Q. Consider the following reactions

Na₂B₄O₇  →_Δ  2X + Y

CuSO₄ + Y  →_{Non-Luminous flame}  Z + SO₃

2Z + 2X + Carbon  →_{Luminous flame}  2Q + Na₂B₄O₇ + CO

The oxidation states of Cu in Z and Q, respectively are :

Correct Answer: +2 and +1

Solution

On heating borax:

Na₂B₄O₇ → 2NaBO₂ + B₂O₃

Hence,

X = NaBO₂,   Y = B₂O₃

In non-luminous flame, copper sulphate reacts with boron trioxide to form copper metaborate:

CuSO₄ + B₂O₃ → Cu(BO₂)₂ + SO₃

Thus,

Z = Cu(BO₂)₂,   Cu is in +2 oxidation state

In luminous flame, carbon reduces Cu(II) to Cu(I) forming cuprous oxide:

Cu(BO₂)₂ → Cu₂O

Hence,

Q = Cu₂O,   Cu is in +1 oxidation state

Therefore, the oxidation states of Cu in Z and Q are +2 and +1.

Related JEE Main Questions