Question:
What is EAN of Mn in \(\mathrm{Mn}(\mathrm{CO})_5\) ?
What is EAN of Mn in \(\mathrm{Mn}(\mathrm{CO})_5\) ?
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EAN equals atomic number minus oxidation state plus electrons donated by ligands.
Updated On: Nov 13, 2026
35
36
34
37
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
\(\text{EAN}=Z-\text{(oxidation state)}+2\times\text{(number of ligands)}\) for neutral two-electron donors.
Step 2: Values
Mn has \(Z=25\); CO is neutral, so the oxidation state of Mn in \(\text{Mn(CO)}_5\) is 0.
Step 3: Calculate
\(\text{EAN}=25-0+5\times2=35\).
Final Answer:
The effective atomic number of Mn in \(\text{Mn(CO)}_5\) is 35. \[ \boxed{\text{(A) }35} \]
\(\text{EAN}=Z-\text{(oxidation state)}+2\times\text{(number of ligands)}\) for neutral two-electron donors.
Step 2: Values
Mn has \(Z=25\); CO is neutral, so the oxidation state of Mn in \(\text{Mn(CO)}_5\) is 0.
Step 3: Calculate
\(\text{EAN}=25-0+5\times2=35\).
Final Answer:
The effective atomic number of Mn in \(\text{Mn(CO)}_5\) is 35. \[ \boxed{\text{(A) }35} \]
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