# NEET Chemistry The p-Block Elements (XII) Questions Solved

BOARD

(a) Complete the following chemical equations :

(i)

(ii)

(b) How would you account for the following ?

(i) The value of electron gain enthalpy with negative sign for sulphur is higher than that for oxygen.

(ii) $N{F}_{3}$ is an exothermic compound but $NC{l}_{3}$ is endothermic compound.

(iii) $CI{F}_{3}$ molecule has a T-shaped structure and not a trigonal planar one.                        5 mark

(a) (i) $6NaOH+3C{l}_{2}\to 5NaCl+1NaCl{O}_{3}+3{H}_{2}O$

(ii) $Xe{F}_{6}\left(s\right)+3{H}_{2}O\left(l\right)\to Xe{O}_{3}\left(s\right)+6HF\left(aq\right)$

(b) (i) Because of enthalpy of dissociation of S-S bond is higher than O-O bond and the hydration energy of ${S}^{2-}$ is less than that of ${O}^{2-}$ ion.

(ii) Due to smaller size of F as compared to Cl, the N-F bond is much stronger than N-Cl bond while bond dissociation energy of ${F}_{2}$ is much lower than that of $C{l}_{2}$. Therefore, energy released during the formation of $N{F}_{3}$ molecule is more than the energy needed to break ${N}_{2}$ and ${F}_{2}$ molecules into individual atoms. In other words, formation of $N{F}_{3}$ is an exothermic reaction.

The energy released during the formation of $NC{l}_{3}$ molecule is less than the energy needed to break ${N}_{2}$ and $C{l}_{2}$ molecule into individual atoms. Thus formation of $NC{l}_{3}$ is an endothermic  reaction.

(iii) The electronic configuration of Cl is  It has only one half filled orbital. But to form three Cl-F bonds, we need three half filled orbitals. To achieve this, one of the $3{P}_{y}^{2}$ electrons gets excited to 3d orbital. The resulting five orbitals of the third shell of Cl in the first excited state undergoes $S{P}^{3}d$ hybridization to give T-shaped structure to $CI{F}_{3}$ molecule. Since Cl does not undergo $S{P}^{2}$ hybridization, therefore $CI{F}_{3}$ does not have trigonal planar structure.

5 mark

Difficulty Level: