The correct bond order for carbon-oxygen bond in following metal carbonyls \([Cr(CO)_6], [V(CO)_6]^-, [Mn(CO)_6]^+\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \([V(CO)_6]^-<[Cr(CO)_6] < [Mn (CO)_6]^+\)
The metal–carbonyl bond is synergic: CO donates its lone pair to the metal (σ bond), and the metal donates electrons from a filled d orbital into the vacant antibonding π* orbital of CO (π back bond). Electrons entering the π* orbital of CO weaken the C–O bond, so more back donation means a lower C–O bond order.
The three species are isoelectronic (CO is neutral, so the oxidation state of the metal equals the charge of the complex):
[V(CO)6]−: V is −1; V (Z = 23, 3d34s2) + 1 electron = 6 d electrons. The negative charge makes the metal the most electron-rich, so back donation is the greatest and the C–O bond order is the lowest.
[Cr(CO)6]: Cr is 0; Cr (Z = 24) has 6 valence electrons, all in d orbitals in the complex. Back donation is intermediate.
[Mn(CO)6]+: Mn is +1; Mn (Z = 25, 3d54s2) − 1 electron = 6 d electrons. The positive charge holds the d electrons more tightly, so back donation is the least and the C–O bond order is the highest.
C–O bond order: [V(CO)6]− < [Cr(CO)6] < [Mn(CO)6]+
Shortcut: the most negative complex has the lowest C–O bond order; only Option (1) starts with \( \ce{[V(CO)6]^{-}} \).
NCERT (2026-27 edition):
Bonding in Metal Carbonyls – …
