For each of the following (real or imaginary) complexes determine a0 the number of electrons in...

For each of the following (real or imaginary) complexes determine a0 the number of electrons in the coordination sphere and b0 the oxidation state of the metal in each of the complexes.

a. (CO)₅Cr=C(OEt)Me

b. NbBr₃(PPh₃)₂(=CHCMe₃)

c. (Me₃CO)₃W(triple bond CCMe₃)

d. [(n⁵-C₅H₅)RuCl(Me)(PMe₃)]^-

e. [(n⁵-C₅H₅)Ir(PMe₃)(=CH₂)]⁺

Solutions

Expert Solution

(a) (CO)5Cr=C(OEt)Me

Electron inside the coordination sphere count

5 CO = 10 e, Cr = 6e, =C = 2e, Total electron 18, and oxidation state +2

(b) NbBr₃(PPh₃)₂(=CHCMe₃)

Electron count inside coordination sphere: Nb = 5e, 3Br = 3, 2PPh₃ = 4e, =C = 2 ; Total electron count = 14 e, Oxidation state of Nb is +5.

Electron count inside coordination sphere: 3e from 3(Me3CO), 6e from W, and 3e from triple bondCCMe₃ . Total electron count =12 e; Oxidation state of metal ion is +6.

(d) [(n⁵-C₅H₅)RuCl(Me)(PMe₃)]^-

Electron count inside coordination sphere: 5e from (n⁵-C₅H₅), 8e from Ru, 1e from Cl, 1e from Me, 2e from PMe₃ ; Total electron count = 17+1(from negative charge) =18 e; Oxidation state of metal ion is +2.

e. [(n⁵-C₅H₅)Ir(PMe₃)(=CH₂)]+

Electron count inside coordination sphere: 5 e from (n⁵-C₅H₅), 9e from Ir, 2e from PMe_3, 2e from (=CH₂ ). Total electron count = 18-1(positive charge) =17 , Oxidation state of metal ion is +4.

queen_honey_blossom answered 1 year ago

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