Chemical Bonding and Molecular Structure - Result Question 104
####106. Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them:
[2008, $2012 M$ ]
(a) $NO<O_2^{-}<C_2^{2-}<He_2^{+}$
(b) $O_2^{-}<NO<C_2^{2-}<He_2^{+}$
(c) $C_2^{2-}<He_2^{+}<O_2^{-}<NO$
(d) $He_2^{+}<O_2^{-}<NO<C_2^{2-}$
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Solution:
- (d) Calculating the bond order of various species.
$O_2^{-}: KK \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_z^{2}$
$\pi 2 p_x^{2} \pi 2 p_y^{2} \pi^{} 2 p_x^{2} \pi^{} 2 p_y^{1}$
Number of electrons in bonding B.O. $=\frac{\text{ Number of electrons in non-bonding }}{2}$ $=\frac{8-5}{2}$ or 1.5
NO: $KK \sigma 2 s^{2} \sigma^{} 2 s^{2} \pi 2 p_x^{2} \approx \pi 2 p_y^{2} \sigma 2 p_z^{2} \pi^{} 2 p_x^{1}$
B.O. $=\frac{N_b-N_a}{2}=\frac{8-2}{2}$ or 2.5
$C_2^{2-}: k k \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_x^{2} \pi 2 p_y^{2} \sigma 2 p_z^{2}$
B.O. $=\frac{N_b-N_a}{2}=\frac{8-2}{2}$ or 3
$He_2^{+}=\sigma 1 s^{2} \sigma^{*} 1 s^{1}$
B. O. $=\frac{N_b-N_a}{2}=\frac{2-1}{2}$ or 0.5
From these values we find the correct order of increasing bond order is
$He_2^{2+}<O_2^{-}<NO<C_2^{2-}$
