Chemistry Q61 – JEE-MAIN 2026

Home Chemistry Coordination CompoundsQuestion #327

Which of the following sequences of hybridisation, geometry and magnetic nature are correct for the given coordination compounds ? A.

[N i C l_{4}]^{2 -} - s p^{3}

, tetrahedral, paramagnetic B.

[N i (N H_{3})_{6}]^{2 +} - s p^{3} d^{2}

, octahedral, paramagnetic C.

[N i (C O)_{4}] - s p^{3}

, tetrahedral, paramagnetic D.

[N i (C N)_{4}]^{2 -} - d s p^{2}

, square planar, diamagnetic Choose the correct answer from the options given below :

Clue 1 of 3: Determine Oxidation State

The first step for any coordination compound is to determine the oxidation state of the central metal atom.

🧠 Full Solution Path

Step 1: Analyze

[N i C l_{4}]^{2 -}

and

[N i (N H_{3})_{6}]^{2 +}

✦ Active

For $[N i C l_{4}]^{2 -}$ : The oxidation state of Ni is $+ 2$ . Electronic configuration of $N i^{2 +}$ is $3 d^{8}$ . $C l^{-}$ is a weak field ligand, so no electron pairing occurs. This leads to $s p^{3}$ hybridization, a tetrahedral geometry, and 2 unpaired electrons, making it paramagnetic. Thus, option A is correct.

For $[N i (N H_{3})_{6}]^{2 +}$ : The oxidation state of Ni is $+ 2$ . Electronic configuration of $N i^{2 +}$ is $3 d^{8}$ . $N H_{3}$ is a strong field ligand, but for $N i^{2 +}$ in octahedral complexes, it typically forms an outer orbital complex, meaning no pairing in the $3 d$ orbitals. This results in $s p^{3} d^{2}$ hybridization, an octahedral geometry, and 2 unpaired electrons, making it paramagnetic. Thus, option B is correct.

Step 2: Analyze

[N i (C O)_{4}]

○ Expand

For $[N i (C O)_{4}]$ : The oxidation state of Ni is $0$ . Electronic configuration of $N i (0)$ is $3 d^{8} 4 s^{2}$ . $C O$ is a very strong field ligand, causing the $4 s$ electrons to pair up and shift into the $3 d$ orbitals, resulting in a $3 d^{10}$ configuration. This leads to $s p^{3}$ hybridization, a tetrahedral geometry, and 0 unpaired electrons, making it diamagnetic. Option C states it is paramagnetic, which is incorrect.

Step 3: Analyze

[N i (C N)_{4}]^{2 -}

○ Expand

For $[N i (C N)_{4}]^{2 -}$ : The oxidation state of Ni is $+ 2$ . Electronic configuration of $N i^{2 +}$ is $3 d^{8}$ . $C N^{-}$ is a very strong field ligand, causing electron pairing in the $3 d$ orbitals, which makes one $3 d$ orbital available for hybridization. This results in $d s p^{2}$ hybridization, a square planar geometry, and 0 unpaired electrons, making it diamagnetic. Thus, option D is correct.

💡 Teacher's Secret Hint

Remember that $d^{8}$ complexes with strong field ligands often adopt a square planar geometry.

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Chemistry Question 61 – JEE-MAIN 2026

Clue 1 of 3: Determine Oxidation State

🧠 Full Solution Path