In $(C H_{3})_{3} N$ nitrogen is $s p^{3}$ hybridized whereas in $(S i H_{3})_{3} N$ it is $s p^{2}$ hybridised, why?

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Solution

$S i$ contains vacant $3 d$ orbitals, the lone pair of $N$ is involved in $p_{π} - d_{π}$ back bonding. Hence $N - S i$ bonds gain partial double bond character i.e. its hybridization becomes $s p^{2}$ . However, $N {(C H_{3})}_{3}$ does not have any such back bonding due to absence of $d$ orbitals. As a result $N$ has normal $s p^{3}$ hybridization with pyramidal shape.

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