A) \[\operatorname{sp} < s{{p}^{2}} < sp\]
B) \[\operatorname{sp} < s{{p}^{2}} > s{{p}^{3}}\]
C) \[\operatorname{sp} > s{{p}^{2}} < s{{p}^{3}}\]
D) \[\operatorname{sp} > s{{p}^{2}} > s{{p}^{3}}\]
Correct Answer: D
Solution :
\[\operatorname{sp} > s{{p}^{2}} > s{{p}^{3}}\] Smaller the size of hybrid orbital, greater is the electronegativity thus correct order of electronegativity is \[\operatorname{sp} > s{{p}^{2}} > s{{p}^{3}}\]
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