Question:

The correct order of enthalpy of hydration for the transition metal ions is

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Across a period, decreasing metal ion radius increases hydration enthalpy.
Updated On: Dec 14, 2025
  • Cr$^{2+}$>Mn$^{2+}$>Co$^{2+}$>Ni$^{2+}$
  • Ni$^{2+}$>Co$^{2+}$>Mn$^{2+}$>Cr$^{2+}$
  • Ni$^{2+}$>Co$^{2+}$>Cr$^{2+}$>Mn$^{2+}$
  • Cr$^{2+}$>Mn$^{2+}$>Ni$^{2+}$>Co$^{2+}$
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The Correct Option is C

Solution and Explanation

Step 1: Hydration enthalpy depends on ionic radius and charge density.
Smaller ionic size and higher charge density give more negative (larger magnitude) hydration enthalpy.
Across a period in transition metals, ionic radius decreases.
Step 2: Compare the given ions.
The size order for M$^{2+}$ ions across the 1st transition series is:
Cr$^{2+}$>Mn$^{2+}$>Co$^{2+}$>Ni$^{2+}$.
Step 3: Hydration enthalpy trend.
Smaller radius → higher hydration enthalpy. Thus, Ni$^{2+}$ (smallest) has highest hydration enthalpy, followed by Co$^{2+}$, Mn$^{2+}$, and Cr$^{2+}$ (largest).
Step 4: Conclusion.
Hence, Ni$^{2+}$>Co$^{2+}$>Mn$^{2+}$>Cr$^{2+}$.
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