Question

For which of the following ions will the magnetic moment ($\mu$) be maximum?

• A. $Zn^{2+}$
• B. $Fe^{2+}$
• C. $Co^{2+}$
• D. $Ni^{2+}$

Hint:

Greater the number of unpaired electrons, greater will be the magnetic moment.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Magnetic moment ($\mu$) depends on the number of unpaired electrons present in the ion.
It is calculated using the spin-only formula:
\[ \mu = \sqrt{n(n+2)} \text{ B.M.} \] where $n$ = number of unpaired electrons.
Step 2: Electronic Configuration and Unpaired Electrons:
$Zn^{2+$:}
Electronic configuration of Zn = $[Ar] 3d^{10}4s^2$
For $Zn^{2+}$: $[Ar] 3d^{10}$
Unpaired electrons = 0
$Fe^{2+$:}
Electronic configuration of Fe = $[Ar] 3d^6 4s^2$
For $Fe^{2+}$: $[Ar] 3d^6$
Unpaired electrons = 4
$Co^{2+$:}
Electronic configuration of Co = $[Ar] 3d^7 4s^2$
For $Co^{2+}$: $[Ar] 3d^7$
Unpaired electrons = 3
$Ni^{2+$:}
Electronic configuration of Ni = $[Ar] 3d^8 4s^2$
For $Ni^{2+}$: $[Ar] 3d^8$
Unpaired electrons = 2
Since $Fe^{2+}$ has the maximum number of unpaired electrons (4), it will have the maximum magnetic moment.
Step 3: Final Answer:
The magnetic moment will be maximum for $Fe^{2+$}.