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

From their Lewis structures, determine the number of \(\sigma\) and \(\pi\) bonds in each of the following molecules or ions: (a) \(\mathrm{CO}_{2}\); (b) cyanogen, \((\mathrm{CN})_{2} ;(\mathrm{c})\) formaldehyde, \(\mathrm{H}_{2} \mathrm{CO} ;\) (d) formic acid, \(\mathrm{HCOOH},\) which has one \(\mathrm{H}\) and two \(\mathrm{O}\) atoms attached to \(\mathrm{C}\).

Short Answer

Expert verified
In conclusion, the number of \(\sigma\) and \(\pi\) bonds for each molecule or ion are: - \(CO_2\): 2 \(\sigma\) bonds and 2 \(\pi\) bonds - \((CN)_2\): 2 \(\sigma\) bonds and 4 \(\pi\) bonds - \(H_2CO\): 3 \(\sigma\) bonds and 1 \(\pi\) bond - \(HCOOH\): 4 \(\sigma\) bonds and 1 \(\pi\) bond

Step by step solution

01

CO2:

For CO2, the central atom is Carbon. Each Oxygen atom forms a double bond with Carbon. The Lewis structure for CO2 is: O=C=O
02

(CN)2:

For cyanogen, two CN molecules are triple-bonded with one another. The Lewis structure for cyanogen is: N≡C-C≡N
03

H2CO:

For formaldehyde (H2CO), the central atom is Carbon. Carbon forms a double bond with Oxygen and single bonds with each of the two Hydrogen atoms. The Lewis structure for H2CO is: H-C=O | H
04

HCOOH:

For formic acid (HCOOH), the central atom is Carbon. Carbon forms a single bond with a Hydrogen atom and a double bond with one of the Oxygen atoms. The other Oxygen atom is single bonded to Carbon and also single bonded to a Hydrogen atom. Thus, the Lewis structure for HCOOH is: O || H-C-O-H ##Step 2: Identify the bond types##
05

CO2:

For CO2, we see two double bonds: C=O and C=O.
06

(CN)2:

For cyanogen, we see two triple bonds: C≡N and C≡N.
07

H2CO:

For H2CO, we observe a double bond C=O and two single bonds: C-H and C-H.
08

HCOOH:

For HCOOH, we observe a double bond C=O, and three single bonds: C-H, C-O, and O-H. ##Step 3: Count the total number of \(\sigma\) and \(\pi\) bonds##
09

CO2:

For CO2, the two double bonds consist of two \(\sigma\) bonds and two \(\pi\) bonds.
10

(CN)2:

For cyanogen, the two triple bonds consist of two \(\sigma\) bonds and four \(\pi\) bonds.
11

H2CO:

For H2CO, there is one \(\sigma\) bond and one \(\pi\) bond in the double bond, and two more \(\sigma\) bonds in the single bonds. This totals to three \(\sigma\) bonds and one \(\pi\) bond.
12

HCOOH:

In HCOOH, the double bond consists of one \(\sigma\) bond and one \(\pi\) bond, and the three single bonds are all \(\sigma\) bonds. This totals to four \(\sigma\) bonds and one \(\pi\) bond. In conclusion, the number of \(\sigma\) and \(\pi\) bonds for each molecule or ion is as follows: - CO2: 2 \(\sigma\) bonds and 2 \(\pi\) bonds - (CN)2: 2 \(\sigma\) bonds and 4 \(\pi\) bonds - H2CO: 3 \(\sigma\) bonds and 1 \(\pi\) bond - HCOOH: 4 \(\sigma\) bonds and 1 \(\pi\) 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!

Key Concepts

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

Sigma and Pi Bonds
Sigma (\f\(\sigma\f\)) and pi (\f\(\pi\f\)) bonds are fundamental concepts in molecular structure and chemical bonding.
\f\(\sigma\f\) bonds are the first bonds formed during the chemical bonding process. They are characterized by head-to-head overlap of atomic orbitals along the internuclear axis and are generally stronger than \f\(\pi\f\) bonds. Each single bond in a molecule is a \f\(\sigma\f\) bond.
\f\(\pi\f\) bonds, on the other hand, arise from the side-to-side overlap of p orbitals. They provide added stability to molecules by sharing electron density above and below the \f\(\sigma\f\) bond axis. Double and triple bonds consist of one \f\(\sigma\f\) bond along with one or two \f\(\pi\f\) bonds, respectively.
\f\(\pi\f\) bonds restrict rotation of the bonded atoms due to their electron cloud distribution, while \f\(\sigma\f\) bonds allow for free rotation. These properties are crucial when studying molecular geometry and reactivity.
Molecular Geometry
Understanding molecular geometry is essential when delving into the three-dimensional arrangement of atoms in a molecule. It is determined by the number of bonding pairs and lone pairs of electrons surrounding the central atom.
VSEPR (Valence Shell Electron Pair Repulsion) theory is often used to predict molecular shapes. This theory asserts that electron pairs around a central atom will arrange themselves as far apart as possible to minimize repulsion.
Common molecular geometries include linear, trigonal planar, tetrahedral, and bent shapes. These geometries affect the molecule's physical properties and reactivity. For example, \f\(\text{CO}_2\f\) is a linear molecule, resulting in a nonpolar molecule despite having polar bonds. This underscores the importance of overall molecular geometry in determining the molecular polarity and its interactions with other substances.
Chemical Bonding Analysis
Chemical bonding analysis examines how atoms combine to form molecules. It encompasses the identification of \f\(\sigma\f\) and \f\(\pi\f\) bonds, understanding molecular geometry, and considering electronegativity to assess bond polarity.
In our exercise, the analysis was focused on counting the \f\(\sigma\f\) and \f\(\pi\f\) bonds in different molecules, where the number of these bonds gave insights into the bond order and strength. High bond orders, as seen in multiple \f\(\pi\f\) bonds in cyanogen, imply stronger and shorter bonds.
Moreover, chemical bonding analysis also looks into resonance structures, hybridization state of atoms, and molecular orbitals – all contributing to a comprehensive picture of how electrons are distributed in a molecule.
Lewis Structure Representation
Lewis structure representation is a valuable tool for visualizing the valence electrons in molecules, showing how atoms are bonded, and the presence of lone pairs of electrons.
Developing a correct Lewis structure starts with determining the total valence electrons, arranging atoms with the least electronegative atom often in the center, and distributing electrons to fill the octet rule (or duet for hydrogen).
In complex molecules, there may be multiple valid Lewis structures, called resonance structures, that represent the true distribution of electrons. The structures from our exercise revealed different bonding situations, from single bonds in molecular hydrogen to the triple bonds in cyanogen, providing a direct way to visualize and count \f\(\sigma\f\) and \f\(\pi\f\) bonds, crucial for understanding the molecule's reactivity and properties.

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

Consider the Lewis structure for glycine, the simplest amino acid: (a) What are the approximate bond angles about each of the two carbon atoms, and what are the hybridizations of the orbitals on each of them? (b) What are the hybridizations of the orbitals on the two oxygens and the nitrogen atom, and what are the approximate bond angles at the nitrogen? (c) What is the total number of \(\sigma\) bonds in the entire molecule, and what is the total number of \(\pi\) bonds?

Consider the bonding in an \(\mathrm{MgH}_{2}\) molecule. (a) Draw a Lewis structure for the molecule, and predict its molecular geometry. (b) What hybridization scheme is used in \(\mathrm{MgH}_{2}\) ? (c) Sketch one of the two-electron bonds between an \(\mathrm{Mg}\) hybrid orbital and an \(\mathrm{H} 1 \mathrm{~s}\) atomic orbital.

(a) Draw Lewis structures for ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\), and acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right) .\) (b) What is the hybridization of the carbon atoms in each molecule? (c) Predict which molecules, if any, are planar. (d) How many \(\sigma\) and \(\pi\) bonds are there in each molecule? (e) Suppose that silicon could form molecules that are precisely the analogs of ethane, ethylene, and acetylene. How would you describe the bonding about Si in terms of hydrid orbitals? Silicon does not readily form some of the analogous compounds containing \(\pi\) bonds. Why might this be the case?

(a) Starting with the orbital diagram of a sulfur atom, describe the steps needed to construct hybrid orbitals appropriate to describe the bonding in \(\mathrm{SF}_{2}\). (b) What is the name given to the hybrid orbitals constructed in (a)? (c) Sketch the large lobes of these hybrid orbitals. (d) Would the hybridization scheme in part (a) be appropriate for \(\mathrm{SF}_{4} ?\) Explain.

Would you expect the nonbonding electron-pair domain in \(\mathrm{NH}_{3}\) to be greater or less in size than for the corresponding one in \(\mathrm{PH}_{3}\) ? Explain.

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