The shape of ${\mathrm{H}}_2\mathrm{O}$ molecule is bent while that of ${\mathrm{CO}}_2$ is linear. The correct explanation for this fact is that dipole moment of:

1.	carbon dioxide is 3.84 D and that of water is zero.
2.	water is zero and that of carbon dioxide is 1.84 D.
3.	carbon dioxide is zero and that of water is 1.84 D.
4.	carbon dioxide is 3.84 D and that of water is 1.84 D.

The application of dipole moment is/are :

The correct order of increasing ionic character in the molecules: LiF, K₂O, ${\mathrm{N}}_2,<mspace\; width="1em"></mspace>{\mathrm{SO}}_2$ and ClF₃ is:

1. ${\mathrm{N}}_2<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><<mspace\; width="1em"></mspace>{\mathrm{ClF}}_3<mspace\; width="1em"></mspace><<mspace\; width="1em"></mspace>{\mathrm{SO}}_2<<mspace\; width="1em"></mspace>{\mathrm{K}}_2\mathrm{O}<mspace\; width="1em"></mspace><<mspace\; width="1em"></mspace>\mathrm{LiF}$

2. ${\mathrm{N}}_2<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><<mspace\; width="1em"></mspace>{\mathrm{ClF}}_3<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><<mspace\; width="1em"></mspace>\mathrm{LiF}<mspace\; width="1em"></mspace><<mspace\; width="1em"></mspace>{\mathrm{SO}}_2<<mspace\; width="1em"></mspace>{\mathrm{K}}_2\mathrm{O}$

3. ${\mathrm{SO}}_2<<mspace\; width="1em"></mspace>{\mathrm{K}}_2\mathrm{O}<mspace\; width="1em"></mspace><<mspace\; width="1em"></mspace>{\mathrm{ClF}}_3<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><<mspace\; width="1em"></mspace>\mathrm{LiF}<mspace\; width="1em"></mspace><<mspace\; width="1em"></mspace>{\mathrm{N}}_2$

4. ${\mathrm{N}}_2<mspace\; width="1em"></mspace><<mspace\; width="1em"></mspace>{\mathrm{SO}}_2<mspace\; width="1em"></mspace><<mspace\; width="1em"></mspace>{\mathrm{ClF}}_3<mspace\; width="1em"></mspace><<mspace\; width="1em"></mspace>{\mathrm{K}}_2\mathrm{O}<mspace\; width="1em"></mspace><<mspace\; width="1em"></mspace>\mathrm{LiF}$

CH₄ does not exhibit square planar geometry because:

 BeH₂ molecule has a zero dipole moment because -

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 :

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:

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.