Q5 / 155 · WBJEE 2026 · Chemistry
The calculated magnetic moment for low spin $[Ru(EDTA)]^-$ is
- A$2.73$ BM
- Correct answerB$1.73$ BM
- C$3.23$ BM
- D$0.00$ BM
Solution
Let the oxidation state of Ru in the complex $[\text{Ru}(\text{EDTA})]^-$ be $x$.
The ligand $\text{EDTA}^{4-}$ has a charge of $-4$.
Therefore, $x + (-4) = -1 \Rightarrow x = +3$.
Ruthenium (Ru) is a $4d$-series element in Group 8. The electronic configuration of neutral Ru is $[\text{Kr}] 4d^7 5s^1$.
For the $\text{Ru}^{3+}$ ion, three electrons are removed, giving the configuration $[\text{Kr}] 4d^5$.
The complex is given as low spin. In an octahedral crystal field, the five $d$-electrons will pair up in the lower energy $t_{2g}$ orbitals.
The electronic arrangement will be $t_{2g}^5 e_g^0$.
This configuration has $1$ unpaired electron ($n = 1$).
The spin-only magnetic moment $\mu$ is calculated using the formula:
$\mu = \sqrt{n(n+2)} \text{ BM}$
$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM}$
Answer: $1.73$ BM
The ligand $\text{EDTA}^{4-}$ has a charge of $-4$.
Therefore, $x + (-4) = -1 \Rightarrow x = +3$.
Ruthenium (Ru) is a $4d$-series element in Group 8. The electronic configuration of neutral Ru is $[\text{Kr}] 4d^7 5s^1$.
For the $\text{Ru}^{3+}$ ion, three electrons are removed, giving the configuration $[\text{Kr}] 4d^5$.
The complex is given as low spin. In an octahedral crystal field, the five $d$-electrons will pair up in the lower energy $t_{2g}$ orbitals.
The electronic arrangement will be $t_{2g}^5 e_g^0$.
This configuration has $1$ unpaired electron ($n = 1$).
The spin-only magnetic moment $\mu$ is calculated using the formula:
$\mu = \sqrt{n(n+2)} \text{ BM}$
$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM}$
Answer: $1.73$ BM
Hence, the correct answer is option: 'B'.