Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

D and F Block Elements question

2023 · 31 Jan · Shift 1 · Q1
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /D and F Block Elements
  5. /2023 · 31 Jan · Shift 1 · Q1

D and F Block Elements question

2023 · 31 Jan · Shift 1 · Q1

JEE MainChemistryD and F Block ElementsMCQ+4 / −1
When Cu2+\mathrm{Cu}^{2+}Cu2+ ion is treated with KI\mathrm{KI}KI, a white precipitate, X\mathrm{X}X appears in solution. The solution is titrated with sodium thiosulphate, the compound Y\mathrm{Y}Y is formed. X\mathrm{X}X and Y\mathrm{Y}Y respectively are :
  1. A
    X=CuI2\mathrm{X=CuI_2}X=CuI2​ Y=Na2S4O6\mathrm{Y=Na_2S_4O_6}Y=Na2​S4​O6​
  2. B
    X=Cu2I2\mathrm{X=Cu_2I_2}X=Cu2​I2​ Y=Na2S4O6\mathrm{Y=Na_2S_4O_6}Y=Na2​S4​O6​
  3. C
    X=CuI2\mathrm{X=CuI_2}X=CuI2​ Y=Na2S2O3\mathrm{Y=Na_2S_2O_3}Y=Na2​S2​O3​
  4. D
    X=Cu2I2\mathrm{X=Cu_2I_2}X=Cu2​I2​ Y=Na2S4O5\mathrm{Y=Na_2S_4O_5}Y=Na2​S4​O5​
View written solutionFree

Correct answer: B

  1. Reaction of Cu2+\mathrm{Cu^{2+}}Cu2+ with KI\mathrm{KI}KI

When Cu2+\mathrm{Cu^{2+}}Cu2+ is treated with iodide ions, iodide reduces Cu2+\mathrm{Cu^{2+}}Cu2+ to Cu+\mathrm{Cu^+}Cu+ and itself gets oxidized to iodine.

The reaction is:

2Cu2++4I−→2CuI↓+I22\mathrm{Cu^{2+}} + 4\mathrm{I^-} \rightarrow 2\mathrm{CuI}\downarrow + \mathrm{I_2}2Cu2++4I−→2CuI↓+I2​

Now, CuI\mathrm{CuI}CuI is a white precipitate. It may also be written as Cu2I2\mathrm{Cu_2I_2}Cu2​I2​ in dimeric form.

So,

X=CuI=Cu2I2X = \mathrm{CuI} = \mathrm{Cu_2I_2}X=CuI=Cu2​I2​
  1. Titration of liberated iodine with sodium thiosulphate

The iodine formed is titrated with sodium thiosulphate:

I2+2Na2S2O3→2NaI+Na2S4O6\mathrm{I_2} + 2\mathrm{Na_2S_2O_3} \rightarrow 2\mathrm{NaI} + \mathrm{Na_2S_4O_6}I2​+2Na2​S2​O3​→2NaI+Na2​S4​O6​

Thus, the product formed is sodium tetrathionate:

Y=Na2S4O6Y = \mathrm{Na_2S_4O_6}Y=Na2​S4​O6​
  1. Match with options
  • Option A: X=CuI2X=\mathrm{CuI_2}X=CuI2​, Y=Na2S4O6Y=\mathrm{Na_2S_4O_6}Y=Na2​S4​O6​ → incorrect because CuI2\mathrm{CuI_2}CuI2​ is not the white precipitate formed.
  • Option B: X=Cu2I2X=\mathrm{Cu_2I_2}X=Cu2​I2​, Y=Na2S4O6Y=\mathrm{Na_2S_4O_6}Y=Na2​S4​O6​ → correct.
  • Option C: X=CuI2X=\mathrm{CuI_2}X=CuI2​, Y=Na2S2O3Y=\mathrm{Na_2S_2O_3}Y=Na2​S2​O3​ → incorrect.
  • Option D: X=Cu2I2X=\mathrm{Cu_2I_2}X=Cu2​I2​, Y=Na2S4O5Y=\mathrm{Na_2S_4O_5}Y=Na2​S4​O5​ → incorrect.

Therefore, the correct answer is:

B\boxed{\text{B}}B​
PreviousNext

More from D and F Block Elements

  • Nd2+ = ​2023 · MCQ
  • The correct order of basicity of oxides of vanadium is :2023 · MCQ
  • The difference in oxidation state of chromium in chromate and dichromate salts is ​.2022 · Numerical
  • Manganese (VI) has ability to disproportionate in acidic solution. The difference in oxidation states of two ions it forms in acidic solution is ​.2022 · Numerical
  • Metals generally melt at very high temperature. Amongst the following, the metal with the highest melting point will be :2022 · MCQ
  • Among Co3+, Ti2+, V2+ and Cr2+ ions, one if used as a reagent cannot liberate H2​ from dilute mineral acid solution, its spin-only magnetic moment in gaseous state is ​ B.M. (Nearest integer)2022 · Numerical
  • Cerium (IV) has a noble gas configuration. Which of the following is correct statement about it?2022 · MCQ
  • Among the following, which is the strongest oxidizing agent?2022 · MCQ