Question

Which one of the following octahedral complexes has the highest spin-only magnetic moment?

• A. [Cr(H$_2$O)$_4$(OH)$_2$]
• B. [V(H$_2$O)$_4$I$_2$]$^+$
• C. [Fe(NH$_3$)$_4$(CN)$_2$]$^+$
• D. [Co(NH$_3$)$_4$Cl$_2$]

Hint:

A quick way to estimate the spin-only magnetic moment:
If the number of unpaired electrons is $n$, the magnetic moment value is always approximately $n.\text{something}$ B.M. (e.g., if $n=4$, $\mu \approx 4.9$ B.M.).
Simply find the complex with the highest number of unpaired electrons!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The spin-only magnetic moment ($\mu_{s}$) depends on the number of unpaired electrons ($n$). Higher $n$ means higher $\mu_{s}$.
Key Formula or Approach:
\[ \mu_{s} = \sqrt{n(n + 2)} \text{ B.M.} \]
We must find the oxidation state and configuration for each metal ion.

Step 2: Detailed Explanation:
$\bullet$ (A) $[Cr(H_{2}O)_{4}(OH)_{2}]$: Cr is $+2$ ($d^{4}$). $H_{2}O/OH^{-}$ are weak ligands.
Configuration: $t_{2g}^{3} e_{g}^{1} \implies n = 4$ unpaired.
$\bullet$ (B) $[V(H_{2}O)_{4}I_{2}]^{+$:} V is $+3$ ($d^{2}$).
Configuration: $t_{2g}^{2} e_{g}^{0} \implies n = 2$ unpaired.
$\bullet$ (C) $[Fe(NH_{3})_{4}(CN)_{2}]^{+$:} Fe is $+3$ ($d^{5}$). $CN^{-}$ is very strong $\rightarrow$ low spin.
Configuration: $t_{2g}^{5} e_{g}^{0} \implies n = 1$ unpaired.
$\bullet$ (D) $[Co(NH_{3})_{4}Cl_{2}]$: Co is $+2$ ($d^{7}$). $Cl^{-}$ is weak $\rightarrow$ high spin.
Configuration: $t_{2g}^{5} e_{g}^{2} \implies n = 3$ unpaired.

Step 3: Final Answer:
Complex (A) has the highest $n$ value (4).
This matches option (A).