question_answer

The\[s{{p}^{3}}\]hybridisation is present on the central atom of

A)  \[HCHO\]          

B)  \[BC{{l}_{3}}\]

C)  \[PC{{l}_{3}}\]            

D)  \[S{{O}_{3}}\]

Correct Answer: C

Solution :

 For\[s{{p}^{3}}\]hybridisation,the number of hybridized orbitals of central atom should be four. Number of hybridised orbitals = number of\[\sigma \]bonds + number of lone pairs? Number of hybridised orbitals in HCHO is 3. \[H\overset{\sigma }{\mathop{-}}\,\overset{\begin{smallmatrix}  H \\  |\sigma  \end{smallmatrix}}{\mathop{C}}\,\overset{\sigma }{\mathop{=}}\,O\] ie, \[3+0=3\] Number of hybridised orbitals in\[BC{{l}_{3}}\]are 3. \[Cl\overset{\sigma }{\mathop{-}}\,\underset{\begin{smallmatrix}  |\sigma  \\  Cl \end{smallmatrix}}{\mathop{B}}\,\overset{\sigma }{\mathop{-}}\,Cl\]ie, \[3+0=3\] Number of hybridised orbitals in\[PC{{l}_{3}}\]are 4. \[Cl-\underset{\begin{smallmatrix}  | \\  Cl \end{smallmatrix}}{\mathop{\overset{\bullet \,\bullet }{\mathop{B}}\,}}\,-Cl\]ie, \[3+1=4\] Number of hybridised orbitals in\[S{{O}_{3}}\]are 3. ie, \[3+0=3\] Therefore, the hybridisation on central atom in\[PC{{l}_{3}}\]is.