(A) Account for the following:
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine, it shows an oxidation state of +4.
(ii) Cr2+ is a strong reducing agent.
(iii) Cu2+ salts are coloured, while Zn2+ salts are white.
(B) Complete the following equations:
(i) 2MnO2 + 4KOH + O2 →
(ii) Cr2O72- + 14H+ + 6e- →

(A)-(i)Element Name and Symbol Atomic Number Common Oxidation States
Scandium (Sc) 21 +3
Titanium (Ti) 22 +4
Vanadium (V) 23 +2, +3, +4, +5
Chromium (Cr) 24 +2, +3, +6
Manganese (Mn) 25 +2, +3, +4, +6, +7
Iron (Fe) 26 +2, +3
Cobalt (Co) 27 +2, +3
So Manganese (Mn) shows the highest number of oxidation states.
Element Sc Ti V Cr Mn Fe Co
M.P. (oC) 1540 1668 1890 1875 1245 1537 1495
(ii) Scandium shows only +3 oxidation state.
Zn has electronic configuration = 3d104s2
Zn2+ = 3d10
Cu = 3d54s1
Cu2+ = 3d6
In the case of Zn, a fully filled d orbital is present; therefore, no d-d transition can be possible in this case, it is colorless. In the case of copper, because of d-d transition electrons emit light in the visible range and hence they are colored compounds
(iii) Cr2+ gets converted to Cr3+ as the +3 oxidation state has half-filled t2g orbitals - thus it is a good reducing agent
(B)
(i) 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
(ii) Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O