MCQ
Hypo is used in photography because it is:
- AA strong reducing agent
- BA strong oxidizing agent
- ✓A strong Complexing agent
- DPhoto sensitive Compound
✓
Answer
Correct option: C.
A strong Complexing agent
Hypo forms the complex $\left[\mathrm{Ag}\left(\mathrm{S}_2 \mathrm{O}_3\right)_2\right]^{3-}$ from insoluble $Ag.$
$\left[\mathrm{Ag}\left(\mathrm{S}_2 \mathrm{O}_3\right)_2\right]^{3-}$ is formed by treatment of $\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 \cdot 5 \mathrm{H}_2 \mathrm{O}$ with $\text{AgBr}.$
Therefore, Hypo is used in photography because it is a strong Complexing agent.
$\left[\mathrm{Ag}\left(\mathrm{S}_2 \mathrm{O}_3\right)_2\right]^{3-}$ is formed by treatment of $\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 \cdot 5 \mathrm{H}_2 \mathrm{O}$ with $\text{AgBr}.$
Therefore, Hypo is used in photography because it is a strong Complexing agent.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Detergents are sodium salts of a long chain:
Ascorbic acid is a
$K_f$ for water is $1.86\, K\, kg\, mol^{-1} .$ If your automobile radiator holds $1.0\, kg$ of water, .......... $gm$ of ethylene glycol $(C_2H_6O_2)$ must you add to get the freezing point of the solution lowered to $-2.8\,^o C$.
→Peptization is a process of:
The mole fraction of a solvent in aqueous solution of a solute is $0.8$. The molality (in $mol\, kg^{-1}$) of the aqueous solution is
→
What would be the product formed when $1\,-$ Bromo $-3-$ chloro cyclobutane reacts with two equivalents of metallic sodium in ether
→Scheme $1$ and $2$ describe the conversion of $P$ to $Q$ and $R$ to $S$, respectively. Scheme $3$ describes the synthesis of $T$ from $Q$ and $S$. The total number of $Br$ atoms in a molecule of $T$ is. . . . . . . .
→Scheme $1$ : (image)
Scheme $2$ :(image)
Scheme $3$ :(image)
Which of the following is correct
→Penta acetate of glucose $\xrightarrow[{\operatorname{Re} agent}]{{Tollen's}}$ $+\,ve$ test $(A)$
Penta acetate of glucose $\xrightarrow{{Benedicts}}$ $+\,ve$ test $(B)$
Penta acetate of glucose $\xrightarrow[{\operatorname{Re} agent}]{{Schiffs}}$ $-\,ve$ test $(C)$
Penta acetate of glucose $\xrightarrow{{NaHS{O_3}}}$ $-\,ve$ test $(D)$
Glucose, on oxidation with $Br_2(aq)$ gives
→Which one of the following statements is true for protein synthesis (translation)