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Chapter 1: Problem 10

What is the most likely electron configuration for a sodium ion? (A) 1\(s^{2} 2 s^{2} 2 p^{5}\) (B) 1\(s^{2} 2 s^{2} 2 p^{6}\) (C) 1\(s^{2} 2 s^{2} 2 p^{6} 3 s^{1}\) (D) 1\(s^{2} 2 s^{2} 2 p^{5} 3 s^{2}\)

Short Answer

Expert verified
Option B, \(1s^{2} 2s^{2} 2p^{6}\), is the most likely electron configuration for a sodium ion.

Step by step solution

01

Identify Sodium's Electron Configuration

Sodium (Na) has 11 electrons in its ground state. The electron configuration for Sodium (Na) is thus \(1s^{2} 2s^{2} 2p^{6} 3s^{1}\).
02

Determine the Electron Configuration of Sodium Ion

When sodium (Na) becomes an ion (Na+), it loses one electron in order to have a stable, full outer energy level. So Na+ lose the single electron in the 3s orbital. The electron configuration becomes \(1s^{2} 2s^{2} 2p^{6}\). This resembles the electron configuration of Neon (Ne) which is a Noble Gas.

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Most popular questions from this chapter

\(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu} \quad E^{\circ}=+0.3 \mathrm{V}\) \(\mathrm{Fe}^{2+}+2 e^{-} \rightarrow \mathrm{Fe} \quad E^{\circ}=-0.4 \mathrm{V}\) Based on the reduction potentials given above, what is the reaction potential for the following reaction? \(\mathrm{Fe}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Fe}+\mathrm{Cu}^{2+}\) (A) \(-0.7 \mathrm{V}\) (B) \(-0.1 \mathrm{V}\) (C) \(+0.1 \mathrm{V}\) (D) \(+0.7 \mathrm{V}\)

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Based on the given reduction potentials, which of the following would lead to a reaction? (A) Placing some \(\operatorname{Cr}(s)\) in a solution containing \(\mathrm{Pb}^{2+}\) ions (B) Placing some \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) ions (C) Placing some \(\mathrm{Cr}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) ions (D) Placing some \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Pb}^{2+}\) ions

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. How much formic acid is necessary to reach the equivalence point? (A) 10.0 mL (B) 20.0 mL (C) 30.0 mL (D) 40.0 mL

Which of the following species is amphoteric? (A) \(\mathrm{H}^{+}\) (B) \(\mathrm{CO}_{3}^{2-}\) (C) \(\mathrm{HCO}_{3}^{-}\) (D) \(\mathrm{H}_{2} \mathrm{CO}_{3}\)

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 $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ Correctly identify the anode and cathode in this reaction as well as where oxidation and reduction are taking place. (A) Cu is the anode where oxidation occurs, and Zn is the cathode where reduction occurs. (B) Cu is the anode where reduction occurs, and Zn is the cathode where oxidation occurs. (C) Zn is the anode where oxidation occurs, and Cu is the cathode where reduction occurs. (D) Zn is the anode where reduction occurs, and Cu is the cathode where oxidation occurs.
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