(a) Account for the following :
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.
(ii) `Cr^(2+)` is a strong reducing agent.
(iii) `Cu^(2+)` salts are coloured while `Zn^(2+)` salts are white.
(b) Complete the following equations :
(i) `2MnO_(2) + 4KOH + O_(2) overset(Delta)(to)`
(ii) `Cr_(2)O_(7)^(2-) + 14H^(+)6I^(-) to`
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(a) (i) Manganese (Mn) shows highest oxidalion state of +7 with oxygen but with fluorin, it shows the highest of +4 because oxygen has the ability to form multiple bonds with Manganese. Hence, oxidation state can be extended to +7.
(ii) `Cr^(2+)` is a strong reducing agent because `Cr^(2+) to Cr^(3+) + e^(-)` resulting in change from `d^(4)` to more stable `d^(4)(t_(2)g^(3))`
(iii) `Cu^(2+)` has 1 unpaired electron which get excited and when come back to ground state, it emits radiation lies in visible region. Hence, coloured whereas, `Zn^(2+)` has completely filled d-orbital `(d^(10))` configuration. Hence, no unpaired electron. Therefore, `Zn^(2+)` salts are white.
(b) (i) ` 2Mn O_(2) + 4KOH + O_(2) to underset("Potssium Manganate")(2K_(2)MnO_(4) + 2H_(2)O)`
(ii) `Cr_(2)O_(7)^(2-) + 14H^(+) + 6I^(-) to 2Cr^(3+) + 7H_(2)O + 3I_(2)`
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