SATHEE: Chemical Bonding and Molecular Structure

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 $N_{2}^{+}$

(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

$\begin{aligned} (b) N O^{+} = σ 1 s^{2} σ^{*} 1 s^{2} σ 2 s^{2} σ * 2 s^{2} σ 2 p_{x}^{2} \\ π 2 p_{y}^{2} = π 2 p_{z}^{2} \\ Bond order of N O^{+} = \frac{1}{2} (N_{b} - N_{a}) \\ = \frac{1}{2} (10 - 4) = \frac{1}{2} \times 6 = 3 \end{aligned}$

Similarly, Bond order of $N O = \frac{1}{2} (10 - 5)$

$= \frac{1}{2} (5) = 2.5$

Bond order of $N O^{-} = \frac{1}{2} (10 - 6) = \frac{1}{2} (4) = 2$

Bond order of $O_{2}^{-} = \frac{1}{2} (10 - 7) = \frac{1}{2} (3) = 1.5$

By above calculation, we get

Decreasing bond order

$N O^{+} > N O > N O^{-} > O_{2}^{-}$

Chemical Bonding and Molecular Structure - Result Question 117

Solution:

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Chemical Bonding and Molecular Structure - Result Question 117

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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