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

What is the hybridization of the carbon atom in phosgene, \(\mathrm{Cl}_{2}\) CO? Give a complete description of the \(\sigma\) and \pi bonding in this molecule.

Short Answer

Expert verified
The carbon in phosgene is \(\mathrm{sp^2}\) hybridized, forming \(\sigma\) bonds with both chlorines and oxygen, and a \(\pi\) bond with oxygen.

Step by step solution

01

Determine Carbon Atom's Hybridization

To find the hybridization of the carbon atom in phosgene (ClyC=O), we examine how many attachment points the carbon atom has. There are two chlorine atoms and one oxygen atom connected to carbon. Count both single bonds and double bonds equally as one attachment each. Therefore, the carbon in phosgene has three total points of attachment, indicating \(\mathrm{sp^2}\) hybridization.
02

Description of \(\sigma\) Bonding

\(\sigma\) bonds are formed by the overlap of hybrid orbitals. In phosgene, the carbon atom is \(\mathrm{sp^2}\) hybridized, so it has three \(\mathrm{sp^2}\) orbitals available. These orbitals overlap with one 3p orbital from each chlorine atom and an \(\mathrm{sp^2}\) or \(\text{lone pair }2p\) orbital from the oxygen. Therefore, all three \(\mathrm{sp^2}\) hybrid orbitals from carbon form \(\sigma\) bonds: two with the chlorine atoms and one with the oxygen atom.
03

Description of \(\pi\) Bonding

\(\pi\) bonds are formed by the lateral overlap of unhybridized p orbitals. In phosgene, the carbon-oxygen double bond consists of one \(\sigma\) bond and one \(\pi\) bond. The \(\pi\) bond forms by the side-by-side overlap of a p orbital reserved on the carbon and a compatible p orbital on the oxygen atom.
04

Electron Geometry Verification

Verify \(\mathrm{sp^2}\) hybridization by examining the molecular geometry of phosgene. The carbon is central, bound to two chlorines and one oxygen, making the electron geometry trigonal planar when considering the three attachment points. This geometry is consistent with \(\mathrm{sp^2}\) hybridization.

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 Bonding
Understanding sigma (\( \sigma \)) and pi (\( \pi \)) bonding is crucial in hybridization. Sigma bonds are the primary bonds formed by direct overlap of orbitals. In phosgene, the carbon atom is \( \mathrm{sp^2} \) hybridized, which means it holds three \( \mathrm{sp^2} \) hybrid orbitals. These orbitals engage in sigma bonding with two chlorine atoms and one oxygen atom.
  • Each \( \mathrm{C-Cl} \) bond is a sigma bond formed by the overlap of a carbon \( \mathrm{sp^2} \) orbital and a chlorine 3p orbital.
  • The \( \mathrm{C=O} \) double bond consists of one sigma bond between a carbon \( \mathrm{sp^2} \) orbital and an oxygen \( \mathrm{sp^2} \) or lone pair 2p orbital.
  • For the pi bond, the unhybridized p orbital of carbon overlaps laterally with a p orbital of oxygen.
Sigma bonds are stronger than pi bonds, as they involve head-on overlapping. Pi bonds, however, provide extra connection strength by adding a sideways overlap. Both bonding types work together to stabilize molecules.
sp2 Hybridization
Hybridization refers to the mixing of atomic orbitals to create new hybrid orbitals. In phosgene, carbon's \( \mathrm{sp^2} \) hybridization results from mixing one 2s and two 2p orbitals, forming three equivalent \( \mathrm{sp^2} \) orbitals. These are arranged around the carbon for optimal bonding, resulting in a trigonal planar formation typical of this hybridization type.
  • The trigonal planar arrangement is due to the equal energy level and geometric spacing of these hybrid orbitals.
  • This hybridization allows carbon to have one remaining unhybridized p orbital, available for \( \pi \) bonding with oxygen.
  • Each \( \mathrm{sp^2} \) hybrid orbital forms a sigma bond with either chlorine or oxygen.
The concept of \( \mathrm{sp^2} \) hybridization helps explain the carbon's ability to form versatile compounds with different atoms.
Trigonal Planar Geometry
Trigonal planar geometry is a structural arrangement where a central atom, like carbon in phosgene, is connected to three atoms positioned at the corners of an equilateral triangle. These atoms, including two chlorine and one oxygen, create a 120-degree angle between each bond.
  • Such a structure is indicative of \( \mathrm{sp^2} \) hybridization.
  • Each peripheral atom in this setup lies in the same plane as the central atom.
  • The consistent angles and planar arrangement are essential for molecular stability and reactivity.
Trigonal planar geometry ensures there is minimal repulsion between electrons in the bonds by maximizing their separation, which is crucial for understanding molecular shapes and electron distribution.
Electron Geometry Verification
Verifying electron geometry involves matching predicted and actual shapes of molecules. For phosgene, this verification confirms \( \mathrm{sp^2} \) hybridization through its trigonal planar electron arrangement.
  • Carbon in phosgene maintains three points of attachment—two chlorines and one oxygen, consistent with \( \mathrm{sp^2} \) hybridization.
  • The overall shape should correspond to the trigonal planar model, validating the calculated hybrid state.
  • Observing bond angles also reinforces symmetry, further confirming the predicted geometry.
Accurate electron geometry verification is vital because it connects theoretical understanding with observable chemical behavior, ensuring correct predictions about molecular interactions 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

Draw a Lewis structure for diimide, \(\mathbf{H}-\mathbf{N}=\mathbf{N}-\mathbf{H}\) Then, using valence bond theory, describe the bonding in this compound. What orbitals overlap to form the bond between nitrogen atoms in this compound?

Consider the following list of small molecules and ions: \(\mathrm{C}_{2}, \mathrm{O}_{2}^{-}, \mathrm{CN}^{-}, \mathrm{O}_{2}, \mathrm{CO}, \mathrm{NO}, \mathrm{NO}^{+}, \mathrm{C}_{2}^{2-}, \mathrm{OF}^{-} .\) Identify (a) all species that have a bond order of 3 (b) all species that are paramagnetic (c) species that have a fractional bond order

Carbon dioxide \(\left(\mathrm{CO}_{2}\right),\) dinitrogen monoxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) the azide ion \(\left(\mathrm{N}_{3}^{-}\right),\) and the cyanate ion (OCN^-) have the same arrangement of atoms and the same number of valence shell electrons. However, there are significant differences in their electronic structures. (a) What hybridization is assigned to the central atom in each species? Which orbitals overlap to form the bonds between atoms in each structure. (b) Evaluate the resonance structures of these four species. Which most closely describe the bonding in these species? Comment on the differences in bond lengths and bond orders that you expect to see based on the resonance structures.

Specify the electron-pair and molecular geometry for each underlined atom in the following list. Describe the hybrid orbital set used by this atom in each molecule or ion. (a) \(\underline{\mathrm{BBr}}_{\mathrm{s}}\) (b) \(\underline{\mathrm{CO}}_{2}\) (c) \(\underline{\mathrm{CH}}_{2} \mathrm{Cl}_{2} \quad\) (d) \(\underline{\mathrm{CO}}_{3}^{2-}\)

Bromine and fluorine react at temperatures higher than \(150^{\circ} \mathrm{C}\) to give a compound that is \(45.69 \%\) Br and \(54.31 \% \mathrm{F}\) (a) What is the empirical formula of the compound? (b) Assuming the molecular formula of the compound is the same as its empirical formula, suggest a structure for the molecule. What is the Br atom hybrid. ization in the molecule? (c) The molecule has a small dipole moment. Does this agree with your structural proposal in (b) above? Why or why not? If it does not agree, can you propose an alternative structure?

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