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Chapter 11: Problem 40

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

Short Answer

Expert verified
The molecular orbital diagram for CaF includes 9 valence electrons filling up σ2s, σ*2s, σ2pz, π2px, and π2py orbitals sequentially. The bond order is 1.5. The bond length of CaF+ would be longer than that of CaF as the removal of an electron decreases bond order, hence increasing bond length.

Step by step solution

01

Identify the valence electrons

Start by identifying the valence electrons in both atoms. Calcium (Ca) is in group 2, hence it has 2 valence electrons. Fluorine (F) is in group 17, hence it has 7 valence electrons. Thus, the total number of valence electrons in CaF is 2 + 7 = 9 electrons.
02

Construct the molecular orbital diagram

Start by drawing the energy levels of atomic orbitals of the individual atoms on the sides, then fill up the molecular orbitals in the center sequentially with the total number of valence electrons identified in step 1. There are 3 types of molecular orbitals formed: sigma, pi and sigma*. Fill up in the sequence σ2s, σ*2s, σ2pz, π2px=π2py, π*2px=π*2py, σ*2pz. Since we have 9 valence electrons, they fill up σ2s(2), σ*2s(2), σ2pz(2), π2px(2), and the remaining 1 electron goes to π2py.
03

Deduce the bond order

First calculate the bond order which is defined as half the difference between number of electrons in bonding and antibonding orbitals. Bond Order = 0.5*(Number of electrons in bonding orbitals - number of electrons in antibonding orbitals) = 0.5*(6-3)=1.5.
04

Compare Bond lengths

When one more electron is removed to form CaF+, it's being removed from a bonding orbital, thus reducing the overall bond order. When bond order decreases, bond length increases. Hence, the bond length of CaF+ will be longer than that of CaF.

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Most popular questions from this chapter

The anion \(I_{4}^{2-}\) is linear, and the anion \(I_{5}^{-}\) is V-shaped, with a \(95^{\circ}\) angle between the two arms of the V. For the central atoms in these ions, propose hybridization schemes that are consistent with these observations.

The Lewis structure of \(\mathrm{N}_{2}\) indicates that the nitrogento-nitrogen bond is a triple covalent bond. Other evidence suggests that the \(\sigma\) bond in this molecule involves the overlap of \(s p\) hybrid orbitals. (a) Draw orbital diagrams for the N atoms to describe bonding in \(\mathrm{N}_{2}\) (b) Can this bonding be described by either \(s p^{2}\) or \(s p^{3}\) hybridization of the \(\mathrm{N}\) atoms? Can bonding in \(\mathrm{N}_{2}\) be described in terms of unhybridized orbitals? Explain.

The paramagnetism of gaseous \(\mathrm{B}_{2}\) has been established. Explain how this observation confirms that the \(\pi_{2 p}\) orbitals are at a lower energy than the \(\sigma_{2 p}\) orbital for \(\mathrm{B}_{2}\)

Consider the molecules \(\mathrm{CO}^{+}\) and \(\mathrm{CN}^{-}\) and use molecular orbital theory to answer the following: (a) Write the molecular orbital configuration of each ion (ignore the 1 s electrons). (b) Predict the bond order of each ion. (c) Which of these ions is paramagnetic? Which is diamagnetic? (d) Which of these ions do you think has the greater bond length? Explain.

Indicate several ways in which the valence-bond method is superior to Lewis structures in describing covalent bonds.
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