The hybridizations of atomic orbitals of nitrogen in , and respectively are [2016]

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Q.
The hybridizations of atomic orbitals of nitrogen in ${NO}_{2}^{+}$ , ${NO}_{3}^{-}$ and ${NH}_{4}^{+}$ respectively are [2016]

1 $s p, s p^{3}$ and $s p^{2}$

2 $s p^{2}, s p^{3}$ and $sp$

3 $s p, s p^{2}$ and $s p^{3}$

4 $s p^{2}, s p$ and $s p^{3}$

Ans.
(3)

$X = \frac{1}{2} (V E + M A - c + a)$

For $N O_{2}^{+},$ $X = \frac{1}{2} (5 + 0 - 1) = 2 i.e., s p hybridisation$

For $N O_{3}^{-},$ $X = \frac{1}{2} (5 + 0 + 1) = 3 i.e., s p^{2} hybridisation$

For $N H_{4}^{+},$ $X = \frac{1}{2} (5 + 4 - 1) = 4 i.e., s p^{3} hybridisation$

Solution

Q.
The hybridizations of atomic orbitals of nitrogen in ${NO}_{2}^{+}$ , ${NO}_{3}^{-}$ and ${NH}_{4}^{+}$ respectively are [2016]

1 $s p, s p^{3}$ and $s p^{2}$

2 $s p^{2}, s p^{3}$ and $sp$

3 $s p, s p^{2}$ and $s p^{3}$

4 $s p^{2}, s p$ and $s p^{3}$

Ans.
(3)

$X = \frac{1}{2} (V E + M A - c + a)$

For $N O_{2}^{+},$ $X = \frac{1}{2} (5 + 0 - 1) = 2 i.e., s p hybridisation$

For $N O_{3}^{-},$ $X = \frac{1}{2} (5 + 0 + 1) = 3 i.e., s p^{2} hybridisation$

For $N H_{4}^{+},$ $X = \frac{1}{2} (5 + 4 - 1) = 4 i.e., s p^{3} hybridisation$

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Solution

Q. The hybridizations of atomic orbitals of nitrogen in NO2+, NO3- and NH4+ respectively are [2016]

1 sp,sp3 and sp2

2 sp2,sp3 and sp

3 sp,sp2 and sp3

4 sp2,sp and sp3

Ans. (3) X=12(VE+MA-c+a) For NO2+, X=12(5+0-1)=2 i.e., sp hybridisation For NO3-, X=12(5+0+1)=3 i.e., sp2 hybridisation For NH4+, X=12(5+4-1)=4 i.e., sp3 hybridisation

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Q.
The hybridizations of atomic orbitals of nitrogen in ${NO}_{2}^{+}$ , ${NO}_{3}^{-}$ and ${NH}_{4}^{+}$ respectively are [2016]

1 $s p, s p^{3}$ and $s p^{2}$

2 $s p^{2}, s p^{3}$ and $sp$

3 $s p, s p^{2}$ and $s p^{3}$

4 $s p^{2}, s p$ and $s p^{3}$

Ans.
(3)

$X = \frac{1}{2} (V E + M A - c + a)$

For $N O_{2}^{+},$ $X = \frac{1}{2} (5 + 0 - 1) = 2 i.e., s p hybridisation$

For $N O_{3}^{-},$ $X = \frac{1}{2} (5 + 0 + 1) = 3 i.e., s p^{2} hybridisation$

For $N H_{4}^{+},$ $X = \frac{1}{2} (5 + 4 - 1) = 4 i.e., s p^{3} hybridisation$