Question

Identify the complex ion with spin only magnetic moment of 4.90 BM.

• A. [Co(NH\(_3\))\(_6\)]\(^{3+}\)
• B. [Cr(NH\(_3\))\(_6\)]\(^{3+}\)
• C. [Mn(CN)\(_6\)]\(^{3-}\)
• D. [MnCl\(_6\)]\(^{3-}\)

Hint:

To solve problems involving spin-only magnetic moment: 1. Calculate the oxidation state} of the central metal ion. 2. Determine the electronic configuration} of the metal ion. 3. Identify the nature of the ligand} (strong field or weak field). Strong field ligands cause electron pairing (low spin complexes for d\(^4\), d\(^5\), d\(^6\), d\(^7\)), while weak field ligands do not (high spin complexes). For d\(^1\), d\(^2\), d\(^3\), d\(^8\), d\(^9\), d\(^{10}\) configurations, the number of unpaired electrons is generally independent of ligand field strength. 4. Fill the d-orbitals} according to the ligand field strength and Hund's rule to find the number of unpaired electrons (n). 5. Calculate the magnetic moment} using the formula $\mu_s = \sqrt{n(n+2)}\operatorname{BM}$. Common strong field ligands include CN\(^-\), CO, en, NH\(_3\). Common weak field ligands include F\(^-\), Cl\(^-\), Br\(^-\), I\(^-\), H\(_2\)O (though H\(_2\)O can be intermediate).

Answer 1

The Correct Option is D

Step 1: Understand the formula for spin-only magnetic moment.
The spin-only magnetic moment ($\mu_s$) is given by the formula:
$\mu_s = \sqrt{n(n+2)}\operatorname{BM}$ where 'n' is the number of unpaired electrons and BM stands for Bohr Magneton. We are given $\mu_s = 4.90\operatorname{BM}$.
Let's find the number of unpaired electrons (n) corresponding to this magnetic moment.
$4.90 = \sqrt{n(n+2)}$
Squaring both sides:
$(4.90)^2 = n(n+2)$
$24.01 = n^2 + 2n$
$n^2 + 2n - 24.01 = 0$
We can approximate $n$ to be 4 since $\sqrt{4(4+2)} = \sqrt{4 \times 6} = \sqrt{24} \approx 4.898$.
So, we are looking for a complex ion with 4 unpaired electrons. 
Step 2: Determine the number of unpaired electrons for each complex ion option.
1. [Co(NH\(_3\))\(_6\)]\(^{3+}\)
Oxidation state of Co: Let it be $x$. $x + 6(0) = +3 \implies x = +3$.
Cobalt (Co) has atomic number 27. Electronic configuration: [Ar] 3d\(^7\) 4s\(^2\).
Co\(^{3+}\) electronic configuration: [Ar] 3d\(^6\).
NH\(_3\) (ammonia) is a strong field ligand.
In the presence of a strong field ligand, the 3d\(^6\) electrons pair up.
$t_{2g}^6 e_g^0$.
Number of unpaired electrons (n) = 0.
$\mu_s = \sqrt{0(0+2)} = 0\operatorname{BM}$. This is not 4.90 BM. 
2. [Cr(NH\(_3\))\(_6\)]\(^{3+}\)
Oxidation state of Cr: Let it be $x$. $x + 6(0) = +3 \implies x = +3$.
Chromium (Cr) has atomic number 24. Electronic configuration: [Ar] 3d\(^5\) 4s\(^1\).
Cr\(^{3+}\) electronic configuration: [Ar] 3d\(^3\).
NH\(_3\) is a strong field ligand. However, for d\(^3\) configuration, irrespective of strong or weak field ligand, the electrons occupy $t_{2g}$ orbitals singly before pairing.
$t_{2g}^3 e_g^0$.
Number of unpaired electrons (n) = 3.
$\mu_s = \sqrt{3(3+2)} = \sqrt{3 \times 5} = \sqrt{15} \approx 3.87\operatorname{BM}$. This is not 4.90 BM. 
3. [Mn(CN)\(_6\)]\(^{3-}\)
Oxidation state of Mn: Let it be $x$. $x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3$.
Manganese (Mn) has atomic number 25. Electronic configuration: [Ar] 3d\(^5\) 4s\(^2\).
Mn\(^{3+}\) electronic configuration: [Ar] 3d\(^4\).
CN\(^-\) (cyanide) is a very strong field ligand.
In the presence of a strong field ligand, the 3d\(^4\) electrons pair up.
$t_{2g}^4 e_g^0$. (This configuration is for low spin complex, one electron pairs up in $t_{2g}$).
Number of unpaired electrons (n) = 2.
$\mu_s = \sqrt{2(2+2)} = \sqrt{2 \times 4} = \sqrt{8} \approx 2.83\operatorname{BM}$. This is not 4.90 BM.
4. [MnCl\(_6\)]\(^{3-}\)
Oxidation state of Mn: Let it be $x$. $x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3$.
Manganese (Mn) has atomic number 25. Electronic configuration: [Ar] 3d\(^5\) 4s\(^2\).
Mn\(^{3+}\) electronic configuration: [Ar] 3d\(^4\). Cl\(^-\) (chloride) is a weak field ligand.
In the presence of a weak field ligand, the 3d\(^4\) electrons will follow Hund's rule and occupy orbitals singly as much as possible before pairing. This forms a high spin complex. $t_{2g}^3 e_g^1$.
Number of unpaired electrons (n) = 4.
$\mu_s = \sqrt{4(4+2)} = \sqrt{4 \times 6} = \sqrt{24} \approx 4.898\operatorname{BM}$. This matches 4.90 BM. 
Step 3: Conclude the complex ion.
The complex ion [MnCl\(_6\)]\(^{3-}\) has 4 unpaired electrons and thus a spin-only magnetic moment of approximately 4.90 BM. 
The final answer is $\boxed{[MnCl}_6]^{3-}}$.