Question

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

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

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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