question_answer

X, Y are anhydrides of sulphurous acid and sulphuric acid, respectively. The hybridization state and the shape of X and Y are

A)

\[X\]	\[Y\]
\[s{{p}^{2}},\] angular	\[s{{p}^{3}},\] tetrahedral

               

B)

\[X\]	\[Y\]
\[s{{p}^{2}},\] angular	\[s{{p}^{2}},\] angular

C)

\[X\]	\[Y\]
\[s{{p}^{2}},\] angular	\[s{{p}^{2}},\] planar triangular

D)

\[X\]	\[Y\]
\[s{{p}^{3}},\]planar	\[s{{p}^{3}},\]planar

Correct Answer: C

Solution :

                 \[\underset{\text{sulphurous}\,\text{acid}}{\mathop{{{H}_{2}}S{{O}_{3}}}}\,\xrightarrow[-2{{H}_{2}}O]{{{P}_{2}}{{O}_{5}}}\underset{\begin{smallmatrix}  \,\,\,\,\,\,\,\,\,\,\,\,\,(X) \\  (anydride\,of\,  \\  sulphurous\,acid) \end{smallmatrix}}{\mathop{S{{O}_{2}}}}\,\]                            \[\Rightarrow \,\,3bp+01p\] Thus, hybridisation is \[s{{p}^{2}}\]and shape is angular due to presence of one unpaired electron. \[{{H}_{2}}S{{O}_{4}}\xrightarrow[-2{{H}_{2}}O]{{{P}_{2}}{{O}_{5}}}\underset{\begin{smallmatrix}  \,\,\,\,\,\,\,\,\,\,\,(Y) \\  anhydride\,of\,  \\  sulphuric\,acid \end{smallmatrix}}{\mathop{S{{O}_{3}}}}\,\]         \[\Rightarrow \] \[3bp+01p\] Thus, in \[S{{O}_{3}}\]hybridisation of S is \[s{{p}^{2}}\]and its shape is triangular planar.