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Step-by-Step Solution
Step 1: Write the electronic configurations of the neutral atoms
• Titanium (Ti), atomic number 22: $[\text{Ar}]\,3d^2\,4s^2$
• Vanadium (V), atomic number 23: $[\text{Ar}]\,3d^3\,4s^2$
• Chromium (Cr), atomic number 24: $[\text{Ar}]\,3d^5\,4s^1$
• Manganese (Mn), atomic number 25: $[\text{Ar}]\,3d^5\,4s^2$
Step 2: Determine the electron loss for each ion in the correct option
The correct option given is: Ti2+, V3+, Cr4+, Mn5+.
• Ti2+ means 2 electrons are removed from Ti.
• V3+ means 3 electrons are removed from V.
• Cr4+ means 4 electrons are removed from Cr.
• Mn5+ means 5 electrons are removed from Mn.
Step 3: Calculate the resulting electron configurations
• Ti2+: Start with $[\text{Ar}]\,3d^2\,4s^2$. Removing 2 electrons (preferentially from the 4s orbital) → $[\text{Ar}]\,3d^2\,4s^0$.
• V3+: Start with $[\text{Ar}]\,3d^3\,4s^2$. Removing 3 electrons (2 from 4s and 1 from 3d) → $[\text{Ar}]\,3d^2\,4s^0$.
• Cr4+: Start with $[\text{Ar}]\,3d^5\,4s^1$. Removing 4 electrons (1 from 4s and 3 from 3d) → $[\text{Ar}]\,3d^2\,4s^0$.
• Mn5+: Start with $[\text{Ar}]\,3d^5\,4s^2$. Removing 5 electrons (2 from 4s and 3 from 3d) → $[\text{Ar}]\,3d^2\,4s^0$.
Step 4: Verify the 3d2 configuration
Each ion in this series ends up having the configuration $[\text{Ar}]\,3d^2\,4s^0$, which is 3d2.
Conclusion
Among the given options, only Ti2+, V3+, Cr4+, Mn5+ all have the 3d2 configuration. Therefore, this is the correct series.
