(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`

Explanation

(a) (i) Mn can form double bonds with oxygen atoms and shows an oxidation state of +7. Fluorine is monovalent, only single bonds can be formed with fluorine and therefore the highest oxidation state shown by Mn with fluorine is +4.
(ii) `Cr^(2+)` is a strong reducing agent.
`Cr^(2+)` has the electronic configuration `d^(4)`. When it acts as a reducing agent, it is itself oxidised and changes to `Cr^(3+)`. The d-orbital configuration becomes `d^(3)`. It has stable half-filled `t_(2g)` configuration.
(iii) The colour of transition metal ions are due to d-d transitions of electrons i.e., from lower d-levels to higher d-levels `(t_(2g)" to "e_(g))`. `Cu^(2+)` ions contain an unpaired electron which on d-d transition gives blue colour. `Zn^(2+)` ions have no unpaired electron. There is no possibility of transition of electrons hence no colour.
(b) (i) `2MnO_(2)+4KOH+O_(2)to2K_(2)MnO_(4)+2H_(2)O`
(ii) `Cr_(2)O_(7)^(2-)+14H^(+)+6I^(-)to2Cr^(3+)+7H_(2)O+3I_(2)`