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Chapter 1: Problem 8

The first ionization energy for a neutral atom of chlorine is 1.25 \(\mathrm{MJ} / \mathrm{mol}\) and the first ionization energy for a neutral atom of argon is 1.52 \(\mathrm{MJ} / \mathrm{mol}\) How would the first ionization energy value for a neutral atom of potassium compare to those values? (A) It would be greater than both because potassium carries a greater nuclear charge then either chlorine or argon. (B) It would be greater than both because the size of a potassium atom is smaller than an atom of either chlorine or argon. (C) It would be less than both because there are more electrons in potassium, meaning they repel each other more effectively and less energy is needed to remove one. (D) It would be less than both because a valence electron of potassium is farther from the nucleus than one of either chlorine or argon.

Short Answer

Expert verified
The first ionization energy value for a neutral Potassium atom would be less than both Chlorine and Argon because a valence electron of potassium is farther from the nucleus than one of either chlorine or argon.

Step by step solution

01

Understanding first ionization energy

First ionization energy is the energy required to remove the most loosely held electron from an atom. It is generally seen that the ionization energy increases from left to right across a period and decreases down a group on the periodic table.
02

Considering Potassium’s position in the periodic table

Potassium is in the first group and fourth period of the periodic table. This means it has one electron in its outermost shell, which is farther from the nucleus because of the addition of a new energy level.
03

Comparing with Chlorine and Argon

Chlorine and Argon lie to the right of Potassium on the same period. This means they not only have a higher nuclear charge (more protons), but their outer electrons are also closer to the nucleus, hence more tightly held.
04

Making the correct choice

From the information gathered, a Potassium atom's first ionization energy must be lower than that of Chlorine or Argon because the outer electron is farther from the nucleus. Hence, it would involve less energy to remove this electron. This is perfectly described in answer (D).

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Most popular questions from this chapter

A bottle of water is left outside early in the morning. The bottle warms gradually over the course of the day. What will happen to the pH of the water as the bottle warms? (A) Nothing; pure water always has a pH of 7.00. (B) Nothing; the volume would have to change in order for any ion concentration to change. (C) It will increase because the concentration of \(\left[\mathrm{H}^{+}\right]\) is increasing. (D) It will decrease because the auto-ionization of water is an endothermic process.

Which of the following pairs of substances would make a good buffer solution? (A) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\) and \(\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\) (B) \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)\) and \(\mathrm{LiOH}(a q)\) (C) \(\mathrm{HCl}(a q)\) and \(\mathrm{KCl}(a q)\) (D) \(\mathrm{HF}(a q)\) and \(\mathrm{NH}_{3}(a q)\)

\(2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{AlCl}_{3}(s)\) The reaction above is not thermodynamically favored under standard conditions, but it becomes thermodynamically favored as the temperature decreases toward absolute zero. Which of the following is true at standard conditions? (A) \(\Delta S\) and \(\Delta H\) are both negative. (B) \(\Delta S\) and \(\Delta H\) are both positive. (C) \(\Delta S\) is negative, and \(\Delta H\) is positive. (D) \(\Delta S\) is positive, and \(\Delta H\) is negative.

A student titrates 20.0 \(\mathrm{mL}\) of 1.0 \(M \mathrm{NaOH}\) with 2.0 \(\mathrm{M}\), \(\mathrm{HCO}_{2} \mathrm{H}\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right) .\) Formic acid is a monoprotic acid. \(\mathrm{CH}_{3} \mathrm{NH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \leftrightarrow \mathrm{OH}^{-}(a q)+\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}(a q)\) The above equation represents the reaction between the base methylamine \(\left(K_{\mathrm{b}}=4.38 \times 10^{-4}\right)\) and water. Which of the following best represents the.concentrations of the various species at equilibrium? (A) \(\left[\mathrm{OH}^{-}\right]>\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]\) (B) \(\left[\mathrm{OH}^{-}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]\) (C) \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]>\left[\mathrm{OH}^{-}\right]>\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]\) (D) \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]>\left[\mathrm{OH}^{-}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}+\right]\)

$$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Cl}_{2}(g) \mapsto 2 \mathrm{NOCl}(g) \Delta G^{\circ}=132.6 \mathrm{kJ} / \mathrm{mol}$$ For the equilibrium above, what would happen to the value of \(\Delta G^{\circ}\) if the concentration of \(\mathrm{N}_{2}\) were to increase and why? (A) It would increase because the reaction would become more themodynamically favored. (B) It would increase because the reaction would shift right and create more products. (C) It would decrease because there are more reactants present. (D) It would stay the same because the value of \(K_{e q}\) would not change.
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