An iodide of -A- element A 2 I n is oxidized to IO 3-and AO- by MnO 4-in acidic medium- If 5 - 10-3 mol of A 2 I n requires 700 mL of 0-02 M KM nO 4 solution for complete oxidation- which of the following statements is-are correct-A- The value of n-2B- Empirical formula of iodide is AIC- 0-15 mole of A 2 I n would require 0-3 mole of K 2 Cr 2 O 7 solution in acidic medium if its oxidised products are IO 3- AOD- - A - may be a metal from first group of periodic table

The correct options are A The value of n=2 B Empirical formula of iodide is AIA_2I_n+MnO_4^ IO_3^ +AO+Mn^2+ Here oxidation of both A and I is taking place so n ...

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Byju's Answer

Standard XII

Chemistry

Stoichiometric Calculations

An iodide of ...

Question

An iodide of " $A$ " element $(A_{2} I_{n})$ is oxidized to $I O_{3}^{-}$ and $A O$ , by $M n O_{4}^{-}$ in acidic medium. If $5 \times 10^{- 3} mol$ of $A_{2} I_{n}$ requires $700 mL$ of $0.02 M K M n O_{4}$ solution for complete oxidation, which of the following statement(s) is/are correct?

The value of

n = 2

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Empirical formula of iodide is

A I

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0.15

mole of

A_{2} I_{n}

would require

0.3

mole of

K_{2} C r_{2} O_{7}

solution in acidic medium if its oxidised products are

I O_{3}^{-}

A O

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^{''} A^{''}

may be a metal from first group of periodic table

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Solution

The correct options are
A The value of $n = 2$
B Empirical formula of iodide is $A I$
$A_{2} I_{n} + M n O_{4}^{-} H^{+} - - \to I O_{3}^{-} + A O + M n^{2 +}$
Here oxidation of both $A$ and $I$ is taking place so n-factor ( $n_{f a c t o r}$ ) of $A_{2} I_{n}$ is sum of the two.
oxidation state of $A$ in $A_{2} I_{n} = n / 2$
$n_{f a c t o r} =$ (change in oxidation state) x (total number of atoms) so,
$n_{f} o f A_{2} I_{n} = n_{f} (I) + n_{f} (A) = 6 \times 2 + (2 - \frac{n}{2}) \times 2 = 16 - n$
and
$n_{f}$ of $M n O_{4}^{-} = 5$
equating the number of equivalents we get:
${equivalents}_{A_{2} I_{n}} = {equivalents}_{M n O_{4}^{-}}$

$moles \times n_{f a c t o r} (A_{2} I_{n}) = moles \times n_{f a c t o r} (M n O_{4}^{-})$

$5 \times 10^{- 3} \times (16 - n) = \frac{700 \times 0.02 \times 5}{1000}$
$n = 2$

so, option (a) is correct and the compound is $A_{2} I_{2}$
and hence option (b) is correct: Empirical formula of the idoide $= A I$

$n_{f}$ of $A_{2} I_{2} = 12 + 2 = 14$

$A_{2} I_{2} n_{f} = 14 + K_{2} C r_{2} O_{7} n_{f} = 6 H^{+} - - \to I O_{3}^{-} + A O + C r^{3 +}$

$e q u i_{A_{2} I_{2}} = e q u i_{K_{2} C r_{2} O_{7}}$

$0.15 \times 14 = n_{K_{2} C r_{2} O_{7}} \times 6$

$n_{K_{2} C r_{2} O_{7}}$ used

$= \frac{0.15 \times 14}{6} = \frac{0.15 \times 7}{3} = 0.35$

hence option (c) is incorrect

option (d) is incorrect, " $A$ " is not a first group metal as it shows variable oxidation state.

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Similar questions

$A_{2} O_{n}$ is oxidized to $A O_{3}^{-}$ by $K M n O_{4}$ in acidic medium. If 1.34 mmol of $A_{2} O_{n}$ requires 32.2 mL of 0.05 M acidified $K M n O_{4}$ solution for complete oxidation, which of the following statement(s) is/are correct?