Which of the following factors are favourable for the formation of an ionic bond?

1.	Low ionization enthalpy, High negative electron gain enthalpy, High lattice energy.
2.	High ionization enthalpy, High electron gain enthalpy, High lattice energy.
3.	Low ionization enthalpy, Low electron gain enthalpy, Low lattice energy.
4.	High ionization enthalpy, Low electron gain enthalpy, High lattice energy.

The correct relationship between bond strength and bond order is :

The bond length can be defined as:

${\mathrm{}}_{}{\mathrm{}}_{}$\(\text{H}_3\text{PO}_3\) 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:

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 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