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Q.

The number of hybrid orbitals in methyl carbocation is x and that in methyl carbanion is y, then the value of x+y is

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answer is 7.

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Detailed Solution

Methyl carbocation $\overset{\oplus}{C} H_{3}$ $s p^{2}$ hybridisation

Methyl carbanion $\bar{C} H_{3} - S P^{3}$ hybridisation

$x + y = 3 + 4 = 7$

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