A) \[s,\,\,{{p}_{x}},\,\,{{p}_{y}},\,\,{{d}_{yz}}\]
B) \[s,\,\,{{p}_{x}},\,\,{{p}_{y}},\,\,{{d}_{{{x}^{2}}-{{y}^{2}}}}\]
C) \[s,\,\,{{p}_{x}},\,\,{{p}_{y,\,\,}}{{d}_{{{z}^{2}}}}\]
D) \[s,\,\,{{p}_{y}},\,\,{{p}_{z}},{{d}_{xy}}\]
Correct Answer: B
Solution :
For square planar geometry hybridization is \[ds{{p}^{2}}\]involving s,\[{{p}_{x}},\,{{p}_{y}}\] and \[{{d}_{{{x}^{2}}-}}_{{{y}^{2}}}\] orbital.
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