Let x be the EAN of [Mo(η⁵-C₅H₅)(CO)₃(CH₃)], y the number of five-membered chelate rings in [Cu(gly)₂], and z the number of metal–carbon bonds in K₄[Fe(CN)₆]. Find x/y − z. (Mo = 42)

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Mo(II) is 40 electrons, plus 6 from Cp⁻, 6 from three CO and 2 from CH₃⁻, giving x = 54. Each gly gives one five-membered ring, so y = 2. The six cyanides bond through C, so z = 6. 54/2 − 6 = 27 − 6 = 21.

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