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Chapter 8: Problem 68

Based on your knowledge of the chemistry of the alkali metals, predict some of the chemical properties of francium, the last member of the group.

Short Answer

Expert verified
Francium, being the last member of the alkali metals, is likely to be the most reactive, have the lowest ionization energy and electronegativity, and the largest atomic size in the group.

Step by step solution

01

Identify Chemical Group Trends

Francium is a member of the alkali metal group in the periodic table. This group is known for certain distinctive trends. The elements in this group typically have a single valence electron, which they tend to lose in reactions, forming positive ions. As you move down the group, the desire to lose an electron increases, making elements more reactive.
02

Predict Francium's Reactivity

Since Francium is the last member of the alkali metal group, it’s at the bottom of the group. This implies that it should be the most reactive of all the alkali metals. This is due to the electron shielding effect, which lessens the pull of the nuclear charge on the valence electrons, making it easier for the atom to lose its valence electron.
03

Predict Francium's Ionization Energy

Also, based on the trend that ionization energy decreases moving down the alkali group, Francium's outermost electron should be the easiest to remove amongst alkali metals.
04

Predict Francium's Electronegativity

Furthermore, the electronegativity of alkali metals tends to decrease as we move down the group. Hence, Francium should be the least electronegative of the alkali metals.
05

Predict Francium's Atomic Size

Lastly, the atomic size increases moving down the alkali metal group. This means Francium would have the largest atomic size among the alkali metals.

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Most popular questions from this chapter

Write equations representing the following processes: (a) The electron affinity of \(\mathrm{S}^{-}\) (b) The third ionization energy of titanium (c) The electron affinity of \(\mathrm{Mg}^{2+}\) (d) The ionization energy of \(\mathrm{O}^{2-}\)

As discussed in the chapter, the atomic mass of argon is greater than that of potassium. This observation created a problem in the early development of the periodic table because it meant that argon should be placed after potassium. (a) How was this difficulty resolved? (b) From the following data, calculate the average atomic masses of argon and potassium: Ar-36 (35.9675 amu; 0.337 percent), \(\begin{array}{lll}\text { Ar- } 38 & (37.9627 & \text { amu } ; & 0.063 \text { percent), } & \text { Ar-40 }\end{array}\) \((39.9624 \mathrm{amu} ; 99.60\) percent \() ; \mathrm{K}-39(38.9637 \mathrm{amu} ;\) 93.258 percent \(,\) K- 40 ( 39.9640 amu; 0.0117 percent), K-41 \((40.9618\) amu; 6.730 percent).

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