Question

question 2. name the binary ionic compounds indicated by the following formulas: a. bao b. cabr₂ c. aln d. cucl₂ e. fe₂o₃ f. pbs g. zncl₂ options: barium oxygen, barium(ii) oxide, barium oxide, barium oxygen(ii), calcium bromide, calcium(ii) bromide, calcium bromine, calciumide bromine, aluminum nitride, aluminum nitrogen, aluminum(iii) nitride, aluminide nitride, copper(ii) chloride, copper(i) chloride, copper chloride

Explanation:

Step 1: Naming BaO

Barium (Ba) is in group 2, so it has a +2 charge. Oxygen (O) has a -2 charge. For binary ionic compounds with a metal in group 1 or 2 (fixed charge), we use the metal name and non - metal name with -ide suffix. So BaO is Barium oxide.

Step 2: Naming \(CaBr_2\)

Calcium (Ca) is in group 2, charge +2. Bromine (Br) has -1 charge. We name the metal (Calcium) and the non - metal with -ide (bromide). So \(CaBr_2\) is Calcium bromide.

Step 3: Naming AlN

Aluminum (Al) has a +3 charge (group 13, common charge). Nitrogen (N) has -3 charge. We use the metal name (Aluminum) and non - metal with -ide (nitride). So AlN is Aluminum nitride.

Step 4: Naming \(CuCl_2\)

Copper can have multiple charges. In \(CuCl_2\), chlorine (Cl) has -1 charge, so 2 Cl⁻ means Cu has +2 charge. We use the metal name with its charge in parentheses (II) and non - metal with -ide. So \(CuCl_2\) is Copper(II) chloride.

Step 5: Naming \(Fe_2O_3\)

Oxygen (O) has -2 charge, 3 O atoms give a total of -6 charge. So 2 Fe atoms must have a total of +6 charge, meaning each Fe has +3 charge. Iron (Fe) with +3 charge is named as Iron(III), and oxygen becomes oxide. So \(Fe_2O_3\) is Iron(III) oxide.

Step 6: Naming PbS

Lead (Pb) can have multiple charges, but in PbS, sulfur (S) has -2 charge, so Pb has +2 charge. We name the metal (Lead) with its charge? Wait, no, for PbS, if we consider common naming (or if Pb is in +2 state here), but actually, for binary ionic, Lead(II) sulfide? Wait, no, the formula is PbS. Sulfur is -2, so Pb is +2. So name is Lead(II) sulfide? Wait, but in the options? Wait, the original problem's f is PbS. Wait, maybe I missed. Wait, the user's question is about a - g, but the options shown have up to some. Wait, let's re - check. The user's question has a: BaO, b: \(CaBr_2\), c: AlN, d: \(CuCl_2\), e: \(Fe_2O_3\), f: PbS, g: \(ZnCl_2\).
For g: \(ZnCl_2\), Zinc (Zn) has +2 charge, Chlorine -1. So Zinc chloride.

But let's answer each part as per the options:

a. BaO: The correct option is "Barium oxide" (among the options, "Barium oxide" is an option).

b. \(CaBr_2\): The correct option is "Calcium bromide".

c. AlN: The correct option is "Aluminum nitride".

d. \(CuCl_2\): The correct option is "Copper(II) chloride".

e. \(Fe_2O_3\): Iron(III) oxide (but if not in options, maybe the options for e are missing, but based on the given options, for d it's Copper(II) chloride.

Wait, the user's image shows options for a: Barium oxygen, Barium(II) oxide, Barium oxide, Barium oxygen(II); for b: Calcium bromide, Calcium(II) bromide, Calcium bromine, Calciumide bromine; for c: Aluminum nitride, Aluminum nitrogen, Aluminum(III) nitride, Aluminide nitride; for d: Copper(II) chloride, Copper(I) chloride, Copper chloride.

So:

a. BaO: Barium has a +2 charge (group 2), so we don't need to use (II) as it's a fixed charge metal. So correct name is Barium oxide (option "Barium oxide").

b. \(CaBr_2\): Calcium is group 2, +2 charge, bromide. So "Calcium bromide" (option "Calcium bromide").

c. AlN: Aluminum is +3, nitrogen -3. So "Aluminum nitride" (option "Aluminum nitride").

d. \(CuCl_2\): Chlorine is -1, 2 Cl⁻ so Cu is +2. So "Copper(II) chloride" (option "Copper(II) chloride").

e. \(Fe_2O_3\): Oxygen is -2, 3 O: -6, so 2 Fe: +6, each Fe +3. So Iron(III) oxide (but if not in options, maybe the user's options for e are missing, but let's proceed with the given).

f. PbS: Sulfur is -2, so Pb is +2. So Lead(II) sulfide (but options? If in options, but the user's options shown don't have PbS options fully, but let's assume).

g. \(ZnCl_2\): Zinc is +2, Chlorine -1. So Zinc chloride.

But let's answer each sub - question:

a. BaO:

Answer:

a. Barium oxide

b. \(CaBr_2\):