Question

What is the formula for barium fluoride?

Answer 1

Hint:Barium fluoride is an ionic compound which is insoluble in water. It is a colourless solid which occurs in nature as a rare mineral frankdicksonite. It is transparent, generally ranging from ultraviolet to infrared, from \[150-200\;nm\;{\text{to}}11-11.5\;\mu m\].

Complete step by step answer:
As the name suggests, the given compound barium fluoride comprises a metal barium and a non-metal fluorine. We know that any compound containing a metal and a non-metal is considered to be an ionic compound. In ionic compounds, we have to take into account the charges of each of the elements present in the compound. Now let us identify the formula for the given compound i.e. barium fluoride.
From the periodic table, it is clear that barium is a metal having a symbol \[Ba\] and it belongs to group 1 thus, it possesses +2 cationic charge while fluorine is a non-metal having a symbol \[F\] and it possesses -1 anionic charge. It will look like $B{a^{2 + }}{F^{1 - }}$. As barium fluoride is neutral, the net charge on the compound should be zero. So in the present case, $( + 2) + ( - 1) = + 1 \ne 0$. Thus, this can’t be the formula for barium fluoride as one to one ratio of the charges makes the ionic formula. Now, we can change the subscripts in order to balance the charge. We will change the subscript of both metal and non-metal following the criss-cross method to neutralise the charge i.e.${(B{a^{2 + }})_1}{({F^{1 - }})_2}$. Now the charges are balanced as $(1 \times ( + 2)) + (2 \times ( - 1)) = 0$.
Hence, the formula for magnesium iodide is $Ba{F_2}$.

Note: While writing the chemical formula of an ionic compound, never write the subscript '1'. And it should be noted that we can also have one or two polyatomic ions like \[N{H_4}N{O_3}\] in the same compound. In this case, we have to find and write their names as identified from the Common Ion Table.