Answer
a. Zinc chloride: ZnCl2
b. Tin(IV) fluoride: SnF4
c. Calcium nitride: Ca3N2
d. Aluminum sulfide: Al2S3
e. Mercury(I) selenide: Hg2Se
f. Silver iodide: AgI
Work Step by Step
a. Zinc chloride (ZnCl2):
Zinc (Zn) is a metallic element, and chlorine (Cl) is a nonmetallic element. When they combine, the formula for zinc chloride is ZnCl2, where the subscript 2 indicates that there are two chlorine atoms for every one zinc atom.
b. Tin(IV) fluoride (SnF4):
Tin (Sn) is a metallic element, and fluorine (F) is a nonmetallic element. The Roman numeral "IV" indicates that tin has an oxidation state of +4, so the formula for tin(IV) fluoride is SnF4, where the subscript 4 indicates that there are four fluorine atoms for every one tin atom.
c. Calcium nitride (Ca3N2):
Calcium (Ca) is a metallic element, and nitrogen (N) is a nonmetallic element. The formula for calcium nitride is Ca3N2, where the subscripts 3 and 2 indicate that there are three calcium atoms for every two nitrogen atoms.
d. Aluminum sulfide (Al2S3):
Aluminum (Al) is a metallic element, and sulfur (S) is a nonmetallic element. The formula for aluminum sulfide is Al2S3, where the subscripts 2 and 3 indicate that there are two aluminum atoms for every three sulfur atoms.
e. Mercury(I) selenide (Hg2Se):
Mercury (Hg) is a metallic element, and selenium (Se) is a nonmetallic element. The Roman numeral "I" indicates that mercury has an oxidation state of +1, so the formula for mercury(I) selenide is Hg2Se, where the subscript 2 indicates that there are two mercury atoms for every one selenium atom.
f. Silver iodide (AgI):
Silver (Ag) is a metallic element, and iodine (I) is a nonmetallic element. The formula for silver iodide is AgI, where the subscripts are not necessary because there is one silver atom for every one iodine atom.
