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

Write electron-pair structures for each of the following. Include both bonding and nonbonding pairs and predict the preferred shape of the molecule or ion as linear, triangular (planar), angular, tetrahedral, or pyramidal. a. \(\mathrm{CO}_{2}\) b. \(\mathrm{N} \equiv \mathrm{C}-\mathrm{O}^{\ominus}\) c. \(\mathrm{CH}_{2}=\mathrm{C}=\mathrm{O}\) d. \(\mathrm{CH}_{3}^{+}\) e. \(\mathrm{F}_{2} \mathrm{C}=\mathrm{CH}_{2}\) f. \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{N}\) g. \(\mathrm{SiF}_{4}\) h. \(\mathrm{HCOOH}\) i. \(\mathrm{H}_{3} \mathrm{O}^{\ominus}\) j. \(\mathrm{CH}_{3} \mathrm{SH}\) \(\mathbf{k} . * \mathrm{SO}_{3}\)

Short Answer

Expert verified
CO2: Linear; N≡C-O⁻: Linear; CH2=C=O: Linear; CH3⁺: Triangular planar; F2C=CH2: Trigonal planar; CH3C≡N: Linear; SiF4: Tetrahedral; HCOOH: Trigonal planar; H3O⁻: Trigonal pyramidal; CH3SH: Tetrahedral; SO3: Triangular planar.

Step by step solution

01

Lewis Structure for CO2

The central atom is carbon, which forms double bonds with two oxygen atoms (O=C=O). Both oxygen atoms have two lone pairs each. The molecule is linear.
02

Lewis Structure for N≡C-O⁻

The nitrogen atom triple bonds with carbon and carbon single bonds to oxygen, which bears a negative charge. Nitrogen has one lone pair, carbon shares all valence electrons in bonds, and oxygen has three lone pairs. The shape is linear.
03

Lewis Structure for CH2=C=O

Carbon in the middle forms a double bond with another carbon and an oxygen. Each carbon in the double bond uses all valences in bonds, and oxygen has two lone pairs. The shape is linear around the central carbon.
04

Lewis Structure for CH3⁺

Central carbon bonds to three hydrogen atoms with a positive charge since it lacks a lone pair. The molecular shape is triangular planar.
05

Lewis Structure for F2C=CH2

Central carbon (part of the double bond) bonds to two fluorines and a carbon, and that other carbon connects to two hydrogens. The geometry around the double bond is trigonal planar.
06

Lewis Structure for CH3C≡N

Methyl group (CH3) bonds to a carbon triple-bonded to a nitrogen atom. The geometry around carbon with the triple bond is linear.
07

Lewis Structure for SiF4

Central silicon atom forms single bonds with four fluorine atoms. Each fluorine has three lone pairs. The molecular geometry is tetrahedral.
08

Lewis Structure for HCOOH

Carbon forms a double bond with oxygen and a single bond with OH. The remaining oxygen in OH has two lone pairs, and the geometry around carbon is trigonal planar.
09

Lewis Structure for H3O⁻

Oxygen is the central atom bonded to three hydrogens and one lone pair, carrying a negative charge. The shape of the ion is trigonal pyramidal.
10

Lewis Structure for CH3SH

Carbon forms three single bonds with hydrogen and one with sulfur. Sulfur then bonds to a hydrogen and carries two lone pairs. The shape around carbon is tetrahedral.
11

Lewis Structure for SO3

Sulfur is the central atom, forming double bonds with three surrounding oxygen atoms. Each oxygen has two lone pairs, and the molecule is triangular planar.

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.

Molecular Geometry
Understanding molecular geometry helps us visualize the 3D arrangement of atoms in a molecule, which influences its physical and chemical properties.
The geometry is determined by the number of bonding and nonbonding electron pairs around the central atom, which push against each other.
Common molecular geometries include:
  • Linear: Atoms are arranged in a straight line, like in CO$_2$ and N$ upleeqslant$C-O$^-.$
  • Trigonal Planar: Atoms lie flat on a plane, as seen in CH$_3^+$ and SO$_3.$
  • Tetrahedral: Atoms form a shape like a 3D pyramid with four triangle sides, as shown in SiF$_4.$
  • Trigonal Pyramidal: Similar to a tetrahedral, but with one atom replaced by a lone pair, like in H$_3$O$^-.$
  • Angular: Also known as bent, where two atoms form a V-shape, influenced by lone pairs.
Understanding these shapes is crucial for predicting how molecules interact with each other.
Electron-Pair Structure
Electron-pair structures, commonly known as Lewis Structures, show the arrangement of the electrons within a molecule.
Electrons can be
  • Bonding Pairs: Involved in bonding two atoms together
  • Nonbonding Pairs (Lone Pairs): Not involved in bonding; they belong exclusively to one atom
To draw a Lewis Structure, we:
  • Count all valence electrons available for bonding.
  • Place the least electronegative atom in the center.
  • Form bonds by pairing electrons between atoms and complete octets around atoms when possible.
  • Assign any extra electrons as lone pairs.
For example, in HCOOH, the central carbon atom forms a double bond with an oxygen, illustrating how double bonds share more electrons between atoms than single bonds.
VSEPR Theory
VSEPR stands for Valence Shell Electron Pair Repulsion theory, which helps predict the molecular geometry of molecules.
According to VSEPR, electron pairs around a central atom will arrange themselves as far apart as possible to minimize repulsion.
  • This affects both bonding and nonbonding pairs, meaning the exact molecular shape can differ depending on the number of lone pairs.
  • For example, even with four electron pairs, SiF$_4$ is tetrahedral, while H$_3$O$^-
  • is trigonal pyramidal due to one lone pair.
The presence of lone pairs can significantly affect the molecule's overall shape because lone pairs occupy more space than bonding pairs, therefore pushing the bonds closer together, as seen in angular geometries.
Bonding and Nonbonding Pairs
Atoms in a molecule can be linked by sharing electrons, referred to as bonding pairs, which form part of the molecule's structural framework.
Bonding pairs are responsible for the bonds that hold atoms together, like the double bonds in carbon dioxide ( CO$_2$), where each carbon shares four electrons with two oxygens.
In contrast, nonbonding pairs, or lone pairs, do not contribute to the structure of a molecule directly, but they affect its geometry:
  • They occupy space and repel other electron pairs, causing distortions in the expected geometry.
  • In H$_3$O$^{-}$, the lone pair's repulsion causes a shift from tetrahedral to trigonal pyramidal shape.
Analyzing both types of pairs gives insight into why molecules form certain shapes, which is essential for predicting their chemical behavior and reactivity.

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

Write electron-pair structures including bonding and unshared pairs for each of the following compounds. Predict the preferred shape of the molecule as linear, angular, planar and triangular, tetrahedral, or pyramidal. a. \(\oplus \mathrm{NO}_{2}\) b. \(\mathrm{CS}_{2}\) c. \(\mathrm{O}=\mathrm{C}=\mathrm{C}=\mathrm{O}\) d. \(\mathrm{H}_{2} \mathrm{C}=\mathrm{NH}\) e. \(\mathrm{HN}=\mathrm{NH}\) f. \(\mathrm{CH}_{3}\) g. ClNO h. \(\mathrm{NH}_{2}\) i. \(\mathrm{BH}_{4}^{-}\)

Use electron-repulsion arguments to explain the following: a. The \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angle in \(\mathrm{NH}_{4}^{+}\) is larger than in \(\mathrm{NH}_{3}\). b. The \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angle in \(\mathrm{NH}_{3}\left(107.3^{\circ}\right)\) is larger than the \(\mathrm{F}-\mathrm{N}-\mathrm{F}\) bond angles in \(\mathrm{NF}_{3}\left(102.1^{\circ}\right)\). c. The \(\mathrm{Cl}-\mathrm{C}-\mathrm{Cl}\) angle in \(\mathrm{Cl}_{2} \mathrm{C}=\mathrm{O}\) (phosgene, \(111.3^{\circ}\) ) is less than the \(\mathrm{H}-\mathrm{C}-\mathrm{H}\) angle in \(\mathrm{H}_{2} \mathrm{C}=\mathrm{O}\) (methanal, \(118^{\circ}\) ). d. The \(\mathrm{H}-\mathrm{C}-\mathrm{H}\) angle in methanal \(\left(118^{\circ}\right)\) is greater than the \(\mathrm{H}-\mathrm{C}-\mathrm{H}\) angle in ethene \(\left(116.7^{\circ}\right)\).

The \(\mathrm{P}-\mathrm{H}\) bond distance in \(\mathrm{PH}_{3}\) is \(1.42 \mathrm{~A}\) and the \(\mathrm{N}-\mathrm{H}\) bond distance in \(\mathrm{NH}_{3}\) is \(1.01 \mathrm{~A}\). Use the bond angle of \(93^{\circ}\) for \(\mathrm{PH}_{3}\) and \(107.3^{\circ}\) for \(\mathrm{NH}_{3}\) and calculate the distance between the hydrogens for each molecule. Would you expect the repulsion between the hydrogen nuclei in \(\mathrm{PH}_{3}\), to be more, or less, than in \(\mathrm{NH}_{3}\) ?

Draw an atomic-orbital picture of 1,4 -dichlorobutatriene, \(\mathrm{ClCH}=\mathrm{C}=\mathrm{C}=\mathrm{CHCl}\) Examine your diagram carefully and predict the number and kind of stereoisomers possible for this structure.

The boron orbitals in diborane, \(\mathrm{B}_{2} \mathrm{H}_{6}\), overlap with hydrogen \(1 s\) orbitals in such a way to produce a structure having four ordinary \(\mathrm{B}-\mathrm{H}\) bonds, each of which is an electron-pair bond associating two nuclei. The remaining two hydrogens each are bonded to both boron nuclei through an electron-pair bond associated with three atomic nuclei. This type of bond is referred to as a three-center bond. a. Would you expect diborane to be planar or nonplanar? Explain, using electron-repulsion arguments. b. Make an atomic-orbital diagram for diborane. c. Explain why the terminal \(\mathrm{H}-\mathrm{B}-\mathrm{H}\) angle is larger than the internal \(\mathrm{H}-\mathrm{B}-\mathrm{H}\) angle.

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