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.You need to login to perform this action.
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