H3PO3 can be represented by structures 1 and 2 shown below. 

ncert solutions for class 11 chemistry chapter 4

These two structures cannot be taken as the canonical forms of the resonance hybrid, because  : 

1. The positions of the atoms have changed.

2. The positions of the atoms are constant.

3. H3PO3 does not show resonance.

4. Two hydrogen atoms are missing.

Subtopic:  Resonance & Nature of Compounds |
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The shape of H2O molecule is bent while that of CO2 is linear. The correct explanation is -

1. The dipole moment of carbon dioxide is 3.84 D and water is zero

2. The dipole moment of water is zero and carbon dioxide is 1.84 D

3. The dipole moment of carbon dioxide is zero and water is 1.84 D

4. The dipole moment of carbon dioxide is 3.84 D and water is 1.84 D

Subtopic:  Polarity |
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The application of dipole moment is/are -

1. It is used to differentiate polar and non-polar bonds.

2. It is helpful in calculating the percentage ionic character of a molecule.

3. Both '1' and '2'

4. None of the above.

Subtopic:  Polarity |
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The difference between electronegativity and electron gain enthalpy is -

1. An element has a constant value of the electron gain enthalpy but not of electronegativity

2. There is no difference between these two terms

3. An element has a constant value of electronegativity but not of electron gain enthalpy.

4. None of above

Subtopic:  Lattice & Hydration Energy |
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The correct order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3 is -

1. N2  < ClF3 < SO2< K2O < LiF

2. N2  < ClF3  < LiF < SO2< K2O

3. SO2< K2O < ClF3  < LiF < N2

4. N2 < SO2 < ClF3 < K2O < LiF

 

Subtopic:  Polarity |
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CH4 does not exhibit square planar geometry because-

1. For square planar geometry, 5 bonds are required.

2. Carbon does not have d-orbitals to undergo dsp2 hybridization.

3. Due to steric hindrance CH4 does not exhibit square planar geometry.

4. Carbon does not have d-orbitals to undergo d2sphybridization.

Subtopic:  Hybridisation |
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 BeH2 molecule has a zero dipole moment because -

1. Dipole moments of each H–Be bond are unequal and are in opposite directions

2. Dipole moments of each H–Be bond are equal and are in opposite directions

3. Dipole moments of each H–Be bond are equal and are in same directions

4. Dipole moments of each H–Be bond are unequal and are in same directions 

Subtopic:  Polarity |
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The shapes of sp, sp2, sp3 orbitals formed due to hybridization of atomic orbitals would be respectively-

1. Trigonal planar, linear, tetrahedral

2. Linear, trigonal planar, tetrahedral

3. Linear, tetrahedral, trigonal planar

4. Tetrahedral, trigonal planar, linear

Subtopic:  Hybridisation |
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BF3+NH3F3B.NH3

The correct statement about the above reaction is -

1. Hybridization of ‘B’ changes to sp2 from sp3 while there is no change in the hybridization of 'N'

2. Hybridization of ‘N’ changes to sp3 from sp2 while there is no change in the hybridization of 'B'

3. Hybridization of ‘N’ changes to sp2 from sp3 while there is no change in the hybridization of 'B'

4. Hybridization of ‘B’ changes to sp3 from sp2 while there is no change in the hybridization of 'N'.

Subtopic:  Hybridisation |
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The total number of sigma and pi bonds in C2His

1. 6 sigma bonds and 1 pi-bond.

2. 3 sigma bonds and 3 pi-bonds.

3. 5 sigma bonds and 1 pi-bond.

4. 2 sigma bonds and 2 pi-bonds.

Subtopic:  Hybridisation |
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