In the reaction of potassium dichromate, potassium chloride and sulfuric acid (conc.), the oxidation state of the chromium in the product is +

$$K_2Cr_2O_7 + 4KCl + 6H_2SO_4 \xrightarrow{\Delta} 2CrO_2Cl_2 + 6KHSO_4 + 3H_2O$$

Rule: In any neutral molecule, the sum of oxidation states equals zero. Let the oxidation state of chromium in $$CrO_2Cl_2$$ be $$x$$. Then
$$x + 2\times(\text{ox. state of O}) + 2\times(\text{ox. state of Cl}) = 0\quad-(1)$$

We know: Oxygen has oxidation state $$-2$$ and chlorine has oxidation state $$-1$$. Substituting these into $$(1)$$ gives:
$$x + 2(-2) + 2(-1) = 0$$
$$x = 6$$

Therefore, the oxidation state of chromium in the product $$CrO_2Cl_2$$ is +6.