JEE Chemistry 2025 Question with Solution

Given: Match the complexes in LIST-I with their shape and magnetic moment in LIST-II.

Find: The correct correspondence and hence the correct option.

Strong field ligands such as $CO$ and $CN^-$ cause pairing of electrons, while weak field ligands such as $Cl^-$ and $Br^-$ do not cause pairing. Also, magnetic moment is related to the number of unpaired electrons.

1. For $[Ni(CO)_4]$: $CO$ is a strong field ligand. This complex is tetrahedral and diamagnetic, so its magnetic moment is $0 \; \text{BM}$. Hence,

$$[Ni(CO)_4] \rightarrow \text{III}$$

2. For $[Ni(CN)_4]^{2-}$: $CN^-$ is a strong field ligand and causes pairing. The complex is square planar and diamagnetic, so

$$[Ni(CN)_4]^{2-} \rightarrow \text{II}$$

3. For $[NiCl_4]^{2-}$: $Cl^-$ is a weak field ligand, so electrons remain unpaired. The complex is tetrahedral with magnetic moment $2.8 \; \text{BM}$. Hence,

$$[NiCl_4]^{2-} \rightarrow \text{I}$$

4. For $[MnBr_4]^{2-}$: $Br^-$ is a weak field ligand. $Mn^{2+}$ is $d^5$ and remains high spin, giving a tetrahedral complex with magnetic moment $5.9 \; \text{BM}$. Hence,

$$[MnBr_4]^{2-} \rightarrow \text{IV}$$

Therefore, the correct matching is

$$A-III, \; B-II, \; C-I, \; D-IV$$

So, the correct option is B.

Given: The complexes $[Ni(CO)_4]$, $[Ni(CN)_4]^{2-}$, $[NiCl_4]^{2-}$ and $[MnBr_4]^{2-}$.

Find: Their correct geometries and magnetic moments.

• Strong field ligands: $CO, \; CN^-$
• Weak field ligands: $Cl^-, \; Br^-$
• Magnetic moment formula:

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

where $n$ is the number of unpaired electrons.

For $[Ni(CO)_4]$, nickel is in oxidation state $0$ and the complex is tetrahedral with all electrons paired, so magnetic moment is $0 \; \text{BM}$.

For $[Ni(CN)_4]^{2-}$, the strong field ligand $CN^-$ gives a square planar arrangement with paired electrons, so magnetic moment is $0 \; \text{BM}$.

For $[NiCl_4]^{2-}$, the weak field ligand $Cl^-$ gives a tetrahedral high spin complex. With $2$ unpaired electrons,

$$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.8 \; \text{BM}$$

So this matches I.

For $[MnBr_4]^{2-}$, manganese is $Mn^{2+}$, that is $d^5$. With weak field ligand $Br^-$, it remains high spin with $5$ unpaired electrons.

$$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9 \; \text{BM}$$

So this matches IV.

Hence,

$$A-III, \; B-II, \; C-I, \; D-IV$$

Therefore, the correct option is B.