Question

Which of the following does not have a magnetic moment of $1.73\mathrm{B}.\mathrm{M}$?

• A. ${\mathrm{O}}_2^{+}$
• B. ${\mathrm{O}}_2^{-}$
• C. $\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]$
• D. $\mathrm{CuI}$

Answer 1

The correct option is D

$\mathrm{CuI}$

The explanation of the correct option:

Option (D)

1. The magnetic moment is given by the formula $\sqrt{\mathrm{n}(\mathrm{n}+2)}$ where $\mathrm{n}$is the number of unpaired electrons. The magnetic moment is measured in Bohr-Magneton (B.M.).
2. In Copper iodide ($\mathrm{CuI}$), copper is in a +1 oxidation state and its electronic configuration is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^{10}4{\mathrm{s}}^0$. This implies that copper in its +1 oxidation state has no unpaired electrons i.e. $\mathrm{n}=0$. Thus, the magnetic moment comes out to be $\sqrt{0(0+2)}=0\mathrm{B}.\mathrm{M}.$

The explanation of incorrect options:
Option (A): The electronic configuration of ${\mathrm{O}}_2^{+}$is $KK{\left(\sigma 1s\right)}^2{\left({\sigma }^{*}1s\right)}^2{{\left(\sigma 2s\right)}^2{\left({\sigma }^{*}2s\right)}^2\left(\sigma 2p_z\right)}^2{\left(\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}\right)}^2={\left(\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}\right)}^2{\left({\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}\right)}^1={\left({\mathrm{\pi }}_{}^{*}2{\mathrm{p}}_{\mathrm{y}}\right)}^0$ which clearly shows an unpaired electron present in the ${\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{x}}}^{*}$ molecular orbital i.e. $n=1$. Thus, the magnetic moment is $\sqrt{1(1+2)}=\sqrt{2}=1.73\mathrm{B}.\mathrm{M}.$

Option (B): The electronic configuration of ${\mathrm{O}}_2^{-}$ is $KK{\left(\sigma 1s\right)}^2{\left({\sigma }^{*}1s\right)}^2{{\left(\sigma 2s\right)}^2{\left({\sigma }^{*}2s\right)}^2\left(\sigma 2p_z\right)}^2{\left(\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}\right)}^2={\left(\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}\right)}^2{\left({\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}\right)}^2={\left({\mathrm{\pi }}_{}^{*}2{\mathrm{p}}_{\mathrm{y}}\right)}^1$ which clearly shows an unpaired electron present in the ${\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}}^{*}$ molecular orbital i.e $n=1$. Thus, the magnetic moment is $\sqrt{1(1+2)}=\sqrt{2}=1.73B.M.$
Option (C): In $\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]$, copper is in a +2 oxidation state in which its electronic configuration would be $\left[\mathrm{Ar}\right]3{\mathrm{d}}^{9}4{\mathrm{s}}^0$. Thus, there would be 1 unpaired electron in the 3d orbital i.e $n=1$. Thus, the magnetic moment is $\sqrt{1(1+2)}=\sqrt{2}=1.73\mathrm{B}.\mathrm{M}.$

Therefore, the correct answer is option (D).