Chemical Bonding and Molecular Structure - Result Question 51
####52. 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 $NH_2^{-}$
Show Answer
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]
