SATHEE: Chemical Bonding and Molecular Structure

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, $N O_{2}^{-}, N O_{3}^{-}$ , $N H_{2}^{-}, N H_{4}^{+}, S C N^{-}$ ?

[2011]

(a) $N O_{2}^{-}$ and $N O_{3}^{-}$

(b) $N H_{4}^{+}$ and $N O_{3}^{-}$

(d) $N O_{2}^{-}$ and $N H_{2}^{-}$

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Solution:

$N O_{2}^{-}, H = \frac{1}{2} [5 + 0 + 1 - 0] = 3 = s p^{2}$

$N O_{3}^{-}, H = \frac{1}{2} [5 + 0 + 1 - 0] = 3 = s p^{2}$

$N H_{2}^{-}, H = \frac{1}{2} [5 + 2 + 1 + 0] = 4 = s p^{3}$

$N H_{4}^{+}, H = \frac{1}{2} [5 + 4 + 0 - 1] = 4 = s p^{3}$

$S C N^{-}, H = \frac{1}{2} [4 + 0 + 0 + 0] = 2 = s p$

$∴ N O_{2}^{-}$ and $N O_{3}^{-}$ 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]

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Chemical Bonding and Molecular Structure - Result Question 51

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