The shape of Cr(CO)6 molecule is ............
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Octahedral
The shape of a metal carbonyl follows from the number of CO groups around the metal: tetracarbonylnickel(0) is tetrahedral, pentacarbonyliron(0) is trigonal bipyramidal and hexacarbonylchromium(0) is octahedral.
In \( \ce{Cr(CO)6} \) chromium is in the zero oxidation state (CO is a neutral ligand, so x + 6(0) = 0 and x = 0). Cr (Z = 24) is [Ar]3d54s1. CO is a strong field ligand, so all six valence electrons pair up in three 3d orbitals (3d6). This leaves two empty 3d orbitals, which with one 4s and three 4p orbitals form six d2sp3 hybrid orbitals. Each accepts a lone pair from one CO, and the six ligands point to the corners of an octahedron.
Why the other options are incorrect
Option (1) Planar trigonal — Incorrect. Three bonds with sp2 hybridisation (as in \( \ce{BF3} \)); \( \ce{Cr(CO)6} \) has six Cr–C bonds.
Option (2) Two Tetrahedra fused with each other — Incorrect. This describes a species such as \( \ce{Cr2O7^{2-}} \), in which two tetrahedra share one oxygen; \( \ce{Cr(CO)6} \) has a single metal centre.
Option (3) Trigonal bipyramidal — Incorrect. This is the shape of the five-coordinate \( \ce{Fe(CO)5} \).
NCERT (2026-27 edition):
Bonding in Metal Carbonyls
