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Chapter 10: Problem 83

Assume that the third-period element phosphorus forms a diatomic molecule, \(\mathrm{P}_{2}\), in an analogous way as nitrogen does to form \(\mathrm{N}_{2}\). (a) Write the electronic configuration for \(\mathrm{P}_{2}\). Use \(\left[\mathrm{Ne}_{2}\right]\) to represent the electron configuration for the first two periods. (b) Calculate its bond order. (c) What are its magnetic properties (diamagnetic or paramagnetic)?

Short Answer

Expert verified
Part (a) - The electronic configuration for \(\mathrm{P}_2\) is \(\mathrm{[Ne]}\sigma_{2s}^{2}\sigma_{2s*}^{2}\sigma_{2p}^{2}\pi_{2p}^{4}\pi_{2p*}^{2}\). Part (b) - The Bond order for \(\mathrm{P}_2\) is 1. Part (c) - The \(\mathrm{P}_2\) molecule is diamagnetic.

Step by step solution

01

Writing the Electronic configuration of Phosphorus

Phosphorus (P) lies in the third period of the Periodic Table in Group 15. Its atomic number is 15. Thus, excluding the inner (K and L) electronic configuration, the valence electrons' configuration of phosphorus (P) is as follows: \( P: 3s^23p^3 \). Considering two phosphorus atoms, in \(\mathrm{P}_2\), the total number of electrons available for molecular orbital filling is thus \(2 \times (3s^23p^3) = 12.\) Thus, the molecular electronic configuration for \(\mathrm{P}_2\) is \(\mathrm{[Ne]}\sigma_{2s}^{2}\sigma_{2s*}^{2}\sigma_{2p}^{2}\pi_{2p}^{4}\pi_{2p*}^{2}\).
02

Calculating the Bond Order

The Bond Order is defined as the difference in the number of electron pairs residing in the bonding and anti-bonding molecular orbitals. Hence, using the formula for bond order we have, \( Bond Order = \frac{1}{2} \times (number\: of\: electrons\: in\: bonding\: molecular\: orbitals\: -\: number\: of\: electrons\: in\: antibonding\: molecular\: orbitals) = \frac{1}{2} \times (6-4) = 1\). Therefore, \(\mathrm{P}_2\) has a bond order of 1.
03

Determining the Magnetic Properties

The magnetic properties of a molecule are determined based on the presence or absence of unpaired electrons in its molecular orbitals. A molecule will be paramagnetic if it has unpaired electrons and diamagnetic if it does not. From the molecular orbital diagram for \(\mathrm{P}_2\), we can see that all the electrons are paired. Hence, \(\mathrm{P}_2\) is diamagnetic.

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