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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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