(i) Sodium salt of an acid (A) is formed on boiling white phosphorus with NaOH solution.
(ii) On passing chlorine through phosphorus kept fused under water, another acid (B) is formed.
(iii) Phosphorus on treatment with concentrated `HNO_(3)` given an acid (C) which is also formed by the action of dilute `H_(2)SO_(4)` on powdered phosphorite rock.
(iv) (A) on treatment with a solution of `HgCl_(2)` first gives a white precipitate of compound (D) and then a grey precipitate of (E). Identify (A) to (E) and write balanced chemical equation for the reactions at steps (i) to (iv)

Answer

(A) `NaH_(2) PO_(2)`; (B) `H_(3)PO_(3)`; (C) `H_(3)PO_(4)` (D) `Hg_(2)Cl_(2)` (E) Hg

Explanation

The given change are :
(i) `P_(4) + 3NaOH + 3H_(2)O rarr 3NaH_(2)PO_(2)` (sodium hypophosphite) `+ PH_(3)`
Thus acid (A) is `H_(3)PO_(2)` i.e., hypophosphorus acid.
(ii) `2P + 3Cl_(2) + 6H_(2)O rarr 2H_(3) PO_(3)` (phosporic acid) `+6HCl`
Thus, acid (B) is `H_(3)PO_(3)`
(iii)`P_(4) + 20 HNO_(2) rarr 4H_(3) PO_(4) (C) + 20NO_(2) + 4H_(2)O`
`P_(4) + 10H_(2)SO_(4) rarr 4H_(3)PO_(4) (C) ("phosporic acid") + 10 SO_(2) + 4H_(2)O`
Thus, acid (C) is `H_(3)PO_(4)`
(iv) `H_(3)PO_(2) + 2H_(2)O rarr H_(3)PO_(4) + 4H`
`HgCl_(2) + 2H rarr Hg_(2) Cl_(2) (D) ("white") + 2HCl , Hg_(2)Cl_(2) + 2H rarr 2Hg (E) ("grey") + 2HCl`