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: B
Solution :
The correct order regarding the electronegativity of hybrid orbitals of carbon is \[sp>s{{p}^{2}}>s{{p}^{3}}\] because in \[sp,\,\,s{{p}^{2}}\] an \[s{{p}^{3}}\] hybrid orbitals s-orbital character is 50%, 33.3% and 25% respectively and due to higher s-orbital character electron attraction tendency i.e., electronegativity increases.
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