In $P_4$ (Tetrahedral) molecule

1.  Each P is joined to four P

2.  Each P is joined to three P

3.  Each P is joined to two P

4.  $P_4$ does not exist

molecular shapes of ${\mathrm{SF}}_4,<mspace\; width="1em"></mspace>{\mathrm{CF}}_4<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>\mathrm{XeF}$₄ are 

1.  Same with 2, 0 and 1 lone pairs of electrons respectively

2.  The same with 1, 1 and 1 lone pairs of electrons respectively 

3.  Different with 0, 1 and 2 lone pairs of electrons respectively 

4.  Different with 1, 0 and 2 lone pairs of electrons respectively

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 $\mathrm{NO},<mspace\; width="1em"></mspace>{\mathrm{NO}}^{+}<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{NO}}^{-}$ follow the order

1.  $N{O}^{+}<mspace\; width="1em"></mspace>><mspace\; width="1em"></mspace>NO<mspace\; width="1em"></mspace>><mspace\; width="1em"></mspace>N{O}^{-}$

2.  $NO<mspace\; width="1em"></mspace>><mspace\; width="1em"></mspace>N{O}^{+}<mspace\; width="1em"></mspace>><mspace\; width="1em"></mspace>N{O}^{-}$

3.  $N{O}^{-}<mspace\; width="1em"></mspace>><mspace\; width="1em"></mspace>NO<mspace\; width="1em"></mspace>><mspace\; width="1em"></mspace>N{O}^{+}$

4.  $N{O}^{+}<mspace\; width="1em"></mspace>><mspace\; width="1em"></mspace>N{O}^{-}<mspace\; width="1em"></mspace>><mspace\; width="1em"></mspace>NO$

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, ${\mathrm{sp}}^2<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{sp}}^3$ 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.  $2AlC{l}_3<mspace\; width="1em"></mspace>\to A{l}_{2}C{l}_6$

2.  $BeC{l}_2\to BeC{l}_2<mspace\; width="1em"></mspace>\left(solid\right)$

3.  $2lC{l}_3<mspace\; width="1em"></mspace>\to l_{2}C{l}_6$

4.  $2B{H}_3<mspace\; width="1em"></mspace>\to B_2{H}_6$