Un-decomposed silver bromide from developed film, in black and white photography, makes a complex with hypo. The formula of the complex is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \([Ag(S_2O_3)_2]^{3-}\)
In black and white photography, the silver bromide that light did not decompose has to be removed from the developed film. This is done by washing with hypo, sodium thiosulphate, whose \( \ce{S2O3^2-} \) ions form a soluble complex with silver ion:
\( \displaystyle \ce{AgBr + 2Na2S2O3 -> Na3[Ag(S2O3)2] + NaBr} \)
The charge on the complex ion follows from its parts:
\( \displaystyle (+1) + 2 \times (-2) = -3 \)
So the complex is \( \ce{[Ag(S2O3)2]^3-} \), the dithiosulphatoargentate(I) ion.
Why the other options are incorrect
Option (2) \([Ag(S_2O_3)_2]^{2-}\) — Incorrect. The ligands are right but the charge is wrong: one \( \ce{Ag^{+}} \) and two \( \ce{S2O3^2-} \) give −3.
Option (3) \([Ag_2(S_2O_8)_2]^{2-}\) — Incorrect. \( \ce{S2O8^2-} \) is the peroxodisulphate ion, a strong oxidising agent, not the thiosulphate ion of hypo.
Option (4) \([Ag(S_2O_8)_2]^{3-}\) — Incorrect. This again has peroxodisulphate instead of thiosulphate; hypo does not contain \( \ce{S2O8^2-} \).
NCERT (2026-27 edition):
Importance and Applications of Coordination Compounds – …
