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

However, these two structures cannot be considered as canonical forms of a resonance hybrid because:

1.	The positions of the atoms are different in each structure.
2.	The positions of the atoms remain unchanged.
3.	\(\text{H}_3\text{PO}_3\) 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 for this fact is that dipole moment of:

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 -

Among ${\mathrm{NH}}_3$ and ${\mathrm{NF}}_3$,  _____ has higher dipole moment because -

1. NF_{3 ;} the resultant bond moment of the N–H bonds add up with the lone pair while N – F bonds partly cancels the moment of the lone pair.

2. NH_{3 ;} the resultant bond moment of the N–F bonds add up with lone pair while N - H bonds partly cancels the moment of the lone pair.

3. NF_{3 ;} the resultant bond moment of the N–F bonds add up with the lone pair while N – H bonds partly cancels the moment of the lone pair.

4. NH_{3 ;} the resultant bond moment of the N–H bonds add up with lone pair while N – F bonds partly cancels the moment of the lone pair.

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 :