Match the solutions given in Column I and the colours given in Column II.
| Column I (Aqueous solution of salt) |
Column II (Colour) |
||
|---|---|---|---|
| A. | $\mathrm{FeSO_4 . 7H_2O}$ | 1. | Green |
| B. | $\mathrm{NiCl_2 . 4H_2O}$ | 2. | Light pink |
| C. | $\mathrm{MnCl_2 4H_2O}$ | 3. | Blue |
| D. | $\mathrm{CoCl_2 . 6H_2O}$ | 4. | Pale green |
| E. | $\mathrm{Cu_2Cl_2}$ | 5. | Pink |
| 6. | Colourless |
A. $\rightarrow$ (4) B. $\rightarrow$ (1) C. $\rightarrow(2)$ D. $\rightarrow$ (5) E. $\rightarrow$ (6)
| (Aqueous solution of salt) | (Colour) | ||
|---|---|---|---|
| A. | $\mathrm{FeSO_4 . 7H_2O}$ | 1. | Pale green |
| B. | $\mathrm{NiCl_2 . 4H_2O}$ | 2. | Green |
| C. | $\mathrm{MnCl_2 4H_2O}$ | 3. | Light pink |
| D. | $\mathrm{CoCl_2 . 6H_2O}$ | 4. | Pink |
| E. | $\mathrm{Cu_2Cl_2}$ | 5. | Colourless |
Match the property given in Column I with the element given in Column II.
| Column I (Property) |
Column II (Element) |
||
|---|---|---|---|
| A. | Lanthanoid which shows +4 oxidation state | 1. | Pm |
| B. | Lanthanoid which can show +2 oxidation state | 2. | Ce |
| C. | Radioactive lanthanoid | 3. | Lu |
| D. | Lanthanoid which has $4f^7$ electronic configuration in +3 oxidation state | 4. | Eu |
| E. | Lanthanoid which has $4f^{14}$ electronic configuration in +3 oxidation state | 5. | Gd |
| 6. | Dy |
A. $\rightarrow(2)$ B. $\rightarrow$ (4) C. $\rightarrow$ (1) D. $\rightarrow(5)$ E. $\rightarrow$ (3)
A. Lanthanoid which shows +4 oxidation state is cerium.
${ }_{58} \mathrm{Ce}=[\mathrm{Xe}] 4 f^2 5 d^0 6 \mathrm{~s}^2 ;$ Oxidation state $=+3,+4$
B. Lanthanoid which can show +2 oxidation state is europium.
${ }_{63} \mathrm{Eu}=[\mathrm{Xe}] 4 f^7 5 d^0 6 \mathrm{~s}^2$; Oxidation state $=+2,+3$
C. Radioactive lanthanoid is promethium. It is the only synthetic (man-made) radioactive lanthanoid.
D. Lanthnoid which has $4 f^7$ electronic configuration in +3 oxidation state is gadolinium.
${ }_{64} \mathrm{Gd}=[\mathrm{Xe}] 4 f^7 5 d^1 6 s^2$; Oxidation state $=+3$
E. Lanthanoid which has $4 f^{14}$ electronic configuration in +3 oxidation state is lutetium
${ }_{71} \mathrm{Lu}=[\mathrm{Xe}] 4 f^{14} 5 d^1 6 s^2 ;$ Oxidation state $=+3$
Match the properties given in Column I with the metals given in Column II.
| Column I (Property) |
Column II (Metal) |
||
|---|---|---|---|
| A. | Element with highest second ionisation enthalpy | 1. | Co |
| B. | Element with highest third ionisation enthalpy | 2. | Cr |
| C. | M in M(CO)$_6$ is | 3. | Cu |
| D. | Element with highest heat of atomisation | 4. | Zn |
| 5. | Ni |
A. $\rightarrow(3)$ B. $\rightarrow$ (4) C. $\rightarrow(2)$ D. $\rightarrow$ (1)
A. $\mathrm{Cu}^{+}=3 d^{10}$ which is very stable configuration due to full-filled orbitals. Hence, removal of second electron requires very high energy.
B. $\mathrm{Zn}^{2+}=3 d^{10}$ which is very stable configuration. Hence, removal of third electron requires very high energy.
C. Metal carbonyl with formula $\mathrm{M}(\mathrm{CO})_6$ is $\mathrm{Cr}(\mathrm{CO})_6$.
D. Nickel is the element with highest heat of atomisation.
Assertion $(\mathrm{A}) \mathrm{Cu}^{2+}$ iodide is not known.
Reason $(\mathrm{R}) \mathrm{Cu}^{2+}$ oxidises $\mathrm{I}^{-}$to iodine.
Assertion (A) Separation of Zr and Hf is difficult.
Reason (R) Because Zr and Hf lie in the same group of the Periodic Table.