Question

The calculated magnetic moments (spin only value) for species $[FeCl_4]^{2-}$, $[Co(C_2O_4)_3]^{3-}$ and $MnO_4^{2-}$ respectively are :

• A. 4.90, 0 and 1.73 BM
• B. 5.92, 4.90 and 0 BM
• C. 5.82, 0 and 0 BM
• D. 4.90, 0 and 2.83 BM

Hint:

Spin-only magnetic moment $\mu = \sqrt{n(n+2)} \text{ BM}$. A quick way to estimate: if $n=1$, $\mu \approx 1.7$; if $n=4$, $\mu \approx 4.9$.

Answer 1

The Correct Option is A

Step 1: $[FeCl_4]^{2-}$: $Fe^{2+}$ is $d^6$. $Cl^-$ is a weak field ligand. Number of unpaired electrons $(n) = 4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM}$.
Step 2: $[Co(C_2O_4)_3]^{3-}$: $Co^{3+}$ is $d^6$. Oxalate $(ox)$ is a strong field ligand for $Co^{3+}$ (causing pairing). $n = 0$. $\mu = 0 \text{ BM}$.
Step 3: $MnO_4^{2-}$: $Mn$ is in +6 state, so it is $d^1$. $n = 1$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM}$.