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Chapter 8: Q20CQ (page 338)

Question:Figure 8.16 shows that in the Z = 3 to 10 filling of the n = 2 shell (lithium to neon), there is an upward trend in elements' first Ionization energies. Why is there a drop as Z goes from 4 to 5, from beryllium to boron?

Short Answer

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Answer

As the subshell 2p1 is not closed, electrons in this shell could be removed with lesser ionization energy compared to beryllium.

Step by step solution

01

Explanation

It is easier with lesser ionization energy to remove the unpaired electron in the subshell2p1 of Boron (B)than the electron in the closed subshellof Beryllium (Be)in the process of ionization energy from to Li(z = 3 to 10).

02

Drop between 4 to 5

Beryllium has an electronic configuration . The subshells are closed. Electrons in the closed shells are very tightly bound and the positive nuclear charge is very large when compared to the negative charge of the inner shielding electrons. So, these electrons cannot be easily detached. So, higher energy is needed to remove these electrons, whereas, Boron has the electronic configuration 1s22s22p1. As the subshell is not closed, electrons in this shell could be removed with laser ionization energy compared to Beryllium.

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