1. Identify the complex and geometry

The complex is $[Pt(en)(NO_2)_2]$ where:

• $\text{en} =$ ethylenediamine, a bidentate ligand
• $Pt$ is typically square planar in such complexes
• There are two $NO_2^-$ ligands

So we have a square planar complex with one bidentate ligand occupying two adjacent positions, and two monodentate $NO_2^-$ ligands occupying the remaining two positions.

2. Check for geometrical isomerism

Since $en$ is bidentate, in a square planar complex it must occupy adjacent (cis) positions.

That leaves the two $NO_2$ ligands also adjacent to each other. Thus, there is only one geometrical arrangement possible.

So, geometrical isomerism does not increase the count here.

3. Check for linkage isomerism of $NO_2^-$

The ligand $NO_2^-$ is ambidentate:

• It can bind through N: nitro
• It can bind through O: nitrito

Since there are two $NO_2$ ligands, the possibilities are:

1. Both bonded through N: dinitro
2. Both bonded through O: dinitrito
3. One through N and one through O: nitro-nitrito

These are distinct linkage isomers.

Thus, total number of isomers = $3$

4. Evaluate options

• A: 3 ✅
• B: 1 ❌
• C: 4 ❌
• D: 2 ❌

5. Final answer

The number of isomers possible for $[Pt(en)(NO_2)_2]$ is $\boxed{3}$