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Chapter 10: Problem 51

Use molecular orbital theory to explain why the \(\mathrm{Be}_{2}\) molecule does not exist

Short Answer

Expert verified
According to Molecular Orbital Theory, after filling up the Molecular Orbital diagram for Be2, the bond order comes out to be 0 which indicates that no bond forms between the atoms. Hence, Be2 molecule does not exist.

Step by step solution

01

Discuss Be Atomic Orbitals

Beryllium (Be) has an atomic number of 4. Therefore, its electronic configuration is 1s² 2s². For Be2, the outermost orbitals of Be atoms (2s orbitals) overlap to form Molecular Orbitals.
02

Construct the MO diagram

In the MO diagram for this molecule, 2s orbitals from each Be atom combine to form two molecular orbitals, one bonding (marked as \(\sigma_{2s}\)) and one antibonding (marked as \(\sigma_{2s}^{*}\)). In a simple molecular orbital diagram like this, the bonding orbitals are placed below the corresponding antibonding orbitals.
03

Filling of the MOs

Two electrons each from both Be atoms, totaling 4 electrons, are to be filled in the molecular orbitals now. The first two electrons will go to the \(\sigma_{2s}\) orbital. The next two electrons will fill the \(\sigma_{2s}^{*}\) orbital.
04

Calculate Bond Order

The bond order is given by the formula: \[Bond Order = 0.5×(Number of electrons in bonding orbitals – Number of electrons in antibonding orbitals)\], Substituting in the values we obtained, \[Bond order = 0.5×(2-2) = 0\]. A bond order of 0 means that no bond forms between the atoms, indicating that the Be2 molecule does not exist.

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Most popular questions from this chapter

Specify which hybrid orbitals are used by carbon atoms in these species: (a) \(\mathrm{CO},\) (b) \(\mathrm{CO}_{2},\) (c) \(\mathrm{CN}^{-}\).

Define these terms: bonding molecular orbital, antibonding molecular orbital, pi molecular orbital, sigma molecular orbital.

Draw a molecular orbital energy level diagram for each of these species: \(\mathrm{He}_{2}, \mathrm{HHe}, \mathrm{He}_{2}^{+} .\) Compare their relative stabilities in terms of bond orders. (Treat HHe as a diatomic molecule with three electrons.)

The compound 1,2 -dichloroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\right)\) is nonpolar, while cis-dichloroethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2}\right)\) has a dipole moment: The reason for the difference is that groups connected by a single bond can rotate with respect to each other, but no rotation occurs when a double bond connects the groups. On the basis of bonding considerations, explain why rotation occurs in \(1,2-\) dichloroethane but not in \(c i s\) -dichloroethylene.

Sketch the shapes of these molecular orbitals: \(\sigma_{1 s}\), \(\sigma_{1 s}^{\star}, \pi_{2 p},\) and \(\pi_{2 p}^{\star}\) How do their energies compare?
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