Chemical Bonding and Molecular Structure - Result Question 117
####120. The relationship between the dissociation energy of $N_2$ and $N_2{ }^{+}$is :
(a) Dissociation energy of $N_2^{+}>$dissociation energy of $N_2$
(b) Dissociation energy of $N_2=$ dissociation energy of $N_2{ }^{+}$
(c) Dissociation energy of $N_2>$ dissociation energy of $N_2{ }^{+}$
(d) Dissociation energy of $N_2$ can either be lower or higher than the dissociation energy of $N_2{ }^{+}$
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Solution:
- (c) Dissociation energy of any molecules depends upon bond order. Bond order in $N_2$ molecule is 3 while bond order in $N_2^{+}$is 2.5 .
Further we know that more the Bond order, more is the stability and more is the BDE.
121
$ \begin{aligned} & \text{ (b) } NO^{+}=\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma * 2 s^{2} \sigma 2 p_x{ }^{2} \\ & \pi 2 p_y{ }^{2}=\pi 2 p_z{ }^{2} \\ & \text{ Bond order of } NO^{+}=\frac{1}{2}(N_b-N_a) \\ & =\frac{1}{2}(10-4)=\frac{1}{2} \times 6=3 \end{aligned} $
Similarly, Bond order of $NO=\frac{1}{2}(10-5)$
$ =\frac{1}{2}(5)=2.5 $
Bond order of $NO^{-}=\frac{1}{2}(10-6)=\frac{1}{2}(4)=2$
Bond order of $O_2^{-}=\frac{1}{2}(10-7)=\frac{1}{2}(3)=1.5$
By above calculation, we get
Decreasing bond order
$ NO^{+}>NO>NO^{-}>O_2^{-} $
