Question

Which of the following is paramagnetic with bond order $0.5$?

• A. ${\mathrm{O}}_2$
• B. ${\mathrm{N}}_2$
• C. ${\mathrm{He}}_2$
• D. ${\mathrm{H}}_2^{+}$

Answer 1

The correct option is D

${\mathrm{H}}_2^{+}$

The explanation for the correct option

(D)${\mathbf{H}}_{\mathbf{2}}^{\mathbf{+}}$

• The molecule which is paramagnetic should have unpaired electrons in it.
• ${\mathrm{H}}_2^{+}$ has only one electron. So there will be one unpaired electron.
• Thus ${\mathrm{H}}_2^{+}$ is paramagnetic.
• Bond order can be determined using the formula,
• $Bondorder=\frac{Numberofbondingelectrons-Numberofantibondingelectrons}{2}$
• The molecular electronic configuration of ${\mathrm{H}}_2^{+}$ is $\mathrm{\sigma }1{\mathrm{s}}^1$
• Therefore the bond order will be,
• $\frac{1-0}{2}=0.5$
• Hence option D is correct.

The explanation for incorrect options

(A)${\mathrm{O}}_2$

• ${\mathrm{O}}_2$ has total $16$ electrons.
• The molecular electronic configuration is, $\mathrm{\sigma }1{\mathrm{s}}^2\mathrm{\sigma }*1{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{s}}^2\mathrm{\sigma }*2{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2(\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2)(\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{x}}^1=\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{y}}^1)$
• There are two unpaired electrons, thus${\mathrm{O}}_2$ is paramagnetic
• $$\begin{gathered} Bondorder=\frac{10-6}{2} \\ =2 \end{gathered}$$
• Thus option A is incorrect.

(B) ${\mathrm{N}}_2$

• ${\mathrm{N}}_2$ has total $14$ electrons.
• The molecular electronic configuration is $\mathrm{\sigma }1{\mathrm{s}}^2\mathrm{\sigma }*1{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{s}}^2\mathrm{\sigma }*2{\mathrm{s}}^2(\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2)\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2$
• There are no unpaired electrons present.
• Thus option B is incorrect.

(C)${\mathrm{He}}_2$

• ${\mathrm{He}}_2$ has four electrons.
• The molecular electronic configuration is $\mathrm{\sigma }1{\mathrm{s}}^2\mathrm{\sigma }*1{\mathrm{s}}^2$
• No unpaired electrons so not paramagnetic.
• Thus option C is incorrect.

Therefore the correct answer is option (D) ${\mathbf{H}}_{\mathbf{2}}^{\mathbf{+}}$.