Question

How many unpaired electrons are there in lead(I)?

Answer 1

Hint: The number of unpaired electrons is evaluated using Hund's rule and Pauli Exclusion Principle. Lead is an element which is positioned in the p-block elements of the periodic table.

Complete step by step answer:
Lead is an element in the periodic table with atomic number $$82$$ and electronic configuration $$[Xe]4f^{14}5d^{10}6s^{2}6p^2$$. The valence shell of lead is $$6$$ which have a total of $$4$$ electrons. Of the four valence electrons two electrons are present in $$6s$$ and two electrons are present in $$6p$$ orbitals.
As the higher orbital is $$p$$-orbital which contains only two electrons is neither a half-filled nor a full-filled orbital, so lead is termed as a $$p$$-block element. According to Hund’s rule and Pauli Exclusion Principle the two electrons must reside on two orbitals as singly filled. The $$p$$-orbitals contain a total of three orbitals. Thus the lead atom contains two unpaired electrons.

When lead releases an electron and forms lead(I), an electron is lost from the p orbital. Thus lead(I) contains only one electron in the $$p$$-orbital. The number indicates that lead is in $$+1$$ oxidation state. Thus the lead with a single electron is positioned in the $$p$$-orbital has one unpaired electron.

Note:
The number written in the bracket after the name of the element in Roman numerals is called the oxidation state of the element. Hund’s rule of maximum spin multiplicity states that pairing of electrons in the orbitals will not occur until all the orbitals are singly filled. Pauli’s exclusion principle states that none of the two electrons in an orbital will have the same set of four quantum numbers.