Answer to Question #141963 in General Chemistry for gia

Question

Assume that the third-period element phosphorus forms a diatomic molecule, P2, in an analogous way as nitrogen does to form N2. (a) Write the electronic configuration for P2. Use [Ne2] to represent the electron configuration for the first two periods. (b) Calculate its bond order. (c) What are its magnetic properties (diamagnetic or paramagnetic)?

Answer 1

(a) Let's write the electronic configuration of phosphorus:

$[Ne] 3s^2 3p^3$

where the core shell (1s, 2s, 2p) is covered by [Ne] noble gas configuration and 5 valence electrons are located on 3s and 3p levels.

(b) To calculate the bond's order we have to look at the number of unpaired valence electrons. By the Pauli principle, the electrons occupy orbitals by the energy grows. The 3s level can contain maximum 2 electrons and it is completed (look at the electronic configuration (a)), and 3p level contains 3 unpaired electrons which means that there are 3 electrons more necessary to fill the 3p level. This is showing that the most probable bond order is 3.

(c) To check which of the magnetic properties the $P_2$ molecule has, we have to look at the number of unpaired electrons. From (a) and (b) we already know that there are 3 unpaired electrons in P. As soon as we have 2 phosphorus, it means that all the electrons will interact to form a triple bond, thus the number of unpaired electrons is zero. Because there are no unpaired electrons, the $P_2$ molecule is diamagnetic.