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Chapter 1: Problem 16

Use the following information to answer questions 14-16 The radius of atoms and ions is typically measured in Angstroms \((A),\) which is equivalent to \(1 * 10^{-10} \mathrm{m} .\) Below is a table of information for three different elements. TABLE NOT AVAILABLE Which of the following represents the correct electron configuration for the zinc ion, \(\mathrm{Zn}^{2+} ?\) (A) \([\mathrm{Ar}] 3 \mathrm{d}^{10}\) (B) \([\mathrm{Ar}] 4 \mathrm{s}^{2} 3 \mathrm{d}^{8}\) (C) \([\mathrm{Ar}] 4 \mathrm{s}^{2} 4 \mathrm{d}^{8}\) (D) \([\mathrm{Kr}] 4 \mathrm{s}^{2} 3 \mathrm{d}^{8}\)

Short Answer

Expert verified
The correct electron configuration for the zinc ion, \(Zn^{2+}\), is \([\mathrm{Ar}] 3 \mathrm{d}^{10}\).

Step by step solution

01

Understand Atomic Structure

Zinc has an atomic number of 30. This means it has 30 protons and, in a neutral state, also 30 electrons. The electron configuration is a way of displaying how these electrons are distributed in various atomic orbitals. Standard electron configuration first fills the s orbitals (max 2 electrons), then p orbitals (max 6 electrons), then d orbitals (maximum 10 electrons), and so on.
02

Understand Ionization

When an atom becomes an ion, it loses or gains electrons to achieve a more stable electron configuration. A positive charge means it has lost electrons. In this case, \(Zn^{2+}\) has lost two electrons. We need to keep this in mind when writing the electron configuration.
03

Write the Electron Configuration for Zn

First, write the electron configuration for a neutral zinc atom. You should get: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2\). This configuration shows the zinc atom with all 30 electrons.
04

Write the Electron Configuration for \(Zn^{2+}\)

Next, account for the lost electrons in the zinc ion. Zinc loses two electrons from the 4s orbital when it ionizes, so the electron configuration becomes: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10}\). This shows the ionized zinc atom with 28 electrons. Using noble gas shorthand and replacing the electron configuration of Argon ([Ar] represents \(1s^2 2s^2 2p^6 3s^2 3p^6\)), the final electron configuration of \(Zn^{2+}\) is \([\mathrm{Ar}] 3 \mathrm{d}^{10}\).

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Most popular questions from this chapter

\(2 \mathrm{HI}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\mathrm{I}_{2}(g)+\) energy A gaseous reaction occurs and comes to equilibrium, as shown above. Which of the following changes to the system will serve to increase the number of moles of \(\mathrm{I}_{2}\) present at equilibrium? (A) Increasing the volume at constant temperature (B) Decreasing the volume at constant temperature (C) Increasing the temperature at constant volume (D) Decreasing the temperature at constant volume

\(\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{2} \mathrm{O}(s)\) Which of the following is true for the above reaction? (A) The value for \(\Delta S\) is positive. (B) The value for \(\Delta G\) is zero. (C) The value for \(\Delta H\) is positive. (D) The reaction is favored at 1.0 \(\mathrm{atm}\) and 298 \(\mathrm{K}\) .

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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 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. If the formic acid were replaced with a strong acid such as HCl at the same concentration (2.0 M), how would that change the volume needed to reach the equivalence point? (A) The change would reduce the amount as the acid now fully dissociates. (B) The change would reduce the amount because the base will be more strongly attracted to the acid. (C) The change would increase the amount because the reaction will now go to completion instead of equilibrium. (D) Changing the strength of the acid will not change the volume needed to reach equivalence.
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