Question

Which of the following ion is diamagnetic?

• A. ${\mathrm{Zn}}^{+2}$
• B. ${\mathrm{Cr}}^{+3}$
• C. ${\mathrm{Fe}}^{+3}$
• D. ${\mathrm{Mn}}^{+2}$

Answer 1

The correct option is A

${\mathrm{Zn}}^{+2}$

Explanation for the correct option

A. ${\mathrm{Zn}}^{+2}$

• The diamagnetic ions one those that do not have unpaired electrons.
• Atomic number of zinc ($\mathrm{Zn}$) = 30
• Thus, the number of electrons in ${\mathrm{Zn}}^{+2}$$=30-2=28{\mathrm{e}}^{-}$
• Therefore, the electronic configuration of ${\mathrm{Zn}}^{+2}$is - $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{10}$
• As all the orbitals are completely filled in ${\mathrm{Zn}}^{+2}$, that means there is no unpaired electron present. Hence, ${\mathrm{Zn}}^{+2}$ is a diamagnetic ion.

Explanation for the incorrect options:

B. ${\mathrm{Cr}}^{+3}$

1. Atomic number of Chromium ($\mathrm{Cr}$) = 24
2. Thus, the number of electrons in ${\mathrm{Cr}}^{+3}$$=24-3=21{\mathrm{e}}^{-}$
3. Therefore, the electronic configuration of ${\mathrm{Cr}}^{+3}$is - $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^3$
4. As the d orbital in ${\mathrm{Cr}}^{+3}$ is only partially filled with 3 unpaired electrons, that means ${\mathrm{Cr}}^{+3}$ is a paramagnetic ion.

C. ${\mathrm{Fe}}^{+3}$

1. Atomic number of Iron ($\mathrm{Fe}$) = 26
2. Thus, the number of electrons in ${\mathrm{Fe}}^{+3}$$=26-3=23{\mathrm{e}}^{-}$
3. Therefore, the electronic configuration of ${\mathrm{Fe}}^{+3}$is - $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^5$
4. As the d orbital in ${\mathrm{Fe}}^{+3}$ is half with 5 unpaired electrons, that means ${\mathrm{Fe}}^{+3}$ is a paramagnetic ion.

D. ${\mathrm{Mn}}^{+2}$

1. Atomic number of Manganese ($\mathrm{Mn}$) = 25
2. Thus, the number of electrons in ${\mathrm{Mn}}^{+2}$$=25-2=23{\mathrm{e}}^{-}$
3. Therefore, the electronic configuration of ${\mathrm{Mn}}^{+2}$is - $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^5$
4. As the d orbital in ${\mathrm{Mn}}^{+2}$ is half with 5 unpaired electrons, that means ${\mathrm{Mn}}^{+2}$ is a paramagnetic ion.

Hence, the correct option is A, ${\mathrm{Zn}}^{+2}$.