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Chapter 1: Problem 16

In general, do metals or nonmetals from the same period have higher ionization energies? Why? (A) Metals have higher ionization energies because they usually have more protons than nonmetals. (B) Nonmetals have higher ionization energies because they are larger than metals and harder to ionize. (C) Metals have higher ionization energies because there is less electron shielding than there is in nonmetals. (D) Nonmetals have higher ionization energies because they are closer to having filled a complete energy level.

Short Answer

Expert verified
(D) Nonmetals have higher ionization energies because they are closer to having filled a complete energy level.

Step by step solution

01

Understanding Ionization energy

Ionization energy is the energy required to remove an electron from an atom or ion in its gaseous state. As you go from left to right across a period in the periodic table, the ionization energy generally increases.
02

Compare Metals and Nonmetals

On the periodic table, metals are on the left and nonmetals are on the right. When comparing elements in the same period, generally nonmetals will have higher ionization energy than metals. The explanation is due to the electron configuration and atomic structure
03

Choose the Correct Answer

Comparing the options given, the right answer would be a statement that correctly explains why nonmetals have higher ionization energies. The statement should refer to the fact that nonmetals are closer to having a filled energy level. This makes them more stable and thus more difficult (requires more energy) to remove an electron. The option that correctly states this is option D. Hence (D) Nonmetals have higher ionization energies because they are closer to having filled a complete energy level, is the correct answer.

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Most popular questions from this chapter

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ As the reaction progresses, what will happen to the overall voltage of the cell? (A) It will increase as \(\left[\mathrm{Zn}^{2+}\right]\) increases. (B) It will increase as \(\left[\mathrm{Cu}^{+}\right]\) increases. (C) It will decrease as \(\left[\mathrm{Zn}^{2+}\right]\) increases. (D) The voltage will remain constant.

A student added 1 liter of a 1.0\(M\) \(\mathrm{KCl}\) solution to 1 liter of a 1.0 \(M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution. A lead chloride precipitate formed, and nearly all of the lead ions disappeared from the solution. Which of the following lists the ions remaining in the solution in order of decreasing concentration? (A) \(\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{K}^{+}\right]>\left[\mathrm{Pb}^{2+}\right]\) (B) \(\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{Pb}^{2+}\right]>\left[\mathrm{K}^{+}\right]\) (C) \(\left[\mathrm{K}^{+}\right]>\left[\mathrm{Pb}^{2+}\right]>\left[\mathrm{NO}_{3}^{-}\right]\) (D) \(\left[\mathrm{K}^{+}\right]>\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{Pb}^{2+}\right]\)

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ What will happen in the salt bridge as the reaction progresses? (A) The Na' ions will flow to the Cu/Cu' half-cell. (B) The Br' ions will flow to the Cu/Cu' half-cell. (C) Electrons will transfer from the Cu/Cu' half-cell to the Zn/Zn" half cell. (D) Electrons will transfer from the \(\mathrm{Zn} / \mathrm{Zn}^{2+}\) half- cell to the Cu/Cu' half cell.

Use the following information to answer questions 1-5. \(\begin{array}{ll}{\text { Reaction } 1 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=?} \\ {\text { Reaction } 2 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)} & {\Delta H=-37 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 3 : \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=-46 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 4 : \mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{H}_{2}(g)} & {\Delta H=-65 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}}\end{array}\) If reaction 2 were repeated at a higher temperature, how would the reaction's value for \(\Delta G\) be affected? (A) It would become more negative because entropy is a driving force behind this reaction. (B) It would become more positive because the reactant molecules would collide more often. (C) It would become more negative because the gases will be at a higher (D) It will stay the same; temperature does not affect the value for \(\Delta G\) .

The following reaction is found to be at equilibrium at 25°C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) The value for \(K_{\mathrm{c}}\) at \(25^{\circ} \mathrm{C}\) is \(8.1 .\) What must happen in order for the reaction to reach equilibrium if the initial concentrations of all three species was 2.0 \(M\) ? (A) The rate of the forward reactions would increase, and \(\left[\mathrm{SO}_{3}\right]\) would decrease. (B) The rate of the reverse reaction would increase, and \(\left[\mathrm{SO}_{2}\right]\) would decrease. (C) Both the rate of the forward and reverse reactions would increase, and the value for the equilibrium constant would also increase. (D) No change would occur in either the rate of reaction or the concentrations of any of the species.
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