Redox Reactions Test - 6

Because in electrochemical series Cu is above all in sequence I₂, Ag and Br₂, therefore, Cu will reduce Br₂.

Alkaline earth metals have in common an outer s- electron shell which is full; that is, that is why orbital contains its full complement of two electrons, which these elements readily lose to form cations with charge +2, and an oxidation state (oxidation number) of +2.

Halogens have high electronegativity and electron affinity. They have greater tendency to accept electrons or easily reduced, therefore they are strong oxidizing agent.
As the reduction potential decrease down the group,the oxidizing power decrease down the group the order of the oxidizing power will be as under
F2 > Cl2> Br2> I2
¥ The oxidizing power depends on,
Heat of dissociation of halogen molecule.
Electron affinity of atom.
Hydration energy of the ion.
Heat of vaporization
If a halogen has low energy of dissociation, a high electron affinity, and higher hydration of its ion, it will have high oxidizing power.
F has although low electron affinity than Cl but low dissociation energy and have high hydration energy of its ion, therefore Fluorine is strongest oxidizing agent.

NH₄⁺NO₃^-.In NH₄⁺
N is -3 and 
NO₃^- , N is +5.

In CrO^-_{2 ,} Cr is in +3 oxidation state

In ClO^-_{3 ,} Cl is in +5 oxidation state

In CrO^2-_{4 ,} ​​_​​​​ Cr is in +6 oxidation state

In MnO^-₄ , Mn is in +7 oxidation state

In (3d⁵4s²) +7 will be the maximum oxidation state.

Ne is a noble gas, so does not show any positive or negative oxidation state.

F (Fluorine)  is most electronegative element so it always show -1 oxidation state.

Cs is alkali metal so will show positive oxidation state of +1.