Chemical Bonding And Molecular Structure Question 351
The hybridization of atomic orbitals of nitrogen in $ NO_2^{+},NO_2^{-} $ and $ NH_4^{+} $ are
Options:
A) $ sp^{2},sp^{3} $ and $ sp^{2} $ respectively
B) $ sp,sp^{2} $ and $ sp^{3} $ respectively
C) $ sp^{2},sp $ and $ sp^{3} $ respectively
D) $ sp^{2},sp^{3} $ and sp respectively
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Answer:
Correct Answer: B
Solution:
- $ NO_2^{+}=\frac{1}{2}[5+0+0-1]=2sp^3; $
$ NO_2^{-}=\frac{1}{2}[5+0+1-0]=3sp^3; $
$ NH_4^{+}=\frac{1}{2}[5+4+0-1]=4sp^{3} $
BETA
