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JEE Advanced levelSingle correct optionDifficulty 3 of 3cc215

A neutral, mononuclear, diamagnetic carbonyl $M (CO) X_{5}$ of a 3d metal obeys the 18-electron rule. Which metal M and shape fit?

ACr, square pyramidal
BMn, trigonal bipyramidal
CCo, trigonal bipyramidal
DFe, trigonal bipyramidal

Answer: D

Valence electrons:

x + 5 \times 2 = 18

gives

x = 8

, so M is Fe (group 8). Five ligands with

d s p^{3}

hybridisation give a trigonal bipyramid.

Mn (CO) X_{5}

would have 17 electrons and

Co (CO) X_{5}

19, both paramagnetic;

Cr (CO) X_{5}

has 16.

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