your subject teacher.

Question 6.
Take an aluminium tube about 15 cm long and 1 to 1.5 cm in diameter. Cut slice of a medium – sized potato about 2 cm thick. Insert the tube in the slice, press it, and rotate it 2 – 3 times. Remove the tube. You will find a piece of potato fixed in the tube like a piston head. Repeat the same process with the other end of the tube. Now you have the tube with both ends closed by potato pieces with an air column in between.

Take a pencil with one end unsharpened. Place this end at one of the pieces of potato. Press it suddenly to push the potato piece in the tube. Observe what happens. The activity shows rather dramatically how increased air pressure can push things.
Answer:
Do yourself.

Winds, Storms and Cyclones Additional Important Questions

Objective Type Questions

Question 1.
Choose the correct alternative:

Question (i)
A storm is marked by –
(a) Strong winds
(b) Rain
(c) Thunder the lightning
(d) All the above.
Answer:
(d) All the above.

Question (ii)
The moving air is called –
(a) Wind
(b) Strong winds
(c) Storm
(d) None of these.
Answer:
(a) Wind

Question (iii)
The amount of water on the earth remains more or less the same because of –
(a) Thunder
(b) Storm
(c) Flood
(d) Water cycle.
Answer:
(d) Water cycle.

Question (iv)
The word monsoon is derived from –
(a) Arabic word
(b) English word
(c) Hindi word
(d) Urdu word.
Answer:
(a) Arabic word

Question (v)
The diameter of the eye of the cyclone varies from –
(a) 10 km to 15 km
(b) 10 km to 20 km
(c) 10 km to 30 km
(d) 10 km to 40 km.
Answer:
(c) 10 km to 30 km

Question (vi)
A violent tornado can travel at speeds of about –
(a) 200 km/h
(b) 300 km/h
(c) 350 km/h
(d) None of these.
Answer:
(b) 300 km/h

Question 2.
Fill in the blanks :

1. Orissa was hit by a cyclone with wind speed of 200 km/h on …………….
2. On 29 October, 1999, a second cyclone with wind speed of ……………. hit Orissa again.
3. The greater the difference in pressure, the ……………. the air moves.
4. The worm air is lighter than the ……………. air.
5. At the poles, the air is colder than that at latitudes about ……………. degrees.
6. The word monsoon is derived from the Arabic word …………….
7. Clouds bring ……………..
8. Farmers in our country depend mainly on rains for their …………….
9. A large cyclone is a violently rotating mass of ………… in the atmosphere.
10. A tornado is a ……………. funnel shaped cloud that reaches from the sky to the ground.

Answer:

1. 18 October 1999
2. 260 km/h
3. Faster
4. Cold,
5. 60
6. Mausam
7. Rain
8. Harvests
9. Air
10. Dark.

Question 3.
Which of the following statements are true (T) or false (F):

1. The cyclone affected agriculture, communication, transport and electricity supply.
2. On heating the air expands and occupies more space
3. In winter, the direction of the wind flow gets reversed.
4. The winds from the oceans carry water arid bring rain.
5. Water cycle is not a continuous phenomenon.
6. Thunderstorms are caused by violent air current inside the cumulus clouds.
7. The cyclones are called hurricane in America.
8. The storms are called typhoons in China.
9. Uneven heating on the earth is the main cause of wind movement.
10. We must stand under a high-rise building or a tree when caught in a thunderstorm.
11. Tropical cyclones occur throughout the year.
12. Lightning, rains and storms are always harmful for the earth.

Answer:

1. True (T)
2. True (T)
3. True (T)
4. True (T)
5. False (F)
6. True (T)
7. True (T)
8. True (T)
9. True (T)
10. False (F)
11. False (F)
12. False (F)

Question 4.
Match the items in Column A with Column B:

Answer:

(i) (b)
(ii) (c)
(iii) (d)
(iv) (a).

Winds, Storms and Cyclones Very Short Answer Type Questions

Question 1.
What is a wind?
Answer:
The moving air is called wind.

Question 2.
Define the term cycle?
Answer:
A cycle is an event or phenomenon which repeats it selfs after sometime.

Question 3.
Define the term evaporation?
Answer:
The process of changing water from its liquid form to its vapour is known as evaporation.

Question 4.
When you fly a kite, does the wind coming from your back help?
Answer:
Yes.

Question 5.
If you are in a boat, is it easier to row it if there is wind coming from behind you?
Answer:
Yes.

Question 6.
Define tornadoes?
Answer:
A tornado is a dark funnel shaped cloud that reaches from the sky to the ground. In our country tornadoes are not very frequent.

Question 7.
Which region gets maximum sunlight?
Answer:
Regions close to the equator get maximum sunlight.

Question 8.
What do you mean by the “eye” of a storm?
Answer:
The centre of a cyclone is calm area. It is called the eye of the storm.

Question 9.
Can you imagine what would happen if high speed winds blow over the roofs of buildings?
Answer:
If the roofs were weak, they would be lifted and blown away.

Question 10.
What do you mean by “hurricane”?
Answer:
“Hurricane” is the term used for storm in West Indies and America.

Question 11.
When is cyclone alert issued?
Answer:
A cyclone alert or cyclone watch is issued 48 hours in advance of any expected storm.

Question 12.
Which factors contribute to the development of cyclone?
Answer:
Facters like wind speed, wind direction, humidity and temperature contribute to the development of cyclones.

Question 13.
When is cyclone warning issued?
Answer:
A cyclone warning is issued 24 hours in advance.

Question 14.
Why smoke always rises up?
Answer:
Smoke is hotter than air, so it is also lighter than air. That is why smoke always moves up.

Question 5.
How do “high – speed winds” harm us ?
Answer:
High – speed winds accompanying a cyclone can damage houses, telephones and other communication systems, trees, etc. causing tremendous loss of life and property.

Question 16.
What is “beaufort scale”?
Answer:
The number and name of a wind is determined by the speed at which it flows on an internationally accepted scale, called beaufort scale.

Question 17.
Is our body a conductor?
Answer:
Yes.

Question 18.
How are high building protected from lightning?
Answer:
High buildings are protected from lightning by fixing lightning conductor on the building.

Winds, Storms and Cyclones Short Answer Type Questions

Question 1.
How is storm caused?
Answer:
When the wind blows gently, it is called a breeze. But, when it blows very fast it cause storm. Storm may be defined as something taking place in the weather of a violent nature. At sea, a storm may be a strong wind or gale. On land, a storm usually means a weather situation marked by heavy rain and often with strong winds, lightning and thunder.

Question 2.
Explain the structure of a tornado?
Answer:
The diameter of a tornado can be as small as a metre and as large as a kilometer, or even wider. The funnel of a tornado sucks dust, debris and everything near it at the base (due to low pressure) and throws out near the top.

Question 3.
How is lightning useful in nature?
Answer:
Lightning is useful in nature because during lightning in tense heat and high temperature are produced. As a result, nitrogen combines with oxygen to form its oxides. These oxides of nitrogen further get dissolved in water to form a dilute solution of nitric acid that comes to the ground with rain. This is how nature provides nitrogenous compounds to plants that are important for their growth.

Question 4.
How are lightning and thunder caused?
Answer:
When two oppositily charged clouds are near each other, the air between them becomes good conductor because charges begin, to move in air very speedily. The presence of electric charges in very large quantities in the air causes to appear as sleaks of lightning and thunder.

Question 5.
Explain the terms thunderstorms and cyclones.
Answer:
Thunderstorms develop in hot, humid tropical areas like India very frequently. The rising temperatures produce strong upward rising winds. These winds carry water droplets upwards, where they freeze, and fall down again. The swift movement of the falling waterd roplets along with the rising air create lightning and sound. It is this event that we call a thunderstorm.

Question 6.
Suggest precautios if a storm is accompanied by lightning?
Answer:
If a storm is accompanied by lightning, we must take the following precautions:

1. Do not take shelter under an isolated tree. If you are in a forest take shelter under a small tree. Do not lie on the ground,
2. Do not take shelter under an umbrella with a metallic end.
3. Do not sit near a window. Open garages, storage sheds, metal sheds are not safe places to take shelter.
4. A car or a bus is a safe place to take shelter.
5. If you are in water, get out and go inside a building.

Question 7.
Suggest some effective safety measures for cyclone.
Answer:
Some Effective Safety Measures:

1. A cyclone forecast and warning service.
2. Rapid communication of. warnings to the Government agencies, the ports, fishermen, ships and to the general public.
3. Construction of cyclone shelters in the cyclone prone areas, and Administrative arrangements for moving people fast to safer places.

Winds, Storms and Cyclones Long Answer Type Questions

Question 1.
How does a thunderstorm becomes a cyclone?
Answer:
Before cloud formation, water takes up heat from the atmosphere to change into vapour. When water vapour changes back to liquid form as raindrops, this heat is released to the atmosphere. The heat released to the atmosphere warms the air around. The air tends to rise and causes a drop in pressure. More air rushes to the centre of the storm. This cycle is repeated. The chain of events ends with the formation of a very low – pressure system with very high – speed winds revolving around it. It is this weather condition that we call a cyclone. Factors like wind speed, wind direction, temperature and humidity contribute to the development of cyclones.

Question 2.
Define the structure of a cyclone?
Answer:
Structure of a cyclone:
The centre of a cyclone is a calm area. It is called the eye of the storm. A large cyclone is a violently rotating mass of air in the atmosphere, 10 to 15 km high. The diameter of the eye varies from 10 to 30 km. It is a region free of clouds and has light winds. Around this calm and clear eye (See Fig.), there is a cloud region of about 150 km in size.

In this region there are high – speed winds (150-250 km/h) and thick clouds with heavy rain. Away from this region the wind speed gradually decreases. The formation of a cyclone is a very complex process.

Question 3.
With a neat diagram show the formation of a cyclone.
Answer:

Question 4.
Describe the action taken by the people and some precautions if you are staying in a cyclone hit area?
Answer:
Action on the part of the people:

1. We should not ignore the warnings issued by the meteorological department throught TV, radio, or newspapers.
2. We should make necessary arrangements to shift the essential household goods, domestic animals and vehicles, etc. to safer places.
3. We should avoid driving on roads through standing water, as floods may have damaged the roads.
4. We should keep ready the phone numbers of all emergency sendees like police, fire brigade, and medical centres.

Some precautions, if you are staying in a cyclone hit area:

1. Do not drink water that could be contaminted. Always store drinking water for emergencies.
2. Do not touch wet switches and fallen power lines.
3. Do not go out just for the sake of fun.
4. Do not pressurise the rescue force by making undue demands.
5. Cooperate and help your neighbours and friends

Question 5.
Show the phenomena that lead to the formation of clouds and falling of rain and creation of storms and cyclones with the help of a flow diagram or a flow chart.
Answer:

Question 6.
On a map, show the regions near the equator where cyclones form.
Answer:

MP Board Class 7th Science Solutions

MP Board Class 12th Chemistry Important Questions Chapter 4 Chemical Kinetics

Chemical Kinetics Important Questions

Chemical Kinetics Objective Type Questions

Question 1.
Choose the correct answer :

Question 1.
For most of the reactions, the value of temperature coefficient lies in between :
(a) 1 and 3
(b) 2 and 3
(c) 1 and 4
(d) 2 and 4.
Answer:
(b) 2 and 3

Question 2.
For First order reaction value of tm is : (MP 2012 Supp.)
(a) \(\frac { 0.693 }{ { k }_{ 1 } }\)
(b) \(\frac { 2.303 }{ { k }_{ 1 } }\)
(c) \(\frac { 0.303 }{ { k }_{ 1 } }\)
(d) \(\frac {0.693}{t}\)
Answer:
(a) \(\frac { 0.693 }{ { k }_{ 1 } }\)

Question 3.
A first order reaction gets completed to 75% in 32 minutes. How much time would have been required for 50% completion :
(a) 24 minute
(b) 16 minute
(c) 8 minute
(d) 4 minute.
Answer:
(b) 16 minute

Question 4.
Hydrolysis of sucrose to glucose and fructose is :

an example of:
(a) First order reaction
(b) Second order reaction
(c) Third order reaction
(d) Zero order reaction.
Answer:
(a) First order reaction

Question 5.
Plants prepare starch in the process of:
(a) Flash photolysis
(b) Photolysis
(c) Photosynthesis
(d) None of these.
Answer:
(c) Photosynthesis

Question 6.
For first order reaction the specific reaction constant depends upon :
(a) Concentration of reactants
(b) Concentration of products
(c) Time
(d) Temperature.
Answer:
(d) Temperature.

Question 7.
In the reaction between A and B to form C, A represents first order and B represents second order. Rate equation will be written as :
(a) Rate = k [A]²[B]
(b) Rate = k [A] [B]²
(c) Rate = k [A]^1/2[B]
(d) Rate = k [A] [B]^1/2
Answer:
(b) Rate = k [A] [B]²

Question 8.
The rate of chemical reaction depends upon : (MP 2017)
(a) Active mass
(b) Atomic mass
(c) Equivalent weight
(d) Molecular mass.
Answer:
(a) Active mass

Question 9.
Arrhenius equation is :
(a) k = e^-E_aRT
(b) k = \(\frac { { E }_{ a } }{ RT }\)
(c) k = log\(\frac { { E }_{ a } }{ RT }\)
(d) k = Ae^-E_a/RT
Answer:
(d) k = Ae^-E_a/RT

Question 10.
Unit of velocity constant of first order reaction :
(a) mol litre^-1 sec^-1
(b) mol^-1 litre⁺¹ sec^-1
(c) sec^-1
(d) mol litre^-1 sec
Answer:
(c) sec^-1

Question 11.
Unit of velocity constant of zero order reaction :
(a) mol^-1 litre^-1 sec^-1
(b) mol^-1 litre⁺¹ sec^-1
(c) sec^-1
(d) mol litre^-1 sec
Answer:
(a) mol^-1 litre^-1 sec^-1

Question 12.
Rate constant of a reaction increased with the increase of which of the following factor as :
(a) Pressure
(b) Temperature
(c) Concentration of reaction
(d) All of these.
Answer:
(b) Temperature

Question 13.
Factor on which the rate constant of the 1st order reaction does not depend :
(a) Temperature
(b) Catalyst
(c) Activation energy
(d) Concentration of reactant.
Answer:
(d) Concentration of reactant.

Question 14.
Minimum energy required for molecules to react is called :
(a) Potential energy
(b) Kinetic energy
(c) Nuclear energy
(d) Activation energy.
Answer:
(d) Activation energy.

Question 15.
Law of mass action prepared by :
(a) Dalton
(b) Guddberg and Waaje
(c) Hunds and Mulliken
(d) Arrhenius.
Answer:
(b) Guddberg and Waaje

Question 16.
Unit of reaction rate is:
(a) mol litre^-1 sec^-1
(b) mol^-1 litre sec^-1
(c) mol^-1 litre^-1 sec
(d) mol litre sec
Answer:
(a) mol litre^-1 sec^-1

Question 2.
Fill in the blanks :

1. The rate of a reaction does not depends on the concentration of the reacting species, then the reaction is of ……………………
2. Half life period of a radioactive element is 140 days. On taking 1 gm element initially the amount left after 560 days will be ……………………
3. Fast reactions are completed in less than …………………… seconds.
4. Unit of rate constant for third order reaction is ……………………
5. Difference in the minimum and maximum energy state of reactants is called ……………………
6. Reactions which take place by the absorption of radiations are called …………………… (MP 2016)
7. The total number of molecules which participate in a reaction is called ……………………
8. In the mechanism of a reaction, the slowest step is called ……………………

Answer:

1. Zero
2. \(\frac {1}{2}\) gm
3. 10^-9
4. mol^-2 Litre² second^-1
5. Activation energy
6. Photo chemical reaction
7. Molecularity
8. Rate determining step.

Question 3.
Answer in one word / sentence :

1. What is the relation between threshold energy and activation energy?
2. What is the expression of rate constant for first order reaction?
3. Write alternative form of Arrhenius equation.
4. Write an example of zero order reaction.
5. What is Quantum Efficiency?
6. Write the expression of half life period for first order reaction.
7. Explain Threshold energy.

Answer:

7. The Number of photons absorbed minimum amount of energy which the reactant molecules should possess for effective collision.

Chemical Kinetics Very Short Answer Type Questions

Question 1.
What is the effect of rate constant on temperature? How can this effect of temperature be measured quantitatively? (NCERT)
Answer:
Rate of reaction increases with increase in temperature and with 10°C rise in
temperature its value becomes two times. Arrhenius equation expresses the effect of temperature on constant
K = A_e^{-E _/RT}
Where A is frequency factor and Ea is activation energy of the reaction.

Question 2.
The rate of reaction for A + 2B → Product, will be equal to it\(\frac {-d[A]}{dt}\) = k[A] [B]² if B is present in excess amount then what will be the order of reaction?
Answer:
It will be 1^st order of reaction.

Question 3.
Write alternative form of Arrhenius equation.
Answer:

Question 4.
For a zero order reaction t_1/2is proportional to what?
Answer:
Initial concentration of reactant [A].

Question 5.
What is Energy Barrier?
Answer:
The minimum energy achieved by the reactant only after which it can be converted to product is known as energy barrier. Reactant molecules cannot form activated complex till they reach this height (activation energy) and cannot be converted to form product

Question 6.
Explain the rate determining step.
Answer:
Some chemical reactions complete in one or more steps. Rate of reaction is determined by the slowest step which is known as rate determining step.

Question 7.
If unit of rate constant is litre/mole/sec. then what will be the order of reaction?
Answer:
Order of reaction = Second.

Question 8.
Order of a reaction is zero can its molecularity be zero?
Answer:
Molecularity of a reaction is always a whole number. Thus, it cannot be zero.

Question 9.
What is specific reaction rate?
Answer:
Specific reaction rate of a reaction at a given temperature is equal to that rate of a reaction when concentration of each reactant is unity.
Reaction : A_2(g) + B_2(g) → 2AB_g
Rate of reaction ∝ [A₂][B₂]
If [A₂] = [B₂] = 1, then
Rate of reaction = k, where k = Specific reaction rate constant.

Question 10.
When is the average rate of the reaction equal to its instantaneous rate?
Answer:
When value of time interval is nearly zero or when time by infinite form is minute then the average rate of reaction is comparable to its instantaneous rate.

Question 11.
What are pseudo unimolecular reactions?
Answer:
There are some reactions in which molecularity is more than one i.e. two or more molecules are present, but in the chemical reaction concentration of only one reactant molecule is changed and it is only responsible for rate of reaction. Thus, order is one. These reactions are called pseudo unimolecular reactions.
Example:

Above reaction is bimolecular but concentration of water does not affect the rate of reaction, thus, rate of reaction is proportional to the concentration of sucrose only.
Rate = k[C₁₂H₂₂O₁₁]
Thus, inversion of sucrose is a first order reaction. It is known as pseudo unimolecular reaction.

Question 12.
What is temperature coefficient?
Answer:
Generally, rate of any reaction increases due to increase in temperature. On increasing temperature by 10°C, velocity of reaction is increased up to 2 – 3 times. If velocity constant of a reaction is and it is increased to increasing temperature by 10°C, the ratio of these two constant is called temperature coefficient, i.e.,
\(\frac { { k }_{ t+10 } }{ { k }_{ t } }\) ≈ 2 to 3
Thus, at various time the ratio of rate of reaction which differ by 10°C is known as temperature coefficient.

Question 13.
As compared to water, gasoline vaporizes faster. Why?
Answer:
Value of activation energy of vaporization of gasoline is less than the value of activation energy of water.

Chemical Kinetics Short Answer Type Questions

Question 1.
The conversion of molecules x to y follows second order kinetics. If concetration of x is increased to three times how will it affect the rate of formation of y? (NCERT)
Solution:
For the reaction x → y
Rate of reaction (r) = k[x]² …. (1)
If concentration of x is increased three times, now
Rate of reaction (r)¹ = k[3x]² = k[9x² ] …. (2)
Dividing equation (2) by Question (1),
\(\frac { { r }^{ 1 } }{ r }\) = \(\frac { k[9{ x }^{ 2 }] }{ k[{ x }^{ 2 }] }\) = 9
Thus, the rate of reactions will become 9 times.

Question 2.
Write down the expression representing the rate of reaction.
Answer:
Rate of reaction is the rate of change in concentration of reactant or concentration of product in unit time interval.

Unit of rate of reaction depend on the units of concentration and time. If concentration is represented in mole per litre and time in second, then unit of rate of reaction is mol per litre per second. If time is expressed in minute then unit of reaction rate is mole per litre per minute.

Question 3.
What is the meaning of instantaneous rate of reaction?
Answer:
The rate of reaction does not remain constant during the whole time interval because rate of reaction depends upon the concentration of reactants. As the concentration of reactants decreases with time, the rate of reaction also decreases with time.
In order to express the reaction rate as accurately as possible, the instantaneous rate of reaction is expressed. For this the time interval (∆t) is taken as small as possible.
Instantaneous rate of reaction = –[\(\frac {∆A}{∆t}\)_{∆t → o}
= – \(\frac {d[A]}{dt}\)
Where, d [A] is a change in concentration of reactant A.

Question 4.
What do you know about molecularity of any reaction?
Answer:
Molecularity of Reaction:
‘Number of moles of reactant participating in elementary step of chemical reaction is called molecularity of the reaction.’
There are many reactions which proceed through the number of steps and each step being independent is called elementary step. The rate of reaction is determined by slowest step and it is known as rate determining step. Here molecularity can be defined as “Number of molecules/atoms or ions participating in rate determining step is called molecularity.”
Example:
(i) Unimolecular reaction :
O₃ → O₃ + O
(ii) Bimolecular reaction:
NO + O₃ → NO₂ + O₂

Question 5.
Differentiate between molecularity and order of reaction. (MP 2018)
Answer:
Differences between Molecularity and Order of reaction :
Molecularity:

• It is the total number of molecules which participate in the reaction.
• It is a theoretical concept.
• It is always a whole number.
• Its value is never zero.
• It does not provide any information about mechanism of reaction.

Order of reaction:

• It is the number of molecules which participate in reaction and whose concentration is changed.
• Order of reaction is determined by experimentally.
• Fractional values are also possible.
• Zero value is possible.
• It provides information about mechanisms of reaction.

Question 6.
Write any four factors which affects rate of a chemical reaction.
Answer:
The rate of reaction depends upon the following factors:
1. Concentration of reactants:
At constant temperature, the rate of a reaction increases by increasing the concentration of the reactants.

2. Temperature of the system:
If the concentration of the reactants are constant then the rate of reaction increases by increasing temperature. For 10 degree rise of temperature, the reaction rate becomes double or triple.

3. Presence of catalyst:
Positive catalyst increases the rate of reaction and negative catalyst decreases the reaction rate.

4. Nature of the reactants:
Nature of reactants also affect the reaction rate. In any chemical reaction some old bonds are broken and new bonds are formed. Thus, in case of more simple molecules the lesser is the number of bond breaking and the rate of reaction increase whereas in case of complex molecules more bonds are broken and rate decreases.

5. Exposure to radiations:
The rate of some reactions increases due to some special radiations. For example: In the absence of light the reaction between hydrogen and chlorine is slow, but in the presence of light, the reaction proceeds at a faster rate.

Question 7.
How does rate of any reaction depend on temperature? Explain.
Answer:
Generally, rate of any reaction increases due to increase in temperature. On increasing temperature by 10°C, velocity of reaction is increased up to 2 – 3 times. If velocity constant of a reaction is and it is increased to increasing temperature by 10°C, the ratio of these two constant is called temperature coefficient, i.e.,
\(\frac { { k }_{ t+10 } }{ { k }_{ t } }\) ≈ 2 to 3
Arrhenius provide the following relation for showing the effect of temperature on velocity constant.
k = A.e^-E_a

Question 8.
Write a short note on activation energy?
Answer:
According to Arrhenius, any chemical reaction is only possible when reacting molecules are activated with minimum energy which is called threshold energy. Kinetic energy of most of the molecules are less than this minimum energy. The excess energy which is required to activate reactant molecules, is called activation energy.
Activation energy can be determined by the use of Arrhenius equation

Question 9.
Write four differences between Rate of reaction and Rate constant. (MP 2018)
Answer:
Differences between Rate of reaction and Rate constant:
Rate of reaction:

• It is expressed in terms of consumption of reactants or formation of product per unit time.
• It depends on concentration of reactant at particular moment.
• It generally decreases with the progress of reaction.
• Its unit is mol L^-1 cm^-1.

Rate constant:

• It is proportionality constant in differential form in rate law or rate equation.
• It is independent of concentration of reactant.
• It does not depend on the progress of reaction.
• It changes according to order of reaction.

Question 10.
Prove that half – life period is independent of the initial concentration for the firsforder reaction.
Solution:
Half – life period for the reaction is that period in which the initial concentration of reactant is reduced to half.
For first order reaction :

Thus, the half – life period is independent of the initial concentration for the first order reaction.

Question 11.
(i) 2N₂O₅ → 4N02 +02
(ii) H₂ +I₂ → 2HI
Reaction (i) is first order reaction and (ii) is second order reaction. Why?
Answer:
(i) In reaction 2N₂O₅ → 4NO₂ + O₂, when graph is plotted between rate of reaction and then again graph is plotted between rate of reaction and [N₂O₅]², it is shown that in the first graph straight line is obtained i.e.
rate ∝ [N₂O₅]
or rate = k[N₂O₅]
Therefore, 2N₂O₅ → 4NO₂ + O₂ is first order reaction.

(ii) In reaction H₂+I₂ → 2HI when graph is plotted between rate of reaction and (H₂) (I₂), it is seen that straight line is obtained.
Therefore rate ∝ [H₂][I₂]
Hence, this reaction is of 2^nd order reaction.

Question 12.
What do you understand by order of reaction? Give example. (MP 2011,12,16)
Answer:
Order of reaction:
The order of a reaction is defined as the sum of all the powers to which concentration terms in the rate law are raised to express the observecfrate of the reaction. Suppose there is a general reaction,
aA + bB + cC → product
For which the rate law is
Rate = – \(\frac {dx}{dt}\) = k[A]^p[B]^q[C]^r
Then the order of the reaction n = p + q + r, where, p, q and r are the orders with
respect to individual reactants and overall order is the sum of the exponents i.e.,p + q + r.
When n = 1 the reaction is of first order, if n = 2 the reaction is of second order and so on.
For example : Decomposition of ammonium nitrite occurs as follows :
NH₄NO₂ → N₂ +2H₂O
Rate of reaction = – \(\frac {dx}{dt}\) = k [NH₄NO₂]
Thus, order of this reaction will be 1.

Question 13.
Write unit of rate constant k for the zero order, first order and second order reaction (MP 2012)
Answer:

Chemical Kinetics Long Answer Type Questions

Question 1.
Determine the expression for zero order reaction.
Answer:
Reactions in which rate of reaction does not depend on the concentration of the rede tan ts are called zero order reactions. Consider a zero order reaction
A → B
Where, A and B are concentration of reactants and products. Since in this type of reaction, rate of reaction does not depend on the concentration of reactant therefore, rate of change in concentration of reactant remains constant.
Rate of reaction = Constant.
Let initial cone, of reactants be a moles /It and after time‘t’ x moles are converted into product. Then cone, of A after time t will be (a – x) mol /It.
\(\frac {-d[A]}{dt}\) or \(\frac {dx}{dt}\) ∝(a – x)⁰
or \(\frac {dx}{dt}\) = k₀(a – x)(a – x)⁰ … (1)
Where, k₀ is velocity constant of a zero order reaction.
\(\frac {dx}{dt}\) ∝ (a-x)(a – x)⁰ … (2)
dx = k₀dt … (3)
Integrating equation. (2), we get
x = k₀t + c … (4)
Where, c = integration constant
When, t = 0 then x = 0
Substituting it in equation. (4), we get
0 = k₀ x O + C
or C = 0  …(5)
Substituting this in equation. (4), we get
x = k₀t  … (6)
or k₀ = \(\frac {x}{t}\) = … (7)
Thus, equation. (7) is the velocity equation for zero order reaction. Unit of k
k₀ = \(\frac {x}{t}\)
= \(\frac {mole/litre}{second}\) = mole litre^-1 second^-1

Theoretical yield calculator tool makes the calculation faster and it displays the theoretical yield value of the chemical reaction.

Question 2.
Write the Arrhenius equation in the form of equation of straight line. What will be the slope of a graph using this equation? Calculate the activation energy for the decomposition in a decomposition reaction in which the value of slope obtained is – 9920 when log k is plotted against \(\frac {1}{T}\)
Solution:
Arrhenius equation : Arrhenius gave a relation between rate constant of reaction and temperature which is known as Arrhenius equation, i.e.,
k = Ae^-E_a/RT
Where, k = Rate constant of reaction, A = Frequency factor, Ea= Activation energy, R = Gas constant and T = Absolute temperature.

Taking logarithm both sides of above equation, we get
log_e k = – \(\frac { { E }_{ a } }{ RT }\) + log_e A
or log₁₀ k = log₁₀ A – \(\frac { { E }_{ a } }{ 2.303RT }\) .. (i)
This is a straight line equation. If log₁₀ k and \(\frac {1}{T}\) is plotted at different temperatures, a straight line is obtained whose slope = \(\frac { { -E }_{ a } }{ 2.303RT }\) From the graph, value of slope = \(\frac { { E }_{ a } }{ 2.303RT }\) can be calculated and activation energy (E_a ) can be calculated.
Slope = \(\frac { { -E }_{ a } }{ 2.303RT }\) (Activation energy)
Or E_a = Slope x 2.303 x R
Or E_a = – 9920 x 2.303 x (-4.58)
∴ Ea = 104633.5808 calorie per gm molecule

Question 3.
Derive an expression for rate constant of first order reaction.
Answer:
The reaction in which rate of reaction depends upon the concentration of one mole, are called first order reaction.
Let this reaction is
A → product
Suppose intial concentration of A is a gram mole and after t second x mole consumed and remaining concentration is (a – x) gram mole.
So after t time, the rate of reaction will be proportional to (a – x)
\(\frac {dx}{dt}\) ∝(a – x)
or \(\frac {dx}{dt}\) = k (a – x) (k = Velocity constant) … (1)
or \(\frac {dx}{(a – x)}\) = kdt … (2)
on intergating this equation,
∫\(\frac {dx}{(a – x)}\) = ∫ kdt
or -ln(a – x) = kt + c
Where c is intergating constant, …. (3)
If t = 0 then x = 0
Putting the value of c from eqn. (3),
-ln a = c .. (4)
Putting the value of c from eqn. (4), into eqn. (3).
-ln(a – x) = kt – ln a
or ln a – ln (a – x) = kt
ln \(\frac {a}{(a – x)}\) = kt
or k = \(\frac {1}{(t)}\)ln\(\frac {a}{(a – x)}\)
charging the base of log,
k = \(\frac {2.303}{(t)}\) log \(\frac {a}{(a – x)}\)
It is the desired expression for first order of reaction

Chemical Kinetics Numerical Questions

Question 1.
A first order reaction has a rate constant 1.15 x 10^-3 s^-1 How long will 5g of this reactant take to reduce to 3g? (NCERT)
Solution:
According to question,

Applying first order kinetic equation

Question 2.
The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate E_a.
Solution:

Question 3.
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: (NCERT)

Solution:
Given that the reaction between A and B is first order w.r.t. A and zero order w.r.t. B.
Rate = k [A]¹[B]⁰ but [B]⁰ = I.
Rate = k [A]
From experiment I,
2 x 10^-2 = k (0. 1) ⇒ k = 0.2 min^-1
From experiment II,
4 x 10^-2 = 0.2 [A] ⇒[A] = 0.2 min L^-1
From experiment III,
Rate = (0.2) (0.4) = 0 08 mol L^-1 min^-1
From experiment IV,
2 x 10^-2= 0.2 [A] ⇒ [A] = 0.1 mol L^-1.

Question 4.
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. (NCERT)
Solution:

If 90% reaction is getting completed

If 99% reaction is getting completed

Dividing equation (1) by equation. (2),

Question 5.
A first order reaction takes 40 min for 30% decomposition. Calculate tm. [CBSE (Delhi) 2013], (NCERT)
Solution:
30% decomposition means that x = 30% of a = 0.30 a
As reaction is of first order

For a first order reaction,

Question 6.
Show that time required for completing 99.9% of a first order reaction is 10 times of its half life period.
Solution:
If initial concentration of reactant is a, then t = ? for x = 0.999 a
We know that for a first order reaction

MP Board Class 12th Chemistry Important Questions

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 4 Determinants Ex 4.2 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 4 Determinants Ex 4.2

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 4 Determinants Ex 4.4 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 4 Determinants Ex 4.4

In this article, we will share MP Board Class 10th Hindi Solutions पद्य Chapter 3 प्रेम और सौन्दर्य Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Hindi Navneet Solutions पद्य Chapter 3 प्रेम और सौन्दर्य

प्रेम और सौन्दर्य अभ्यास

बोध प्रश्न

प्रेम और सौन्दर्य अति लघु उत्तरीय प्रश्न

प्रश्न 1.
श्रीकृष्ण के हृदय में किसकी माला शोभा पा रही है?
उत्तर:
श्रीकृष्ण के हृदय में गुंजाओं की माला शोभा पा रही है।

प्रश्न 2.
गोपाल के कुंडलों की आकृति कैसी है?
उत्तर:
गोपाल के कुंडलों की आकृति मछली जैसी है।

प्रश्न 3.
श्रद्धा का गायन-स्वर किस तरह का है?
उत्तर:
श्रद्धा का गायन-स्वर मधुकरी (भ्रामरी) जैसा है।

प्रश्न 4.
‘मधुर विश्रांत और एकान्त जगत का सुलझा हुआ रहस्य’-सम्बोधन किसके लिए है?
उत्तर:
यह सम्बोधन मनु के लिए है। श्रद्धा कहती है कि तुम्हें देखकर ऐसा लगता है मानो तुमने संसार के रहस्य को सुलझा लिया है, इसलिए तुम निश्चित होकर बैठे हो।

प्रश्न 5.
माथे पर लगे टीके की तुलना किससे की है?
उत्तर:
माथे पर लगे टीके की तुलना सूर्य से की गयी है।

प्रेम और सौन्दर्य लघु उत्तरीय प्रश्न

प्रश्न 1.
गोपाल के गले में पड़ी गुंजों की माला की तुलना किससे की गई है?
उत्तर:
गोपाल के गले में पड़ी गुंजों की माला की तुलना दावानल की ज्वाला से की गई है।

प्रश्न 2.
श्रीकृष्ण के ललाट पर टीका की समानता किससे की गई है?
उत्तर:
श्रीकृष्ण के ललाट पर शोभित टीके की समानता सूर्य से की गयी है।

प्रश्न 3.
मनु को हर्ष मिश्रित झटका-सा क्यों लगा?
उत्तर:
श्रद्धा की वाणी सुनते ही मनु को एक हर्ष मिश्रित झटका लगा और वे मोहित होकर यह देखने लगे कि यह संगीत से मधुर वचन कौन कह रहा है।

प्रेम और सौन्दर्य दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
पीताम्बरधारी श्रीकृष्ण के सौन्दर्य का वर्णन कीजिए।
उत्तर:
श्याम वर्ण पर पीताम्बर धारण किए हुए श्रीकृष्ण का स्लौन्दर्य ऐसा लग रहा है मानो नीलमणि के पर्वत पर प्रात:कालीन सूर्य की किरणें पड़ रही हैं।

प्रश्न 2.
‘सरतरु की मनु सिन्धु में,लसति सपल्लव डार’ का आशय स्पष्ट कीजिए।
उत्तर:
नायक कहता है कि इस नायिका का झिलमिली नामक आभूषण अपार चमक के साथ झीने पट में झलक रहा है। उसे देखकर ऐसा लगता है मानो कल्प वृक्ष की पत्तों सहित डाल समुद्र में विलास कर रही है।

प्रश्न 3.
अरुण रवि मण्डल उनको भेद दिखाई देता हो, छवि धाम का भावार्थ लिखिए।
उत्तर:
कवि कहता है कि श्रद्धा का मुख ऐसा सुन्दर दिखाई दे रहा था जैसे सूर्य अस्त होने से पहले छिप गया हो परन्तु जब लाल सूर्य उन नीले मेघों को चीर कर दिखाई देता है तो वह अत्यन्त सुन्दर दिखाई देता है। श्रद्धा के मुख का सौन्दर्य वैसा ही था।

प्रश्न 4.
अधोलिखित पद्यांशों की सप्रसंग व्याख्या कीजिए
(अ) तो पै वारौ उरबसी …………. खै उरबसी समान।
उत्तर:
हे चतुर राधिका! सुन, तू तो इतनी सुन्दर है कि तुझ पर मैं इन्द्र की अप्सरा उर्वशी को भी न्योछावर कर दूँ। हे राधा! तू तो मोहन के उर (हृदय) में उरबसी आभूषण के समान बसी हुई है। अतः दूसरों की बात सुनकर तुम मौन धारण मत करो।

(ब) हृदय की अनुकृति ………….. सौरभ संयुक्त।
उत्तर:
पूर्ववत्।
व्याख्या-कवि कहता है कि मनु ने वह सुन्दर दृश्य देखा जो नेत्रों को जादू के समान मोहित कर देने वाला था। श्रद्धा उस समय फूलों की शोभा से युक्त लता के समान लग रही थी। श्रद्धा चाँदनी से घिरे हुए काले बादल के समान लग रही थी। श्रद्धा ने नीली खाल का वस्त्र पहन रखा था इस कारण वह बादल के समान दिखाई दे रही थी। किन्तु उसकी शारीरिक कान्ति उसके परिधान के बाहर भी जगमगा रही थी। श्रद्धा हृदय की भी उदार थी और उसी के अनुरूप वह देखने में उदार लग रही थी, उसका कद लम्बा था और उससे स्वच्छन्दता झलक रही थी। वायु के झोंकों में वह ऐसी लगती थी मानो बसन्त की वायु से हिलता हुआ कोई छोटा साल का पेड़ हो और वह सुगन्धि में डूबा हो।

प्रेम और सौन्दर्य काव्य सौन्दर्य

प्रश्न 1.
अधोलिखित काव्यांश में अलंकार पहचान कर लिखिए
(अ) धस्यौ मनो हियगढ़ समरू ड्योढ़ी लसत निसान।’
(ब) ‘विश्व की करुण कामना मूर्ति’।।
उत्तर:
(अ) उत्प्रेक्षा अलंकार
(ब) रूपक अलंकार।

प्रश्न 2.
फिरि-फिरि चित त ही रहतु, टुटी लाज की लाव।
अंग-अंग छवि झऔर में, भयो भौंर की नाव।”
में छन्द पहचान कर उसके लक्षण लिखिए।
उत्तर:
इसमें दोहा छन्द है जिसका लक्षण इस प्रकार है-
दोहा-लक्षण-दोहा चार चरण का मात्रिक छन्द है। इसकी प्रत्येक पंक्ति में 24 मात्राएँ होती हैं। पहले तथा तीसरे चरणों में 13-13 मात्राएँ और दूसरे तथा चौथे चरणों में 11-11 मात्राएँ। होती हैं।

प्रश्न 3.
संयोग श्रृंगार का एक उदाहरण रस के विभिन्न: अंगों सहित लिखिए।
उत्तर:
संयोग शृंगार :
जहाँ प्रेमी-प्रेमिका की संयोग दशा में प्रेम का अंकन, माधुर्यमय वार्ता, स्पर्श, दर्शन आदि का वर्णन हो वहाँ संयोग श्रृंगार होता है। इसमें स्थायी भाव-रति, विभाव-प्रेमी-प्रेमिका, अनुभाव-परिहास, कटाक्ष, स्पर्श, आलिंगन आदि तथा संचारी भाव-हर्ष, उत्सुकता, मद आदि होते हैं।

अंगों सहित उदाहरण :
“जा दिन ते वह नन्द को छोहरा, या वन धेनु चराई गयी है। मोहिनी ताननि गोधन, गावत, बेनु बजाइ रिझाइ गयौ है।। वा दिन सो कछु टोना सो कै, रसखानि हियै में समाइ गयौ है। कोऊन काहू की कानि करै, सिगरो, ब्रज वीर, बिकाई गयौ है।”

स्पष्टीकरण :
यहाँ पर आश्रय गोपियाँ तथा आलम्बन श्रीकृष्ण हैं। वन में गाय चराना, वंशी बजाना, गाना आदि उद्दीपन हैं। मोहित होना, लोक लज्जा न मानना आदि अनुभाव हैं। ‘स्मृति’ | संचारी भाव है। इस प्रकार रति स्थायी भाव संयोग-शृंगार में परिणत हुआ है।

प्रश्न 4.
वियोग श्रृंगार को परिभाषित करते हुए उदाहरण रस के अंगों सहित लिखिए।
उत्तर:
वियोग श्रृंगार-जहाँ प्रेमी-प्रेमिका की वियोग दशा में प्रेम का अंकन, विरह वेदना का सरस वर्णन हो वहाँ वियोग श्रृंगार होता है। इसे विप्रलम्भ श्रृंगार भी कहते हैं। इसमें स्थायी भाव-रति, विभाव-प्रेमी-प्रेमिका, अनुभाव-प्रस्वेद, अश्रु, कम्पन, रुदन आदि तथा संचारी भाव-स्मृति, विषाद, आवेग,आदि होते हैं।

अंगों सहित उदाहरण :
“ऊधौ मोहिं ब्रज बिसरत नाहीं।
वृन्दावन गोकुल वन उपवन, सघन कुंज की छाहीं।
प्रात समय माता जसुमति अरु, नन्द देख सुख पावत।
माखन रोटी दह्यौ सजायौ, अति हित साथ खवावत।
गोपी ग्वाल बाल संग खेलत, सब दिन हँसत सिरात॥”

स्पष्टीकरण :
इसमें आश्रय श्रीकृष्ण तथा आलम्बन गोकुल की वस्तुएँ, नन्द किशोर आदि हैं। वृन्दावन, वन उपवन, रोटी, दही आदि उद्दीपन हैं। आँसू मलिनता आदि अनुभाव हैं। स्मृति संचारी भाव है तथा रति स्थायी भाव है।

प्रेम और सौन्दर्य महत्त्वपूर्ण वस्तुनिष्ठ प्रश्न

बहु-विकल्पीय प्रश्न

प्रश्न 1.
श्रीकृष्ण के हृदय पर किसकी माला शोभा पा रही है?
(क) मणियों की
(ख) मोतियों की
(ग) गुंजों की
(घ) फूलों की।
उत्तर:
(ग) गुंजों की

प्रश्न 2.
श्रीकृष्ण के कानों में किस प्रकार के कुण्डल सुशोभित हैं?
(क) मकर की आकृति के
(ख) मछलियों की आकृति के
(ग) भौरों की आकृति के
(घ) हिरणों की आकृति के।
उत्तर:
(ख) मछलियों की आकृति के

प्रश्न 3.
प्रसादजी ने श्रद्धा सर्ग में किस प्रकार का चित्रण किया है?
(क) मनु और श्रद्धा के प्रेम का
(ख) प्रकृति का चित्रण
(ग) नारी सौन्दर्य का
(घ) उपर्युक्त सभी।
उत्तर:
(घ) उपर्युक्त सभी।

प्रश्न 4.
सृष्टि में प्रलय के पश्चात् मनु ने सबसे पहले किसको देखा?
(क) पर्वतों को
(ख) वायु के झोंके
(ग) वृक्षों को
(घ) श्रद्धा को।
उत्तर:
(घ) श्रद्धा को।

रिक्त स्थानों की पूर्ति

1. सोहत ओढ़े पीतु पटु स्याम ………… गात।
2. गोपाल के गले में ………… माला थी।
3. नील परिधान बीच सुकुमार खुल रहा मृदुल …………. अंग।
4. श्रद्धा के अपूर्व सौन्दर्य को देख ………… आकर्षित होते हैं।

उत्तर:

1. सलौने
2. गुंजों
3. अधखिला
4. मनु।

सत्य/असत्य

1. बिहारी के दोहों में राधा-कृष्ण के प्रेम का सुन्दर वर्णन है।
2. ‘कामायनी’ आधुनिक काल का श्रेष्ठ महाकाव्य है। (2009)
3. ‘नील घनशावक से सुकुमार सुधा भरने को विधु के पास’ पंक्ति प्रसाद की लहर’ कविता की है।
4. प्रलयकाल के बाद जब मनु अकेले रह जाते हैं तब उन्हें श्रद्धा के दर्शन होते हैं।

उत्तर:

1. सत्य
2. सत्य
3. असत्य
4. सत्य।

सही जोड़ी मिलाइए

उत्तर:
1. → (घ)
2. → (ग)
3. → (क)
4. → (ख)

एक शब्द/वाक्य में उत्तर

1. बिहारी के काव्य में किस रस की प्रधानता है?
2. प्राचीन काल में कवि राजाओं के दरबार में रहकर उनकी प्रशंसा में कविता क्यों करते थे?
3. श्रद्धा की मधुर वाणी सुनकर किसकी समाधि भंग हुई?
4. श्रीकृष्ण के हृदय पर किसकी माला शोभा पा रही है? (2009)

उत्तर:

1. श्रृंगार रस
2. जीविकोपार्जन के लिए
3. मनु की
4. गुंजों की।

सौन्दर्य-बोध भाव सारांश

रीतिकालीन कवि बिहारीजी ने श्रीकृष्ण की वेश-भूषा एवं सुन्दर छवि का वर्णन किया है। उनका पीताम्बर प्रात:काल की पीली धूप की भाँति सुशोभित है। श्रीकृष्ण का सौन्दर्य इतना अपूर्व है कि कोई भी कवि उनके अलौकिक सौन्दर्य के वर्णन में सफल नहीं हो सकता। श्रीकृष्ण के मस्तक पर लगा लाल रंग का टीका सूर्यमण्डल में प्रवेश करके उनके सौन्दर्य में चार-चाँद लगा रहा है।

श्रीकृष्ण के हृदय में राधा निवास करती हैं। उनका सौन्दर्य उर्वशी अप्सरा से भी अधिक है। नायिका का मन नायक के प्रेम के कारण सांसारिक कार्यों में नहीं लगता है। उसका चंचल मन नायक के ध्यान में निमग्न रहता है। नायिका को अपने चारों ओर नायक श्रीकृष्ण की छवि दिखाई देती रहती है।

प्रेम और सौन्दर्य संदर्भ-प्रसंगसहित व्याख्या

सोहत औ. पीतु पटु, स्याम सलौने गात।
मनौं नीलमनि-सैल पर, आतपु परयौ प्रभात॥ (1)

शब्दार्थ :
सोहत = शोभा दे रहा है। सलौने = सुन्दर चमकीले। गात = शरीर। सैल = पर्वत। आतपु = घाम, धूप।

सन्दर्भ :
प्रस्तुत दोहा बिहारी द्वारा रचित ‘सौन्दर्य-बोध’ शीर्षक से लिया गया है।

प्रसंग :
यहाँ पर पीताम्बरधारी श्यामसुन्दर श्रीकृष्ण के स्वरूप का मोहक अंकन किया गया है।

व्याख्या :
कविवर बिहारी कहते हैं कि नायक के श्याम सलौने शरीर पर पीला वस्त्र ओढ़ने से ऐसा लगता है मानो नीलमणि के,पर्वत पर प्रात:कालीन धूप पड़ रही हो।

विशेष :

1. इस दोहे में नायक के शारीरिक सौन्दर्य का वर्णन हुआ है।
2. छेकानुप्रास एवं उत्प्रेक्षा अलंकारों का प्रयोग।
3. दोहा छन्द।

सखि सोहत गोपाल के, उर गुंजनु की माल।
बाहिर लसति मनौ पिए, दावानल की ज्वाल॥ (2)

शब्दार्थ :
उर = वक्षस्थल पर। गुंजनु की माल = गुंजा रत्नों की माला। लसति = शोभित हो रही है। दावानल = जंगल की आग।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
नायिका ने गुंजों के कुंज में नायक से मिलने का वायदा किया था पर किसी कारण वह वहाँ न पहुँच सकी इससे नायक दुःखित हो जाता है। इसी भाव को यहाँ व्यक्त किया गया है।

व्याख्या :
नायिका अपनी सखी से जो उसके वृत्तांत से परिचित है, कहती है कि श्रीकृष्ण के हृदय पर गुंज माला मुझे ऐसी लगती है मानो उस कुंज में मुझे न पाकर इनको जो दुःखरूपी दावानल का पान करना पड़ा, उसी की ज्वाला बाहर निकल रही है, इसे देखकर मेरी आँखों को बड़ा ताप होता है।

विशेष :

1. गुरुजनों के मध्य नायिका अपनी बात को संकेतों के माध्यम से कहना चाहती है।
2. अनुप्रास एवं उत्प्रेक्षा अलंकारों का प्रयोग।
3. छन्द दोहा।

लिखन बैठि जाकी सबी, गहि-गहि गरव गरूर।
भए न केते जगत के, चतुर चितेरे कूर॥ (3)

शब्दार्थ :
सबी (सबीह) = यथार्थ चित्र। गरव गरूर = घमण्ड के साथ। केते = कितने। जगत = संसार के। चितेरे = चित्र बनाने वाले। कूर = विदलित, बुरी तरह विकृत।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
इस दोहे में नायिका की सखी नायक से उसके। क्षण-क्षण पर बढ़ते हुए यौवन तथा शरीर की कान्ति की प्रशंसा कर रही है।

व्याख्या :
नायिका की सखी नायक से कह रही है कि भला मैं बेचारी उस नायिका की प्रतिक्षण विकसित होती हुई शोभा। का वर्णन कैसे कर सकती हूँ? उसका यथार्थ चित्र लिखने के लिए घमण्ड में भर-भरकर बैठे जगत के कितने चतुर मूढ़ मति (क्रूर) नहीं हुए।

विशेष :

1. नायिका के शरीर की, प्रतिक्षण बदलती रूप छवि का वर्णन किया गया है।
2. सबी’ (सबीह) अरबी भाषा का शब्द है जिसका अर्थ यथार्थ चित्र से है।
3. गहि-गहि में पुनरुक्तिप्रकाश अलंकार है।

मकराकृति गोपाल कैं, सोहत कुंडल कान।
धस्यो मनौ हियगढ़ समरु, ड्योढ़ी लसत निसान॥ (4)

शब्दार्थ :
मकराकृति = मछली की आकृति के। सोहत = शोभा देते हैं। धस्यौ = धस गया, अपने अधिकार में किया। हियगढ़ = हृदय रूपी गढ़ में। लसत = शोभा दे रहे हैं। निसान = ध्वजा।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
सखी द्वारा नायिका का वर्णन करने पर नायक के ऊपर। जो कामदेव का प्रभाव पड़ा है, उसी का वर्णन यहाँ किया गया है।

व्याख्या :
हे सखी। गोपाल के कानों में मछली की आकृति के कुंडल ऐसे सुशोभित हैं, मानो हृदय रूपी गढ़ (किले) पर कामदेव ने विजय प्राप्त कर ली है। इसी कारण उसके ध्वज मकान की ड्योढ़ी पर फहरा रहे हैं।

विशेष :

1. नायक पर कामदेव के प्रभाव का वर्णन किया गया है।
2. दूसरी पंक्ति में उत्प्रेक्षा अलंकार।
3. शृंगार रस का वर्णन।

नीको लसत लिलार पर, टीको जरित जराय।
छविहिं बढ़वत रवि मनौ, ससि मंडल में आय॥ (5)

शब्दार्थ :
नीको= अच्छा। लसत= शोभा देता है। लिलार पर = माथे पर। जरित जराय = जड़ा हुआ। रवि = सूर्य ससि = चन्द्रमा। टीको = माथे पर पहनने वाला आभूषण।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
नायिका के जड़ाऊ टीके को माथे पर देखकर उससे रीझकर नायक कहता है।

व्याख्या :
नायक कहता है कि उसके ललाट पर जड़ाऊ काम से जड़ा हुआ टीका ऐसा अच्छा लग रहा है मानो चन्द्र-मंडल में आकर सूर्य उसकी छवि बढ़ा रहा हो।

विशेष :

1. माथे पर पहने हुए टीके की शोभा का वर्णन
2. द्वितीय पंक्ति में उत्प्रेक्षा अलंकार।
3. शृंगार रस।

झीनैं पट मैं झिलमिली, झलकति ओप अपार।
सुरतरु की मनु सिंधु में, लसति सपल्लव डार।। (6)

शब्दार्थ :
ओप = चमक। सुरतरु = देवताओं का वृक्ष अर्थात् कल्पवृक्ष। लसति = शोभित होती है। सपल्लव = पत्तों सहित। डार = डाल।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
झीने पट में फूटकर निकली हुई नायिका के शरीर की झलक पर रीझकर नायक कहता है।

व्याख्या :
नायक कहता है ओहो! इसकी झिलमिली (एक प्रकार का कानों में पहने जाने वाला आभूषण) अपार चमक के साथ झीने पट में झलक रही है। मानो कल्प वृक्ष की पत्तों सहित डाल समुद्र में विलास कर रही हो।

विशेष :

1. झीने पट में झिलमिली की झलक का सुन्दर वर्णन है।
2. दूसरी पंक्ति में उत्प्रेक्षा अलंकार।
3. शृंगार रस।

त्यौं-त्यौं प्यासेई रहत, ज्यौं-ज्यौं पियत अघाई।
सगुन सलोने रूप की, जुन चख तृषा बुझाई॥ (7)

शब्दार्थ :
प्यासेई रहत = प्यास नहीं बुझती है। अघाई = छककर पीने पर भी। चख = नेत्र। तृषा = प्यास।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
नायिका नायक को बार-बार देखती है फिर भी उसकी प्यास नहीं बुझती है, इसी का यहाँ वर्णन है।

व्याख्या :
नायिका कहती है कि मेरे प्यासे नेत्र ज्यों-ज्यों उस सगुन सलोने रूप को अघाकर पीते हैं त्यों-त्यों ही वे प्यासे बने रहते हैं क्योंकि सगुन एवं सलोने (खारे) पानी से आँखों की प्यास बुझती नहीं है अपितु बढ़ती ही जाती है।

विशेष :

1. इसमें कवि ने यह भाव प्रकट किया है कि जैसे खारे पानी से प्यास तृप्त नहीं होती है उसी तरह नायक के सलोने रूप को देखने की इच्छा भी पूरी नहीं होती है।
2. चख-तृषा में रूपक।
3. श्रृंगार रस।

तो पर वारौं उरबसी, सुनि राधिके सुजान।
तू मोहन के उर बसी है, उरबसी समान॥ (8)

शब्दार्थ :
उरबसी = उर्वशी नामक अप्सरा। सुजान = चतुर। उरबसी = हृदय में बस गयी है। उरबसी = हृदय पर पहनने का आभूषण।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
जब राधिका जी ने यह सुना कि श्रीकृष्ण किसी अन्य नायिका से प्रेम करने लगे हैं तो उन्होंने मौन धारण कर लिया है। सखी इसी मौन को छुड़ाने के निमित्त कहती है

व्याख्या :
हे चतुर राधिका! सुन, तू तो इतनी सुन्दर है कि तुझ पर मैं इन्द्र की अप्सरा उर्वशी को भी न्योछावर कर दूँ। हे राधा! तू तो मोहन के उर (हृदय) में उरबसी आभूषण के समान बसी हुई है। अतः दूसरों की बात सुनकर तुम मौन धारण मत करो।

विशेष :

1. नायिका के मौन को छुड़ाने का प्रयास है।
2. उरबसी’ में श्लेष, उरबसी-उरबसी में यमक अलंकार।
3. शृंगार रस।

फिरि-फिरि चित उत ही रहतु, टूटी लाज की लाव।
अंग-अंग-छवि-झौर में, भयो भौर की नाव॥ (9)

शब्दार्थ :
लाव = नाव बाँधने की रस्सी। लाज की लाव = लज्जा रूपी लाव। झौर = झूमर (झुंड) भौंर = भँवर।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
इस दोहे में नायिका अपनी सखी से अपनी स्नेह दशा का वर्णन करती है।

व्याख्या :
हे सखी! नायक के अंग-प्रत्यंग की छवि के झूमर में फँसकर मेरा चित्त (मन) भँवर में पड़ी हुई नाव जैसा बनकर, बार-बार घूमकर पुनः उसी अर्थात् नायक की ओर ही लगा रहता है। वह किसी दूसरी तरफ नहीं जाता है क्योंकि उसकी लज्जारूपी रस्सी टूट गई है।

विशेष :

1. नायक के प्रति विशेष आकर्षण ने उसकी लोक-लज्जा को मिटा दिया है।
2. फिरि-फिरि, अंग-अंग में पुनरुक्तिप्रकाश अलंकार, सम्पूर्ण में अनुप्रास।
3. शृंगार रस।

जहाँ-जहाँ ठाढ़ौ लख्यौ, स्याम सुभग-सिरमौरू।
उनहूँ बिन छिन गहि रहतु, दृगनु अजौं वह ठौरू॥ (10)

शब्दार्थ :
लख्यौ = देखा। सुभग = भाग्यवानों का। सिरमौरू = शिरोमणि। अजौं = अब भी। ठौरू = स्थान।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
श्रीकृष्ण के गोकुल से मथुरा चले जाने पर ब्रजबालाएँ आपस में जो वार्तालाप कर रही हैं, उसी का यहाँ वर्णन है।

व्याख्या :
ब्रज की युवतियाँ परस्पर कह रही हैं कि जिन-जिन स्थानों पर हमने प्यारे श्रीकृष्ण को खड़ा देखा था उन स्थानों में उनके कारण ऐसी सुन्दरता आ गई है कि उनके यहाँ न रहने पर भी हमारे नेत्रों को ऐसे प्रिय लगते हैं कि हम अब भी अपने आपको श्याम सुन्दर के संग पाती हैं।

विशेष :

1. सुभग सिरमौरु का अर्थ सुन्दर पुरुषों में शिरोमणि (श्रीकृष्ण) के लिए आया है।
2. श्रीकृष्ण के अलौकिक प्रभाव की छटा।
3. जहाँ-जहाँ में पुनरुक्ति, स्याम सुभग-सिरमौर में अनुप्रास अलंकार।

श्रद्धा भाव सारांश

जयशंकर प्रसाद ने श्रद्धा तथा मनु की प्राचीन कथा के माध्यम से मानव मन की तर्क प्रधान बुद्धि का अंकन किया है।

श्रद्धा मनु से प्रश्न करती है आप कौन हैं जो इस एकान्त स्थान को अपनी आभा से आलोकित किये हो? संसार रूपी सागर की लहरों के समीप इस निर्जन स्थान पर मौन होकर क्यों बैठे हो?

आपको देखकर ऐसा आभास होता है कि आपने दुनिया के रहस्य का निदान कर लिया है। आप करुणा की सजल मूर्ति हो। मनु ने श्रद्धा के अपूर्व सौन्दर्य को निहारा। वह उनके नेत्रों को उलझाने वाला सुन्दर माया-जाल था। श्रद्धा का शरीर पुष्पों से आच्छादित कोमल लता के समान अथवा बादलों के मध्य चन्द्र की चाँदनी जैसा दृष्टिगोचर हो रहा था।

श्रद्धा का शरीर बसन्त की शीतल-मन्द-सुगन्धित वायु से झूमते हुए सुरभित शाल के लघु वृक्ष की भाँति सुशोभित हो रहा था। श्रद्धा की केशराशि उसके कंधों तक पीछे की ओर बिखरी हुई थी। बालों की आभा उसके मुखमण्डल की शोभा को और भी अधिक बढ़ा रही थी। श्रद्धा की मुस्कान कोमल कली पर उगते हुए सूर्य की उज्ज्वल किरण अलसाई होने के कारण सुन्दर लग रही थी।

श्रद्धा की सुन्दरता इस प्रकार दृष्टिगोचर हो रही थी मानो पुष्पों का वन लेकर मन्द-मन्द बहती हुई वायु उसके आँचल को सुगन्ध से परिपूर्ण करके कृतज्ञता ज्ञापन कर रही हो।

श्रद्धा संदर्भ-प्रसंगसहित व्याख्या

(1) “कौन तुम? संसृति-जलनिधि तीर-तरंगों से फेंकी मणि एक,
कर रहे निर्जन का चुपचाप प्रभा की धारा से अभिषेक?
मधुर विश्रांत और एकांत-जगत का सुलझा हुआ रहस्य,
एक करुणामय सुन्दर मौन और चंचल मन का आलस्य।”

शब्दार्थ :
संसृति = संसार। जलनिधि = सागर। निर्जन = एकान्त। प्रभा = कान्ति। अभिषेक = तिलक करना, सुशोभित करना। मधुर = आकर्षक। विश्रांत = थके हुए। . सन्दर्भ-प्रस्तुत छन्द महाकवि जयशंकर प्रसाद द्वारा रचित ‘कामायनी’ महाकाव्य के ‘श्रद्धा’ सर्ग से लिया गया है।

प्रसंग :
इस छन्द में प्रलय काल के पश्चात् श्रद्धा का मनु से साक्षात्कार होता है। श्रद्धा मनु से पूछती है।

व्याख्या :
श्रद्धा मनु से पूछती है कि तुम कौन हो? जिस प्रकार सागर की लहरों द्वारा किनारे पर फेंकी गई मणि सूनेपन को अपनी ज्योति से सुशोभित करती है, उसी प्रकार तुम भी इस संसार रूपी सागर के किनारे बैठे हुए मणि के समान इस एकान्त स्थान को अपनी कान्ति से सुशोभित कर रहे हो। तुम्हारा रूप मधुर है, तुम थके हुए से प्रतीत हो रहे हो और इस एकान्त सूने स्थान पर बैठे हो। तुम्हें देखकर ऐसा लगता है मानो तुमने संसार के रहस्य को सुलझा लिया है इसलिए तुम यहाँ निश्चित होकर बैठे हो। तुम्हारे मुख पर करुणा भी है और तुम्हारा मौन बड़ा आकर्षक प्रतीत होता है। तुम्हारी यह शान्ति सदैव चंचल रहने वाले मन की शिथिलता के समान है।

विशेष :

1. कविता का आरम्भ नाटकीय ढंग से होता है।
2. रूपक एवं उपमा अलंकार का प्रयोग।
3. श्रृंगार रस।

(2) सुना यह मनु ने मधु गुंजार मधुकरी-सा जब सानन्द,
किए मुख नीचा कमल समान प्रथम कवि का ज्यों सुन्दर छन्द,
एक झिटका-सा लगा सहर्ष, निरखने लगे लुटे से कौन,
गा रहा यह सुन्दर संगीत? कुतूहल रहन सका फिर मौन।

शब्दार्थ :
मधु-गुंजार = मनोहर शब्द। मधुकरी = भँवरी। सानन्द = आनन्द सहित। झिटका = झटका।

सन्दर्भ एवं प्रसंग :
पूर्ववत्।

व्याख्या :
कवि कहता है कि उस समय मनु नीचे झुके हुए कमल के समान अपना मुख नीचा किए बैठे थे। उन्होंने भँवरी की गुंजार के समान श्रद्धा की यह मधुर वाणी बड़े हर्ष से सुनी। उस समय वे अकेले थे; किसी अन्य की मधुर वाणी सुनकर उनका प्रसन्न होना स्वाभाविक था। श्रद्धा द्वारा कहे गये ये शब्द मनु के लिए आदि कवि वाल्मीकि के प्रथम सुन्दर छन्द के समान थे।

वाल्मीकि कवि के प्रथम छन्द में करुणा का भाव समाया हुआ था। इधर श्रद्धा के वचनों में भी करुणा है। श्रद्धा की वाणी सुनते ही मनु को एक झटका-सा लगा और वे मोहित होकर यह देखने लगे कि कौन यह संगीत से मधुर वचन कह रहा है? जब मनु ने श्रद्धा को अपने सामने देखा तो कुतूहल के कारण वह शान्त न रह सके।

विशेष :

1. आदि कवि वाल्मीकि के मुख से जो प्रथम छन्द निकला था, उसमें करुणा मौजूद थी।
2. अनुप्रास एवं उपमा अलंकार का प्रयोग।
3. भाषा खड़ी बोली।

(3) और देखा वह सुन्दर दृश्य नयन का इन्द्रजाल अभिराम,
कुसुम-वैभव में लता समान चन्द्रिका से लिपटा घनश्याम।
हृदय की अनुकृति बाह्य उदार एक लम्बी काया उन्मुक्त,
मधु-पवन, क्रीड़ित ज्यों शिशु साल, सुशोभित हो सौरभ-संयुक्त।

शब्दार्थ :
इन्द्रजाल = जादू। अभिराम = सुन्दर। कुसुम-वैभव = फलों का ऐश्वर्य। चन्द्रिका = चाँदनी। घनश्याम = काला बादल। अनुकृति = अनुरूप। बाह्य = देखने में। उन्मुक्त = स्वछंद। मधु-पवन = बसन्त की वाय शिशु-साल = साल का छोटा वृक्ष। सौरभ-संयुक्त = सुगन्धि से युक्त।

सन्दर्भ एवं प्रसंग :
पूर्ववत्।

व्याख्या :
कवि कहता है कि मनु ने वह सुन्दर दृश्य देखा जो नेत्रों को जादू के समान मोहित कर देने वाला था। श्रद्धा उस समय फूलों की शोभा से युक्त लता के समान लग रही थी। श्रद्धा चाँदनी से घिरे हुए काले बादल के समान लग रही थी। श्रद्धा ने नीली खाल का वस्त्र पहन रखा था इस कारण वह बादल के समान दिखाई दे रही थी। किन्तु उसकी शारीरिक कान्ति उसके परिधान के बाहर भी जगमगा रही थी। श्रद्धा हृदय की भी उदार थी और उसी के अनुरूप वह देखने में उदार लग रही थी, उसका कद लम्बा था और उससे स्वच्छन्दता झलक रही थी। वायु के झोंकों में वह ऐसी लगती थी मानो बसन्त की वायु से हिलता हुआ कोई छोटा साल का पेड़ हो और वह सुगन्धि में डूबा हो।

विशेष :

1. श्रद्धा के अनुपम सौन्दर्य का वर्णन।
2. उपमा तथा उत्प्रेक्षा अलंकार।
3. भाषा-खड़ी बोली।

(4) मसूण, गांधार देश के नील रोम वाले मेषों के चर्म,
ढंक रहे थे उसका वपु कांत बन रहा था वह कोमल वर्म।
नील परिधान बीच सुकुमार खुला रहा मृदुल अधखुला
अंग, खिला हो ज्यों बिजली का फूल मेघवन बीच गुलाबी रंग।

शब्दार्थ :
मसृण = चिकने। गांधार देश = कंधार देश (अफगानिस्तान देश वर्तमान में)। रोम = रोयें। मेष = मेंढ़ा। चर्म = खाल। वपु = शरीर। कान्त = सुन्दर। वर्म = आवरण। परिधान = वस्त्र। मृदुल = कोमल।

सन्दर्भ एवं प्रसंग :
पूर्ववत्।

व्याख्या :
कवि कहता है कि कंधार देश के रोयें वाले मेंढ़ों की कोमल खाल से उसका सुन्दर शरीर ढका हुआ था। वही खाल (चमड़ा) उसका कोमल आवरण (वस्त्र) बन रहा था।
उस नीले आवरण के बीच से उसका कोमल अंग दिखाई दे रहा था। ऐसा लग रहा था मानो मेघ-वन के बीच में गुलाबी रंग का बिजली का फूल खिला हो।

विशेष :

1. श्रद्धा की वेशभूषा और सुन्दरता का वर्णन है।
2. उपमा तथा उत्प्रेक्षा अलंकार।
3. भाषा-खीड़ी बोली।

(5) आज, वह मुख! पश्चिम के व्योम बीच जब घिरते हों घनश्याम,
अरुण रवि-मण्डल उनको भेद दिखाई देता हो छविधाम।
या कि, नव इंद्रनील लघु शृंग फोड़कर धधक रही हो कांत,
एक लघु ज्वालामुखी अचेत माधवी रजनी में अश्रान्त।

शब्दार्थ :
व्योम= आकाश। अरुणलाल। रवि-मण्डल = सूर्य मण्डल। छविधाम = सुन्दर। इन्द्रनील = नीलम। लघु श्रृंग = छोटी चोटी। माधवी रजनी = बसन्त की रात। अश्रान्त = निरन्तर।
सन्दर्भ एवं प्रसंग-पूर्ववत्।
व्याख्या-कवि कहता है कि आह! वह मुख बहुत ही। सुन्दर था। सन्ध्या के समय पश्चिम दिशा में जब काले बादल आ जाते हैं और सूर्य अस्त होने से पहले छिप जाता है किन्तु जब
लाल सूर्य उन मेघों को चीर कर दिखाई देता है तो वह अत्यन्त सुन्दर दिखाई देता है। श्रद्धा के मुख का सौन्दर्य भी वैसा ही था।
श्रद्धा के मुख की सुन्दरता का वर्णन करते हुए आगे कवि। कहता है कि नीलम की नन्ही-सी चोटी हो और बसन्त ऋतु की मधुर रात्रि में एक छोटा-सा ज्वालामुखी उस नीलम की चोटी
को फोड़कर धधक रहा हो तो जैसी उसकी शोभा होगी, वैसी ही। शोभा श्रद्धा के मुख की थी।

विशेष :

1. श्रद्धा द्वारा पहना गया वस्त्र नीला है इस कारण नीले मेघों के बीच सूर्य की कल्पना की गई है।
2. नीलम की चोटी कल्पना भी इसी नीले आवरण के कारण की गयी है।
3. उत्प्रेक्षा अलंकार।

(6) घिर रहे थे घुघराले बाल अंस अवलम्बित मुख के पास,
नील घनशावक से सुकुमार सुधा भरने को विधु के पास।
और उस मुख पर वह मुसकान। रक्त किसलय पर ले विश्राम,
अरुण की एक किरण अम्लान अधिक अलसाई हो अभिराम।

शब्दार्थ :
अंस. = कंधा। अवलम्बित = सहारे। घन-शावक = बादल के बच्चे। सुधा = अमृत। विधु = चन्द्रमा। रक्त किसलय = लाल कोंपल। अरुण = सूर्य। अम्लान = कान्तिमान। अभिराम = सुन्दर।

सन्दर्भ एवं प्रसंग :
पूर्ववत्।

व्याख्या :
श्रद्धा के मुख के पास उसके कंधे पर धुंघराले बाल बिखरे हुए थे। उन्हें देखकर ऐसा लग रहा था मानो मेघों के बालक अर्थात् छोटे बादल चन्द्रमा के पास अमृत भरने को आये हों और श्रद्धा के मुख पर मुस्कराहट ऐसी शोभा दे रही थी मानो कोई सूर्य की कान्तिमान किरण लाल कोंपलों पर विश्राम करके अलसा रही हो।

विशेष :

1. श्रद्धा के बाल नीले मेघों के समान हैं और मुख = चन्द्रमा के समान, अतः उत्प्रेक्षा अलंकार।
2. श्रद्धा के ओंठ लाल कोंपल के समान हैं और मुस्कराहट सूर्य की किरण के समान। अतः उत्प्रेक्षा।
3. श्रृंगार रस।

(7) नित्य-यौवन छवि से ही दीप्त विश्व की करुण कामना मूर्ति,
स्पर्श के आकर्षण से पूर्ण प्रकट करती ज्यों जड़ में स्फूर्ति।
उषा की पहिली लेखा कांत, माधुरी से भींगी भर मोद,
मद भरी जैसे उठे सलज्ज भोर की तारक-द्युति की गोद॥

शब्दार्थ :
यौवन की छवि = यौवन की शोभा। दीप्ति = शोभित। करुण = दयावान। कामना मूर्ति = इच्छा की मूर्ति। स्पर्श-पूर्ण = श्रद्धा को देखकर उसे स्पर्श करने की इच्छा होती थी। स्फूर्ति = चेतना। लेखा कांत = सुन्दर किरण। माधुरी = सुषमा। मोद = हर्ष। मदभरी = मस्ती से भरी हुई। भोर = प्रात:काल। तारक द्युति की गोद = तारों की शोभा की छाया में।

सन्दर्भ एवं प्रसंग :
पूर्ववत्।

व्याख्या :
कवि कहता है कि श्रद्धा अनन्त यौवन की शोभा से दीप्त थी। वह संसार भर की सदय इच्छा की मूर्ति थी। उसे देखकर उसे स्पर्श करने की तीव्र इच्छा उत्पन्न होती थी। ऐसा लगता था मानो उसका सौन्दर्य जड़ वस्तुओं में भी चेतना भर देता था।

श्रद्धा उषा की पहली सुन्दर किरण के समान है। उसमें माधुर्य, आनन्द, मस्ती एवं लज्जा है। जिस प्रकार उषा की प्रथम किरण अन्धकार को दूर करती हुई निकल जाती है उसी प्रकार श्रद्धा के दर्शन से मनु के हृदय में छाया निराशा का अंधकार भी दूर होने लगा।

विशेष :

1. उषा की प्रथम किरण का मानवीकरण है।
2. श्रद्धा को पाकर मनु की निराशा कुछ कम होने लगी है।
3. उपमा एवं उत्प्रेक्षा अलंकार।

(8) कुसुम कानन अंचल में मंद-पवन प्रेरित सौरभ साकार,
रचित-परमाणु-पराग-शरीर, खड़ा हो, ले मधु का आधार।
और पड़ती हो उस पर शुभ्र नवल मधु राका मन की साथ,
हँसी का मद विह्वल प्रतिबिम्ब मधुरिमा खेला सदृश अबाध।

शब्दार्थ :
कानन-अंचल = जंगल के बीच। मंद-पवन = धीरे-धीरे चलने वाली वायु। सौरभ साकार = सुगन्धि की साकार मूर्ति। परमाणु – पराग = पराग के परमाणु। मधु = पुष्प रस। शुभ्र = स्वच्छ। नवल = नवीन। मधु राका = बसन्त की पूर्णिमा। मद विह्वल = मस्ती से भरी हुई। मधुरिमा खेला सदृश अबाध = हँसी में अक्षय माधुर्य भरा है।

सन्दर्भ एवं प्रसंग :
पूर्ववत्।

व्याख्या :
कवि कहता है कि श्रद्धा फूलों से भरे हुए वन के बीच सौरभ की मूर्ति के समान दिखाई देती है, जिससे मन्द पवन खेल रहा है। वह सौरभ की मूर्ति फलों के पराग के परमाणुओं से बनी है। इस पराग निर्मित मूर्ति पर मन की कामना रूपी नवीन बसन्त की पूर्णिमा की चाँदनी पड़ रही हो तो जैसी शोभा होगी, वैसी ही शोभा श्रद्धा की हो रही थी। उस समय श्रद्धा की मस्त हँसी निरन्तर माधुर्य से खेला करती थी।

विशेष :

1. श्रद्धा को पराग के परमाणुओं से निर्मित मूर्ति के समान दिखाकर उसके अनुपम सौन्दर्य का वर्णन किया है।
2. उपमा अलंकार, अनुप्रास की छटा।
3. शृंगार रस।