Question

If the magnetic moment of a dioxygen species is 1.73 B.M, it may be:

• A. ${\mathrm{O}}_2,{{\mathrm{O}}_2}^{–},\mathrm{or}{{\mathrm{O}}_2}^{+}$
• B. ${{\mathrm{O}}_2}^{–},\mathrm{or}{{\mathrm{O}}_2}^{+}$
• C. ${\mathrm{O}}_2\mathrm{or}{{\mathrm{O}}_2}^{–}$
• D. ${\mathrm{O}}_2,{{\mathrm{O}}_2}^{+}$

Answer 1

The correct option is B

${{\mathrm{O}}_2}^{–},\mathrm{or}{{\mathrm{O}}_2}^{+}$

Explanation for the correct option:-

Option (B) ${{\mathrm{O}}_2}^{–},\mathrm{or}{{\mathrm{O}}_2}^{+}$

Step 1: Calculating the number of unpaired electrons:

We know,

$$\begin{gathered} \mathrm{\mu }=\sqrt{\mathrm{n}(\mathrm{n}+2)}=1.73\mathrm{B}.\mathrm{M} \\ \therefore \sqrt{\mathrm{n}(\mathrm{n}+2)}=1.73 \\ \Rightarrow {\left(\sqrt{\mathrm{n}(\mathrm{n}+2)}\right)}^2={\left(1.73\right)}^2 \\ \Rightarrow \mathrm{n}(\mathrm{n}+2)=3 \\ \Rightarrow {\mathrm{n}}^2+2\mathrm{n}=3 \\ \mathrm{n}=1 \end{gathered}$$

Thus there is 1 unpaired electron

Step 2: Writing the electronic configuration

${{\mathrm{O}}_2}^{-}$has a total of 17 electrons

${{\mathrm{O}}_2}^{-}=\mathrm{\sigma }1{\mathrm{s}}^2{\mathrm{\sigma }}^{\ast }1{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{s}}^2{\mathrm{\sigma }}^{\ast }2{\mathrm{s}}^2\mathrm{\sigma }2{{\mathrm{p}}_{\mathrm{z}}}^2\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{x}}}^2=\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^2{\mathrm{\pi }}^{\ast }2{{\mathrm{p}}_{\mathrm{x}}}^2={\mathrm{\pi }}^{\ast }2{{\mathrm{p}}_{\mathrm{y}}}^1$

we can see that it has only 1 unpaired electron.

${{\mathrm{O}}_2}^{+}$has a total of 15 electrons

${{\mathrm{O}}_2}^{+}=\mathrm{\sigma }1{\mathrm{s}}^2{\mathrm{\sigma }}^{\ast }1{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{s}}^2{\mathrm{\sigma }}^{\ast }2{\mathrm{s}}^2\mathrm{\sigma }2{{\mathrm{p}}_{\mathrm{z}}}^2\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{x}}}^2=\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^2{\mathrm{\pi }}^{\ast }2{{\mathrm{p}}_{\mathrm{x}}}^1$

we can see that it has only 1 unpaired electron.

As we already calculated for the magnetic moment to be $1.73,\mathrm{n}=1$

Thus ${{\mathrm{O}}_2}^{–},\mathrm{or}{{\mathrm{O}}_2}^{+}$both have a magnetic moment of 1.73 B.M

Hence it is the correct option

Explanation for the incorrect options:-

Option (A) ${\mathrm{O}}_2,{{\mathrm{O}}_2}^{–},\mathrm{or}{{\mathrm{O}}_2}^{+}$

• ${\mathrm{O}}_2$ has 16 electrons, and its electronic configuration is $\mathrm{\sigma }1{\mathrm{s}}^2{\mathrm{\sigma }}^{\ast }1{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{s}}^2{\mathrm{\sigma }}^{\ast }2{\mathrm{s}}^2\mathrm{\sigma }2{{\mathrm{p}}_{\mathrm{z}}}^2\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{x}}}^2=\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^2{\mathrm{\pi }}^{\ast }2{{\mathrm{p}}_{\mathrm{x}}}^1={\mathrm{\pi }}^{\ast }2{{\mathrm{p}}_{\mathrm{y}}}^1$
• Thus we can see that it has 2 unpaired electrons present, that is $\mathrm{n}=2$

$$\begin{gathered} \therefore \mathrm{\mu }=\sqrt{\mathrm{n}(\mathrm{n}+2)} \\ \Rightarrow \mathrm{\mu }=\sqrt{2(2+2)} \\ \Rightarrow \mathrm{\mu }=\sqrt{8} \\ \Rightarrow \mathrm{\mu }=2.83\mathrm{B}.\mathrm{M} \end{gathered}$$

• Hence it is the incorrect option.

Option (C) ${\mathrm{O}}_2\mathrm{or}{{\mathrm{O}}_2}^{–}$

• ${\mathrm{O}}_2$ has a magnetic moment of 2.83 B.M
• Hence it is the incorrect option

Option (D) ${\mathrm{O}}_2,{{\mathrm{O}}_2}^{+}$

• ${\mathrm{O}}_2$ has a magnetic moment of 2.83 B.M
• Hence it is the incorrect option

Therefore, option (B) is the correct answer.