Construct the molecular orbital diagram for CF. Would you expect the bond length of \(\mathrm{CF}^{+}\) to be longer or shorter than that of CF?

Short Answer

Expert verified
Yes, the bond length of \(\mathrm{CF}^{+}\) is expected to be longer than that of CF.

Step by step solution

01

Identify the Atomic Orbitals (AOs)

Starting with the carbon atom which is in the second period, Carbon (C) has 1s, 2s, and 2p atomic orbitals. Fluorine (F), also in the second period, has the same 1s, 2s, and 2p atomic orbitals.
02

Overlap the AOs to form MOs

Overlap these atomic orbitals to form molecular orbitals. Atomic s orbitals overlap to form sigma molecular orbitals, while p orbitals can form both sigma and pi orbitals. Remember that each pairwise overlap generates one bonding and one antibonding orbital. This would give us \(\sigma_{2s}\), \(\sigma_{2s}^*\), \(\sigma_{2p}\), \(\pi_{2p}\), \(\pi_{2p}^*\), and \(\sigma_{2p}^*\) molecular orbitals.
03

Fill the MOs with Electrons

Using the Aufbau principle (fill from lowest orbital up) and the Pauli exclusion principle (maximum of two electrons per orbital with opposite spins), populate the molecular orbitals. CF has 6 valence electrons from C and 7 from F, for a total of 13 valence electrons (\(C:[He]2s^{2}2p^{2}\), F:\([Ne]2s^{2}2p^{5}\)). In CF+\, we have one less electron, so that's 12 valence electrons. The order of filling would be \(\sigma_{2s}\), \(\sigma_{2s}^*\), \(\sigma_{2p}\), \(\pi_{2p}\), \(\pi_{2p}^*\), and \(\sigma_{2p}^*\).
04

Interpret the Impact on Bond Length

A decrease in the number of bonding electrons or an increase in the number of antibonding electrons would increase bond length. Here, we notice that losing an electron from CF to form CF+ involves removing a bonding electron, implying a lesser bond attraction and therefore a longer bond length in CF+ compared to CF.

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!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free