Which one of the following octahedral complexes has the highest spin-only magnetic moment?
Show 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!
Step 1: Understanding the Question:
This question asks us to identify the octahedral coordination complex that possesses the highest spin-only magnetic moment ($\mu_s$). Step 2: Key Formula or Approach:
The spin-only magnetic moment ($\mu_s$) is directly related to the number of unpaired electrons ($n$) in the central metal ion:
\[ \mu_s = \sqrt{n(n + 2)} \text{ B.M.} \]
Therefore, the complex with the highest number of unpaired d-electrons ($n$) will have the highest spin-only magnetic moment.
To find $n$, we must determine the oxidation state, the d-electron configuration of the metal ion, and whether the ligands are weak-field (high-spin) or strong-field (low-spin). Step 3: Detailed Explanation:
Let's analyze each complex individually:
• (A) [Cr(H$_2$O)$_4$(OH)$_2$]:
- Water ($\text{H}_2\text{O}$) is a neutral ligand (charge = 0), and hydroxide ($\text{OH}^-$) has a charge of $-1$.
- Let the oxidation state of Chromium be $x$. Thus, $x + 4(0) + 2(-1) = 0 \implies x = +2$.
- $\text{Cr}^{2+}$ has a $d^4$ valence configuration.
- Since $\text{H}_2\text{O}$ and $\text{OH}^-$ are relatively weak-field ligands, they do not cause pairing. This results in a high-spin octahedral state: $t_{2g}^3 e_g^1$.
- This configuration contains 4 unpaired electrons ($n = 4$).
• (B) [V(H$_2$O)$_4$I$_2$]$^+$:
- Let the oxidation state of Vanadium be $y$. Thus, $y + 4(0) + 2(-1) = +1 \implies y = +3$.
- $\text{V}^{3+}$ has a $d^2$ configuration.
- For a $d^2$ system, the electrons occupy the $t_{2g}$ orbitals: $t_{2g}^2 e_g^0$.
- This configuration contains 2 unpaired electrons ($n = 2$).
• (C) [Fe(NH$_3$)$_4$(CN)$_2$]$^+$:
- Let the oxidation state of Iron be $z$. Thus, $z + 4(0) + 2(-1) = +1 \implies z = +3$.
- $\text{Fe}^{3+}$ has a $d^5$ configuration.
- Here, cyanide ($\text{CN}^-$) is a very strong-field ligand, and ammonia ($\text{NH}_3$) is a moderate-to-strong-field ligand, prompting a low-spin configuration: $t_{2g}^5 e_g^0$.
- This low-spin configuration contains only 1 unpaired electron ($n = 1$).
• (D) [Co(NH$_3$)$_4$Cl$_2$]:
- Let the oxidation state of Cobalt be $w$. Thus, $w + 4(0) + 2(-1) = 0 \implies w = +2$.
- $\text{Co}^{2+}$ has a $d^7$ configuration.
- Because chloride ($\text{Cl}^-$) is a weak-field ligand, this complex is typically high-spin: $t_{2g}^5 e_g^2$.
- This high-spin configuration contains 3 unpaired electrons ($n = 3$). Step 4: Final Answer:
Comparing the number of unpaired electrons:
- [Cr(H$_2$O)$_4$(OH)$_2$] has $n = 4$
- [V(H$_2$O)$_4$I$_2$]$^+$ has $n = 2$
- [Fe(NH$_3$)$_4$(CN)$_2$]$^+$ has $n = 1$
- [Co(NH$_3$)$_4$Cl$_2$] has $n = 3$
The complex [Cr(H$_2$O)$_4$(OH)$_2$] has the highest number of unpaired electrons ($n = 4$), resulting in the highest spin-only magnetic moment. This corresponds to option (A).
