Permanganate (VII) ion, $M n O_{4}^{-}$ oxidies $I^{-}$ ion to $I_{2}$ and gives manganese (IV) oxide $M n O_{2}$ in basic medium. The skeletal ionic equation is given as
$p M n {O_{4}}_{(a q)}^{-} + x H_{2} O (l) \to r M n O_{2} (s) + s I_{2} (s) + y O H_{(a q)}^{-}$ The values of p,q,r and s are:

p (1), q (2), r (8), s (4)

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p (2), q (6), r (2), s (3)

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p (2), q (4), r (2), s (8)

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p (1), q (4), r (8), s (2)

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Solution

The correct option is B p (2), q (6), r (2), s (3)
$p M n {O_{4}}^{-} (a q) + q I^{-} + x H_{2} O (l) \to r M n O_{2} (s) + s I_{2} (s) + y O H^{-} (a q)$
in a balanced chemical reaction number of all atoms at right side should be equal to the left side of the reaction.
Balancing a chemical reaction as:
Step 1: Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atom as:
$+ 7 M n {O_{4}}^{-} (a q) +^{-} + H_{2} O (l) \to + 4 M n O_{2} (s) +_{2} (s) + O H^{-} (a q)$
Step 2: Identify the atoms that are oxidized and those that are reduced as:
Reduction: $+ 7 M n {O_{4}}^{-} (a q) \to + 4 M n O_{2} (s)$
Oxidation: $^{-} \to_{2} (s)$
Step 3: oxidation-number change is:
Reduction: $+ 7 M n {O_{4}}^{-} (a q) \to + 4 M n O_{2} (s)$ Gain of 3 electron per Mn atom
Oxidation: $^{-} \to_{2} (s)$ : Loss of total 2 electrons
Step 4: Balance the total change in oxidation number as:
Reduction: $+ 7 M n {O_{4}}^{-} (a q) \to + 4 M n O_{2} (s) \times 2$ : Gain of 6 electron total
Oxidation: $^{-} \to_{2} (s) \times 3$ : Loss of 6 electron
$2 + 7 M n {O_{4}}^{-} (a q) \to 2 + 4 M n O_{2} (s)$
$6^{-} \to 3_{2} (s)$ :
Step 5: Balance O atoms in reduction reaction by adding $H_{2} O$ and then balance H by $H^{+}$ as:
$2 + 7 M n {O_{4}}^{-} (a q) + 8 H^{+} \to 2 + 4 M n O_{2} (s) + 4 H_{2} O$
Step 6: For base catalysed reaction add $O H^{-}$ to both side to neutralize $H^{+}$ as:
$2 + 7 M n {O_{4}}^{-} (a q) + 8 H^{+} + 8 O H^{-} \to 2 + 4 M n O_{2} (s) + 4 H_{2} O + 8 O H^{-}$
or, $2 + 7 M n {O_{4}}^{-} (a q) + 4 H_{2} O \to 2 + 4 M n O_{2} (s) + 8 O H^{-}$
Thus overall reaction is:
$2 + 7 M n {O_{4}}^{-} (a q) + 6^{-} + 4 H_{2} O \to 2 + 4 M n O_{2} (s) + 3_{2} (s) + 8 O H^{-}$
$p + 7 M n {O_{4}}^{-} (a q) + q^{-} + x H_{2} O \to r + 4 M n O_{2} (s) + s_{2} (s) + y O H^{-}$
thus $p = 2, q = 6, r = 2, s = 3, x = 4, y = 8$

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Permanganate (VII) ion,MnO−4 oxidies I− ion to I2 and gives manganese (IV) oxide MnO2 in basic medium. The skeletal ionic equation is given aspMnO4−(aq)+xH2O(l)→rMnO2(s)+sI2(s)+yOH−(aq)The values of p,q,r and s are:

Permanganate (VII) ion, $M n O_{4}^{-}$ oxidies $I^{-}$ ion to $I_{2}$ and gives manganese (IV) oxide $M n O_{2}$ in basic medium. The skeletal ionic equation is given as
$p M n {O_{4}}_{(a q)}^{-} + x H_{2} O (l) \to r M n O_{2} (s) + s I_{2} (s) + y O H_{(a q)}^{-}$ The values of p,q,r and s are: