Match the complex ions given in Column I with the hybridisation and number of unpaired electrons given in Column II and assign the correct code.
| Column I (Complex ion) |
Column II (Hybridisation, number of unpaired electrons) |
||
|---|---|---|---|
| A. | $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ | 1. | $dsp^2,1$ |
| B. | $\left[\mathrm{Co}(\mathrm{CN})_4\right]^{2-}$ | 2. | $sp^3 d^2,5$ |
| C. | $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}$ | 3. | $d^2 sp^3, 3$ |
| D. | $\left[\mathrm{MnF}_6\right]^{4-}$ | 4. | $sp^3 d^2, 2$ |
Match the complex species given in Column I with the possible isomerism given in Column II and assign the correct code.
| Column I (Complex species) |
Column II (Isomerism) |
||
|---|---|---|---|
| A. | $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}$ | 1. | Optical |
| B. | cis $-\left[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2\right]^{+}$ | 2. | Ionisation |
| C. | $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right)\right] \mathrm{Cl}_2$ | 3. | Coordination |
| D. | $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]\left[\mathrm{Cr}(\mathrm{CN})_6\right]$ | 4. | Geometrical |
Match the compounds given in Column I with the oxidation state of cobalt present in it (given in column II) and assign the correct code.
| Column I (Compound) |
Column II (Oxidation state of Co) |
||
|---|---|---|---|
| A. | $\left[\mathrm{Co}(\mathrm{NCS})\left(\mathrm{NH}_3\right)_5\right]\left(\mathrm{SO}_3\right)$ | 1. | +4 |
| B. | $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{SO}_4$ | 2. | 0 |
| C. | $\mathrm{Na}_4\left[\mathrm{Co}\left(\mathrm{S}_2 \mathrm{O}_3\right)_3\right]$ | 3. | +2 |
| D. | $\left[\mathrm{CO}_2(\mathrm{CO})_8\right]$ | 4. | +3 |
Assertion (A) Toxic metal ions are removed by the chelating ligands.
Reason (R) Chelate complexes tend to be more stable.
Assertion (A) $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}_6\right)\right] \mathrm{Cl}_2$ and $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_2$ are reducing in nature.
Reason (R) Unpaired electrons are present in their d-orbitals.