Question

Explain, which one of the following pairs has higher ionization enthalpy:

(i) Be and B (ii) Cu and K​

Answer 1

Answer:

1)Be and B

2)CuandK

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Answer 2

The element which will have higher ionization enthalpy is:

(i) Beryllium (Be)

(ii) Copper (Cu)

Explanation:

(i) Be and B

• The ionization enthalpy (I.E) is defined as the minimum amount of energy (E) required to remove the valence electron (e) from an isolated gaseous atom.
• The electronic configuration of Beryllium is $1s^{2} 2s^{2}$ and that of boron is $1s^{2}2s^{2}2p^{1}$, it is easier to remove an unpaired electron from valence shell.
• It will be easier to remove the one unpaired electron from 2p orbital of boron, so it will have less ionization energy (I.E) than that of beryllium which has completely filled orbitals.
• The values of I.E are: Be=9.32eV, B=8.29eV
• Ionization energy of Be > Ionization energy of B.

(ii) Cu and K

• The electronic configuration of Potassium is $[Ar]4s^{1}$ and that of Copper is $[Ar]4s^{1}3d^{10}$.
• Along a period the size of atoms decreases as the effective nucelar charge increases.
• Hence it will be difficult to remove the valence electron of copper as compared to that of potassium.
• So, ionization enthalpy of copper > ionization enthalpy of potassium.
• I.E of K = 4.34eV and I.E of Cu= 7.72eV.