Correct regarding edges and faces in $l{F}_7$ respectively 

1.  15, 10

2.  10, 10

3.  7, 15

4.  15, 7

Electronegativity of A and B is 1 and 4 respectively. The percentage of the covalent character of AB is

1.  79.5%

2.  8%

3.  20.5%

4.  25%

The bond energies in $\mathrm{NO}, {\mathrm{NO}}^{+} \mathrm{and} {\mathrm{NO}}^{-}$ follow the order

1.  $N{O}^{+} > NO > N{O}^{-}$

2.  $NO > N{O}^{+} > N{O}^{-}$

3.  $N{O}^{-} > NO > N{O}^{+}$

4.  $N{O}^{+} > N{O}^{-} > NO$

The element which readily forms an ionic bond has the electronic configuration

1. $1_{s^2}{2}_{s^2}{2}_{p^3}$

2. $1_{s^2}{2}_{s^2}{2}_{p^1}$

3. $1_{s^2}{2}_{s^2}{2}_{p^2}$

4.$1_{s^2}{2}_{s^2}2p^6{3}_{s^1}$

The correct of $\mathrm{F}\hat{\mathrm{X}}\mathrm{F}$ bond angle (X = central atom) in ${\mathrm{CF}}_4, {\mathrm{POF}}_3 \mathrm{and} {\mathrm{SO}}_2{\mathrm{F}}_2$ is

1.  $S{O}_2{F}_2 < PO{F}_3 < C{F}_4$

2.  $PO{F}_3 < C{F}_4< S{O}_2{F}_2$

3.  $C{F}_4 < S{O}_2{F}_2 < PO{F}_3$

4.  $PO{F}_3 < S{O}_2{F}_2 < C{F}_4$

In the cyanide ion, the negative charge is on 

1.  C

2.  N

3.  Both C and N

4.  Resonate between C and N

The order of electronegativity of carbon in sp, ${\mathrm{sp}}^2 \mathrm{and} {\mathrm{sp}}^3$ hybridized states follows

1. $\mathrm{sp}>{\mathrm{sp}}^2>{\mathrm{sp}}^3$

2. ${\mathrm{sp}}^3>{\mathrm{sp}}^2>\mathrm{sp}$

3. $\mathrm{sp}>{\mathrm{sp}}^3>{\mathrm{sp}}^2$

4. ${\mathrm{sp}}^2>\mathrm{sp}>{\mathrm{sp}}^3$

In which of the dimerization process, the achievement of the octet is not the driving force.

1.  $2AlC{l}_3 \to A{l}_{2}C{l}_6$

2.  $BeC{l}_2\to BeC{l}_2 \left(solid\right)$

3.  $2lC{l}_3 \to l_{2}C{l}_6$

4.  $2B{H}_3 \to B_2{H}_6$

Anhydrous $AlC{l}_3$ is covalent. If the lattice energy for${\mathrm{AlCl}}_3 = 5137 \mathrm{kJ} {\mathrm{mol}}^{-1},$${∆}_{hydration}$ for ${\mathrm{Al}}^{3+} = -4665 \mathrm{kJ} {\mathrm{mol}}^{-1} \mathrm{and} {∆}_{\mathrm{Hydration}} \mathrm{for} {\mathrm{Cl}}^{-} = -381 \mathrm{kJ} {\mathrm{mol}}^{-1}$, will it remain covalent or become ionic in aqueous solution?

1.  It will remain covalent

2.  It will remain ionic

3.  It may or may not become ionic

4.  Being ionic or covalent of any molecule is independent of ionisation energy

The melting point of $Al{F}_3$ is $104°C$ and that of $Si{F}_4$ is $-77°C$ (it sublimes) because:

1.  There is a very large difference in the ionic character of the Al - F and Si - F bonds

2.  In $Al{F}_3, A{l}^{3+}$ interacts very strongly with the neighboring $F^{-}$ ions to give a three-dimensional structure but in $Si{F}_4$ no such interaction is possible

3.  The silicon ion in the tetrahedral $Si{F}_4$ is not shielded effectively from the fluoride ions whereas in $Al{F}_3, the A{l}^{+3}$ ion is shielded on all sides

4.  The attractive forces between the $Si{F}_4$ molecules are strong whereas those between the $Al{F}_3$ molecule are weak