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  • Arrange the following species in decreasing order of oxidation numbers:<img alt="\mathrm{Cl^{-},Mn^{7+},Cr^{6+},Ba^{2+},K^{+},S^{2-}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm

Arrange the following species in decreasing order of oxidation numbers:\mathrm{Cl^{-},Mn^{7+},Cr^{6+},Ba^{2+},K^{+},S^{2-}}

Option: 1

\mathrm{Cl ^{-}, S ^{2-}, K ^{+}, B a^{2+}, C r^{6+}, M n^{7+}}


Option: 2

\mathrm{S^{2-}, Cl ^{-}, K ^{+}, B a^{2+}, C r^{6+}, M n^{7+}}


Option: 3

\mathrm{M n^{7+}, C r^{6+}, B a^{2+}, K^{+}, C l^{-}, S^{2-}}


Option: 4

\mathrm{S ^{2-}, Cl ^{-}, B a^{2+}, C r^{6+}, M n^{7+}, K ^{+}}


Answers (1)

best_answer

For ions composed of only one atom, the oxidation number is equal to the charge on the ion.

Decreasing Order: +7>+6>+2>+1>-1>-2

So,  decreasing the order of oxidation numbers of species 

\mathrm{Mn^{7+},Cr^{6+},Ba^{2+},K^{+},Cl^{-},S^{2-}}

Hence, the correct answer is Option (3)

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Sayak

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