Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Consider octahedral . The nitrite can bond through N or O. The thiocyanate can bond through S or N. How many linkage isomers are there, if you ignore cis and trans? How many isomers in all, when the cis and trans forms of each are counted too?
Cis ligands are apart and trans ligands apart. Tetrahedral complexes have no geometrical isomers. Octahedral has fac and mer forms. To count the isomers of a monodentate octahedral complex, write the ligands as three trans pairs and count the different sets. The net dipole of two bonds with dipole at an angle is .
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Linkage. Nitrite has ways to bond and thiocyanate has . The choices multiply: linkage isomers.
Step 2. Geometry. The complex is of the type , which has geometrical isomers: cis and trans.
Step 3. All together. Each linkage isomer can be cis or trans: .
Answer: linkage isomers and isomers in all.
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