In black and white photography, the developed film is fixed by washing it with hypo (sodium thiosulphate) solution. Which of the following equations correctly represents the reaction of hypo with the undecomposed silver bromide?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \ce{AgBr + 2Na2S2O3 -> Na3[Ag(S2O3)2] + NaBr} \)
Silver bromide does not dissolve in water. In fixing, the hypo solution dissolves the AgBr that was not decomposed on the film by forming a soluble complex ion, [Ag(S2O3)2]3−, in which two thiosulphate ions are ligands on Ag+.
Charge on the complex: Ag is +1 and each S2O32− is −2, so +1 + 2(−2) = −3; three Na+ ions balance it, giving Na3[Ag(S2O3)2].
Two Na2S2O3 supply the two ligands and four Na+: three go with the complex and one with Br− as NaBr.
\( \displaystyle \ce{AgBr(s) + 2Na2S2O3(aq) -> Na3[Ag(S2O3)2](aq) + NaBr(aq)} \)
The complex is washed away, so no light-sensitive AgBr is left on the film.
Why the other options are incorrect
Option (1) \( \ce{AgBr + Na2S2O3 -> Na[Ag(S2O3)] + NaBr} \) — Incorrect. The complex formed has two thiosulphate ligands, [Ag(S2O3)2]3−, not one.
Option (3) \( \ce{2AgBr + Na2S2O3 -> Ag2S2O3 + 2NaBr} \) — Incorrect. This is a simple exchange of ions with no complex formed; silver bromide is removed only because it becomes the soluble complex ion.
Option (4) \( \ce{2AgBr + 2Na2S2O3 -> 2Ag + Na2S4O6 + 2NaBr} \) — Incorrect. This shows silver bromide being reduced to metallic silver. In fixing, silver is not reduced: it stays in the +1 state in the complex and goes into solution.
NCERT (2026-27 edition):
5.7 Importance and Applications of Coordination Compounds – “In black and white photography, the developed film is fixed by washing with hypo solution which dissolves the undecomposed AgBr to form a complex ion, [Ag(S2O3)2]3–.”