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The heptacity of allyl and Cp and the ligation mode of NO in the thermodynamically stable complexes

[(ηx-allyl)Ru(CO)2(NO)] and [(ηy-Cp)Ru(CO)2(NO)],

respectively, are

(The heptacity of allyl and Cp are denoted by ηx and ηy, respectively.)

  • A.(η\(^{3}\), NO-bent) and (η\(^{5}\), NO-linear)
  • B.(η\(^{3}\), NO-linear) and (η\(^{5}\), NO-bent)
  • C.(η\(^{1}\), NO-bent) and (η\(^{3}\), NO-bent)
  • D.(η\(^{1}\), NO-bent) and (η\(^{5}\), NO-linear)

Answer: B

The answer is (η3, NO-linear) and (η5, NO-bent).

Count electrons on the neutral-ligand convention, with Ru contributing 8, each CO 2, linear NO 3 and bent NO 1:

  • (η3-allyl)Ru(CO)2(NO): 8 + 3 + 4 + 3 (linear NO) = 18.
  • (η5-Cp)Ru(CO)2(NO): with linear NO, 8 + 5 + 4 + 3 = 20, too many; with bent NO, 8 + 5 + 4 + 1 = 18.

Each complex settles into its 18-electron form. An η1-allyl or bent NO in the first would leave it short of 18 electrons, and a linear NO with η5-Cp would exceed 18, so the other combinations are wrong.

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