Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Which is the most polar bond? (a) \({\rm{C - C}}\) (b) \({\rm{C - H}}\) (c) \({\rm{N - H}}\) (d) \({\rm{O - H}}\) (e) \({\rm{Se - H}}\) .

Short Answer

Expert verified

The most polar bond is: \({\rm{O - H}}\).

Step by step solution

01

Define Chemical Bonding

A chemical bond is a long-term attraction between atoms, ions, or molecules that allows chemical compounds to form.

02

Explanation

The polarity of a bond is determined by the electronegativity difference between two atoms creating it.

We'll start by looking at the electronegativity values of each element in fig. \({\rm{7}}{\rm{.6}}\), and then compute the electronegativity difference.

03

Finding which one is most polar bond?

a. As both atoms have the same electronegativity, the difference in electronegativity is \({\rm{0}}\).

Therefore, it is not the most polar bond.

04

Finding which one is most polar bond?

b.

\({\rm{C}}\) has an electronegativity of \({\rm{2}}{\rm{.5}}\) and \({\rm{H}}\) has an electronegativity of \({\rm{2}}{\rm{.1}}\), so the difference is \({\rm{0}}{\rm{.4}}\).

Therefore, it is not the most polar bond.

05

Finding which one is most polar bond?

c.

\({\rm{N}}\) has an electronegativity of \({\rm{3}}{\rm{.0}}\) and \({\rm{H}}\) has an electronegativity of \({\rm{2}}{\rm{.1}}\), so the difference is \({\rm{0}}{\rm{.9}}\).

Therefore, it is not the most polar bond.

06

Finding which one is most polar bond?

d.

\({\rm{O}}\) has an electronegativity of \({\rm{3}}{\rm{.5}}\) and \({\rm{H}}\) has an electronegativity of \({\rm{2}}{\rm{.1}}\), the difference is \({\rm{1}}{\rm{.4}}\).

Therefore, it is the most polar bond.

07

Finding which one is most polar bond?

e.

\({\rm{Se }}\)has an electronegativity is \({\rm{2}}{\rm{.4}}\) and \({\rm{H}}\) has an electronegativity is \({\rm{2}}{\rm{.1}}\), so the difference is \({\rm{0}}{\rm{.3}}\).

Therefore, it is not the most polar bond.

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

Explain how a molecule that contains polar bonds can be nonpolar.

Question: When a molecule can form two different structures, the structure with the stronger bonds is usually the more stable form. Use bond energies to predict the correct structure of the hydroxylamine molecule:

Correct the following statement: โ€œThe bonds in solid\({\text{PbC}}{{\text{I}}_{\text{2}}}\)are ionic; the bond in a\({\text{HCl}}\)molecule is covalent. Thus, all of the valence electrons in\({\text{PbC}}{{\text{I}}_{\text{2}}}\)are located on the\({\text{C}}{{\text{I}}^{\text{ - }}}\)ions, and all of the valence electrons in a\({\text{HCl}}\)molecule are shared between the\({\text{H}}\)and\({\text{CI}}\)atoms.โ€

Identify the atoms that correspond to each of the following electron configurations. Then, write the Lewis symbol for the common ion formed from each atom: (a)\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{5}}}\)(b)\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}\)(c)\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{d}}^{{\text{10}}}}\)(d)\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{d}}^{{\text{10}}}}{\text{4}}{{\text{p}}^{\text{4}}}\)(e)\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{d}}^{{\text{10}}}}{\text{4}}{{\text{p}}^{\text{1}}}\).

Write the Lewis symbols of the ions in each of the following ionic compounds and the Lewis symbols of the atom from which they are formed: (a) \({\rm{MgS}}\) (b) \({\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}\) (c) \({\rm{GaC}}{{\rm{l}}_{\rm{3}}}\) (d) \({{\rm{K}}_{\rm{2}}}{\rm{O}}\) (e) \({\rm{L}}{{\rm{i}}_{\rm{3}}}{\rm{N}}\) (f) \({\rm{KF}}\) .

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