Question

What is meant by lanthanoid contraction? (ii) Why do transition metals form coloured compounds? (iii) Why are $E^\circ_{M^{2+}/M}$ values for Mn and Zn more negative than expected? (iv) Which is the most stable oxidation state of Cu and why? (v) Why is Ce$^{4+}$ in aqueous solution a good oxidising agent?

Hint:

Important facts: \[ \boxed{4f\text{ electrons show poor shielding effect}} \] causing: \[ \boxed{\text{Lanthanoid contraction}} \] Colour in transition compounds: \[ \boxed{d-d\text{ transition}} \] Stable configurations: \[ \boxed{d^5=\text{half-filled, }d^{10}=\text{fully filled}} \] Both provide extra stability.

Answer 1

(i) Lanthanoid contraction Lanthanoids have electrons entering the $4f$ orbitals. The shielding effect of $4f$ electrons is very poor. As atomic number increases, nuclear charge increases, but the added $4f$ electrons cannot effectively shield the outer electrons. Therefore, the effective nuclear charge increases and the size of atoms and ions decreases gradually. This phenomenon is called lanthanoid contraction. \[ \boxed{ La^{3+}\gt Ce^{3+}\gt Pr^{3+}\gt ...\gt Lu^{3+} } \] in terms of ionic size.

(ii) Colour formation in transition metal compounds Transition metals generally have partially filled $d$ orbitals. In the presence of ligands, the five $d$ orbitals split into different energy levels. When visible light falls on these compounds, electrons absorb certain wavelengths and jump from lower energy $d$ orbitals to higher energy $d$ orbitals. This process is called: \[ \boxed{d-d\text{ transition}} \] The remaining transmitted or reflected light gives the compound its colour. Example: \[ CuSO_4 \] appears blue due to $d-d$ transition.

(iii) Negative electrode potential of Mn and Zn The standard electrode potential represents the tendency of metal ions to get reduced. For manganese: \[ Mn:[Ar]3d^54s^2 \] After ionisation: \[ Mn^{2+}:[Ar]3d^5 \] The half-filled $d^5$ configuration is highly stable. Similarly: \[ Zn:[Ar]3d^{10}4s^2 \] forms: \[ Zn^{2+}:[Ar]3d^{10} \] which has a completely filled $d^{10}$ configuration. Due to these stable configurations, removal of electrons from Mn and Zn metals is difficult, resulting in more negative electrode potentials.

(iv) Stable oxidation state of copper Copper has electronic configuration: \[ Cu=[Ar]3d^{10}4s^1 \] It can show: \[ Cu^+ \] and \[ Cu^{2+} \] oxidation states. Although Cu$^+$ has a stable $3d^{10}$ configuration, it undergoes disproportionation: \[ 2Cu^+\rightarrow Cu^{2+}+Cu \] Cu$^{2+}$ is more stable in aqueous solution because of its higher hydration energy. Therefore: \[ \boxed{Cu^{2+}\text{ is the most stable oxidation state of copper}} \]

(v) Oxidising nature of Ce$^{4+}$ Cerium commonly exists in $+3$ and $+4$ oxidation states. The reaction: \[ Ce^{4+}+e^-\rightarrow Ce^{3+} \] has a high positive reduction potential. Therefore, Ce$^{4+}$ readily accepts electrons and oxidises other substances. Hence: \[ \boxed{Ce^{4+}\text{ is a strong oxidising agent}} \]

Final Answer:

(i) \[ \boxed{\text{Lanthanoid contraction is the gradual decrease in size of lanthanoids from La to Lu.}} \]

(ii) \[ \boxed{\text{Transition metal compounds are coloured due to }d-d\text{ transitions.}} \]

(iii) \[ \boxed{\text{Mn and Zn have stable }d^5\text{ and }d^{10}\text{ configurations, causing more negative electrode potentials.}} \]

(iv) \[ \boxed{Cu^{2+}\text{ is the most stable oxidation state of copper.}} \]

(v) \[ \boxed{Ce^{4+}\text{ is a good oxidising agent because it reduces easily to stable }Ce^{3+}.} \]