JEE Advance - Chemistry (2017 - Paper 2 Offline - No. 15)
The order of the oxidation state of the phosphorous atom in
$${H_3}P{O_2},{H_3}P{O_4},{H_3}P{O_3}$$ and $${H_4}{P_2}{O_6}$$ is
$${H_3}P{O_2},{H_3}P{O_4},{H_3}P{O_3}$$ and $${H_4}{P_2}{O_6}$$ is
$${H_3}P{O_3} > {H_3}P{O_2} > {H_3}P{O_4} > {H_4}{P_2}{O_6}$$
$${H_3}P{O_4} > {H_3}P{O_2} > {H_3}P{O_3} > {H_4}{P_2}{O_6}$$
$${H_3}P{O_4} > {H_4}{P_2}{O_6} > {H_3}P{O_3} > {H_3}P{O_2}$$
$${H_3}P{O_2} > {H_3}P{O_3} > {H_4}{P_2}{O_6} > {H_3}P{O_4}$$
Explanation
Let oxidation states of phosphorus in H3PO2, H3PO4, H3PO3 and H4P2O6 be p, q, r and s respectively.
Oxidation state of hydrogen = + 1
Oxidation state of oxygen = $$-$$2
Thus, in H3PO2 :
3 $$\times$$ (+1) + p + 2 $$\times$$ ($$-$$2) = 0 $$\therefore$$ p = +1
In H3PO4 :
3 $$\times$$ (+1) + q + 4 $$\times$$ ($$-$$2) = 0 $$\therefore$$ q = +5
In H3PO3 :
3 $$\times$$ (+1) + r + 3 $$\times$$ ($$-$$2) = 0 $$\therefore$$ r = + 3
In H4P2O6 :
4 $$\times$$ (+1) + 2s + 6 $$\times$$ ($$-$$2) = 0 $$\therefore$$ s = +4
Thus, the order of oxidation state is :
H3PO4 > H4P2O6 > H3PO3 > H3PO2
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