The correct order regarding electronegativity of hybrid orbitals of carbon is:

s p < s p^{2} < s p^{3}

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s p < s p^{2} > s p^{3}

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s p > s p^{2} < s p^{3}

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s p > s p^{2} > s p^{3}

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Solution

The correct option is D $s p > s p^{2} > s p^{3}$
Electronegativity increases as s-character in hybrid orbital increases. Percentage of s character in $s p, s p^{2}$ and $s p^{3}$ are 50%, 33.33% and 25% respectively.
So, the correct option is D.

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Question 03

The types of hybrid orbitals of nitrogen in $N O_{2}^{+}, N O_{3}^{-} a n d N H_{4}^{+}$ respectively are expected to be

(a) $s p, s p^{3} a n d s p^{2}$

(b) $s p, s p^{2} a n d s p^{3}$

(d) $s p^{2}, s p^{3} a n d s p$

C H_{4}, C_{2} H_{6}, a n d C_{2} H_{4}

N O_{3}^{-}, N O_{2}^{+} a n d N H_{4}^{+}

s p^{3}, s p^{2}, s p^{2}, s p^{3}, s p^{2}, s p^{2}, s p, a n d s p

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