What is the oxidation state of cobalt in #"Na"["Co"("NH"_2)("en")("NO"_2)("S"_2"O"_3)]#?

Cobalt is a #+3# oxidation state.

I assume you mean of #"Co"# atom. You'll have to have already memorized these ligands.

• #"NH"_2^(-)#, the amido ligand, binds as #"H"_2"N"=# with a #-1# charge.
• #"en"# is ethylenediamine, a bidentate ligand that is neutral and connects cis to the metal.
• #"NO"_2^(-)#, the nitro ligand, may bind via one oxygen, or possibly via both (likely one), and is a #-1# charge.
• #"S"_2"O"_3^(2-)#, the thiosulfato ligand, is known from general chemistry. It can actually bind at the sulfur, monodentate.

Regardless of the mono/polydentate nature of these ligands, what we care about in this case is their contributed charge. The oxidation state of cobalt can then be found by conservation of charge.

#"OS"_(Co) + "OS"_(NH_2^-) + "OS"_(en) + "OS"_(NO_2^(-)) + "OS"_(S_2O_3^(2-)) = -1#

since sodium forms a #+1# cation. Therefore:

#"OS"_(Co) - 1 + 0 - 1 - 2 = -1#

#color(blue)("OS"_(Co) = +3)#

This makes cobalt a #d^6# metal, in this compound. This compound is called sodium amidoethylenediaminenitrothiosulfatocobaltate(III).