Chemical Bonding And Molecular Structure Ques 76
76. Which of the two ions from the list given below that have the geometry that is explained by the same hybridization of orbitals, $NO_2^{-}, NO_3{ }^{-}$, $NH_2^{-}, NH_4^{+}, SCN^{-}$?
[2011]
(a) $NO_2{ }^{-}$and $NO_3{ }^{-}$
(c) $SCN^{-}$and $NH_2^{-}$
(b) $NH_4^{+}$and $NO_3^{-}$
(d) $NO_2^{-}$and $NH_2^{-}$
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Solution:
- (a)
$NO_2{ }^{-}, \quad H=\frac{1}{2}[5+0+1-0]=3=s p^{2}$
$NO_3{ }^{-}, \quad H=\frac{1}{2}[5+0+1-0]=3=s p^{2}$
$NH_2{ }^{-}, \quad H=\frac{1}{2}[5+2+1+0]=4=s p^{3}$
$NH_4{ }^{+}, \quad H=\frac{1}{2}[5+4+0-1]=4=s p^{3}$
$SCN^{-}, \quad H=\frac{1}{2}[4+0+0+0]=2=s p$
$\therefore NO_2{ }^{-}$and $NO_3{ }^{-}$have same hybridisation.
Hybridisation $=\frac{1}{2}$ [No. of valence electrons of central atom + no. of monovalent atoms attached to it + Negative charge if any - positive charge if any]
BETA
