Question
Cu forms coloured salts while Zn forms colourless salts.
✓
Answer
- Electronic configuration of ${ }_{29} CU$ is $[ Ar ] 3 d^{10} 4 s^1$ while that of Zn is $[ Ar ] 3 d^{10} 4 s^2$.
- Electronic configuration of Cu in its +1 oxidation state is $[ Ar ] 3 d^{10}$ while that in +2 oxidation state is $[ Ar ] 3 d^9$.
- Therefore, Cu contains partially filled d orbitals in +2 oxidation state and thus, $Cu ^{2+}$ salts are coloured.
- However, Zn has completely filled d orbital which is highly stable and hence, it does not form coloured ions.
Hence, Cu forms coloured salts while Zn forms colourless salts.
- Electronic configuration of Cu in its +1 oxidation state is $[ Ar ] 3 d^{10}$ while that in +2 oxidation state is $[ Ar ] 3 d^9$.
- Therefore, Cu contains partially filled d orbitals in +2 oxidation state and thus, $Cu ^{2+}$ salts are coloured.
- However, Zn has completely filled d orbital which is highly stable and hence, it does not form coloured ions.
Hence, Cu forms coloured salts while Zn forms colourless salts.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Write a note on Heisenberg’s uncertainty principle.
How is ammonia manufactured by Haber process?
Draw structures of cis-trans isomers for the following:
Dipole moment of $H_2O$ is higher than that of $NH_3.$ Explain.
→Discuss five applications of radioactivity for peaceful purpose.
Write note on ‘Principal Quantum number.
A produces $B$ by $\alpha$ - emission. If $B$ is in the group 16 of periodic table, what is the group of $A$ ?
→Describe the properties of carbon nanotubes.
Which molecules are polar?
$H-I, H-O-H, H-Br, Br_2, N_2, I_2, NH_3$
→$H-I, H-O-H, H-Br, Br_2, N_2, I_2, NH_3$
Match the pairs :
| Column A | Column B | ||
| i. | Castner-Kellner cell | a. | Na2CO |
| ii. | Slaked lime | b. | CaCO3 |
| iii. | Solvay process | c. | NaOH |
| iv. | Limestone | d. | Ca(OH)2 |