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Chapter 10: Problem 77

Write the ground-state electron configuration for \(\mathrm{B}_{2}\). Is the molecule diamagnetic or paramagnetic?

Short Answer

Expert verified
The ground-state electron configuration for \(\mathrm{B}_{2}\) is \(\sigma_{2s}^2\), \(\sigma_{2s}^*^2\), \(\pi_{2p}^6\). The molecule is diamagnetic.

Step by step solution

01

Identify the number of electrons for the Boron molecule

A single Boron atom has 5 electrons. Thus, a B2 molecule, consisting of two Boron atoms, will have 10 electrons (5 from each atom).
02

Arrange the electrons in molecular orbitals

To determine the ground-state electron configuration of B2, we need to fill the molecular orbitals in order of increasing energy level until all 10 electrons are placed. According to the Aufbau principle, we start from the lowest-energy orbital. For B2, the order is as follows: \(\sigma_{2s}\), \(\sigma_{2s}^*\), \(\pi_{2p}\), \(\sigma_{2p}\). The \(\sigma^*_{2s}\) orbital denotes an antibonding orbital. Two of the 10 electrons will occupy the \(\sigma_{2s}\) orbital and the next two electrons will fill the antibonding \(\sigma^*_{2s}\) orbital. The remaining 6 electrons then fill the \(\pi_{2p}\) orbitals completely.
03

Determine diamagnetic or paramagnetic

Substances that have no unpaired electrons are diamagnetic, while those that have unpaired electrons are paramagnetic. After arranging all the electrons of B2, no electrons are unpaired. Therefore, the B2 molecule is diamagnetic.

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Most popular questions from this chapter

Explain why the bond order of \(\mathrm{N}_{2}\) is greater than that of \(\mathrm{N}_{2}^{+},\) but the bond order of \(\mathrm{O}_{2}\) is less than that of \(\mathrm{O}_{2}^{+}\)

The allene molecule \(\mathrm{H}_{2} \mathrm{C}=\mathrm{C}=\mathrm{CH}_{2}\) is linear (the three \(\mathrm{C}\) atoms lie on a straight line). What are the hybridization states of the carbon atoms? Draw diagrams to show the formation of sigma bonds and pi bonds in allene.

Describe the hybridization state of arsenic in arsenic pentafluoride \(\left(\mathrm{AsF}_{5}\right)\).

Arrange these species in order of increasing stability: \(\mathrm{Li}_{2}, \mathrm{Li}_{2}^{+}, \mathrm{Li}_{2}^{-}\). Justify your choice with a molecular orbital energy level diagram.

The geometries discussed in this chapter all lend themselves to fairly straightforward elucidation of bond angles. The exception is the tetrahedron, because its bond angles are hard to visualize. Consider the \(\mathrm{CCl}_{4}\) molecule, which has a tetrahedral geometry and is nonpolar. By equating the bond moment of a particular \(\mathrm{C}-\mathrm{Cl}\) bond to the resultant bond moments of the other three \(\mathrm{C}-\mathrm{Cl}\) bonds in opposite directions, show that the bond angles are all equal to \(109.5^{\circ}\)
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