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Chapter 9: Problem 81

The most negative electron affinity of the following elements is that of (a) \(\mathrm{Br} ;\) (b) \(\mathrm{Sn} ;\) (c) \(\mathrm{Ba} ;\) (d) Li.

Short Answer

Expert verified
The element with the most negative electron affinity among the given options is Bromine (Br).

Step by step solution

01

Identify the trend in electron affinity across the periodic table

The electron affinity generally increases across a period from left to right and decreases down a group on the periodic table. This is because as you move left to right across a period, the number of protons increases, which means there's a stronger pull on the extra electron, releasing more energy. As you move down a group, the outer electrons are further from the nucleus and are less attracted to it, so less energy is released when an extra electron is added.
02

Apply the trend to given elements

Now, trace the location of each element on the periodic table and apply the above trend to estimate their electron affinities. We have Br (bromine) which is a Halogen and in period 4, Sn (Tin) which is a post-transition metal in period 5, Ba (Barium) which is an alkaline earth metal in period 6, and Li (Lithium) which is an alkali metal in period 2.
03

Identify the element with the most negative electron affinity

Given the trend identified, one can see that halogens which are on the right-most side of the periodic table would have a higher electron affinity compared to other families. Among all these elements, Br (bromine) is the halogen and it's in a higher period compared to Li, so Bromine should have the most negative electron affinity.

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Most popular questions from this chapter

Give the symbol of the element (a) in group 14 that has the smallest atoms; (b) in period 5 that has the largest atoms; (c) in group 17 that has the lowest first ionization energy.

Complete and balance the following equations. If no reaction occurs, so state. (a) \(\operatorname{Rb}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1) \longrightarrow\) (b) \(\mathrm{I}_{2}(\mathrm{s})+\mathrm{Na}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq}) \longrightarrow\) (c) \(\operatorname{SrO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1) \longrightarrow\) (d) \(\mathrm{SO}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(1) \longrightarrow\)

Sketch a periodic table that would include all the elements in the main body of the table. How many "numbers" wide would the table be?

The \(\mathrm{Na}^{+}\) ion and the Ne atom are isoelectronic. The ease of loss of an electron by a gaseous Ne atom, \(I_{1}\) has a value of \(2081 \mathrm{kJ} / \mathrm{mol} .\) The ease of loss of an electron from a gaseous \(\mathrm{Na}^{+}\) ion, \(I_{2}\), has a value of \(4562 \mathrm{kJ} / \mathrm{mol} .\) Why are these values not the same?

Compare the elements \(\mathrm{Al}, \mathrm{Si}, \mathrm{S},\) and \(\mathrm{Cl}.\) (a) Place the elements in order of increasing ionization energy. (b) Place the elements in order of increasing electron affinity.
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