JEE Main 2018 — Coordination Compounds Question with Solution

Assume oxidation state of Cr in all the compounds = x

(i) In [Cr(H₂O)₆] Cl₃ oxidation state of Cr is

x + 0 $$\times$$ 6 + ($$-$$1 $$\times$$ ) = O

$$\Rightarrow$$ x + 0 $$-$$ 3 = O

$$\Rightarrow$$ x = + 3

(ii) [Cr (C₆ H₆)₂] oxidation state of Cr is

x + 0 $$\times$$ 2 = 0

$$\Rightarrow$$ x = 0

(iii) In K₂ [ Cr(CN)₂ (O)₂ (O₂) (NH₃)] oxidation state of Cr is

1 $$\times$$ 2 + x + ($$-$$ 1 $$\times$$ 2) + ($$-$$2 $$\times$$ 2) + ($$-$$2) + 0 = 0

$$\Rightarrow$$ x = + 6

$$\therefore$$ + 3, 0 and + 6 is the correct answer.

Note :

O₂ molecule can have 0, $$-$$ 1, $$-$$ 2 oxidation state but in K₂ [ Cr (CN)₂ (O)₂ (O₂) NH₃ ] if we choose zero as the oxidation state of O₂ then for Cr oxidation state will be $$+$$ 4. But + 4 oxidation state of Cr is unstable and $$+$$ 6 is most stable that is why we choose $$-$$ 2 oxidation state of O₂.