Exercise 8 | The Chemistry of the Transition Metals

Predict the number of unpaired electrons for the following octahedral complex ions:


[Cr(CN)6]4-

[FeF6]3-

[Rh(NH3)6]3+ (low spin)

[Cr(CN)6]4-: oxidation state of Cr = charge – 6 x oxid. state CN- = - 4 + 6 = + 2

Cr (II) complex: [Ar] 3d4 ⇒ 4 d-electrons

CN- ⇒ low spin ⇒ electrons will first fill the t2g orbitals before occupying the higher-energy eg orbitals ⇒ t2g4 eg0

⇒ 2 unpaired electrons


[FeF6]3-: oxidation state of Fe = charge – 6 x oxid. state F- = - 3 + 6 = + 3

Fe (III) complex: [Ar] 3d5 ⇒ 5 d-electrons

F- ⇒ high spin ⇒ d electrons occupy the eg orbitals before they pair up in the t2g orbitals ⇒ t2g3 eg2

⇒ 5 unpaired electrons


[Rh(NH3)6]3+: oxidation state of Rh = charge – 6 x oxid. state of NH3 = + 3 – 0 = + 3

Rh (III) complex: [Kr] 4d6 ⇒ 6 d-electrons

low spin ⇒ electrons will first fill the t2g orbitals before occupying the higher-energy eg orbitals ⇒ t2g6 eg0

⇒ no unpaired electrons