The bond length can be defined as -

1.	The equilibrium distance between the nuclei of two bonded atoms in a molecule.
2.	The farthest distance between the nuclei of two bonded atoms in a molecule.
3.	The shortest distance between the nuclei of two bonded atoms in a molecule.
4.	None of the above.

${\mathrm{H}}_3{\mathrm{PO}}_3$ can be represented by structures 1 and 2 shown below. 

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.

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

The application of dipole moment is/are :

The difference between electronegativity and electron gain enthalpy is :

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:

 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 -