The metal ions that have the calculated spin-only magnetic moment value of 4.9 B.M. are : A. Cr2+ B. Fe2+ C. Fe3+ D. Co2+ E. Mn3+ Choose the correct answer from the options given below: A, B and E Only A, D and E Only B and E Only A, C and E Only

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q62 – JEE-MAIN 2025

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Home Chemistry d- & f-Block ElementsQuestion #369

The metal ions that have the calculated spin-only magnetic moment value of

4.9

B.M. are : A.

{Cr}^{2 +}

{Fe}^{2 +}

{Fe}^{3 +}

{Co}^{2 +}

{Mn}^{3 +}

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 transition metal ion is directly related to the number of unpaired electrons in its d-orbitals.

🥷

Ninja Strategy•Calculate and Eliminate

First, determine the required number of unpaired electrons (n=4). Then, quickly calculate n for each ion and eliminate options that include ions with incorrect n values.

🧠 Full Solution Path

Step 1: Calculate number of unpaired electrons from magnetic moment✦ Active

The spin-only magnetic moment is given by the formula $μ = \sqrt{n (n + 2)}$ B.M., where $n$ is the number of unpaired electrons. We are given $μ = 4.9$ B.M. Let's find $n$ :

4.9 \approx \sqrt{n (n + 2)}

Squaring both sides: (4.9)^{2} \approx n (n + 2)

24.01 \approx n^{2} + 2 n

If n = 4, μ = \sqrt{4 (4 + 2)} = \sqrt{4 \times 6} = \sqrt{24} \approx 4.899 B.M.

Thus, a magnetic moment of $4.9$ B.M. corresponds to $n = 4$ unpaired electrons.

Step 2: Determine unpaired electrons for each ion○ Expand

We need to find the number of unpaired electrons for each given metal ion:

\begin{array}{ll} A. {Cr}^{2 +} & Cr: [Ar] 3 d^{5} 4 s^{1} \Rightarrow {Cr}^{2 +} : [Ar] 3 d^{4} (n = 4) unpaired electrons B. {Fe}^{2 +} & Fe: [Ar] 3 d^{6} 4 s^{2} \Rightarrow {Fe}^{2 +} : [Ar] 3 d^{6} (n = 4) unpaired electrons C. {Fe}^{3 +} & Fe: [Ar] 3 d^{6} 4 s^{2} \Rightarrow {Fe}^{3 +} : [Ar] 3 d^{5} (n = 5) unpaired electrons D. {Co}^{2 +} & Co: [Ar] 3 d^{7} 4 s^{2} \Rightarrow {Co}^{2 +} : [Ar] 3 d^{7} (n = 3) unpaired electrons E. {Mn}^{3 +} & Mn: [Ar] 3 d^{5} 4 s^{2} \Rightarrow {Mn}^{3 +} : [Ar] 3 d^{4} (n = 4) unpaired electrons \end{array}

💡 Teacher's Secret Hint

Remember to consider the high-spin configuration for free ions or weak-field ligands, which is typically assumed unless specified otherwise.

Step 3: Identify ions with 4 unpaired electrons○ Expand

The ions with 4 unpaired electrons are ${Cr}^{2 +}$ , ${Fe}^{2 +}$ , and ${Mn}^{3 +}$ . These correspond to options A, B, and E. Therefore, the correct combination is A, B and E Only.

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Chemistry Question 62 – JEE-MAIN 2025

Clue 1 of 3: Magnetic Moment and Unpaired Electrons

🧠 Full Solution Path