Question

zirconium(iv) fluoride is used to make special glass for fiber - optic cables. write the chemical formula of zirconium(iv) fluoride. refer to a periodic table and a list of polyatomic ions.

Explanation:

Step1: Determine ion charges

Zirconium(IV) has a charge of +4 ($Zr^{4 +}$), and fluoride ion has a charge of - 1 ($F^{-}$).

Step2: Balance charges

To balance the charges, we need 4 fluoride ions for every 1 zirconium(IV) ion.

Answer:

$ZrF_{4}$