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JIPMER 2010
Exam
JIPMER
Year
2010
Subject
Chemistry
Chapter
Chemical Bonding and Molecular Structure

JIPMER 2010 · Chemistry · Chemical Bonding and Molecular Structure

Which of the following has the highest bond order ?

Which of the following has the highest bond order ?

Options

  1. AN2\mathrm{N}_2
  2. BO2\mathrm{O}_2
  3. CHe2\mathrm{He}_2
  4. DH2\mathrm{H}_2

Correct answer

A. N2\mathrm{N}_2

Step-by-step solution

N2(7+7=14e−)=KK∗,σ2s2,σ⋆2s2,π2px2≈π2py2,σ2pz2BO=12(8−2)=3O2(8+8=16e−)=KK∗,σ2s2,σσ∗2s2,σ2pz2,π2px2≈π2py2,π∗2px1≈π2py1BO=12(8−4)=2He2(2+2=4e−)=σ1s2,σ∗1s2BO=12(2−2)=0\begin{aligned} \mathrm{N}_2\left(7+7=14 e^{-}\right)= & K K^*, \sigma 2 s^2, \stackrel{\star}{\sigma} 2 s^2, \pi 2 p_x^2 \\ \approx & \pi 2 p_y^2, \sigma 2 p_z^2 \\ \mathrm{BO}= & \frac{1}{2}(8-2)=3 \\ \mathrm{O}_2\left(8+8=16 e^{-}\right)= & K K^*, \sigma 2 s^2, \sigma_\sigma^* 2 s^2, \sigma 2 p_z^2, \\ & \pi 2 p_x^2 \approx \pi 2 p_y^2, \stackrel{*}{\pi} 2 p_x^1 \approx \pi 2 p_y^1 \\ \mathrm{BO}= & \frac{1}{2}(8-4)=2 \\ \mathrm{He}_2\left(2+2=4 e^{-}\right)= & \sigma 1 s^2, \sigma^* 1 s^2 \\ \mathrm{BO}= & \frac{1}{2}(2-2)=0\end{aligned}H2(1+1=2e−)=σ1s2 BO =12(2−0)=1\begin{aligned} \mathrm{H}_2\left(1+1=2 e^{-}\right) & =\sigma 1 s^2 \\ \text{ BO } & =\frac{1}{2}(2-0)=1 \end{aligned}Hence,Hence,\mathrm{N}_2$ has the highest bond order.
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