The correct order of size is
(A) \(Cr^{+3} < Cr^{+4} < Cr^{+5}\)
(B) \(C^{4-} > N^{3-} > 0^{2-}\),
(C) \(Fe > Fe^{+2} > Fe^{+3}\)
(D) \(He^{+}< Li^{+2} < Be^{+3}\)

Mohr's salt on hydrolysis gives. (A) \(Fe^{2+}\) and \(SO_{4}^{2-}\) ions (B) \(Fe^{2+}\) and \(NH_{4}^{+}\) ions (C) \(Fe^{2+}\),\(NH_{4}^{+}\) and \(SO_{4}^{2-}\) ions (D) \([Fe(NH_{3})_{6}]^{2+}\) and \(SO_{4}^{2-}\) ions

The colourless ion among the following is (A) Mn^(2+) , B) Cu^(+) (C) Cr^(3+) (D) Fe^(2+)

The correct order of the number of unpaired electrons in the ions Cu^(2+), Ni^(2+), Fe^(3+) , and Cr^(3+) is

(i) On the basis of the standard electrode potential values stated for acid solutions, predict whether Ti^(4+) species may be used to oxidise Fe(II) to Fe(III) {:(Ti^(4+) + e^(-) to Ti^(3+), E^(@) = +0.01V), (Fe^(3+) + e^(-) to Fe^(2+), E^(@)= +0.77V):} (ii) Based on the data arrange Fe^(3+), Mn^(2+) " and " Cr^(2+) in the increasing order of stability of +2 oxidation state. (Give a brief reason) E_(Cr^(3+)//Cr^(2+))^(@) = -0.4V E_(Mn^(3+)//Mn^(2+))^(@) = +1.5V E_(Fe^(3+)//Fe^(2+))^(@) = +0.8V

Standard electrode potentials are Fe^(2+)//Fe, E^(@) = -0.44 V Fe^(3+)//Fe^(2+), E^(@) = -0.77 V If Fe^(3+), Fe^(2+) , and Fe block are kept together, then