Identify the complex type

The complex is $[\mathrm{Cu}(\mathrm{en})_2(\mathrm{SCN})_2]$ where:

• $\mathrm{en}$ = ethylenediamine, a bidentate ligand
• $\mathrm{SCN}^-$ = thiocyanate, an ambidentate ligand

Coordination number of Cu is $6$:

• two $\mathrm{en}$ ligands contribute $4$ donor atoms
• two $\mathrm{SCN}^-$ ligands contribute $2$ donor atoms

So the complex is octahedral.

Geometrical arrangement of $[M(AA)_2B_2]$

For an octahedral complex of type $[M(AA)_2B_2]$ where $AA$ is bidentate and $B$ is monodentate, two geometrical isomers are possible:

• cis
• trans

Thus, ignoring linkage for the moment, we get 2 geometrical possibilities.

Optical activity consideration

• The cis form of $[M(AA)_2B_2]$ is optically active and exists as a pair of enantiomers: $\Delta$ and $\Lambda$.
• The trans form has a plane/center of symmetry and is optically inactive.

Therefore, for a normal monodentate ligand $B$, total stereoisomers would be: $2 \text{ (from cis optical pair)} + 1 \text{ (trans)} = 3$

Effect of ambidentate ligand $\mathrm{SCN}^-$

Thiocyanate can bind through:

• $S$ : thiocyanato-$S$
• $N$ : isothiocyanato-$N$

However, the question asks for relatively more stable isomer(s).

For $\mathrm{Cu}^{2+}$, a borderline-soft metal ion, the more stable bonding mode with $\mathrm{SCN}^-$ is generally through nitrogen rather than sulfur in such coordination compounds considered at JEE level. So we count only the more stable linkage form, not all linkage isomers.

Hence linkage isomerism does not increase the count here under the phrase “relatively more stable isomer(s)”.

Total relatively more stable isomers

Therefore, the stable stereoisomers are:

• cis-$\Delta$
• cis-$\Lambda$
• trans

So total number is $3$

Comparison with stored answer

Derived answer = $3$

Stored correct answer = $3$

Hence they agree.