${H_2}{S_2}{O_7}$માં સલ્ફર અને ${K_4}Fe{(CN)_6}$માં આયર્નની અનુક્રમે ઓક્સિડેશન અવસ્થા શું હશે?

Question

AIIMS 2000, Medium

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Solution

a
(a) $\mathop {{H_2}{S_2}{O_7}}\limits^{ * \,} $

$2 \times ( + 1) + 2 \times x + 7 \times ( - 2) = 0$

$ + 2 + 2x - 14 = $$2x = 14 - 2 = 12$

$x = \frac{{12}}{2} = + 6\,\,{\rm{for}}\,\,S$

${K_4}\mathop {Fe}\limits^{ * \,\,} {(CN)_6}$

$4 \times ( + 1)\,x + 6 \times ( - 1) = $$4 + x - 6 = 0$

$x = 6 - 4 = + 2\,\,{\rm{for}}\,\,Fe$.

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