Question:
The sum of coordination number and oxidation number of M in \([\text{M(en)}_2\text{C}_2\text{O}_4]\text{Cl}\) is
The sum of coordination number and oxidation number of M in \([\text{M(en)}_2\text{C}_2\text{O}_4]\text{Cl}\) is
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en and oxalate are both bidentate. Find CN and oxidation number.
Updated On: Oct 1, 2026
- \(9\)
- \(8\)
- \(7\)
- \(6\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
The coordination number is the total number of donor atoms attached to the central metal. The oxidation number is the charge the metal would carry if all ligands were removed with their pairs.
Step 2: Coordination number:
Ethylenediamine (en) is bidentate, so 2 en give 4 donor atoms. Oxalate (\(C_2O_4^{2-}\)) is also bidentate, so it gives 2. Total coordination number \(= 4 + 2 = 6\).
Step 3: Oxidation number:
The complex is \([M(en)_2C_2O_4]^+\), because one \(Cl^-\) is outside. Let M be \(x\). en is neutral and oxalate is \(-2\).
\[ x + 0 + (-2) = +1 \Rightarrow x = +3 \]
Step 4: Sum:
\(6 + 3 = 9\), option (A).
Final Answer:
Coordination number 6 plus oxidation number 3 gives 9. \[ \boxed{9} \]
The coordination number is the total number of donor atoms attached to the central metal. The oxidation number is the charge the metal would carry if all ligands were removed with their pairs.
Step 2: Coordination number:
Ethylenediamine (en) is bidentate, so 2 en give 4 donor atoms. Oxalate (\(C_2O_4^{2-}\)) is also bidentate, so it gives 2. Total coordination number \(= 4 + 2 = 6\).
Step 3: Oxidation number:
The complex is \([M(en)_2C_2O_4]^+\), because one \(Cl^-\) is outside. Let M be \(x\). en is neutral and oxalate is \(-2\).
\[ x + 0 + (-2) = +1 \Rightarrow x = +3 \]
Step 4: Sum:
\(6 + 3 = 9\), option (A).
Final Answer:
Coordination number 6 plus oxidation number 3 gives 9. \[ \boxed{9} \]
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