/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Ques 53 E Calculate the formal charge of c... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Ques 53 E (page 398)

Calculate the formal charge of chlorine in the molecules \({\rm{C}}{{\rm{l}}_{\rm{2}}}\), \({\rm{BeC}}{{\rm{l}}_{\rm{2}}}\), and \({\rm{Cl}}{{\rm{F}}_{\rm{5}}}\).

Short Answer

Expert verified

The formal charge of \({\rm{Cl}}\) in the compound \({\rm{C}}{{\rm{l}}_{\rm{2}}}\) is \(0\), \({\rm{Be}}\) and \({\rm{Cl}}\) in the compound \({\rm{BeC}}{{\rm{l}}_{\rm{2}}}\) is \(0\)and, \({\rm{Cl}}\) and \({\rm{F}}\) in the compound \({\rm{Cl}}{{\rm{F}}_{\rm{5}}}\) is \(0\).

Step by step solution

01

Concept Introduction

A formal charge (\({\rm{F}}{\rm{.C}}{\rm{.}}\)or\({\rm{q}}\)) is the charge on an atom/molecule in covalent bonding, assuming electrons are shared equally in bonds made.

02

Formal Charge of Chlorine

To find the Formal charge use the formula 鈥�

Formal Charge \({\rm{ = }}\) \({\rm{(}}\)Number of valence electrons on atom) \({\rm{ - }}\) (non-bonded electrons\({\rm{ + }}\) Bonding electrons\({\rm{)}}\).

Formal Charge for \({\rm{Cl}}\) is:

\(\begin{aligned}&= \left( 7 \right){\rm{ - }}\left( {{\rm{6 + 1}}} \right)\\&= 0\end{aligned}\)

The formal charge for\({\rm{C}}{{\rm{l}}_{\rm{2}}}\) is found below 鈥�

Element

Bonding Electrons

Non-bonded Electrons

Valence Electrons

Formal Charge

\({\rm{Cl}}\)

\(1\)

\(6\)

\(7\)

\(0\)

Therefore, the formal charge for \({\rm{Cl}}\)is \(0\).

03

Formal Charge of Beryllium Chloride

To find the Formal charge use the formula 鈥�

Formal Charge\({\rm{ = }}\) \({\rm{(}}\)Number of valence electrons on atom) \({\rm{ - }}\) (non-bonded electrons \({\rm{ + }}\) Bonding electrons\({\rm{)}}\).

Formal Charge for \({\rm{Be}}\) is:

\(\begin{aligned}&= \left( {\rm{2}} \right){\rm{ - }}\left( {{\rm{0 + 2}}} \right)\\&= \end{aligned}\)

Formal Charge for \({\rm{Cl}}\) is:

\(\begin{aligned}&= \left( 7 \right){\rm{ - }}\left( {{\rm{6 + 1}}} \right)\\&= 0\end{aligned}\)

The formal charge for \({\rm{BeC}}{{\rm{l}}_{\rm{2}}}\)is found below 鈥�

Element

Bonding Electrons

Non-bonded Electrons

Valence Electrons

Formal Charge

\({\rm{Be}}\)

\(2\)

\(0\)

\(2\)

\(0\)

\({\rm{Cl}}\)

\(1\)

\(6\)

\(7\)

\(0\)

Therefore, the formal charge for\({\rm{Be}}\) and\({\rm{Cl}}\)is\(0\).

04

Formal Charge of Chlorine Pentafluoride

To find the Formal charge use the formula 鈥�

Formal Charge \({\rm{ = }}\) \({\rm{(}}\)Number of valence electrons on atom)\({\rm{ - }}\) (non-bonded electrons \({\rm{ + }}\) Bonding electrons\({\rm{)}}\).

Formal Charge for\({\rm{Cl}}\) is:

\(\begin{aligned} &= \left( 7 \right){\rm{ - }}\left( {{\rm{2 + 5}}} \right)\\&= 0\end{aligned}\)

Formal Charge for \({\rm{F}}\) is:

\(\begin{aligned}&= \left( 7 \right){\rm{ - }}\left( {{\rm{6 + 1}}} \right)\\&= 0\end{aligned}\)

The formal charge for\({\rm{Cl}}{{\rm{F}}_{\rm{5}}}\) is found below 鈥�

Element

Bonding Electrons

Non-bonded Electrons

Valence Electrons

Formal Charge

\({\rm{Cl}}\)

\(5\)

\(2\)

\(7\)

\(0\)

\({\rm{F}}\)

\(1\)

\(6\)

\(7\)

\(0\)

Therefore, the formal charge for \({\rm{Cl}}\) and\({\rm{F}}\)is \(0\).

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 91影视!

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

Identify the molecules with a dipole moment:

  1. \({\rm{S}}{{\rm{F}}_{\rm{4}}}\)
  2. \({\rm{C}}{{\rm{F}}_{\rm{4}}}\)
  3. \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{CCB}}{{\rm{r}}_{\rm{2}}}\)
  4. \({\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{Cl}}\)
  5. \({{\rm{H}}_{\rm{2}}}{\rm{CO}}\)

For which of the following substances is the least energy required to convert one mole of the solid into separate ions?

(a) \({\rm{MgO}}\)

(b) \({\rm{SrO}}\)

(c) \({\rm{KF}}\)

(d) \({\rm{CsF}}\)

(e) \({\rm{Mg}}{{\rm{F}}_{\rm{2}}}\)

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. \({\rm{SiO}}_{\rm{4}}^{{\rm{4 - }}}\)

Methanol,\({{\text{H}}_{\text{3}}}{\text{COH}}\), is used as the fuel in some race cars. Ethanol,\({{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH}}\), is used extensively as motor fuel in Brazil. Both methanol and ethanol produce\({\text{C}}{{\text{O}}_{\text{2}}}\)and\({{\text{H}}_{\text{2}}}{\text{O}}\)when they burn. Write the chemical equations for these combustion reactions using Lewis structures instead of chemical formulas.

From the labels of several commercial products, prepare a list of six ionic compounds in the products. For each compound, write the formula. (You may need to look up some formulas in a suitable reference.)
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Physical 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.