Search any question & find its solution
Question:
Answered & Verified by Expert
Arrange the following in the correct order of their bond orders.
Options:
Solution:
1625 Upvotes
Verified Answer
The correct answer is:
I $>$ III $>$ II $>$ IV
Bond order is the number of chemical bonds present between two atoms of a molecule.
I. $\mathrm{N}_2=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2=\pi 2 p_y^2, \sigma 2 p_z^2$
Bond order $=\frac{10-4}{2}=3$
II. $\mathrm{O}_2=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2$
$\approx 2 p_y^2, \pi^* 2 p_x^1 \approx \pi^* 2 p_y^1$
Bond order $=\frac{10-6}{2}=2$
III. $\mathrm{O}_2^{+}=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \pi 2 p_y^2$,$\pi^* 2 p_x^1$
Bond order $=\frac{10-5}{2}=2.5$
IV. $\mathrm{O}_2^{-}=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \pi 2 p_y^2$,$\pi^* 2 p_x^2 \approx \pi^* 2 p_y^1$
Bond order $=\frac{10-7}{2}=1.5$
Hence, bond order can be arranged as
I $>$ III $>$ II $>$ IV
I. $\mathrm{N}_2=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2=\pi 2 p_y^2, \sigma 2 p_z^2$
Bond order $=\frac{10-4}{2}=3$
II. $\mathrm{O}_2=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2$
$\approx 2 p_y^2, \pi^* 2 p_x^1 \approx \pi^* 2 p_y^1$
Bond order $=\frac{10-6}{2}=2$
III. $\mathrm{O}_2^{+}=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \pi 2 p_y^2$,$\pi^* 2 p_x^1$
Bond order $=\frac{10-5}{2}=2.5$
IV. $\mathrm{O}_2^{-}=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \pi 2 p_y^2$,$\pi^* 2 p_x^2 \approx \pi^* 2 p_y^1$
Bond order $=\frac{10-7}{2}=1.5$
Hence, bond order can be arranged as
I $>$ III $>$ II $>$ IV
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.