The hybridizations of atomic orbitals of nitr | vedclass

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Question

Here's a nice short trick.

$1)$ Count and add all the electrons in valence shell

$2)$ Divide it by $8$

$3)$ On dividing by $8$ you will g

Vedclass · Accepted Answer

$sp, sp^2$ and $sp^3$

Vedclass · Answer

Here's a nice short trick.

$1)$ Count and add all the electrons in valence shell

$2)$ Divide it by $8$

$3)$ On dividing by $8$ you will get a remainder and quotient, add quotient $+$ (remainder $/ 2$)

$4)$ Now get the hybridization corresponding to the number what you got

If $2$ its $sp$, if $3$ its $sp ^2$, if $4$ its $sp ^3$, if $5$ its $sp ^3 d$ and so on.

Ex: $NO ^{2+}$

$ightarrow$ Total electrons in valence shell : $5+6 	imes 2-1=16$.

$ightarrow 16 / 8=2$. Quotient $=2$ and Remainder $=0$

$ightarrow 2+0 / 2=2$ i.e $sp$

$N H ^{4+}$

$ightarrow$ Total electrons in valence shell : $5+7 	imes 4-1=32$.

$ightarrow 32 / 8=4$. Quotient $=4$ and Remainder $=0$

$ightarrow 4+0 / 2=4$ i.e $sp ^3$

$NO ^{3-}$

$ightarrow$ Total electrons in valence shell : $5+6 	imes 3+1=24$.

$ightarrow 24 / 8=3$. Quotient $=3$ and Remainder $=0$

$ightarrow 3+0 / 2=3$ i.e $sp ^2$

The hybridizations of atomic orbitals of nitrogen in $NO_2^+, NO_3^-$ and $NH_4^+$ respectively are

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