Question 31 of 44intermediate🔧 ApplyNumerical3 marks

Calculate the number of unpaired electrons in the following gaseous ions: Mn3+Mn^{3+}, Cr3+Cr^{3+}, V3+V^{3+} and Ti3+Ti^{3+}. Which one of these is the most stable in aqueous solution?

Correct Answer

(most_stable) Cr³⁺ is the most stable in aqueous solution.

(unpaired_electrons) [object Object]

Exercise: EXERCISES | Q: 4.24 | (Page 28)
For More Understanding

Explanation

To find unpaired electrons, we determine the electronic configuration of each ion by removing 3 electrons from the neutral atom (first from 4s, then from 3d). Ti (Z=22): [Ar]3d²4s² → Ti³⁺: [Ar]3d¹ (1 unpaired). V (Z=23): [Ar]3d³4s² → V³⁺: [Ar]3d² (2 unpaired). Cr (Z=24): [Ar]3d⁵4s¹ → Cr³⁺: [Ar]3d³ (3 unpaired). Mn (Z=25): [Ar]3d⁵4s² → Mn³⁺: [Ar]3d⁴ (4 unpaired). Cr³⁺ is most stable due to d³ configuration having maximum exchange energy with all three electrons in parallel spins occupying separate orbitals, as mentioned in the context regarding Hund's rule and exchange energy stabilization.

Solution Steps

  1. Step 1: Write electronic configuration of neutral atoms: Ti[Ar]3d²4s², V[Ar]3d³4s², Cr[Ar]3d⁵4s¹, Mn[Ar]3d⁵4s²

  2. Step 2: Remove 3 electrons to form M³⁺ ions (first from 4s, then from 3d)

  3. Step 3: Ti³⁺: [Ar]3d¹ (1 unpaired), V³⁺: [Ar]3d² (2 unpaired), Cr³⁺: [Ar]3d³ (3 unpaired), Mn³⁺: [Ar]3d⁴ (4 unpaired)

  4. Step 4: Identify most stable ion: Cr³⁺ with d³ configuration has maximum exchange energy stabilization as all three electrons have parallel spins in separate orbitals (Hund's rule)

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