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.

Be₂ molecule does not exist. The best explanation to explain this on the basis of the molecular orbital theory is -

1. The bond order of Be₂ is one.

2. The bond order of Be₂ is negative.

3. The bond order of Be₂ is zero.

4. The bond order of Be₂ is two.

The axial bonds are longer as compared to equatorial bonds in PCl₅ because -

1. Axial bond pairs suffer more repulsion from the equatorial bond pairs

2. Equatorial bond pairs suffer more repulsion from the axial bond pairs

3. Both 1, and 2

4. None of the above

The relation between hydrogen bond and the van der Waals forces is -

1. Both the bonds have equal strength

2. Hydrogen bonds are weaker than van der Waals forces

3. Hydrogen bonds are stronger than van der Waals forces

4. None of the above