JEE Chemistry 2026 Question with Solution

Given: We must arrange the spin-only magnetic moment values of $[\ce{MnBr4}]^{2-}$ (A), $[\ce{Cu(H2O)6}]^{2+}$ (B), $[\ce{Ni(CN)4}]^{2-}$ (C) and $[\ce{Ni(H2O)6}]^{2+}$ (D).

Find: The correct increasing order of magnetic moments.

Concept: Spin-only magnetic moment is given by

$$\mu = \sqrt{n(n+2)} \; \text{BM}$$

where $n$ is the number of unpaired electrons.

Step 1: Analyze each complex.

For $[\ce{MnBr4}]^{2-}$ (A): Mn is in the $+2$ oxidation state with $d^5$ configuration. $\ce{Br^-}$ is a weak field ligand, so the complex is high spin. Number of unpaired electrons $= 5$.

$$\mu = \sqrt{35} \approx 5.92 \; \text{BM}$$

For $[\ce{Cu(H2O)6}]^{2+}$ (B): Cu is in the $+2$ oxidation state with $d^9$ configuration. Number of unpaired electrons $= 1$.

$$\mu = \sqrt{3} \approx 1.73 \; \text{BM}$$

For $[\ce{Ni(CN)4}]^{2-}$ (C): Ni is in the $+2$ oxidation state with $d^8$ configuration. $\ce{CN^-}$ is a strong field ligand, so the complex is low spin. Number of unpaired electrons $= 0$.

$$\mu = 0 \; \text{BM}$$

For $[\ce{Ni(H2O)6}]^{2+}$ (D): Ni is in the $+2$ oxidation state with $d^8$ configuration. $\ce{H2O}$ is a weak field ligand, so the complex is high spin. Number of unpaired electrons $= 2$.

$$\mu = \sqrt{8} \approx 2.83 \; \text{BM}$$

Step 2: Arrange in increasing order.

$$[\ce{Ni(CN)4}]^{2-} < [\ce{Cu(H2O)6}]^{2+} < [\ce{Ni(H2O)6}]^{2+} < [\ce{MnBr4}]^{2-}$$

So, the final order is C < B < D < A.

Therefore, the correct option is D.

Given: The complexes and their labels are A, B, C, and D.

Find: Their increasing order of spin-only magnetic moments.

The magnetic moment depends only on the number of unpaired electrons through

$$\mu = \sqrt{n(n+2)}$$

Hence we only need the order of $n$.

• A: $[\ce{MnBr4}]^{2-}$ has $\ce{Mn^{2+}}$, so it is $d^5$. With weak field $\ce{Br^-}$, it remains high spin, giving $5$ unpaired electrons.
• B: $[\ce{Cu(H2O)6}]^{2+}$ has $\ce{Cu^{2+}}$, so it is $d^9$, giving $1$ unpaired electron.
• C: $[\ce{Ni(CN)4}]^{2-}$ has $\ce{Ni^{2+}}$, so it is $d^8$. Strong field $\ce{CN^-}$ causes pairing, giving $0$ unpaired electrons.
• D: $[\ce{Ni(H2O)6}]^{2+}$ has $\ce{Ni^{2+}}$, so it is $d^8$. With weak field $\ce{H2O}$, it has $2$ unpaired electrons.

Thus the increasing order of unpaired electrons is

$$0 < 1 < 2 < 5$$

which gives

$$\text{C} < \text{B} < \text{D} < \text{A}$$

Therefore, the correct option is D.