Question

Which of the following ions have highest magnetic moment?

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

Answer 1

The correct option is A

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

Explanation for the correct option

A: ${\mathrm{Mn}}^{2+}$

• The formula to calculate the magnetic moment is
• $\mathrm{\mu }=\sqrt{\mathrm{n}(\mathrm{n}+2)}$
• where $\mathrm{n}$ is the number of unpaired electrons.
• Atomic number of Manganese, $\mathrm{Mn}$= 25
• Number of electrons in Manganese ion, ${\mathrm{Mn}}^{2+}$= 23
• Electronic configuration of ${\mathrm{Mn}}^{2+}$is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^5$.
• Thus, ${\mathrm{Mn}}^{2+}$has 5 unpaired electrons and its magnetic moment is $\mathrm{\mu }=\sqrt{5(5+2)}=5.91$.

Explanation for the incorrect options:

B. ${\mathrm{Fe}}^{2+}$

1. Number of electrons in Iron ion, ${\mathrm{Fe}}^{2+}$= 24
2. Electronic configuration of ${\mathrm{Fe}}^{2+}$is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^6$.
3. Thus, ${\mathrm{Fe}}^{2+}$has 4 unpaired electrons and its magnetic moment is $\mathrm{\mu }=\sqrt{4(4+2)}=4.90$.

C. ${\mathrm{Ti}}^{2+}$

1. Number of electrons in Titanium ion, ${\mathrm{Ti}}^{2+}$= 20
2. Electronic configuration of ${\mathrm{Ti}}^{2+}$is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^2$.
3. Thus, ${\mathrm{Ti}}^{2+}$has 4 unpaired electrons and its magnetic moment is $\mathrm{\mu }=\sqrt{2(2+2)}=2.83$.

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

1. Number of electrons in Iron ion, ${\mathrm{Cr}}^{2+}$= 22
2. Electronic configuration of ${\mathrm{Cr}}^{2+}$is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^4$.
3. Thus, ${\mathrm{Cr}}^{2+}$has 4 unpaired electrons and its magnetic moment is $\mathrm{\mu }=\sqrt{4(4+2)}=4.90$.

Hence, the correct option is A).