Question

Consider the following complex ions: Cr(Cn)₆^3- (Metal to Ligand pi-bonding), CrCl₆^3- (Ligand to Metal pi-bonding), Cr(NH₃)₆³⁺ (no pi-bonding), and Cu(NH₃)₄²⁺ (tetrahedral, no pi-bonding)

A) For each complex ion, determine whether the HOMO is bonding, anitbonding, or nonbonding

B) In each case, what effect would a one electron oxidation have on the bond order?

Answer 1

[Cr(CN)6]^3-, Cr(NH₃)₆³⁺ CrCl₆3-  

Cr^3+ electronic configuration is [Ar]3d3

number of electrons from 6 CN-/Cl-/NH3 ligands = 12

Total electrons = 3+12 =15

According to MOT, electronic configuration of octahedral complex is written as

(a1g2)(t1u6)(eg4)(t2g3)(eg*)(a1g*0)(t1u*0)

HOMO is non-bonding (t2g3)

For Cu(NH₃)₄²⁺

Cu^2+ electronic configuration is [Ar]3d9

number of electrons from 4 NH3 ligands = 8

Total electrons = 9+8 =17

According to MOT, electronic configuration of tetrahedral complex is written as

(t2^6)(a1^2)(e^4)(t2*^5)(a1**0)(t2*0)

HOMO is anti-bonding (t2*^5)

B) If one electron is removed from octahedral complex from non-bonding orbital, bond order increases

If one electron is removed from tetrahedral complex from anti-bonding orbital, bond order decreases