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Chapter 4: Problem 23

The first ionization energies of \(\mathrm{Li}, \mathrm{Be}, \mathrm{B}\) and \(\mathrm{C}\) are in the order: (a) \(\mathrm{Li}>\mathrm{Be}<\mathrm{B}<\mathrm{C}\) (b) \(\mathrm{Li}<\mathrm{Be}>\mathrm{B}<\mathrm{C}\) (c) \(\mathrm{Li}>\mathrm{Be}>\mathrm{B}>\mathrm{C}\) (d) \(\mathrm{Li}<\mathrm{Be}>\mathrm{B}>\mathrm{C}\)

Short Answer

Expert verified
Answer: The correct order is Li < Be > B < C.

Step by step solution

01

Option A

\(\mathrm{Li}>\mathrm{Be}<\mathrm{B}<\mathrm{C}\) does not follow the trend, as it states that the ionization energy of Li is higher than Be, which is incorrect since ionization energy increases across a period.
02

Option B

\(\mathrm{Li}<\mathrm{Be}>\mathrm{B}<\mathrm{C}\) follows the periodic table trend, as it shows the ionization energy increasing from Li to Be and then from B to C. There is a slight decrease from Be to B, but this is due to Be having a stable electron configuration of \(1s^2 2s^2\), making it harder to remove the outermost electron. This option seems to be correct.
03

Option C

\(\mathrm{Li}>\mathrm{Be}>\mathrm{B}>\mathrm{C}\) also contradicts the trend, as it shows ionization energy decreasing across the period. This option is incorrect.
04

Option D

\(\mathrm{Li}<\mathrm{Be}>\mathrm{B}>\mathrm{C}\) again contradicts the general trend of ionization energy increasing across the period. This option is incorrect. Based on our analysis, Option B, \(\mathrm{Li}<\mathrm{Be}>\mathrm{B}<\mathrm{C}\) is the correct order of the first ionization energies for the given elements.

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