/*! 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} Problem 49 Which one of these statements ab... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Which one of these statements about formal charge is true? (a) Formal charge is the same as oxidation number. (b) To draw the best Lewis structure, you should minimize formal charge. (c) Formal charge takes into account the different electronegativities of the atoms in a molecule. (d) Formal charge is most useful for ionic compounds. (e) Formal charge is used in calculating the dipole moment of a diatomic molecule.

Short Answer

Expert verified
(b) is true: Minimize formal charge for the best Lewis structure.

Step by step solution

01

Understanding Formal Charge

Formal charge is a concept in chemistry used to determine the sharing of electrons in a molecule. It assumes equal sharing of electrons between atoms, ignoring differences in electronegativity. Unlike oxidation states, formal charge is not an indication of electron density or atom oxidation levels, but rather it helps in determining the most appropriate Lewis structure by minimizing these charges.
02

Checking Each Statement

Let's examine each statement in the question: (a) "Formal charge is the same as oxidation number." This is false. While both are used in chemistry, formal charges assume equal sharing of electrons, unlike oxidation numbers which consider electronegativity differences. (b) "To draw the best Lewis structure, you should minimize formal charge." This is true. Minimizing formal charges generally leads to the most stable and plausible Lewis structure. (c) "Formal charge takes into account the different electronegativities of the atoms in a molecule." This is false. Formal charges ignore electronegativity. (d) "Formal charge is most useful for ionic compounds." This is false. Formal charge is particularly useful for covalent compounds, not ionic. (e) "Formal charge is used in calculating the dipole moment of a diatomic molecule." This is false. Formal charge is not directly used in dipole moment calculations, which depend on bond polarity.
03

Identifying the Correct Answer

Among the given options, (b) is the only true statement. It is crucial to consider and minimize formal charges when drawing Lewis structures to ensure they are as accurate and representative of the actual molecular structure as possible.

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lewis Structure
Lewis structures are diagrams that represent the bonds between atoms in a molecule and the arrangement of lone pairs of electrons. They are essential for understanding how molecules are constructed.
To draw a Lewis structure, follow these steps:
  • Count the total number of valence electrons for all atoms involved.
  • Place the least electronegative atom in the center (except H, which is always terminal).
  • Use pairs of electrons to form bonds between the central atom and surrounding atoms.
  • Distribute remaining electrons to complete octets on the outer atoms first and then the central atom if needed.
  • Adjust for overall charge by adding or removing electrons accordingly.
Minimizing formal charges in a Lewis structure is key to finding the most stable structure. While drawing Lewis structures, remember that they only depict probable distribution of electrons, not the actual electronic arrangement in molecules.
Oxidation Number
The oxidation number (or oxidation state) of an atom in a compound provides insight into the electron transfer processes. It indicates the degree of oxidation of an atom, ideally reflecting electrons that are not equally shared in a chemical bond. When determining oxidation numbers, consider these rules:
  • The oxidation number of a free element (uncombined) is always 0.
  • For monoatomic ions, the oxidation number is equal to the ion's charge.
  • In compounds, hydrogen is assigned an oxidation state of +1, and oxygen is usually -2, with exceptions like peroxides.
  • The sum of oxidation states for all atoms in a neutral molecule must be zero, while in ions, it equals the ion charge.
Oxidation numbers are particularly useful for identifying the electron flow in redox reactions and determining how oxidation and reduction have occurred in chemical processes.
Electronegativity
Electronegativity is a measure of an atom's ability to attract shared electrons when forming a chemical bond. This tendency is crucial for predicting the nature of bonds between atoms. Some key points about electronegativity include:
  • Elements with higher electronegativities tend to attract electrons more strongly, forming polar covalent bonds.
  • The most electronegative element is fluorine, followed by oxygen, nitrogen, and chlorine.
  • Differences in electronegativity can explain the type of bond formation: nonpolar covalent (similar electronegativities), polar covalent (moderate differences), and ionic (large differences).
Understanding electronegativity aids in predicting molecular behavior, reactivity, as well as the distribution of partial charges within molecules, which directly impacts physical properties and chemical reactions.

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

Illustrated are four ions \(-\mathrm{A}, \mathrm{B}, \mathrm{X},\) and \(\mathrm{Y}\) - showing their relative ionic radii. The ions shown in red carry positive charges: a \(2+\) charge for \(\mathrm{A}\) and a \(1+\) charge for \(\mathrm{B}\). Ions shown in blue carry negative charges: a \(1-\) charge for \(X\) and a \(2-\) charge for \(Y\). (a) Which combinations of these ions produce ionic compounds where there is a 1:1 ratio of cations and anions? (b) Among the combinations in part (a), which leads to the ionic compound having the largest lattice energy? [Section 8.2]

Consider the formate ion, \(\mathrm{HCO}_{2}^{-}\), which is the anion formed when formic acid loses an \(\mathrm{H}^{+}\) ion. The \(\mathrm{H}\) and the two \(\mathrm{O}\) atoms are bonded to the central C atom. (a) Draw the best Lewis structure(s) for this ion. (b) Are resonance structures needed to describe the structure? (c) Would you predict that the \(\mathrm{C}-\mathrm{O}\) bond lengths in the formate ion would be longer or shorter relative to those in \(\mathrm{CO}_{2}\) ?

Draw the Lewis structure for \(\mathrm{NO}^{+} .\) Is the nitrogenoxygen bond in \(\mathrm{NO}^{+}\) longer, shorter, or the same length as the nitrogen-oxygen bond in NO? Explain.

What is the Lewis symbol for each of the following atoms or ions? \((\mathbf{a}) \mathrm{Be},(\mathbf{b}) \mathrm{Rb},(\mathbf{c}) \mathrm{I}^{-},(\mathbf{d}) \mathrm{Se}^{2-} .\)

The compound chloral hydrate, known in detective stories as knockout drops, is composed of \(14.52 \% \mathrm{C}, 1.83 \% \mathrm{H},\) \(64.30 \% \mathrm{Cl}\), and \(13.35 \% \mathrm{O}\) by mass, and has a molar mass of \(165.4 \mathrm{~g} / \mathrm{mol} .(\mathbf{a})\) What is the empirical formula of this substance? \((\mathbf{b})\) What is the molecular formula of this substance? (c) Draw the Lewis structure of the molecule, assuming that the Cl atoms bond to a single \(\mathrm{C}\) atom and that there are a \(\mathrm{C}-\mathrm{C}\) bond and two \(\mathrm{C}-\mathrm{O}\) bonds in the compound.

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.