Question:
The increasing order of acidic strength for the aqua complexes $[\text{V}(\text{H}_2\text{O})_6]^{2+}$, $[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$, $[\text{Fe}(\text{H}_2\text{O})_6]^{3+}$, and $[\text{Co}(\text{H}_2\text{O})_6]^{3+}$ is:
The increasing order of acidic strength for the aqua complexes $[\text{V}(\text{H}_2\text{O})_6]^{2+}$, $[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$, $[\text{Fe}(\text{H}_2\text{O})_6]^{3+}$, and $[\text{Co}(\text{H}_2\text{O})_6]^{3+}$ is:
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For metal aqua complexes, oxidation state is the dominant factor determining acidity. If oxidation states are equal, acidity increases from left to right across a period as ionic radius decreases.
Updated On: Jun 16, 2026
- $[\text{V}(\text{H}_2\text{O})_6]^{2+} \lt [\text{Fe}(\text{H}_2\text{O})_6]^{2+} \lt [\text{Fe}(\text{H}_2\text{O})_6]^{3+} \lt [\text{Co}(\text{H}_2\text{O})_6]^{3+}$
- $[\text{Fe}(\text{H}_2\text{O})_6]^{2+} \lt [\text{Fe}(\text{H}_2\text{O})_6]^{3+} \lt [\text{V}(\text{H}_2\text{O})_6]^{2+} \lt [\text{Co}(\text{H}_2\text{O})_6]^{3+}$
- $[\text{Co}(\text{H}_2\text{O})_6]^{3+} \lt [\text{Fe}(\text{H}_2\text{O})_6]^{3+} \lt [\text{Fe}(\text{H}_2\text{O})_6]^{2+} \lt [\text{V}(\text{H}_2\text{O})_6]^{2+}$
- $[\text{Fe}(\text{H}_2\text{O})_6]^{2+} \lt [\text{Fe}(\text{H}{_2}\text{O})_6]^{3+} \lt [\text{Co}(\text{H}_2\text{O})_6]^{3+} \lt [\text{V}(\text{H}_2\text{O})_6]^{2+}$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The question asks us to arrange four transition metal aqua complexes in order of increasing acidic strength (ability to donate a proton from a coordinated water molecule).
Step 2: Key Formula or Approach:
The acidity of a metal aqua complex depends on the charge density of the central metal ion:
\[ \text{Acidity} \propto \text{Charge Density} = \frac{\text{Ionic Charge (z)}}{\text{Ionic Radius (r)}} \]
A higher charge density polarizes the O-H bonds of the coordinated water molecules more strongly, facilitating the release of a proton ($\text{H}^+$).
Step 3: Detailed Explanation:
• First, let us separate the complexes by their oxidation state:
itemize
• $+2$ complexes: $[\text{V}(\text{H}_2\text{O})_6]^{2+}$ and $[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$
• $+3$ complexes: $[\text{Fe}(\text{H}_2\text{O})_6]^{3+}$ and $[\text{Co}(\text{H}_2\text{O})_6]^{3+}$
Because $+3$ metal ions have a significantly higher charge and smaller ionic radii compared to $+2$ metal ions, $+3$ complexes are much stronger acids than $+2$ complexes.
Next, let us compare the $+2$ complexes: $[\text{V}(\text{H}_2\text{O})_6]^{2+}$ vs $[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$.
Going from left to right across the 3d transition series, the effective nuclear charge increases, causing a contraction in ionic radius (transition metal contraction).
Since iron is further to the right than vanadium, the ionic radius of $\text{Fe}^{2+}$ is smaller than that of $\text{V}^{2+}$.
Therefore, $\text{Fe}^{2+}$ has a higher charge density and is more acidic:
\[ [\text{V}(\text{H}_2\text{O})_6]^{2+} \lt [\text{Fe}(\text{H}_2\text{O})_6]^{2+} \]
Finally, let us compare the $+3$ complexes: $[\text{Fe}(\text{H}_2\text{O})_6]^{3+}$ vs $[\text{Co}(\text{H}_2\text{O})_6]^{3+}$.
Cobalt lies to the right of iron in the periodic table, so the ionic radius of $\text{Co}^{3+}$ is smaller than that of $\text{Fe}^{3+}$.
This higher charge density on $\text{Co}^{3+}$ makes $[\text{Co}(\text{H}_2\text{O})_6]^{3+}$ more acidic than $[\text{Fe}(\text{H}_2\text{O})_6]^{3+}$:
\[ [\text{Fe}(\text{H}_2\text{O})_6]^{3+} \lt [\text{Co}(\text{H}_2\text{O})_6]^{3+} \]
Combining these trends yields the complete increasing order of acidity.
itemize
Step 4: Final Answer:
Therefore, the correct increasing order of acidic strength is Option A.
The question asks us to arrange four transition metal aqua complexes in order of increasing acidic strength (ability to donate a proton from a coordinated water molecule).
Step 2: Key Formula or Approach:
The acidity of a metal aqua complex depends on the charge density of the central metal ion:
\[ \text{Acidity} \propto \text{Charge Density} = \frac{\text{Ionic Charge (z)}}{\text{Ionic Radius (r)}} \]
A higher charge density polarizes the O-H bonds of the coordinated water molecules more strongly, facilitating the release of a proton ($\text{H}^+$).
Step 3: Detailed Explanation:
• First, let us separate the complexes by their oxidation state:
itemize
• $+2$ complexes: $[\text{V}(\text{H}_2\text{O})_6]^{2+}$ and $[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$
• $+3$ complexes: $[\text{Fe}(\text{H}_2\text{O})_6]^{3+}$ and $[\text{Co}(\text{H}_2\text{O})_6]^{3+}$
Because $+3$ metal ions have a significantly higher charge and smaller ionic radii compared to $+2$ metal ions, $+3$ complexes are much stronger acids than $+2$ complexes.
Next, let us compare the $+2$ complexes: $[\text{V}(\text{H}_2\text{O})_6]^{2+}$ vs $[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$.
Going from left to right across the 3d transition series, the effective nuclear charge increases, causing a contraction in ionic radius (transition metal contraction).
Since iron is further to the right than vanadium, the ionic radius of $\text{Fe}^{2+}$ is smaller than that of $\text{V}^{2+}$.
Therefore, $\text{Fe}^{2+}$ has a higher charge density and is more acidic:
\[ [\text{V}(\text{H}_2\text{O})_6]^{2+} \lt [\text{Fe}(\text{H}_2\text{O})_6]^{2+} \]
Finally, let us compare the $+3$ complexes: $[\text{Fe}(\text{H}_2\text{O})_6]^{3+}$ vs $[\text{Co}(\text{H}_2\text{O})_6]^{3+}$.
Cobalt lies to the right of iron in the periodic table, so the ionic radius of $\text{Co}^{3+}$ is smaller than that of $\text{Fe}^{3+}$.
This higher charge density on $\text{Co}^{3+}$ makes $[\text{Co}(\text{H}_2\text{O})_6]^{3+}$ more acidic than $[\text{Fe}(\text{H}_2\text{O})_6]^{3+}$:
\[ [\text{Fe}(\text{H}_2\text{O})_6]^{3+} \lt [\text{Co}(\text{H}_2\text{O})_6]^{3+} \]
Combining these trends yields the complete increasing order of acidity.
itemize
Step 4: Final Answer:
Therefore, the correct increasing order of acidic strength is Option A.
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