How would you account for the following?
(i) Transition metals exhibits variable oxidation state.
(ii) `Zr(Z=40)` and `Hf(Z=72)` have almost identical radii.
(iii) Transition metals and their compounds act as catalyst.
OR
Complete the following chemical equation:
(i)`Cr_(2)O_(7)^(2-)+6Fe^(2+)+14H^(+)rarr`
(ii) `CrO_(4)^(2-)+2H^(+)rarr`
(iii) `2MnO_(4)^(2-)+5C_(2)O_(4)^(2-)+16H^(+)rarr`

Answer

### Step-by-Step Solution

**Part (i): Transition metals exhibit variable oxidation states.**

1. **Understanding d-electrons**: Transition metals have a large number of d-electrons, which can be either paired or unpaired. The presence of unpaired d-electrons allows these metals to participate in various oxidation states.

2. **Energy levels of orbitals**: The energy difference between the (n-1)d subshell and the ns subshell is relatively low. This means that electrons can be easily removed from both the d and s orbitals.

3. **Conclusion**: Due to the availability of unpaired d-electrons and the low energy difference between the d and s orbitals, transition metals can lose different numbers of electrons, resulting in variable oxidation states.

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**Part (ii): Zr (Z=40) and Hf (Z=72) have almost identical radii.**

1. **Lanthanoid Contraction**: The phenomenon of lanthanoid contraction occurs due to the poor shielding effect of the f-electrons in the lanthanide series. As you move from lanthanum (La) to lutetium (Lu), the effective nuclear charge increases, causing the size of the atoms to decrease.

2. **Position of Zr and Hf**: Zirconium (Zr) is located above Hafnium (Hf) in the periodic table. The lanthanoid contraction affects the size of Hf, making it smaller than expected.

3. **Conclusion**: As a result of lanthanoid contraction, both Zr and Hf exhibit similar ionic radii despite the significant difference in their atomic numbers.

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**Part (iii): Transition metals and their compounds act as catalysts.**

1. **Vacant d-orbitals**: Transition metals have vacant d-orbitals that allow them to form coordination complexes with reactants, facilitating the reaction process.

2. **Variable oxidation states**: The ability of transition metals to exhibit multiple oxidation states enables them to participate in various reaction mechanisms, forming intermediates that lower the activation energy.

3. **Conclusion**: The combination of vacant d-orbitals and the ability to adopt different oxidation states allows transition metals to act as effective catalysts, increasing the rate of reactions without being consumed.

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**Part (iv): Completing the chemical equations.**

1. **Equation (i)**:
\[ \text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^{+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O} \]

2. **Equation (ii)**:
\[ 2\text{CrO}_4^{2-} + 2\text{H}^{+} \rightarrow \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O} \]

3. **Equation (iii)**:
\[ 2\text{MnO}_4^{-} + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^{+} \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O} \]

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