Predict whether each of the following compounds is molecular or ionic: (a) \(\mathrm{B}_{2} \mathrm{H}_{6},\) (b) \(\mathrm{CH}_{3} \mathrm{OH},\) (c) \(\mathrm{LiNO}_{3},\) (d) \(\mathrm{Sc}_{2} \mathrm{O}_{3}\), (e) \(\mathrm{CsBr}\), (f) \(\mathrm{NOCl},(\mathrm{g}) \mathrm{NF}_{3}\) (h) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\)

Short Answer

Expert verified
(a) B2H6: molecular, (b) CH3OH: molecular, (c) LiNO3: ionic, (d) Sc2O3: ionic, (e) CsBr: ionic, (f) NOCl: molecular, (g) NF3: molecular, (h) Ag2SO4: ionic

Step by step solution

01

(a) B2H6

Elements B (boron) and H (hydrogen) are both non-metals, and their electronegativity difference is 2.06 - 1.57 = 0.49, which is less than 1.7, indicating that B2H6 is a molecular compound.
02

(b) CH3OH

Elements C (carbon), H (hydrogen), and O (oxygen) are all non-metals, and the electronegativity differences between C-H (0.35), C-O (1.43), and H-O (1.24) are all less than 1.7, indicating that CH3OH is a molecular compound.
03

(c) LiNO3

Elements Li (lithium) and N (nitrogen), O (oxygen) are metal and non-metals, respectively. The electronegativity difference between Li and N (1.94), Li and O (2.08) demonstrate that LiNO3 is an ionic compound.
04

(d) Sc2O3

Elements Sc (scandium) is a metal, and O (oxygen) is a non-metal. The electronegativity difference between Sc and O (1.61) is close to 1.7. In this case, Sc2O3 is considered an ionic compound as it consists of a metal and non-metal.
05

(e) CsBr

Elements Cs (cesium) is a metal, and Br (bromine) is a non-metal. The electronegativity difference between Cs and Br (2.36) is greater than 1.7, indicating that CsBr is an ionic compound.
06

(f) NOCl

Elements N (nitrogen), O (oxygen), and Cl (chlorine) are all non-metals, and their electronegativity differences (N-O: 0.51, N-Cl: 0.64, O-Cl: 0.13) are all less than 1.7, indicating that NOCl is a molecular compound.
07

(g) NF3

Elements N (nitrogen) and F (fluorine) are non-metals, and their electronegativity difference (0.98) is less than 1.7, indicating that NF3 is a molecular compound.
08

(h) Ag2SO4

Elements Ag (silver) is a metal, and S (sulfur), O (oxygen) are non-metals. The electronegativity difference between Ag and S (1.45), Ag and O (1.75), indicates that Ag2SO4 is an ionic compound. In summary, compounds (a), (b), (f), and (g) are molecular compounds, while (c), (d), (e), and (h) are ionic compounds.

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.

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

The first atoms of seaborgium (Sg) were identified in 1974 . The longest-lived isotope of Sg has a mass number of \(266 .\) (a) How many protons, electrons, and neutrons are in an \({ }^{266} \mathrm{Sg}\) atom? (b) Atoms of Sg are very unstable, and it is therefore difficult to study this element's properties. Based on the position of Sg in the periodic table, what element should it most closely resemble in its chemical properties?

Name the following ionic compounds: (a) \(\mathrm{Li}_{2} \mathrm{O},\) (b) \(\mathrm{FeCl}_{3}\), (c) \(\mathrm{NaClO}\) (d) \(\mathrm{CaSO}_{3}\), (e) \(\mathrm{Cu}(\mathrm{OH})_{2},\) (f) \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}\), (g) \(\mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\), (h) \(\mathrm{Cr}_{2}\left(\mathrm{CO}_{3}\right)_{3}\) (i) \(\mathrm{K}_{2} \mathrm{CrO}_{4},(\mathrm{j})\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)

Rubidium has two naturally occurring isotopes, rubidium- 85 (atomic mass \(=84.9118\) amu; abundance \(=72.15 \%)\) and rubidium- 87 (atomic mass \(=86.9092\) amu; abundance = \(27.85 \%) .\) Calculate the atomic weight of rubidium.

Millikan determined the charge on the electron by studying the static charges on oil drops falling in an electric field (Figure 2.5). A student carried out this experiment using several oil drops for her measurements and calculated the charges on the drops. She obtained the following data: $$ \begin{array}{cc} \hline \text { Droplet } & \text { Calculated Charge (C) } \\ \hline \text { A } & 1.60 \times 10^{-19} \\ \text {B } & 3.15 \times 10^{-19} \\ \text {C } & 4.81 \times 10^{-19} \\ \text {D } & 6.31 \times 10^{-19} \end{array} $$ (a) What is the significance of the fact that the droplets carried different charges? (b) What conclusion can the student draw from these data regarding the charge of the electron? (c) What value (and to how many significant figures) should she report for the electronic charge?

Because many ions and compounds have very similar names, there is great potential for confusing them. Write the correct chemical formulas to distinguish between (a) calcium sulfide and calcium hydrogen sulfide, (b) hydrobromic acid and bromic acid, (c) aluminum nitride and aluminum nitrite, (d) iron(II) oxide and iron(III) oxide, (e) ammonia and ammonium ion, (f) potassium sulfite and potassium bisulfite, (g) mercurous chloride and mercuric chloride, (h) chloric acid and perchloric acid.

See all solutions

Recommended explanations on Chemistry Textbooks

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