The number of isomers possible for $$[Pt(en)(NO_2)_2]$$ is:

We need to find the number of isomers possible for $$[Pt(en)(NO_2)_2]$$.

Here, $$Pt$$ is platinum(II), which forms square planar complexes with coordination number 4. The ligands are:

1. $$en$$ (ethylenediamine) — a bidentate ligand that occupies two adjacent coordination sites.

2. $$NO_2^-$$ (nitrite ion) — an ambidentate ligand, meaning it can bind to the metal through different atoms.

Since $$en$$ is bidentate and occupies two adjacent positions in the square planar geometry, the two $$NO_2^-$$ ligands must occupy the remaining two adjacent positions. There is only one possible geometric arrangement (no cis/trans isomerism is possible because $$en$$ forces adjacent coordination).

However, $$NO_2^-$$ is ambidentate — it can coordinate through:

(a) The nitrogen atom: this is called the nitro ($$-NO_2$$) linkage.

(b) The oxygen atom: this is called the nitrito ($$-ONO$$) linkage.

With two ambidentate $$NO_2^-$$ ligands, the possible linkage isomers are:

1. Both $$NO_2^-$$ bind through nitrogen: $$[Pt(en)(NO_2)_2]$$ (dinitro)

2. Both $$NO_2^-$$ bind through oxygen: $$[Pt(en)(ONO)_2]$$ (dinitrito)

3. One binds through nitrogen and the other through oxygen: $$[Pt(en)(NO_2)(ONO)]$$ (nitro-nitrito)

Therefore, there are 3 isomers in total.

The answer is Option D: 3.