(a) Complete the following equations : (i) Cr_(2)O_(7)^(2-)+2OH^(-)to (ii) MnO_(4)^(-)+14H^(+)+3e^(-)to (b) Account for the following: (i) Zn is not considered a transition element. (ii) Transition metals form a large number of complexes. (iii) The E^(@) value for the Mn^(3+)//Mn^(2+) couple is much more positive than that for Cr^(3+)//Cr^(2+) couple.

(a) Complete the following equations : (i) Cr_(2)O_(7)^(2-)+2OH^(-)to

1.	(a) Complete the following equations : (i) Cr_(2)O_(7)^(2-)+2OH^(-)to (ii) MnO_(4)^(-)+14H^(+)+3e^(-)to (b) Account for the following: (i) Zn is not considered a transition element. (ii) Transition metals form a large number of complexes. (iii) The E^(@) value for the Mn^(3+)//Mn^(2+) couple is much more positive than that for Cr^(3+)//Cr^(2+) couple.
Answer» Solution :(a) (i) `Cr_(2)O_(7)^(2-)+2OH^(-)to 2CrO_(4)^(2-)+H_(2)O` (ii) `MnO_(4)^(-)+4H^(+)+3e^(-)to MnO_(2)+2H_(2)O` (b) (i) ZN is not considered as a transition element. Zn ATOM has completely filled d-orbitals `(3d^(10))` in its ground as well as in its oxidised state. Therefore, it is not regarded as a transition element. (ii) Transition metals form a large number of complexes because of the FOLLOWING: They have comparatively smaller sizes of the metal ions. They have high ionic charges. d-orbitals are available for bond formation. (III) `E^(@)` value for `Mn^(3+)//Mn^(2+)` couple is much more positive i.e., this REDUCTION takes place with greater tendency because `Mn^(2+)` has `d^(5)`, half-filled stable configuration. In the `Cr^(3+)//Cr^(2+)` couple the change is from `d^(3)` to `d^(4)` which takes place with smaller tendency, hence `E^(@)` value is less positive (in fact, it is negative).