The possible number of geometrical isomers for the complex [CoL2Cl2 ] - (L = H2NCH2CH2O- ) is (are)

Answer: 5 geometrical isomers

L = \(\mathrm{H_2NCH_2CH_2O^-}\) is an unsymmetrical bidentate ligand (it binds through N and O; call them A and B). The complex is octahedral, of the type \(\mathrm{[M(AB)_2X_2]}\) with X = Cl. (Charge check: Co is +3; \(+3 - 2 - 2 = -1\).) Each chelate must span two cis positions.

Case 1: the two Cl trans to each other. Both chelates lie in the square plane. The two N atoms can be trans to each other (then the two O atoms are also trans), or cis to each other. That gives 2 isomers.

Case 2: the two Cl cis to each other. Each Cl has one donor atom trans to it, and the third trans pair is made of the remaining two donor atoms. The possibilities are:

• both N trans to Cl, and O trans to O;
• both O trans to Cl, and N trans to N;
• one N and one O trans to Cl, and N trans to O.

That gives 3 isomers. (The cis forms are chiral, so the number of stereoisomers including optical isomers is larger, but the question asks only for geometrical isomers.)

Total geometrical isomers = 2 + 3 = 5.