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NCERT Exemplar · Q22

Q.Calculate the oxidation number of phosphorus in the following species.

(a) HPO3^2- and
(b) PO4^3-
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Oxidation numbers are assigned using electronegativity rules: oxygen is -2, hydrogen is +1, and the sum of oxidation numbers equals the ion's charge. For (a) HPO32−\text{HPO}_3^{2-}, P is +3; for (b) PO43−\text{PO}_4^{3-}, P is +5.

The idea behind oxidation numbers is a bookkeeping system. We pretend every bond is ionic, giving the shared electrons to the more electronegative atom. This lets us track electron "ownership" and see how many electrons an atom has gained or lost relative to its elemental state. For phosphorus in oxyanions, oxygen always pulls electrons away (it's very electronegative), and hydrogen, being less electronegative than phosphorus, gives its electron away. The net charge on the ion then tells us the phosphorus's oxidation state.

Let's apply this systematically.

  1. Set up the equation for (a) HPO32−\text{HPO}_3^{2-}.

    Let the oxidation number of phosphorus be xx. We know the standard values: hydrogen is +1+1, oxygen is −2-2. The ion has a total charge of −2-2. The sum of all oxidation numbers must equal this charge.

  2. Write and solve the equation.

    There is 1 hydrogen atom, 1 phosphorus atom, and 3 oxygen atoms.

(+1)+(x)+3(−2)=−2(+1) + (x) + 3(-2) = -2

1+x−6=−21 + x - 6 = -2

x−5=−2x - 5 = -2

x=+3x = +3

  1. Now for (b) PO43−\text{PO}_4^{3-}. Again, let phosphorus be xx. Oxygen is −2-2, and there are 4 oxygen atoms. The ion charge is −3-3.

x+4(−2)=−3x + 4(-2) = -3

x−8=−3x - 8 = -3

$$x = +5$$ …

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