Question:medium

The oxidation state of Cr in \([Cr(H_2O)_6]Cl_3\), \([Cr(C_6H_6)_2]\), \(K_2[Cr(CN)_2(O)_2(O_2)(NH_3)]\) respectively are:

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Always separate neutral and charged ligands before calculating oxidation state.
Updated On: Apr 17, 2026
  • \(+3, +4, +6\)
    % option (B) \(+3, +2, +4\)
  • \(+3, 0, +6\)
  • \(+3, 0, +4\)
  • \(+3, 6, +8\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
Oxidation state is the formal charge an atom would bear if all ligands were removed along with the electron pairs that were shared with the central atom. Neutral ligands have zero charge.
Step 2: Key Formula or Approach:
Sum of oxidation states \(= \text{Total charge on the complex/molecule}\).
: Detailed Explanation:
1. \([Cr(H_{2}O)_{6]Cl_{3}\)}:
Water (\(H_{2}O\)) is neutral. The three \(Cl^{-}\) ions give a \(-3\) charge outside the bracket.
\(x + 6(0) + 3(-1) = 0 \Rightarrow x = +3\).
2. \([Cr(C_{6}H_{6})_{2]\) (Dibenzenechromium)}:
Benzene (\(C_{6}H_{6}\)) is a neutral \(\pi\)-ligand.
\(x + 2(0) = 0 \Rightarrow x = 0\).
3. \(K_{2}[Cr(CN)_{2(O)_{2}(O_{2})(NH_{3})]\)}:
Potassium (\(K^{+}\)) is \(+1\), Cyanide (\(CN^{-}\)) is \(-1\), Oxo (\(O^{2-}\)) is \(-2\), Peroxo (\(O_{2}^{2-}\)) is \(-2\), and Ammonia (\(NH_{3}\)) is neutral.
\(2(+1) + x + 2(-1) + 2(-2) + 1(-2) + 0 = 0\)
\(2 + x - 2 - 4 - 2 = 0\)
\(x - 6 = 0 \Rightarrow x = +6\).
Step 3: Final Answer:
The oxidation states are \(+3, 0, +6\).
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