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.You need to login to perform this action.
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