Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 3

Why can a molecule with the structure of NBr_ not exist? (A) Nitrogen only has two energy levels and is thus unable to expand its octet. (B) Bromine is much larger than nitrogen and cannot be a terminal atom in this molecule. (C) It is impossible to complete the octets for all six atoms using only valence electrons. (D) Nitrogen does not have a low enough electronegativity to be the central atom of this molecule.

Short Answer

Expert verified
The reason a molecule with the structure of NBr_ cannot exist is because it is impossible to complete the octets for all six atoms using only the available valence electrons. The correct choice is (C).

Step by step solution

01

Determine the Electron Configurations of Nitrogen and Bromine

By looking at their positions in the periodic table, Nitrogen (N), which belongs to group 15, has 5 valence electrons. Bromine (Br), which belongs to group 17, has 7 valence electrons. So, nitrogen cannot bond with more than 3 bromine atoms without violating the octet rule.
02

Analyze Each Answer Choice

(A) This choice incorrectly assumes nitrogen cannot expand its octet due to it having only two energy levels. It is true that nitrogen is in the second period of the periodic table and elements in this period generally do not expand their octets. But, nitrogen can actually form five bonds (for instance, in ammonium ion, NH4+). (B) This choice inaccurately assigns physical size as a deterrent to bond formation, which is irrelevant as long as bonding rules are obeyed. (C) This is the correct choice, because it is indeed impossible to complete the octets for all six atoms using the available valence electrons. (D) This choice incorrectly focuses on the electronegativity of nitrogen, which concerns its electron attracting ability, not its potential to form bonds.
03

Determine the Best Answer

By considering the electron configurations of nitrogen and bromine, and examining the answer choices, we find that the best answer is (C). This is because this response correctly points out the impossibility to complete the octets for all six atoms using only the available valence electrons.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ When the cell is connected, which of the following reactions takes place at the anode? (A) \(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}(\mathrm{s})\) (B) \(\mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+}+2 e^{-}\) (C) \(2 \mathrm{H}^{+}+2 e^{-} \rightarrow \mathrm{H}_{2}(g)\) (D) \(\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{H}^{+}+2 e^{-}\)

Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ What is the required voltage to make this cell function? (A) 0.34 V (B) 0.42 V (C) 0.76 V (D) 1.10 V

A gaseous mixture at \(25^{\circ} \mathrm{C}\) contained 1 mole of \(\mathrm{CH}_{4}\) and 2 moles of \(\mathrm{O}_{2}\) and the pressure was measured at 2 \(\mathrm{atm}\) . The gases then underwent the reaction shown below. \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) What was the pressure in the container after the reaction had gone to completion and the temperature was allowed to return to \(25^{\circ} \mathrm{C} ?\) (A) 1 atm (B) 2 \(\mathrm{atm}\) (C) 3 \(\mathrm{atm}\) (D) 4 \(\mathrm{atm}\)

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.

In which of the following reactions is entropy increasing? (A) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)\) (B) \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\) (C) \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{HCl}(g)\) (D) \(2 \mathrm{NO}_{2}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free