Which of the following statements is correct ?
(1) The highst oxidation state of a metal is either xhibited in its oxide or its fluoride
(2) `Cr^(2+)` is a stronger reducing agent than `Fe^(2+)`
(3) The stability of the +6 state for Group 6 elements `W^(6+) lt Mo^(6+) lt Cr^(6+)`
(4) `Cu^(+)` ion is not stable in aqueous solutions

Options

A. (i), (iii), (iv)
B. (i), (ii), (iii), (iv)
C. (i), (ii), (iv)
D. (ii), (iii), (iv)

Answer

C

Explanation

(i) The highest oxidation states are often stabilised in the oxide and fluoride compounds. In these compounds `O^(2+)` and `F^(-)` are difficult to be oxized by the central metal because O and F are strong oxidized agents.
(ii) Reducing agent reduces the other and gets oxidized. `Cr^(2+)` is a stronger reducing agent than `Fe^(2+)` because `E^((-))(Cr^(3+)//Cr^(2+))` is negative `(-0.41V)` whereas `E^(Theta)(Fe^(3+)//Fe^(2+))` is positive `(+0.77)`. Thus, `Cr^(2+)` is easily oxidized to `Cr^(3+)` whilt `Fe^(+)` can't be easily oxidized to `Fe^(3+)`.
(iii) The trend in the stability of oxidation states within the group is different for d-block elements and the main group elements (s and p-block elements). For the main group elements, the higher oxidation state become less stable elements, the higher oxidation state become less stable going down a group (inert pair effect). However, for the d-block elements the stability of the higher oxidation states increases going down a group. Chromium in the `+6` oxidation state `(K_(2)Cr_(2)O_(7))` is a good oxidizing agent forming `Cr^(3+)` as the product. This implies that `Cr(+3)` is more stable than `Cr(+6)`. In contrast, molybdenum and tungsten are not easily reduced when they are in `+6` oxidation state `(K_(2)MoO_(4)` and `K_(2)WO_(4))`. The most important states for Mo and W are `(+5)` and `(+6)`. Whilst `Cr(+6)` is strongly oxidizing, `Mo(+6)` and `W(+6)` are stable. Similarly `Cr(+3)` is stable but `Mo(+3)` and `W(+3)` are strongly reducing. The correct stability order of the `+6` state for Group 6 elements is `W^(6+) gt Mo^(6+) gt Cr^(6+)`. We find the same trend in Ground 4 which is composed of Ti, Zr and Hf. For all the three elements, most stable oxidation state is `+4`. However Ti `(+2)` and Ti `(+3)` can be formed form Ti `(+4)` by using good reducing agents but lower oxidation states of Zr and Hf are extremely difficult to prepare.
(iv) `Cu^(2+)` (aq) is much more stable than `Cu^(+)` (aq). Although second ionization enrthalpy of Cu is quite large but `Delta_(hyd)` for `Cu^(2+)` (aq) is much more negative than that for `Cu^(+)` (aq). It more than compensates for the second ionization ehthalpy of Cu. Consequently, many `Cu^(+)` compounds are unstabe in aqueous solution and undergo disprortionation as follows:
`2Cu^(+)` (aq) `rarr` `Cu^(2+)` (aq) `+ Cu (s)`
Going form left to right across 3d-series, `M^(2+)` (aq) ions are known for the last 8 elements form V to Zn while `M^(3+)` (aq) ions are known for the first 7 elements form Sc to Co. Thus, there is an overall increase in stability of `M^(2+)` (aq) with respest to oxidation as one moves across the series. However. In case of iron, `Fe^(2+)` (aq) is less stable than `Fe^(3+)` (aq) on account of extra stability associated with half-filled `(d^(5))` subshell in the case of `Fe^(3+)` (aq).