Consider the following statements about chromates and dichromates.(A)…

The correct answer is Option (1) → (B) and (D) only

Potassium dichromate is a strong oxidising agent. In acidic solution chromium is reduced from +6 to +3, and the reducing agent gives up the electrons:

    \( \displaystyle \ce{Cr2O7^2- + 14H+ + 6e^- -> 2Cr^3+ + 7H2O} \)

(A) The chromate ion is tetrahedral, whereas the dichromate ion consists of two tetrahedra sharing one corner. — Correct statement. In Cr₂O₇²⁻ two CrO₄ tetrahedra share one oxygen atom, with a Cr–O–Cr bond angle of 126°.

(B) In acidic solution, each dichromate ion takes up three electrons when it is reduced to Cr³⁺ ions. — Incorrect statement. Each dichromate ion takes up six electrons. It has two chromium atoms, and each goes from +6 to +3 (three electrons each): 2 × 3 = 6.

(C) Acidified potassium dichromate oxidises tin(II) salts to tin(IV). — Correct statement. Three Sn²⁺ ions supply the six electrons taken up by one dichromate ion:

    \( \displaystyle \ce{3Sn^2+ -> 3Sn^4+ + 6e^-} \)

(D) Acidified potassium dichromate oxidises hydrogen sulphide to sulphate ions. — Incorrect statement. Hydrogen sulphide is oxidised only to sulphur (sulphur goes from −2 to 0), not to sulphate:

    \( \displaystyle \ce{3H2S -> 6H+ + 3S + 6e^-} \)

NCERT (2026-27 edition):
4.4.1 Oxides and Oxoanions of Metals – “…The chromate ion is tetrahedral whereas the dichromate ion consists of two tetrahedra sharing one corner with Cr–O–Cr bond angle of 126°.” … “In acidic solution, its oxidising action can be represented as follows: Cr₂O₇^2– + 14H⁺ + 6e^– → 2Cr³⁺ + 7H₂O (E^o = 1.33V) Thus, acidified potassium dichromate will oxidise iodides to iodine, sulphides to sulphur, tin(II) to tin(IV) and iron(II) salts to iron(III).”