$\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ is blue in colour while $\mathrm{CuSO}_4$ is colourless. Why?
In $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$, water acts as ligand and causes crystal field splitting. Hence, $d-d$ transition is possible thus $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ is coloured. In the anhydrous $\mathrm{CuSO}_4$ due to the absence of water (ligand), crystal field splitting is not possible and hence, it is colourless.
Name the type of isomerism when ambidentate ligands are attached to central metal ion. Give two examples of ambidentate ligands.
Ligand having more than one different binding position are known as ambidentate ligand. e.g., SCN has two different binding positions S and N . Coordination compound containing ambidentate ligands are considered to show linkage isomerism due to presence of two different binding positions.
e.g., (i) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{SCN}\right]^{3+}$ and (ii) $\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right)\right]^{3+}$
Match the complex ions given in Column I with the colours given in Column II and assign the correct code.
| Column I (Complex ion) |
Column II (Colour) |
||
|---|---|---|---|
| A. | $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ | 1. | Violet |
| B. | $\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ | 2. | Green |
| C. | $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ | 3. | Pale blue |
| D. | $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_4(\mathrm{en})\right]^{2+}(\mathrm{aq})$ | 4. | Yellowish orange |
Match the coordination compounds given in Column I with the central metal atoms given in Column II and assign the correct code.
| Column I (Coordination compound) |
Column II (Central metal atom) |
||
|---|---|---|---|
| A. | Chlorophyll | 1. | Rhodium |
| B. | Blood pigment | 2. | Cobalt |
| C. | Wilkinson catalyst | 3. | Magnesium |
| D. | Vitamin B$_{12}$ | 4. | Iron |
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. | $sp^3 d^2, 2$ |