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The equivalent weight of `MnSO_(4)` is half its molecular weight when it is converted toA. (a)`Mn_(2)O_(3)`B. (b)`MnO_(2)`C. (c )`MnO_(4)^(-)`D. (d)`MnO_(4)^(2-)` |
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Answer» Correct Answer - B Equivalent `"weight"=("molecular weight")/("valency factor")` If valency is 2, then equivalent weight will be equal to its molecular weight. In `MnSO_(4)`, the oxidation state of Mn is +II In `Mn_(2)O_(3)`, the oxidation state of Mn is +III In `MnO_(2)`, the oxidation state of Mn is +IV In `MNO_(4)`, the oxidation state of Mn is +VII In `Mn_(4)^(2-)`, the oxidation state of Mn is +VI Thus, when `MnSO_(4)` is converted into `MnO_(2)`, then the valency factor is 2, and the equivalent weight of `MnSO_(4)` will be half of its molecular weight. |
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