Question

(a) First ionization enthalpy of Cr is lower than that of Zn. Why ?

(b) Mention any one property of the transition elements which makes them good catalysts.

Hint:

Cr becomes stable half-filled 3d5 on losing its 4s electron, but Zn has a fully filled 4s orbital. Variable oxidation states help catalysis.

Answer 1

Step 1: Write the outer electronic configurations for part (a).
First ionization enthalpy is the energy needed to remove the outermost electron from a gaseous atom.
The outer configuration of Cr is \( \mathrm{3d^5\,4s^1} \).
The outer configuration of Zn is \( \mathrm{3d^{10}\,4s^2} \).

Step 2: Compare what is lost in each atom.
Cr loses its single 4s electron and reaches \( \mathrm{3d^5} \), a stable half-filled d subshell.
Zn must lose an electron from a completely filled 4s orbital. Its \( \mathrm{3d^{10}\,4s^2} \) arrangement is very stable.
A fully filled configuration holds its electrons more tightly than a half-filled one.
Zn also has a higher effective nuclear charge pulling on the 4s electrons.
So less energy is needed for Cr than for Zn.
\[ \Delta_i H_1(\mathrm{Cr}) = 653\ \mathrm{kJ\ mol^{-1}} < \Delta_i H_1(\mathrm{Zn}) = 906\ \mathrm{kJ\ mol^{-1}} \]

Step 3: Answer part (b) with one property.
Transition elements show variable oxidation states.
This lets them form unstable intermediate compounds with the reactants.
The intermediate then breaks down to give the products and regenerates the catalyst.
Example: \( \mathrm{Fe^{3+}} \) oxidises iodide ions to iodine in the reaction of persulphate with iodide.
Other accepted properties are the ability to form complexes and the presence of vacant d orbitals, which give a large surface for adsorption.

Final Answer:
(a) Cr loses one 4s electron to get the stable half-filled \( \mathrm{3d^5} \) set, while Zn has to lose an electron from a fully filled 4s orbital, so Cr has the lower first ionization enthalpy.
(b) Variable oxidation states (they form intermediate compounds) make them good catalysts.