A) \[sp<s{{p}^{2}}>s{{p}^{3}}\]
B) \[sp<s{{p}^{2}}<s{{p}^{3}}\]
C) \[sp>s{{p}^{2}}<s{{p}^{3}}\]
D) \[sp>s{{p}^{2}}>s{{p}^{3}}\]
Correct Answer: D
Solution :
The electronegativity of the hybrid orbitals decreases as the percentage of s-character decreases. Hence, the correct order is \[sp>s{{p}^{2}}>s{{p}^{3}}\]
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