(a)(i) oxidant, (ii) acid, (iii) base and (iv) reductant
(b)(i) oxidant, (ii) base, (iii) acid and (iv) reductant
(c)(i) acid, (ii) oxidant, (iii) reductant and (iv) base
(d)(i) base, (ii) reductant, (iii) oxidant and (iv) base
Answer
Answer (as printed): A
Explanation
In reaction (i) Water acting as an oxidant oxidizing $\mathrm{Ca}$ to $\mathrm{Ca^{2+}}$ getting itself reduced to $\mathrm{H_2}$.In reaction (ii) Water act as Lewis acid accepting electron from $\mathrm{Cl^{-}}$ which act as a ligand.In reaction (iii) Water act as Lewis base donating electron to $\mathrm{Mg^{2+}}$.In reaction (iv) Water acting as an reductant reducing $\mathrm{F}$ to $\mathrm{F^{-}}$.i) $\mathrm{H_2O + Cl^{-} = [Cl(H_2O)_n]^{-}}$ Where, $\mathrm{Cl}$ = Base, $\mathrm{H_2O}$ = Acidiii) $\mathrm{6H_2O + Mg^{2+}} \rightarrow$ octahedral complex shown below