To determine the number of unpaired electrons in gaseous species of Mn³⁺, Cr³⁺, and V³⁺, follow these steps:

Step 1: Write the Electronic Configurations of Neutral Atoms

• Manganese (Mn, Z = 25):
  Mn: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s² → [Ar] 3d⁵ 4s²
• Chromium (Cr, Z = 24):
  Cr: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹ → [Ar] 3d⁵ 4s¹ (due to half-filled stability)
• Vanadium (V, Z = 23):
  V: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s² → [Ar] 3d³ 4s²

Step 2: Remove Electrons for the Given Ion

• Mn³⁺: Remove 3 electrons from Mn → first from 4s² and one from 3d⁵
  Mn³⁺: [Ar] 3d⁴
• Cr³⁺: Remove 3 electrons from Cr → first from 4s¹ and two from 3d⁵
  Cr³⁺: [Ar] 3d³
• V³⁺: Remove 3 electrons from V → first from 4s² and one from 3d³
  V³⁺: [Ar] 3d²

Step 3: Count Unpaired Electrons

• Mn³⁺ ([Ar] 3d⁴): 4 unpaired electrons
• Cr³⁺ ([Ar] 3d³): 3 unpaired electrons
• V³⁺ ([Ar] 3d²): 2 unpaired electrons

Final Answer:

The number of unpaired electrons in Mn³⁺, Cr³⁺, and V³⁺ are 4, 3, and 2, respectively.