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Chapter 10: Problem 95

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 the \(B_2\) molecule is \(1s^2 1s^*2 2s^2 2s^*2 2p^4\). The molecule is paramagnetic because it has two unpaired electrons.

Step by step solution

01

Determine the Atomic Electron Configurations

\[ \text{Boron (B) is the 5th element on the periodic table, and thus has 5 electrons in atomic state. In the ground state, Boron's electron configuration is \(1s^2 2s^2 2p^1\). The \(B_{2}\) molecule therefore consists of two Boron atoms and has a total of 10 electrons.} \]
02

Construct the Molecular Orbital Diagram for B2

\[ \text{Following the Aufbau principle, fill the molecular orbitals beginning from the lowest energy level. First, fill the 1s and \(1s^*\) (sigma and sigma-star) orbitals. Each of these orbitals can hold two electrons, so far four electrons have been placed (2 in 1s and 2 in \(1s^*\)). The remaining six electrons will go into the 2s and 2s* orbitals in the same way. Up to this point, 8 electrons have been placed. The two remaining electrons will go into the degenerate \(2p\) orbitals. One electron goes into each of the degenerate orbitals.} \]
03

Determine Magnetic Nature

\[ \text{The \(B_2\) molecule is considered to be paramagnetic if it has unpaired electrons, and diamagnetic if all electrons are paired. For the \(B_2\) molecule, there are two unpaired electrons in the degenerate \(2p\) orbitals, and therefore, the molecule is paramagnetic.} \]

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