A) \[C{{H}_{4}}\]
B) \[Xe{{F}_{4}}\]
C) \[{{H}_{2}}O\]
D) \[N{{H}_{3}}\]
Correct Answer: B
Solution :
[a] \[C{{H}_{4}}\to \] \[\begin{align} & =4\sigma \,bond \\ & \Rightarrow s{{p}^{3}}hybridisation \\ \end{align}\] [b] \[Xe{{F}_{4}}\to \] \[=4\sigma \]bond +2 lone pairs \[\Rightarrow \]\[s{{p}^{3}}{{d}^{2}}\]hybridization [c] \[{{H}_{2}}O\to \] \[=2\sigma \]bond + 2 lone pairs \[\Rightarrow \]\[s{{p}^{3}}\]hydridisation [d] \[N{{H}_{3}}\to \] \[=3\sigma \]bond + 1 lone pair \[\Rightarrow \]\[s{{p}^{3}}\]hybridisationYou need to login to perform this action.
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