Question

Poor shielding of the nuclear charge by d or f-orbital electrons is responsible for which of the following facts?
(A) Atomic radius of \[\text{Nb}\] (4-1 series) is comparable to that of \[\text{Ta}\] (5-1 series)
(B) The ${{\text{1}}^{\text{st}}}$ ionization energy of copper is less than that of zinc
(C) The value of electron gain enthalpy is more negative for sulphur than for oxygen.
(D) The ionization energy for gold is greater than that of silver.

Answer 1

Hint: the attraction between the nucleus and the outermost electrons is reduced by the inner shell electrons is called the shielding effect. The s has the highest shielding power followed by the p orbital, d, and then f, d orbital cannot shield the nucleus effectively due to its shape, and therefore the last electrons are very easy to knock out. The elements are always in a state to attain a stable electronic configuration: ${{\text{d}}^{\text{0}}}\text{ , }{{\text{d}}^{\text{5}}}\text{ , }{{\text{d}}^{\text{10}}}$.

Complete step by step solution:
The first ionization enthalpies of the d-block elements are higher than those of the s-block elements but are lesser than those of p-block elements. The ionization enthalpies increase as we move across each series, though not quite regularly.
The increase in ionization enthalpy along a given transition series is attributed to the effect of increasing nuclear charge which would tend to attract the outer electron cloud with the greater force. Consequently, ionization enthalpy is expected to increase.
However, in the case of transition of d-block elements, the addition of an electron takes place to the last but one i.e. $\text{(n-1)}$ d-subshell, and this also increases the screening effect. With the increase in the electrons in the $\text{(n-1)}$d-subshell, the outer electrons in the $\text{ns}$-subshell is shielded more and more.
Thus, the effect of the increasing nuclear charge is opposed by the additional screening effect of the nucleus and consequently, ionization enthalpy increases from the left to right but quite slowly among d-block elements.
The irregular trend in the first ionization enthalpy of the first transition series elements is because the removal of one electron alters the relative energies of $\text{4s}$ and $\text{3d}$ orbitals. The second and third ionization enthalpies are increasing along the period too. However, the magnitude of increase in the second and third ionization enthalpies for the successive elements is much higher.
The zinc has the electronic configuration as follows,
$\text{Zn = }\left[ \text{Ar} \right]\text{3}{{\text{d}}^{\text{10}}}\text{4}{{\text{s}}^{\text{2}}}$
The zinc has a stable electronic configuration. The zinc requires a higher amount of energy to knock out the last shell electronic configuration. Therefore, the first ionization energy for the zinc is higher $\text{906 kJ mo}{{\text{l}}^{\text{-1}}}$ than the copper $\text{745 kJ mo}{{\text{l}}^{\text{-1}}}$.
The electronic configuration of copper is as follows,
\[\text{Cu = }\left[ \text{Ar} \right]\text{3}{{\text{d}}^{\text{10}}}\text{4}{{\text{s}}^{1}}\]
To attain the stable electronic configuration \[\text{3}{{\text{d}}^{\text{10}}}\] one of the electrons is promoted to the d orbital leaves an electron in the $\text{4s}$ orbital. Therefore, it is easier for the copper to lose the last shell electronic configuration. Therefore, the ionization enthalpy for the copper is less than that of the zinc.
Therefore, the first ionization enthalpy of the copper is less than that of the zinc due to the poor shielding of the nuclear charge by d or f-orbital electrons

Hence, (B) is the correct option.

Note: The chromium and copper have the exceptionally high ionization enthalpies value compared to those of their neighbouring elements. This is because of the extra stability of the half-filled ${{\text{d}}^{\text{5}}}$ in the chromium and filled in copper ${{\text{d}}^{10}}$. The value of the second and third enthalpy for zinc is correspondingly low because the ionization involves the removal of an electron resulting in a stable \[\text{3}{{\text{d}}^{\text{10}}}\] configuration.