Q10(b). Identify M from the CO stretching frequencies
The stretching frequency values of both starting material and the products are given below in the reversible reaction. Find M which belongs to the principal quantum number n=5 and draw both the structures: [(η⁵-C₅H₅)M(CO)₃]₂ ⇌ [(η⁵-C₅H₅)M(CO)₂]₂ + 2CO; ν=1960,1915 cm⁻¹ (reactant) and 1889,1859 cm⁻¹ (product).
Answer in brief
M =Mo (a 4d metal, valence shell n=5). [CpMo(CO)₃]₂ has a Mo–Mo single bond; [CpMo(CO)₂]₂ has a Mo≡Mo triple bond; both have only terminal CO.
Key points
- each M must reach 18e: [CpM(CO)₃]₂: x+5+6+1 (M–M)=18⇒ x=6: group 6. A principal quantum number n=5 (valence 5s, 4d series) gives Mo; Cr is n=4, W is n=6.
- two 18-electron CpMo(CO)₃ units joined by a Mo–Mo single bond (3.2 Å), each Mo having three terminal CO. ν_(CO)=1960, 1915 cm⁻¹: terminal carbonyls (above 1850).
- loss of 2 CO leaves [CpMo(CO)₂]₂. Each Mo: 6+5+4=15 e, so it needs three bonds: Mo≡Mo (2.45 Å, bond order 3) to reach 18 e. The two CpMo(CO)₂ units are twisted, with the carbonyls terminal.
- ν_(CO) drops from 1960/1915 to 1889/1859 cm⁻¹ because each Mo now has two instead of three CO: the same metal electron density is back-donated into fewer π^* orbitals, so each C–O bond is weakened more (bond order smaller).
The full model answer
Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.
- Identify M: each M must reach 18e: [CpM(CO)₃]₂: x+5+6+1 (M–M)=18⇒ x=6: group 6. A principal quantum number n=5 (valence 5s, 4d series) gives Mo; Cr is n=4, W is n=6.
- Reactant: two 18-electron CpMo(CO)₃ units joined by a Mo–Mo single bond (3.2 Å), each Mo having three terminal CO. ν_(CO)=1960, 1915 cm⁻¹: terminal carbonyls (above 1850).
- Product: loss of 2 CO leaves [CpMo(CO)₂]₂. Each Mo: 6+5+4=15 e, so it needs three bonds: Mo≡Mo (2.45 Å, bond order 3) to reach 18 e. The two CpMo(CO)₂ units are twisted, with the carbonyls terminal.
- IR shift: ν_(CO) drops from 1960/1915 to 1889/1859 cm⁻¹ because each Mo now has two instead of three CO: the same metal electron density is back-donated into fewer π^* orbitals, so each C–O bond is weakened more (bond order smaller).
- Two bands each: the two CO groups on each Mo give symmetric and antisymmetric stretches; the pair 1960/1915 is the reactant, 1889/1859 the product, which is seen to grow on heating in solution (the reaction is reversible with CO).
[CpMo(CO)₂]₂ adds alkynes across the Mo≡Mo triple bond (a μ-η²:η²-alkyne complex), and [CpMo(CO)₃]₂ is the source of the Mo radical CpMo(CO)₃•.
Coversmetal–metal bonds from IRIdentify M from the CO stretching frequencies
All parts of Question 10
- 10(a) Radiocarbon age of a piece of wood10 marks
- 10(b) Identify M from the CO stretching frequencies10 marks
- 10(c) Why Al cannot be titrated directly with EDTA10 marks