The spin-only magnetic moment ($\mu_s$) is calculated using the formula:

$ \mu_s = \sqrt{n(n+2)} \, \text{Bohr magneton (BM)} $

where '$n$' is the number of unpaired electrons.

Determining Unpaired Electrons

First, find the number of unpaired electrons for each ion:

• E. $Sc^{3+}$: Configuration is $[Ar] 3d^0$. Number of unpaired electrons, $n=0$.
• B. $Ti^{3+}$: Configuration is $[Ar] 3d^1$. Number of unpaired electrons, $n=1$.
• C. $Ti^{2+}$: Configuration is $[Ar] 3d^2$. Number of unpaired electrons, $n=2$.
• A. $Co^{2+}$: Configuration is $[Ar] 3d^7$. Number of unpaired electrons, $n=3$.
• D. $Fe^{2+}$: Configuration is $[Ar] 3d^6$. Number of unpaired electrons, $n=4$.

Calculating Magnetic Moments

Now, calculate the magnetic moment for each ion:

• E. $Sc^{3+}$: $n=0 \implies \mu_s = \sqrt{0(0+2)} = 0 \, \text{BM}$
• B. $Ti^{3+}$: $n=1 \implies \mu_s = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \, \text{BM}$
• C. $Ti^{2+}$: $n=2 \implies \mu_s = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \, \text{BM}$
• A. $Co^{2+}$: $n=3 \implies \mu_s = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \text{BM}$
• D. $Fe^{2+}$: $n=4 \implies \mu_s = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \text{BM}$

Ordering Ions by Magnetic Moment

Arrange the ions in increasing order of their spin-only magnetic moments:

$ 0 \, \text{BM} < 1.73 \, \text{BM} < 2.83 \, \text{BM} < 3.87 \, \text{BM} < 4.90 \, \text{BM} $

This corresponds to the order:

E < B < C < A < D