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Question

The hybridizations of atomic orbitals of nitrogen in $$NO_2^+ ,\ N{ O }_3^-$$ and $$NH_4^+$$ respectively are :


A
sp, sp2 and sp3
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B
sp2, sp and sp3
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C
sp, sp3 and sp2
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D
sp2, sp3 and sp
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Solution

The correct option is A $$sp,\ sp^2 $$ and $$sp^3$$
$$NO_2^{+}$$ contains 2 sigma bonds so hybridization is $$sp$$ and shape is linear.

$$N{ O }_3^{-}$$ contains 3 sigma bonds so hybridization is $$sp^{2}$$ and shape is trigonal.

$$NH_4^{+}$$ contains 4 sigma bonds so hybridization is $$sp^{3}$$ and shape is tetrahedral.

Hence, option A is correct.

Chemistry

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