Question

Why is the nomenclature for Mercury (I) Fluoride $H{g_2}{F_2}$ and not $HgF$ ?

Answer 1

Metals are the chemical elements that can easily undergo loss of electrons. Non-metals are the elements that can attract electrons to attain the nearest inert gas configuration. Mercury is a non-metal that combines with fluorine which is a non-metal forms a compound with molecular formula of $H{g_2}{F_2}$ , the nomenclature can be written based on the oxidation state of central metal atom.

Complete answer:
Chemical elements are of two types based on nature. These are metals, and nonmetals. Metals can easily undergo loss of electrons, and non-metals can try to attract or gain electrons to attain the nearest inert gas configuration. As the inert gases have stable configuration.
Mercury is a non-metal with the chemical symbol $Hg$, it is having less electronegativity compared to fluorine, where fluorine is the most electronegative element in the periodic table.
Mercury combines with fluorine to form different compounds like $H{g_2}{F_2}$ and $HgF$ . but mostly mercury exists as the $H{g_2}^{2 + }$ only. In the compound $H{g_2}{F_2}$ , the cation is $H{g_2}^{2 + }$ .
The empirical formula of $H{g_2}{F_2}$ is $HgF$ .
In $H{g_2}{F_2}$ mercury is the least electronegative element and occupies the central position and fluorine occupies the terminal positions.
The oxidation state of mercury in $H{g_2}{F_2}$ will be $2x + 2\left( { - 1} \right) = 0 \Rightarrow x = + 1$
Thus, the oxidation state of mercury is represented as (I) which means the representation of oxidation number in roman numerals within brackets.
Thus, the nomenclature with Mercury (I) Fluoride is $H{g_2}{F_2}$ but not $HgF$

Note:
Molecular formula is the representation of all the atoms arranged in the compound. Whereas empirical formula is the shortest representation of the atoms present in the molecular formula. In $H{g_2}{F_2}$ the atoms are in the ratio $2:2$ , in $HgF$ the atoms are in $1:1$ ratio, which is the shortest mole ratio.