Which of the following are paramagnetic? A. [NiCl4]2- B. Ni(CO)4 C. [Ni(CN)4]2- D. [Ni(H2O)6]2+ E. Ni(PPh3)4

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q51 – NEET-UG 2025

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Home Chemistry Coordination CompoundsQuestion #330

Which of the following are paramagnetic? A.

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

N i (C O)_{4}

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

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

N i (P P h_{3})_{4}

Clue 1 of 3: Paramagnetism

Paramagnetic substances contain unpaired electrons, which are attracted to an external magnetic field. Diamagnetic substances have all electrons paired and are repelled by a magnetic field.

🧠 Full Solution Path

Step 1: Determine Oxidation State and d-electron Configuration✦ Active

For each complex, determine the oxidation state of Nickel (Ni) and its corresponding d-electron configuration. Nickel (atomic number 28) has an electronic configuration of $[A r] 3 d^{8} 4 s^{2}$ .

A. $[N i C l_{4}]^{2 -}$ : Ni is in $+ 2$ oxidation state ( $N i^{2 +}$ ), so it has a $3 d^{8}$ configuration.

B. $N i (C O)_{4}$ : Ni is in $0$ oxidation state ( $N i (0)$ ), so it has a $3 d^{8} 4 s^{2}$ configuration.

C. $[N i (C N)_{4}]^{2 -}$ : Ni is in $+ 2$ oxidation state ( $N i^{2 +}$ ), so it has a $3 d^{8}$ configuration.

D. $[N i (H_{2} O)_{6}]^{2 +}$ : Ni is in $+ 2$ oxidation state ( $N i^{2 +}$ ), so it has a $3 d^{8}$ configuration.

E. $N i (P P h_{3})_{4}$ : Ni is in $0$ oxidation state ( $N i (0)$ ), so it has a $3 d^{8} 4 s^{2}$ configuration.

Step 2: Analyze Ligand Field Strength, Geometry, and Electron Pairing○ Expand

Determine the geometry and ligand field strength for each complex to predict the number of unpaired electrons. Paramagnetic complexes have unpaired electrons.

A. $[N i C l_{4}]^{2 -}$ : $N i^{2 +}$ ( $d^{8}$ ), $C l^{-}$ is a weak field ligand. It forms a tetrahedral complex. In a weak field tetrahedral $d^{8}$ complex, there are 2 unpaired electrons ( $e^{4} t_{2}^{4}$ ). Thus, it is paramagnetic.

B. $N i (C O)_{4}$ : $N i (0)$ ( $3 d^{8} 4 s^{2}$ ), $C O$ is a strong field ligand. It forms a tetrahedral complex. Strong field ligands cause pairing, leading to a $3 d^{10}$ configuration (all electrons paired). Thus, it is diamagnetic.

C. $[N i (C N)_{4}]^{2 -}$ : $N i^{2 +}$ ( $d^{8}$ ), $C N^{-}$ is a strong field ligand. It forms a square planar complex. Strong field $d^{8}$ square planar complexes have all electrons paired. Thus, it is diamagnetic.

D. $[N i (H_{2} O)_{6}]^{2 +}$ : $N i^{2 +}$ ( $d^{8}$ ), $H_{2} O$ is a weak field ligand. It forms an octahedral complex. In a weak field octahedral $d^{8}$ complex, there are 2 unpaired electrons ( $t_{2 g}^{6} e_{g}^{2}$ ). Thus, it is paramagnetic.

E. $N i (P P h_{3})_{4}$ : $N i (0)$ ( $3 d^{8} 4 s^{2}$ ), $P P h_{3}$ is a strong field ligand. It forms a tetrahedral complex. Similar to $N i (C O)_{4}$ , strong field $P P h_{3}$ causes pairing, leading to a $3 d^{10}$ configuration (all electrons paired). Thus, it is diamagnetic.

💡 Teacher's Secret Hint

Remember that $N i (0)$ complexes with strong field ligands like $C O$ and $P P h_{3}$ are typically tetrahedral and diamagnetic due to $d^{10}$ configuration.

Step 3: Identify Paramagnetic Complexes and Select the Correct Option○ Expand

Based on the analysis, complexes A ( $[N i C l_{4}]^{2 -}$ ) and D ( $[N i (H_{2} O)_{6}]^{2 +}$ ) are paramagnetic.

Comparing this with the given options:

(1) A and C only

(2) B and E only

(3) A and D only

(4) A, D and E only

The correct option is (3).

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Chemistry Question 51 – NEET-UG 2025

Clue 1 of 3: Paramagnetism

🧠 Full Solution Path