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Chapter 1: Problem 5

Why does an ion of phosphorus, \(\mathrm{P}^{3-}\) , have a larger radius than a neutral atom of phosphorus? (A) There is a greater Coulombic attraction between the nucleus and the electrons in \(\mathrm{P}^{3}\) . (B) The core electrons in \(\mathrm{P}^{3-}\) exert a weaker shielding force than those of a neutral atom. (C) The nuclear charge is weaker in \(\mathrm{P}^{3-}\) than it is in P. (D) The electrons in \(\mathrm{P}^{3-}\) have a greater Coulombic repulsion than those in the neutral atom.

Short Answer

Expert verified
The correct answer is (D) The electrons in \(\mathrm{P}^{3-}\) have a greater Coulombic repulsion than those in the neutral atom.

Step by step solution

01

Identify the ion and the atom we are comparing

In this exercise, we are comparing a phosphorus ion, \(\mathrm{P}^{3-}\) and a neutral phosphorus atom. Phosphorus atom has 15 protons, 15 electrons and on average 16 neutrons. When phosphorus gains three electrons to become \(\mathrm{P}^{3-}\), it has 15 protons and now 18 electrons.
02

Understand what happens when an atom becomes an ion

When phosphorus atom gains three electrons to become an ion, those extra electrons will be added to the outermost shell of the atom. These extra electrons increase the electron electron repulsion in that outermost shell, causing them to spread out more and therefore increase the size of the atom.
03

Identify the statement that explains why an ion of phosphorus, \(\mathrm{P}^{3-}\) , has a larger radius than a neutral atom of phosphorus

Given that the electron-electron repulsion has increased in \(\mathrm{P}^{3-}\) as compared to neutral phosphorus atom causing an increase in the size of the ion, the correct statement is the one that states, The electrons in \(\mathrm{P}^{3-}\) have a greater Coulombic repulsion than those in the neutral atom. So, the answer is (D) The electrons in \(\mathrm{P}^{3-}\) have a greater Coulombic repulsion than those in the neutral atom.

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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 $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ If the \(\mathrm{Cu}^{+}+e^{-} \rightarrow \mathrm{Cu}(s)\) half-reaction has a standard reduction potential of \(+0.52 \mathrm{V},\) what is the standard reduction potential for the \(\mathrm{Zn}^{2+}+2 e^{-} \rightarrow \mathrm{Zn}(s)\) half-reaction? (A) \(+0.76 \mathrm{V}\) (B) \(-0.76 \mathrm{V}\) (C) \(+0.24 \mathrm{V}\) (D) \(-0.24 \mathrm{V}\)

A 2.0 L flask holds 0.40 g of helium gas. If the helium is evacuated into a larger container while the temperature is held constant, what will the effect on the entropy of the helium be? (A) It will remain constant because the number of helium molecules does not change. (B) It will decrease because the gas will be more ordered in the larger flask. (C) It will decrease because the molecules will collide with the sides of the larger flask less often than they did in the smaller flask. (D) It will increase because the gas molecules will be more dispersed in the larger flask.

Consider the following reaction showing photosynthesis: $$\begin{array}{c}{6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g)} \\ {\Delta H=+2800 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ Which of the following is true regarding the thermal energy in this system? (A) It is transferred from the surroundings to the reaction. (B) It is transferred from the reaction to the surroundings. (C) It is transferred from the reactants to the products. (D) It is transferred from the products to the reactants.

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 $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ How many moles of electrons must be transferred to create 127 g of copper? (A) 1 mole of electrons (B) 2 moles of electrons (C) 3 moles of electrons (D) 4 moles of electrons

$2 \mathrm{ClF}(g)+\mathrm{O}_{2}(g) \leftrightarrow \mathrm{Cl}_{2} \mathrm{O}(g)+\mathrm{F}_{2} \mathrm{O}(g) \Delta H=167 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}$ During the reaction above, the product yield can be increased by increasing the temperature of the reaction. Why is this effective? (A) The reaction is endothermic; therefore adding heat will shift it to the right. (B) Increasing the temperature increases the speed of the molecules, meaning there will be more collisions between them. (C) The reactants are less massive than the products, and an increase in temperature will cause their kinetic energy to increase more than that of the products. (D) The increase in temperature allows for a higher percentage of molecular collisions to occur with the proper orientation to create the product.
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