The ionization energy of copper is higher than that of potassium, though both have a 4s1 configuration, because the d-electrons in copper
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → form a poor shield, making copper smaller
Shielding means that inner electrons partly cancel the pull of the nucleus on an outer electron. d electrons shield poorly, because d orbitals are spread out in space and do not cover the nucleus well.
Configuration of K: Potassium (Z = 19) is [Ar]4s1.
Configuration of Cu: Copper (Z = 29) is [Ar]3d104s1.
Copper has 10 more protons, and the 10 electrons added with them go into the 3d orbitals. These 3d electrons screen the extra nuclear charge only poorly, so the 4s electron of copper feels a much higher effective nuclear charge. It is pulled closer, the copper atom is much smaller than the potassium atom, and the 4s electron is harder to remove. So copper has the higher ionisation enthalpy.
Why the other options are incorrect
Option (2) form a poor shield, making copper bigger — Incorrect. Poor shielding means a stronger pull of the nucleus, which makes the atom smaller, not bigger. A bigger atom would also have a lower ionisation enthalpy.
Option (3) are strongly shielded, making copper smaller — Incorrect. d electrons are poor shields. Strong shielding would lower the effective nuclear charge and make the 4s electron easier to remove.
Option (4) are strongly shielded, making copper bigger — Incorrect. Both parts are wrong: the shielding is weak, and a bigger atom would hold its outer electron less tightly, giving a lower ionisation enthalpy, not a higher one.
NCERT (2026-27 edition):
4.3.2 Variation in Atomic and Ionic Sizes of Transition Metals – “… It may be recalled that the shielding effect of a d electron is not that effective, hence the net electrostatic attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases. The same trend is observed in the atomic radii of a given series.”