JEE MAIN - Chemistry (2023 - 8th April Evening Shift - No. 15)

The observed magnetic moment of the complex $$\left.\left[\operatorname{Mn}(\underline{N} C S)_{6}\right)\right]^{x^{-}}$$ is $$6.06 ~\mathrm{BM}$$. The numerical value of $$x$$ is __________.
Answer
4

Explanation

The complex is given as $[\mathrm{Mn}(\mathrm{NCS})_{6}]^{x-}$. Here, $\mathrm{NCS}^{-}$ acts as a ligand. Each $\mathrm{NCS}^{-}$ has a charge of -1 and since there are six of them, they contribute a total charge of -6 to the complex.

Manganese (Mn) in this complex is in the +2 oxidation state. We know this because in its ground state, Mn has 5 electrons in its 3d orbitals and 2 electrons in its 4s orbital. But in the Mn²⁺ cation, the 2 electrons from the 4s orbital have been removed, leaving 5 unpaired electrons in the 3d orbitals. This matches the given magnetic moment of 6.06 BM, which corresponds to 5 unpaired electrons.

So, the overall charge of the complex is -4: the Mn²⁺ ion contributes a charge of +2 and the six $\mathrm{NCS}^{-}$ ligands contribute a total charge of -6. When you add these together, you get -4. Hence, $x = -4$.

In summary, the $[\mathrm{Mn}(\mathrm{NCS})_{6}]^{x-}$ complex has an overall charge of -4, meaning x = -4.

So the numerical value of x will be 4.

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