Bridging Carbonyl Ligands in Transition Metal Carbonyl Complexes

The sum of bridging carbonyls in W(CO)_6 and Mn_2(CO)_10 is

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Question type: Numerical
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

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How does each Mn satisfy the 18-electron rule without bridging carbonyls?

Mn has 7 valence electrons. Five terminal CO ligands contribute 10 electrons ( $5 \times 2$ ), and one electron comes from the single $M n - M n$ bond: $7 + 10 + 1 = 18$ electrons per Mn atom.

Answer

Accepted answer: 0

The sum of bridging carbonyls in $W (C O)_{6}$ and $M n_{2} (C O)_{10}$ is 0, as both complexes possess only terminal carbonyl ligands.

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Step-by-step solution

Given: Given complexes: $W (C O)_{6}$ and $M n_{2} (C O)_{10}$ .
Find: Find the total sum of bridging carbonyl ( $μ -CO$ ) ligands present in both complexes.
Strategise: Recall the geometry and coordination mode of CO in each complex: $W (C O)_{6}$ is mononuclear octahedral with all terminal CO ligands. $M n_{2} (C O)_{10}$ consists of two octahedral $M n (C O)_{5}$ units joined by a single $M n - M n$ covalent bond with no bridging carbonyl ligands ( $D_{4 d}$ staggered conformation).
Execute: Number of bridging CO in $W (C O)_{6} = 0$ . Number of bridging CO in $M n_{2} (C O)_{10} = 0$ . Total sum = $0 + 0 = 0$ .
Verify: Unlike $F e_{2} (C O)_{9}$ (which has 3 bridging COs) or $C o_{2} (C O)_{8}$ in solid state (which has 2 bridging COs), $M n_{2} (C O)_{10}$ has zero bridging COs, supported by IR spectroscopy showing only terminal $ν_{C O}$ bands. The sum is 0.

Bridging Carbonyl Ligands in Transition Metal Carbonyl Complexes

What feels right?

The sum of bridging carbonyls in $W (C O)_{6}$ and $M n_{2} (C O)_{10}$ is 0, as both complexes possess only terminal carbonyl ligands.

Why each option works or fails

Solution

01Given

02Find

03Strategise

04Execute

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