Which of the following chemical reaction(s) is (are) involved in the developing of photographic plate?

C_{6} H_{4} (O H)_{2} + 2 A g B r \to 2 A g + C_{6} H_{4} O_{2} + 2 H B r

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A g B r + 2 N a_{2} S_{2} O_{3} \to N a_{3} [A g S_{2} O_{3})_{2}] + N a B r

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A g B r + 2 N H_{3} (a q) \to [A g (N H_{3})_{2}] B r

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2 A g B r + N a_{2} S_{2} O_{3} \to A g_{2} S_{2} O_{3} + 2 N a B r

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Solution

The correct options are
B $C_{6} H_{4} (O H)_{2} + 2 A g B r \to 2 A g + C_{6} H_{4} O_{2} + 2 H B r$
D $A g B r + 2 N a_{2} S_{2} O_{3} \to N a_{3} [A g S_{2} O_{3})_{2}] + N a B r$
Development of the latent image in photography:
To make the latent image visible, the film is passed through a reducing bath containing hydroquinone (an organic reducing agent). More $A g B r$ is reduced and the rate of reduction depends on the intensity of illumination during exposure. Thus, parts of the film which were most strongly illuminated become the darkest.
$2 A g B r + C_{6} H_{4} (O H)_{2} \to 2 A g + 2 H B r + C_{6} H_{4} O_{2}$
This process is carried out in dark to prevent general darkening of the picture. Here the image becomes visible, but the shade is negative in relation to that of the object, i.e., the brighten parts will be darker and vice-versa. Hence, it is called negative plate or film. But at this stage, we cannot expose it to daylight.
Fixing:
To make the negative film to be handled in daylight, the film is dipped in a solution of sodium thiosulphate (hypo). It dissolves away the unreacted AgBr as a complex but leaves the metallic silver unchanged.
$A g B r + 2 N a_{2} S_{2} O_{3} \to N a_{3} [A g (S_{2} O_{3})_{2}] + N a B r$
Since no $A g B r$ remains in the film it can be handled in daylight.

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