The shape of ${\mathrm{H}}_2\mathrm{O}$ molecule is bent while that of ${\mathrm{CO}}_2$ 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

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.

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

The correct order of increasing ionic character in the molecules: LiF, K₂O, ${\mathrm{N}}_2, {\mathrm{SO}}_2$ and ClF₃ is -

1. ${\mathrm{N}}_2 < {\mathrm{ClF}}_3 < {\mathrm{SO}}_2< {\mathrm{K}}_2\mathrm{O} < \mathrm{LiF}$

2. ${\mathrm{N}}_2 < {\mathrm{ClF}}_3 < \mathrm{LiF} < {\mathrm{SO}}_2< {\mathrm{K}}_2\mathrm{O}$

3. ${\mathrm{SO}}_2< {\mathrm{K}}_2\mathrm{O} < {\mathrm{ClF}}_3 < \mathrm{LiF} < {\mathrm{N}}_2$

4. ${\mathrm{N}}_2 < {\mathrm{SO}}_2 < {\mathrm{ClF}}_3 < {\mathrm{K}}_2\mathrm{O} < \mathrm{LiF}$

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 d²sp³ hybridization.

 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 

The shapes of sp, ${\mathrm{sp}}^2$, ${\mathrm{sp}}^3$ 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

${\mathrm{BF}}_3+{\mathrm{NH}}_3\to {\mathrm{F}}_3\mathrm{B}.{\mathrm{NH}}_3$

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'.

The total number of sigma and pi bonds in C₂H₄ is

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.

The lone pairs of electrons can be defined as -

1. Electron pairs that participate in bonding.

2. Electron pairs  that do not participate in bonding.

3. Electron pairs that are present in inner most shell.

4.Electron pairs that are present in valence shell of ions.