Complete the following chemical equation:
(i) `MnO_(4)^(-)(aq)+s_(2)O_(3)^(2-)(aq)+H_(2)O(I)`to
(ii) `MnO_(4)^(-) +NO_(2)^(-)+H^(+)`to
State reason for the following :
(i) `Cu(I)` ion is not stable in an aqueous solution.
(ii) Unlike `Cr^(3+),Mn^(2+),Fe^(3+)` and the subsequently other `M^(2+)` ions of the `3d` series of elements, the `4d` and the `5d` series metals generally do not form stable cationic species.

Answer

### Step-by-Step Solution

#### Part (i): Completing the Chemical Equations

1. **First Reaction**:
- Given: `MnO4^(-)(aq) + S2O3^(2-)(aq) + H2O(l)`
- To find the products, we recognize that `MnO4^(-)` acts as an oxidizing agent and `S2O3^(2-)` acts as a reducing agent.
- The balanced equation is:
\[ 8 \, \text{MnO}_4^{-} + 3 \, \text{S}_2\text{O}_3^{2-} + 8 \, \text{H}_2\text{O} \rightarrow 8 \, \text{MnO}_2 + 6 \, \text{SO}_4^{2-} + 2 \, \text{OH}^- \]
- Here, manganese is reduced from +7 to +4, and sulfur is oxidized from +2 to +6.

2. **Second Reaction**:
- Given: `MnO4^(-) + NO2^(-) + H^(+)`
- The products formed are `NO3^(-)`, `MnO2`, and `H2O`.
- The balanced equation is:
\[ 2 \, \text{MnO}_4^{-} + 3 \, \text{NO}_2^{-} + 2 \, \text{H}^+ \rightarrow 3 \, \text{NO}_3^{-} + 2 \, \text{MnO}_2 + 2 \, \text{H}_2\text{O} \]
- In this reaction, manganese is again reduced from +7 to +4, while nitrogen is oxidized from +3 to +5.

#### Part (ii): Reasons for Stability of Ions

1. **Reason for the instability of `Cu(I)` in aqueous solution**:
- The `Cu(I)` ion tends to disproportionate in aqueous solution to form `Cu(II)` and metallic copper:
\[ 2 \, \text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu} \]
- This occurs because the hydration energy of `Cu(II)` is greater than the second ionization energy of copper, making the formation of `Cu(II)` more favorable.

2. **Reason for the stability of `3d` series ions compared to `4d` and `5d` series**:
- The `3d` series ions like `Cr^(3+)`, `Mn^(2+)`, and `Fe^(3+)` have a more symmetrical distribution of d-electrons, which allows them to form stable ionic complexes.
- In contrast, the atomic radii of `4d` and `5d` series elements are larger, which makes it less favorable for them to form stable cationic species.