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

Dipole moment of \(\mathrm{CO}\) is very small \((0.1 \mathrm{D})\) (inspite of the larger difference in electronegativity between \(\mathrm{C}\) and \(\mathrm{O}\) ) and negative end of the dipole lies near the less electronegative C-atom. It is due to: (a) Occupation of antibonding molecular orbitals (b) Occupation of the lone pairs as well as empty do bitals (c) Occupation of lone-pairs (d) Occupation of empty d-orbitals

Short Answer

Expert verified
The correct answer is (c) Occupation of lone-pairs.

Step by step solution

01

Understanding the Dipole Moment

The dipole moment is a vector quantity indicating the separation of charge in a molecule. A greater difference in electronegativity typically leads to a higher dipole moment.
02

Examining the Molecule CO

Carbon monoxide (CO) has a small dipole moment of 0.1 D. Normally, oxygen, being more electronegative than carbon, should attract electrons more, resulting in a dipole moment towards O.
03

Identifying Charge Localization

The negative end of the dipole being near the carbon atom suggests a charge localization opposite of what is expected based on electronegativity alone.
04

Analyzing Orbital Occupations

This unusual dipole placement can be explained by the electronic configuration; the lone pair of electrons on carbon contributes to the charge distribution, localizing electron density more on carbon.

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.

Carbon Monoxide (CO)
Carbon Monoxide, often abbreviated as CO, is a simple diatomic molecule consisting of one carbon atom and one oxygen atom. Despite its simplicity, the molecule is quite interesting due to its chemical properties. Oxygen is more electronegative than carbon, meaning it tends to attract electrons more strongly. However, what makes CO particularly peculiar is its dipole moment, which is very small at approximately 0.1 Debye.

Usually, we would expect the electron density to be pulled toward the oxygen atom due to its higher electronegativity. But in CO, the small dipole moment indicates a charge localization anomaly. In this molecule, the negative end of the dipole is closer to the carbon atom, not the oxygen. This unusual localization is due to how the molecule's electrons are distributed, influenced by the molecular orbitals they occupy.
Electronegativity
Electronegativity describes the ability of an atom in a molecule to attract shared electrons. The higher an atom's electronegativity, the stronger its pull on electrons. In the periodic table, electronegativity increases from left to right and decreases down a group. In carbon monoxide, oxygen's electronegativity is greater than carbon's, which you'd expect would lead to a dipole moment with the negative pole directed toward oxygen.

However, actual measurements reveal a very small dipole toward carbon instead. This counterintuitive observation can be explained by the molecular orbital theory. Electrons follow orbitals that are influenced by their energy states and how atoms hybridize in a molecule. Hence, electronegativity, while a valuable indicator, doesn't account for every aspect of electronic distribution.
Molecular Orbitals
Molecular orbital theory offers a deeper understanding of how electrons are distributed in molecules like carbon monoxide. In this theory, atomic orbitals combine to form molecular orbitals—these can be bonding, antibonding, or nonbonding, influencing electron distribution and, consequently, molecular properties.

In CO, the molecular orbitals result from the combination of carbon and oxygen's atomic orbitals. Some electrons occupy pi and sigma bonding orbitals, but crucially, the lone pair on carbon occupies a non-bonding molecular orbital. This lone pair plays a significant role in charge localization, causing the electron density to tilt more toward carbon, despite the oxygen's greater electronegativity.
  • Bonding orbitals stabilize the molecule by holding atoms together.
  • Antibonding orbitals work against bond formation.
  • Non-bonding orbitals have no direct effect on bond strength but can affect properties like dipole moments.
Understanding molecular orbitals thus helps clarify why carbon monoxide exhibits such distinct 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

\(\mathrm{BCl}_{3}\) does not exist as dimmer because (a) B has low electronegativity (b) B has small size (c) B has no vacant orbitals (d) B has low electron affinity

Aluminium chloride in acidified aqueous solution forms: (a) Octahedral \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{+3}\) ion (b) Tetrahedral \(\left.[\mathrm{A}]\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{+3}\) ion (c) Octahedral \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{+3}\) ion (d) Tetrahedral \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{+3}\) ion

Which of the following statements about diborane is correct? (1) It contains two 3-centre-2-electron bonds (2) The B atoms in it are sp \(^{3}\) -hybridised (3) The molecule is non-planar (4) All B - H bond lengths are equal due to resonance (a) \(1,2,3\) (b) \(2,3,4\) (c) 2,3 (d) 1,4

The aqueous solution of an inorganic compound \((X)\) gives white precipitate with \(\mathrm{NH}_{4} \mathrm{OH}\) which does not dissolve in excess of \(\mathrm{NH}_{4} \mathrm{OH}\). This aqueous solution also gives white precipitate with \(\mathrm{AgNO}_{3}\) and the precipitate is soluble in dilute \(\mathrm{HNO}_{3} .\) Here \((\mathrm{X})\) is: (a) \(\mathrm{AlCl}_{3}\) (b) \(\mathrm{AlBr}_{3}\) (c) \(\mathrm{AlN}\) (d) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\)

Boron halides behave as Lewis acids because they (a) Have only six electron in the valence shell (b) Have a lone pair of electrons on the \(\mathrm{B}\) atom (c) Are covalent compounds (d) Are ionic compounds

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