CH 301 Lecture Notes - Lecture 13: Sodium Chloride, London Dispersion Force, Hydrogen Bond

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19 Oct 2016

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Sigma bond vs sigma antibond: antibonding orbitals have an , bonding orbital: constructed by adding the mathematical functions for the two 1s atomic orbitals, antibonding molecular orbital: formed by subtracting one of these functions from the other. The number of mos = the number of atomic orbitals used: 2. For every bonding mo (lower energy) there is an anti-bonding orbital (higher energy: 3. Electrons enter the lowest orbital available (aufbau: 4. The maximum # of electrons in an orbital is 2 (pauli exclusion: 5. Electrons spread out before pairing up (hund"s rule) Bond order = bonding electrons antibonding electrons / 2. Off axis or noble plane = means a pi bond. Intermolecular forces: inter: forces between molecules, intra: bonds within a molecule holding it together, intermolecular forces are weaker than intramolecular forces, keys to imf: Polarizability: all affect physical properties such as mp or bp. Intermolecular forces are generally important for condensed phases.

Related Questions

Which of the following statements relating to Molecular Orbital (MO) Theory is incorrect?

Combination of two 2p orbitals may result in either sigma() or pi() MOs.
A species with a bond order of zero will not be stable.
In a stable molecule having an even number of electrons, all electrons must be paired.
Combination of two atomic orbitals produces one bonding and one antibonding MO.
A bonding MO is lower in energy than the two atomic orbitals from which it is formed.

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The molecular orbitals associated with the Ï-bonding MOs insquare cyclobutadiene (C4H4) are, like benzene, form by overlap ofa single p-orbital on each carbon atom that is not involved in thesigma-bonding network of the molecule. These p-orbitals pointperpendicularly with respect to the plane defined by the carbonatoms. There is no central atom, so the only interactions thatmatter are between the p-orbitals among themselves. These MOs haveexactly the same phase relationship as the SALCs you just made inquestion 1. i) Use your results from question 1 to draw the fourÏ-orbitals of C4H4. ii) Use the idea that the energy of a MOincreases with number of new nodes made during bonding (i.e. nodesnot already present in initial AO functions) to sketch a molecularorbital energy level diagram for the Ï-molecular orbitals of C4H4.Remember that, by definition, degenerate orbitals are identicalexcept for trivial rotation in space, and have identical energy.iii) Fill the MOs with the electrons from the atomic p-orbitalsthat you used to form the Ï-MOs of C4H4. Use your diagram todetermine the overall bond order by considering each doublyoccupied bonding MO to contribute 1, each nonbonding orbital 0 andeach antibonding orbital -1. The result would be divided by 4 C-Cbonds to get the average contribution to each C-C bond fromÏ-bonding, which then should be added to a Ï-bonding contributionof 1.0. iv) Repeat step (iii) for the dication C4H4 2+ to determinethe net bond order for this species. Why do you think you got thisvalue? v) Predict which of C4H4 and C4H4 2+ could be observed byESR spectroscopy. vi) Which of C4H4 2+ and C4H4 is more likely tobe susceptible to electrophilic attack.

CH 301 Lecture Notes - Lecture 13: Sodium Chloride, London Dispersion Force, Hydrogen Bond

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