Question #3c8ff

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Question #3c8ff

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Answer 1 · 2017-04-27T01:38:50

The unhybridized p-orbital would contain 'zero' electrons.

Explanation:

C[He]2 $s^{2} p_{x}^{2} p_{y}^{0} p_{z}^{0}$ $s^2p_x^2p_y^0p_z^0$ => ( C:[He]2 $\frac{↑}{s p^{2}} \frac{↑}{s p^{2}} \frac{↑}{s p^{2}} 2 (\frac{0}{p_{z}}))^{+}$ $uarr/(sp^2)uarr/(sp^2)uarr/(sp^2)2(0/p_z))^+$ + $e^{-}$ $e^-$

The carbocation will combine with 3H( $\frac{↑}{1 s}$ $uarr/(1s)$ )
=> $C H_{3}^{+}$ $CH_3^+$ [He]2((3 $σ) (1 p_{z}^{0}))^{+}$ $sigma)(1p_z^0))^+$

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