Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Q21 E (page 392)

Identify the more polar bond in each of the following pairs of bonds: (a) \({\rm{HF or HCl}}\) (b) \({\rm{NO or CO}}\) (c) \({\rm{SH or OH}}\) (d) \({\rm{PCl or HCl}}\) (e) \({\rm{CH or NH}}\) (f) \({\rm{SO or PO}}\) (g) \({\rm{CN or NN}}\) .

Short Answer

Expert verified

The more polar bond in each is:

  1. \({\rm{HF}}\) is more polar bond.
  2. \({\rm{CO}}\)is more polar bond.
  3. \({\rm{OH}}\)is more polar bond.
  4. \({\rm{PCl}}\)is more polar bond.
  5. \({\rm{NH}}\)is more polar bond.
  6. \({\rm{PO}}\)is more polar bond.
  7. \({\rm{CN}}\) is more polar bond.

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

Identifying the most polar bond in each pair

(a)

\({\rm{HF}}\): Difference in electronegativity is: \({\rm{4}}{\rm{.0 - 2}}{\rm{.1 = 1}}{\rm{.9}}\).

\({\rm{HCl}}\): Difference in electronegativity is:\({\rm{3}}{\rm{.0 - 2}}{\rm{.1 = 0}}{\rm{.9}}\).

Therefore, \({\rm{HF}}\) is more polar than \({\rm{HCl}}\).

04

Identifying the most polar bond in each pair

(b)

\({\rm{NO}}\): The difference in electronegativity is:\({\rm{3}}{\rm{.5 - 3}}{\rm{.0 = 0}}{\rm{.5}}\).

\({\rm{CO}}\): The difference in electronegativity is:\({\rm{3}}{\rm{.5 - 2}}{\rm{.5 = 1}}{\rm{.0}}\).

Therefore, \({\rm{CO}}\) has a higher polarity than \({\rm{NO}}\).

05

Identifying the most polar bond in each pair

(c)

\({\rm{SH}}\): Difference in electronegativity is:\({\rm{2}}{\rm{.5 - 2}}{\rm{.1 = 0}}{\rm{.4}}\).

\({\rm{OH}}\): Difference in electronegativity is:\({\rm{3}}{\rm{.5 - 2}}{\rm{.1 = 1}}{\rm{.4}}\).

Therefore, \({\rm{OH}}\) is more polar than \({\rm{SH}}\).

06

Identifying the most polar bond in each pair

(d)

\({\rm{PCl}}\): Difference in electronegativity is: \({\rm{3}}{\rm{.0 - 2}}{\rm{.1 = 0}}{\rm{.9}}\).

\({\rm{SCl}}\): Difference in electronegativity is:\({\rm{3}}{\rm{.0 - 2}}{\rm{.5 = 0}}{\rm{.5}}\).

Therefore, \({\rm{PCl}}\) is more polar than \({\rm{SCl}}\).

07

Identifying the most polar bond in each pair

(e)

\({\rm{CH}}\): Difference in electronegativity is: \({\rm{2}}{\rm{.5 - 2}}{\rm{.1 = 0}}{\rm{.4}}\).

\({\rm{NH}}\): Difference in electronegativity is:\({\rm{3}}{\rm{.0 - 2}}{\rm{.1 = 0}}{\rm{.9}}\).

Therefore, \({\rm{NH}}\) is more polar than \({\rm{CH}}\).

08

Identifying the most polar bond in each pair

(f)

\({\rm{SO}}\): Difference in electronegativity is: \({\rm{3}}{\rm{.5 - 2}}{\rm{.5 = 1}}{\rm{.0}}\).

\({\rm{PO}}\): Difference in electronegativity is:\({\rm{3}}{\rm{.5 - 2}}{\rm{.1 = 1}}{\rm{.4}}\).

Therefore, \({\rm{PO}}\) is more polar than \({\rm{SO}}\).

09

Identifying the most polar bond in each pair

(g)

\({\rm{CN}}\): Difference in electronegativity is: \({\rm{3}}{\rm{.0 - 2}}{\rm{.5 = 0}}{\rm{.5}}\).

\({\rm{NN}}\): As, both atoms have the same electronegativity, the difference in electronegativity is zero.

Therefore, \({\rm{CN}}\) is more polar than \({\rm{NN}}\).

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

Predict the electron pair geometry and the molecular structure of each of the following:

  1. \({\rm{IO}}{{\rm{F}}_{\rm{5}}}\)(I is the central atom)
  2. \({\rm{POC}}{{\rm{l}}_{\rm{3}}}\)(P is the central atom)
  3. \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{SeO}}\)(Se is the central atom)
  4. \({\rm{ClS}}{{\rm{O}}^{\rm{ + }}}\)(S is the central atom)
  5. \({{\rm{F}}_{\rm{2}}}{\rm{SO}}\)(S is the central atom)
  6. \({\rm{N}}{{\rm{O}}_{\rm{2}}}^{\rm{ - }}\)
  7. \({\rm{SiO}}_{\rm{4}}^{{\rm{4 - }}}\)

From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity: (a)\({\rm{As, H, N, P, Sb}}\)(b)\({\rm{Cl, H, P, S, Si}}\)(c)\({\rm{Br, Cl, Ge, H, Sr}}\)(d)\({\rm{Ca, H, K, N, Si}}\)(e)\({\rm{Cl, Cs, Ge, H, Sr}}\).

Use the simulation (http://openstaxcollege.org/l/16MolecPolarity) to perform the following exercises for a two-atom molecule: (a) Adjust the electronegativity value so the bond dipole is pointing toward B. Then determine what the electronegativity values must be to switch the dipole so that it points toward A. (b) With a partial positive charge on A, turn on the electric field and describe what happens. (c) With a small partial negative charge on A, turn on the electric field and describe what happens. (d) Reset all, and then with a large partial negative charge on A, turn on the electric field and describe what happens.

Predict the electron pair geometry and the molecular structure of each of the following molecules or ions:

(a)\({\rm{S}}{{\rm{F}}_{\rm{6}}}\)

(b)\({\rm{PC}}{{\rm{l}}_{\rm{5}}}\)

(c)\({\rm{Be}}{{\rm{H}}_{\rm{2}}}\)

(d) \({\rm{CH}}_{\rm{3}}^{\rm{ + }}\)

Which compound in each of the following pairs has the larger lattice energy? Note: \({\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}\) and \({{\rm{K}}^{\rm{ + }}}\) have similar radii; \({{\rm{S}}^{{\rm{2 - }}}}\) and \({\rm{C}}{{\rm{l}}^{\rm{ - }}}\) have similar radii. Explain your choices.

(a) \({{\rm{K}}_{\rm{2}}}{\rm{O}}\) or \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{O}}\)

(b) \({{\rm{K}}_{\rm{2}}}{\rm{S}}\) or \({\rm{BaS}}\)

(c) \({\rm{KCl}}\) or \({\rm{BaS}}\)

(d) \({\rm{BaS}}\) or \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

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