Chemistry Q66 – NEET-UG 2026

Home Chemistry Coordination CompoundsQuestion #357

Among the species given below, the spin-only magnetic moment is highest for (Given : Atomic number of Ti = 22, Mn = 25, Fe = 26 and Co = 27)

Clue 1 of 3: Magnetic Moment and Unpaired Electrons

The spin-only magnetic moment is directly related to the number of unpaired electrons in the central metal ion. A higher number of unpaired electrons leads to a higher magnetic moment.

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Ninja Strategy•Maximize Unpaired Electrons

The spin-only magnetic moment is directly proportional to the number of unpaired electrons; therefore, identify the complex with the highest number of unpaired electrons by considering the metal's oxidation state and the ligand's field strength.

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Step 1: Determine Oxidation State and

d

-electron Configuration✦ Active

For each complex, calculate the oxidation state of the central metal ion and write its electronic configuration to determine the number of $d$ -electrons.

1. [M n (C N)_{6}]^{3 -} : M n^{3 +} (d^{4})

2. [F e (C N)_{6}]^{3 -} : F e^{3 +} (d^{5})

3. [C o (N H_{3})_{6}]^{3 +} : C o^{3 +} (d^{6})

4. [T i (H_{2} O)_{6}]^{3 +} : T i^{3 +} (d^{1})

Step 2: Identify Ligand Field Strength and Electron Distribution○ Expand

Classify the ligands as strong field (SFL) or weak field (WFL). Strong field ligands cause pairing of electrons in lower energy orbitals ( $t_{2 g}$ for octahedral complexes), while weak field ligands do not. Based on the $d$ -electron configuration and ligand field strength, determine the number of unpaired electrons ( $n$ ) for each complex.

1. [M n (C N)_{6}]^{3 -} : C N^{-} (S F L), d^{4} \Rightarrow t_{2 g}^{4} e_{g}^{0}, n = 2

2. [F e (C N)_{6}]^{3 -} : C N^{-} (S F L), d^{5} \Rightarrow t_{2 g}^{5} e_{g}^{0}, n = 1

3. [C o (N H_{3})_{6}]^{3 +} : N H_{3} (S F L), d^{6} \Rightarrow t_{2 g}^{6} e_{g}^{0}, n = 0

4. [T i (H_{2} O)_{6}]^{3 +} : H_{2} O (W F L), d^{1} \Rightarrow t_{2 g}^{1} e_{g}^{0}, n = 1

💡 Teacher's Secret Hint

Remember that $C N^{-}$ and $N H_{3}$ are strong field ligands, while $H_{2} O$ is a weak field ligand. The effect of ligand field strength is crucial for $d^{4}$ to $d^{7}$ configurations.

Step 3: Calculate Spin-Only Magnetic Moment and Compare○ Expand

Use the formula $μ = \sqrt{n (n + 2)}$ B.M. to calculate the spin-only magnetic moment for each complex and identify the one with the highest value.

1. [M n (C N)_{6}]^{3 -} : n = 2 \Rightarrow μ = \sqrt{2 (2 + 2)} = \sqrt{8} \approx 2.83 B.M.

2. [F e (C N)_{6}]^{3 -} : n = 1 \Rightarrow μ = \sqrt{1 (1 + 2)} = \sqrt{3} \approx 1.73 B.M.

3. [C o (N H_{3})_{6}]^{3 +} : n = 0 \Rightarrow μ = \sqrt{0 (0 + 2)} = 0 B.M.

4. [T i (H_{2} O)_{6}]^{3 +} : n = 1 \Rightarrow μ = \sqrt{1 (1 + 2)} = \sqrt{3} \approx 1.73 B.M.

Comparing the magnetic moments, $[M n (C N)_{6}]^{3 -}$ has the highest value.

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Chemistry Question 66 – NEET-UG 2026

Clue 1 of 3: Magnetic Moment and Unpaired Electrons

🧠 Full Solution Path