Chemistry Q62 – JEE-MAIN 2026

Home Chemistry Coordination CompoundsQuestion #321

Match the LIST-I with LIST-II

\begin{array}{ll} List-I & List-II \\ Complex ion & Calculated spin only \\ magnetic moment (BM) \\ A. [{Cr(H}_{2} {O)}_{6}]^{2 +} & I. 3.87 \\ B. [{Co(H}_{2} {O)}_{6}]^{2 +} & II. 5.92 \\ C. [{Cu(H}_{2} {O)}_{6}]^{2 +} & III. 4.90 \\ D. [{Mn(H}_{2} {O)}_{6}]^{2 +} & IV. 1.73 \end{array}

Choose the correct answer from the options given below:

Clue 1 of 3: Magnetic Moment and Unpaired Electrons

The spin-only magnetic moment of a coordination complex is determined by the number of unpaired electrons in the d-orbitals of the central metal ion.

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Ninja Strategy•Extreme Value Check

Quickly calculate the magnetic moments for the ions with the fewest ( $C u^{2 +}$ , n=1) and most ( $M n^{2 +}$ , n=5) unpaired electrons to rapidly eliminate incorrect options.

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Step 1: Determine Electronic Configuration and Number of Unpaired Electrons (n)✦ Active

For each complex ion, identify the central metal, its oxidation state, and its d-electron configuration. Since $H_{2} O$ is a weak-field ligand, high-spin complexes will be formed (electrons fill orbitals singly before pairing).

A. $[C r (H_{2} O)_{6}]^{2 +}$ : $C r^{2 +}$ is $3 d^{4}$ . In a high-spin configuration, there are 4 unpaired electrons ( $n = 4$ ).

B. $[C o (H_{2} O)_{6}]^{2 +}$ : $C o^{2 +}$ is $3 d^{7}$ . In a high-spin configuration ( $t_{2 g}^{5} e_{g}^{2}$ ), there are 3 unpaired electrons ( $n = 3$ ).

C. $[C u (H_{2} O)_{6}]^{2 +}$ : $C u^{2 +}$ is $3 d^{9}$ . There is 1 unpaired electron ( $n = 1$ ).

D. $[M n (H_{2} O)_{6}]^{2 +}$ : $M n^{2 +}$ is $3 d^{5}$ . In a high-spin configuration, there are 5 unpaired electrons ( $n = 5$ ).

Step 2: Calculate Spin-Only Magnetic Moment and Match○ Expand

Use the formula $μ_{s} = \sqrt{n (n + 2)}$ BM to calculate the magnetic moment for each value of 'n' and match it with the values in List-II.

A. For

n = 4

μ_{s} = \sqrt{4 (4 + 2)} = \sqrt{24} \approx 4.90

BM. (A

\to

III)

B. For

n = 3

μ_{s} = \sqrt{3 (3 + 2)} = \sqrt{15} \approx 3.87

BM. (B

\to

C. For

n = 1

μ_{s} = \sqrt{1 (1 + 2)} = \sqrt{3} \approx 1.73

BM. (C

\to

IV)

D. For

n = 5

μ_{s} = \sqrt{5 (5 + 2)} = \sqrt{35} \approx 5.92

BM. (D

\to

II)

Step 3: Select the Correct Option○ Expand

The correct set of matches is: A-III, B-I, C-IV, D-II. Comparing this with the given options, we find that option 4 is the correct answer.

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Chemistry Question 62 – JEE-MAIN 2026

Clue 1 of 3: Magnetic Moment and Unpaired Electrons

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