‘Spin only’ magnetic moment is same for which of the following ions? A. Ti3+ B. Cr2+ C. Mn2+ D. Fe2+ E. Sc3+ Choose the most appropriate answer from the options given below:

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q72 – NEET-UG 2024

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Home Chemistry d- & f-Block ElementsQuestion #365

‘Spin only’ magnetic moment is same for which of the following ions? A.

T i^{3 +}

C r^{2 +}

M n^{2 +}

F e^{2 +}

S c^{3 +}

Choose the most appropriate answer from the options given below:

Clue 1 of 3: Magnetic Moment Concept

The 'spin only' magnetic moment of an ion depends on the number of unpaired electrons present in its d-orbitals.

🧠 Full Solution Path

Step 1: Determine Electronic Configurations and Unpaired Electrons✦ Active

To find the spin-only magnetic moment, we need to determine the number of unpaired electrons ( $n$ ) for each ion. The general electronic configuration for transition metals is $[A r] 3 d^{x} 4 s^{y}$ . When forming cations, electrons are removed first from the $4 s$ orbital, then from the $3 d$ orbital.

- A. $T i^{3 +}$ (Ti: $[A r] 3 d^{2} 4 s^{2}$ ): $T i^{3 +}$ is $[A r] 3 d^{1}$ . Number of unpaired electrons ( $n$ ) = 1.

- B. $C r^{2 +}$ (Cr: $[A r] 3 d^{5} 4 s^{1}$ ): $C r^{2 +}$ is $[A r] 3 d^{4}$ . Number of unpaired electrons ( $n$ ) = 4.

- C. $M n^{2 +}$ (Mn: $[A r] 3 d^{5} 4 s^{2}$ ): $M n^{2 +}$ is $[A r] 3 d^{5}$ . Number of unpaired electrons ( $n$ ) = 5.

- D. $F e^{2 +}$ (Fe: $[A r] 3 d^{6} 4 s^{2}$ ): $F e^{2 +}$ is $[A r] 3 d^{6}$ . Number of unpaired electrons ( $n$ ) = 4.

- E. $S c^{3 +}$ (Sc: $[A r] 3 d^{1} 4 s^{2}$ ): $S c^{3 +}$ is $[A r]$ . Number of unpaired electrons ( $n$ ) = 0.

💡 Teacher's Secret Hint

Remember the exceptions in electronic configurations for transition metals like Chromium and Copper. Also, electrons are removed from the outermost s-orbital first when forming cations.

Step 2: Compare Number of Unpaired Electrons○ Expand

The 'spin only' magnetic moment ( $μ_{s}$ ) is given by the formula $μ_{s} = \sqrt{n (n + 2)}$ BM. Therefore, ions with the same number of unpaired electrons ( $n$ ) will have the same spin-only magnetic moment.

Comparing the number of unpaired electrons calculated in Step 1:

- $T i^{3 +}$ : $n = 1$

- $C r^{2 +}$ : $n = 4$

- $M n^{2 +}$ : $n = 5$

- $F e^{2 +}$ : $n = 4$

- $S c^{3 +}$ : $n = 0$

We observe that $C r^{2 +}$ and $F e^{2 +}$ both have 4 unpaired electrons.

💡 Teacher's Secret Hint

The magnetic moment is directly proportional to the number of unpaired electrons. More unpaired electrons mean a higher magnetic moment.

Step 3: Identify the Correct Option○ Expand

Since $C r^{2 +}$ (Option B) and $F e^{2 +}$ (Option D) both have 4 unpaired electrons, their spin-only magnetic moments will be identical. Therefore, the most appropriate answer is the option that includes B and D.

This corresponds to option (4).

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Chemistry Question 72 – NEET-UG 2024

Clue 1 of 3: Magnetic Moment Concept

🧠 Full Solution Path