A) \[C{{O}_{2}}\]
B) \[S{{O}_{2}}\]
C) \[{{N}_{2}}O\]
D) \[CO\]
Correct Answer: B
Solution :
When the number of hybrid orbitals is 3, hybridisation is \[s{{p}^{2}}\]. Number of hybrid orbitals,\[H=\frac{1}{2}[V+M-C+A]\][where, V == no. of valence electrons, M = monovalent atoms, C and A = positive and negative charge] \[C{{O}_{2}}\]\[\Rightarrow \]\[H=\frac{1}{2}[4+0-0+0]=2\] \[\therefore \] sp hybridisation. \[S{{O}_{2}}\]\[\Rightarrow \]\[H=\frac{1}{2}[6+0-0+0]=3\] \[\therefore \] \[s{{p}^{2}}\] hybridisation. \[{{N}_{2}}O\]\[\Rightarrow \] \[H=\frac{1}{2}[5+0-0+0]=2.5\] \[\therefore \] sp hybridisation. \[CO\]\[\Rightarrow \]\[H=\frac{1}{2}[4+0-0+0]=2\] \[\therefore \] sp hybridisation. Thus, \[S{{O}_{2}}\] has \[s{{p}^{2}}\] hybridisation.
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