NCERT Class XI Chemistry Classification of Elements and Periodicity in Properties Solutions
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Question : 16
Total: 40
Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne. Explain why?
(i) Be has higherΔ i
(ii) O has lower Δi H than N and F?
(i) Be has higher
(ii) O has lower Δi H than N and F?
Solution:
(i) An s-electron is attracted to the nucleus more than a p-electron.
In beryllium, the electron removed during the ionization is an s-electron whereas the electron removed during ionization of boron is a p-electron.
The penetration of a 2s-electron to the nucleus is more than that of a 2p-electron; hence the 2p electron of boron is more shielded from the nucleus by the inner core of electrons than the 2s electrons of beryllium.
Therefore, it is easier to remove the 2p-electron from boron as compared to the removal of a 2s-electron from beryllium. Thus, boron has a smaller first ionization enthalpy than beryllium.
(ii) O has lower ionisation energy than N because N (1 s 2 2 s 2 2 p x 1 2 p y 1 2 p z 1 1 s 2 2 s 2 2 p x 2 2 p y 1 2 p z 1
In beryllium, the electron removed during the ionization is an s-electron whereas the electron removed during ionization of boron is a p-electron.
The penetration of a 2s-electron to the nucleus is more than that of a 2p-electron; hence the 2p electron of boron is more shielded from the nucleus by the inner core of electrons than the 2s electrons of beryllium.
Therefore, it is easier to remove the 2p-electron from boron as compared to the removal of a 2s-electron from beryllium. Thus, boron has a smaller first ionization enthalpy than beryllium.
(ii) O has lower ionisation energy than N because N (
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