The oxidation state of Cr in $$[\text{Cr}(\text{NH}_3)_4 \text{Cl}_2]^+$$ is

Let the oxidation state of chromium be $$x$$.

Rules used:
1. $$\text{NH}_3$$ is a neutral ligand ⇒ contributes $$0$$ to charge.
2. $$\text{Cl}^-$$ is a monodentate anionic ligand ⇒ contributes $$-1$$ each.

In the complex ion $$\left[\text{Cr}(\text{NH}_3)_4\text{Cl}_2\right]^+$$:
  • Number of $$\text{NH}_3$$ ligands = 4 ⇒ total charge $$0$$.
  • Number of $$\text{Cl}^-$$ ligands = 2 ⇒ total charge $$2(-1) = -2$$.
  • Overall charge on the complex ion = $$+1$$.

Writing the oxidation-state balance:
$$x + 0 + (-2) = +1$$

Solving, $$x = +3$$.

Therefore, the oxidation state of chromium is $$+3$$.

Option A which is: +3