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Electronic Configuration

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Question

What is the increasing order of equivalent weights of the following oxidising agents?
(i) $K M n O_{4} ⟶ M n^{2 +}$
(ii) $K_{2} C r_{2} O_{7} ⟶ C r^{3 +}$
(iii) $K C l O_{3} ⟶ C l^{-}$
[Atomic wt.:
$K = 39, C r = 52, M n = 55, C l = 35.5, O = 16$ ]

K C l O_{3} < K M n O_{4} < K_{2} C r_{2} O_{7}

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K_{2} C r_{2} O_{7} < K M n O_{4} < K C l O_{3}

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K M n O_{4} < K_{2} C r_{2} O_{7} < K C l O_{3}

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K C l O_{3} < K_{2} C r_{2} O_{7} < K M n O_{4}

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Solution

The correct option is A $K C l O_{3} < K M n O_{4} < K_{2} C r_{2} O_{7}$
$(i) K 7 + M n O_{4} ⟶ M n^{2 +}$ $\Rightarrow E_{1} = \frac{39 + 55 + 64}{5} = 31.6$
$(i i) K_{2} 6 + C r_{2} O_{7} ⟶ C r^{3 +}$ $\Rightarrow E_{2} = \frac{39 \times 2 + 52 \times 2 + 16 \times 7}{6} = 49$
$(i i i) K 5 + C l O_{3} ⟶ C l^{-}$
$\Rightarrow E_{3} = \frac{39 + 35.5 + 48}{6} = 20.4$

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Similar questions

K M n O_{4} ⟶ M n^{2 +}

K_{2} C r_{2} O_{7} ⟶ C r^{3 +}

K C l O_{3} ⟶ C l^{-}

K = 39, C r = 52, M n = 55, C l = 35.5, O = 16

K M n O_{4} ⟶ M n^{2 +}

K_{2} C r_{2} O_{7} ⟶ C r^{3 +}

K C l O_{3} ⟶ C l^{-}

K = 39, C r = 52, M n = 55, C l = 35.5, O = 16

a) What is percentage composition of potassium chlorate (KClO₃).

Given:Atomic mass Of K=39,Cl=35.5andO=16

K M n O_{4}

40 g

K = 39; M n = 55; O = 16

20

K C l O_{3}

[K = 39, C l = 35.5, O = 16]

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