Read the passage carefully and answer the questions.
Nearly all the transition elements display typical metallic properties such as high tensile strength, ductility, malleability, high thermal and electrical conductivity and metallic lustre. The transition metals (with the exception of Zn, Cd and Hg) are very hard and have low volatility. Their melting and boiling points are high. They have high enthalpies of atomization. Ions of the same charge in a given series show progressive decrease in radius with increasing atomic number. However, the variation within a series is quite small. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called Lanthanoid contraction which essentially compensates for the expected increase in atomic size with increasing atomic number. The net result of the lanthanoid contraction is that the second and the third d series exhibit similar radii (e.g., Zr 160 pm, Hf 159 pm) and have very similar physical and chemical properties. There is an increase in ionisation enthalpy along each series of the transition elements from left to right. The first ionisation enthalpy, in general, increases, but the magnitude of the increase in the second and third ionisation enthalpies for the successive elements, is much higher along a series. One of the notable features of a transition elements is the great variety of oxidation states these may show in their compounds.
Atomic number of Fe, Co, Ni and Cu are 26, 27, 28 and 29 respectively. Which among them is expected to possess the highest value for the second ionization enthalpy?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Cu
The second ionisation enthalpy is the energy needed to remove an electron from the \( \ce{M+} \) ion. It is very high when this electron must come from a stable, completely filled or half-filled set, because such a configuration resists losing an electron. So first write each atom from its atomic number, then the \( \ce{M+} \) ion (the 4s electron is lost first), and see where the second electron comes from.
Fe (Z = 26): \( \mathrm{[Ar]\,3d^{6}\,4s^{2}} \); \( \ce{Fe+} \) is \( \mathrm{[Ar]\,3d^{6}\,4s^{1}} \). The second electron comes from 4s.
Co (Z = 27): \( \mathrm{[Ar]\,3d^{7}\,4s^{2}} \); \( \ce{Co+} \) is \( \mathrm{[Ar]\,3d^{7}\,4s^{1}} \). The second electron comes from 4s.
Ni (Z = 28): \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \); \( \ce{Ni+} \) is \( \mathrm{[Ar]\,3d^{8}\,4s^{1}} \). The second electron comes from 4s.
Cu (Z = 29): \( \mathrm{[Ar]\,3d^{10}\,4s^{1}} \); \( \ce{Cu+} \) is \( \mathrm{[Ar]\,3d^{10}} \). The second electron must be taken from the completely filled 3d10 set.
The second ionisation enthalpies follow this reasoning (in kJ mol−1):
\( \displaystyle \ce{Fe}\,(1561) \lt \ce{Co}\,(1646) \lt \ce{Ni}\,(1753) \lt \ce{Cu}\,(1958) \)
So the highest second ionisation enthalpy is expected for Cu.
NCERT (2026-27 edition):
Ionisation Enthalpies – “… The lowest common oxidation state of these metals is +2. To form the M2+ ions from the gaseous atoms, the sum of the first and second ionisation enthalpy is required in addition to the enthalpy of atomisation. The dominant term is the second ionisation enthalpy which shows unusually high values for Cr and Cu where M+ ions have the d5 and d10 configurations respectively. The value for Zn is correspondingly low as the ionisation causes the removal of one 4s electron which results in the formation of stable d10 configuration.”