In the given compound which carbon atom will show maximum electronegativity?
$C 1 H_{3} - C 2 H = C 3 = C 4 H_{2}$

Fourth

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First

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Third

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Second

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Solution

The correct option is D Third
Higher is the $s$ character, higher is the electronegativity.
The carbon atoms 1, 2, 3 and 4 are $s p^{3}$ , $s p^{2}$ , $s p$ and $s p^{2}$ hybridized respectively.
The percentage s character decreases in the order $s p (50) > s p^{2} (33)$ and $s p^{3} (25)$ .
Thus, the third carbon will show maximum electronegativity.

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