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The number of unpaired valence electrons in an atom of phosphorus is:

Last updated date: 24th Jul 2024
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Hint: We know that we shall write the electronic configuration of each of these atoms one by one, then we can figure out the number of unpaired electrons by drawing the box diagram for the outer electronic configuration of each element.

Complete answer:
Let us begin with phosphorus. Its atomic number is \[15.\] So its electronic configuration will be $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{3}}$ The p-subshell has \[3\] orbitals and the electrons fill in accordance with Hund’s rule of maximum multiplicity, that is, each degenerate orbital (meaning in cases of orbitals having same energy) is singly occupied by an electron and only after all the orbitals are singly occupied, electron pairing occurs. When we look at the outer electronic configuration of phosphorus, it is $3{{s}^{2}}3{{p}^{3}}$.
Also we know that phosphorus also has a vacant \[3d-\]orbital. In an excited state, electrons are excited from lower energy level to higher energy level. Here in phosphorus in an excited state an electron is excited from \[3s\] to \[3d\] orbit. The electron configuration, the \[3p\] sublevel has room for three more electrons. The orbital filling diagram and the electron dot diagram show the empty spaces for three more electrons and how there are three electrons that aren't paired.

Therefore, the correct answer is option B.

Remember that Phosphorus belongs to the nitrogen family and it can form five bonds or its valency is five in an excited state. But nitrogen forms a maximum of four bonds that are in $NH_{4}^{+}$ . This is because nitrogen belongs to the second period and it does not have vacant d-orbital in its valence shell.