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Sodium Thiosulphate

Consider the ...

Question

Consider the following statements and arrange in the order of true/false as given in the codes.
$S_{1} : 2 S e_{2} C l_{2} \to S e C l_{4} + 3 S e$
$S_{2}$ : Dioxides like $M n O_{2}, P b O_{2}$ do not form $H_{2} O_{2}$ with dilute acids but they evolve $O_{2}$ with conc. $H_{2} S O_{4}$ .
$S_{3}$ : Sodium thiosulphate with $F e C l_{3}$ solution develops a pink or violet colour which soon vanishes.
$S_{4}$ : White precipitate of $P b S_{2} O_{3}$ gets soluble when boiled with water.

F T T T

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T F T T

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F F T T

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T T T F

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Solution

The correct option is C T T T F
$S_{1} : S e_{2} C l_{2} \to S e C l_{4} + 3 S e$ (disproportionation)
$S_{2} : 2 P b O_{2} + 2 H_{2} S O_{4} \to 2 P b S O_{4} + 2 H_{2} O + O_{2}$
$S_{3} : F e^{3 +} + 2 S_{2} O_{3}^{2 -} \to [F e (S_{2} O_{3})_{2}]^{-}; F e^{+ 3} + [F e (S_{2} O_{3})_{2}]^{-}$ (pink or violet) $\to 2 F e^{2 +} + S_{4} O_{6}^{2 -} (c o l o u r l e s s)$
$S_{4} : P b S_{2} O_{3} + H_{2} O Δ - \to P b S ↓ (b l a c k) + H_{2} S O_{4}$ .

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Similar questions

Q. Consider the following statements and arrange in the order of true/false as given in the codes (T/F) :

S_{1}

S r^{2 +}

ions in very dilute solution do not give yellow precipitate with

C r O_{4}^{2 -}

ions.

S_{2}

C a^{2 +}

ions give a yellow precipitate in presence of ammonia with potassium ferrocyanide solution.

S_{3}

M g^{2 +}

ions in solution gives a gelatinous white precipitate of magnesium hydroxide with ammonia solution and the precipitate is readily soluble in solution and aqua regia.

S_{4}

H g S

forms soluble complex in excess of sulphide.

Q. Consider the following statements and arrange in the order of true/false as given in the codes.

S_{1}

: Like

C O_{3}^{- 2}, S O_{3}^{2 -}

ion also gives test with baryta water,

B a (O H)_{2}

S_{2}

A g_{2} S O_{3}

is insolube in dilute

H N O_{3}

S_{3}

: Triodide ions

(I_{3}^{2 -})

produced by the reaction of

C u^{2 +}

and KI solution are not reduced by excess of sodium thiosulphate solution.

S_{4}

: Black precipitate of

C u (S C N)_{2}

turns into white precipitate, when it reacts with saturated solution of

S O_{2}

in water.

Q. Consider the following statements and arrange in the order of true/false as given in the codes.

S_{1}

: The addition of chlorine water dropwise to a solution of a bromide liberates free bromine, which colours the solution orange red.

S_{2}

: Chlorine gas bleaches the filter paper moistened with yellow dyestuff, fluorescein.

S_{3}

: Nitrogen gas is evolved when a solution of nitrate is boiled with zinc dust and sodium hydroxide solution.

Q. Consider the following statements and arrange in the order of true/false as given in the codes.

S_{1}

: When copper sulphate is added to

K_{2} [H g l_{4}]

solution and

S O_{2}

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S_{2}

: When copper sulphate solution is boiled with glucose in presence of caustic soda, a blue precipitate is obtained.

S_{3}

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C u_{2} C l_{2}

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Q. Consider the following statements and arrange in the order of true/false as given in the codes.

S_{1} : N_{2} O

with sodium metal in liquid ammonia forms sodium azide and nitrogen gas is liberated.

S_{2}

: Ammonium dichromate on heating decomposes to give nitrogen and a green coloured compound.

S_{3}

C a N C N

on hydrolysis produces a white precipitate and a gas which turns filter paper moistened with copper sulphate solution deep blue.

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The correct option is C T T T FS1:Se2Cl2→SeCl4+3Se (disproportionation)S2:2PbO2+2H2SO4→2PbSO4+2H2O+O2S3:Fe3++2S2O2−3→[Fe(S2O3)2]−;Fe+3+[Fe(S2O3)2]− (pink or violet) →2Fe2++S4O2−6(colourless)S4:PbS2O3+H2OΔ−→PbS↓(black)+H2SO4.