Question

classify each compound by the number of chlorine atoms indicated in the chemical formula.
1 cl atom 2 cl atoms 3 cl atoms 4 cl atoms 5 cl atoms 6 cl atoms
answer bank
cr(clo₃)₃ hclo₂ ca(clo₃)₂ licl ccl₄ ccl₂f₂ bacl₂ pcl₅

Explanation:

Step1: Analyze each compound

• For $\ce{HClO2}$: The formula has 1 Cl atom (since there's no subscript on Cl, so 1).
• For $\ce{LiCl}$: The formula has 1 Cl atom (subscript 1, implied).
• For $\ce{BaCl2}$: The formula has 2 Cl atoms (subscript 2).
• For $\ce{CCl2F2}$: The formula has 2 Cl atoms (subscript 2 on Cl).
• For $\ce{Ca(ClO3)2}$: The subscript outside the parentheses is 2, and inside is 1 Cl per $\ce{ClO3^-}$, so total Cl atoms: $1\times2 = 2$? Wait, no: $\ce{Ca(ClO3)2}$: the number of Cl is 2 (because the subscript 2 multiplies the Cl in $\ce{ClO3^-}$, which has 1 Cl each, so 21 = 2? Wait, no, wait: $\ce{Cr(ClO3)3}$: Cr has (ClO3) with subscript 3, so Cl atoms: 31 = 3. $\ce{Ca(ClO3)2}$: 2*1 = 2? Wait, no, let's re - evaluate:
• $\ce{HClO2}$: 1 Cl (no subscript, so 1)
• $\ce{LiCl}$: 1 Cl (subscript 1, written as LiCl)
• $\ce{BaCl2}$: 2 Cl (subscript 2)
• $\ce{CCl2F2}$: 2 Cl (subscript 2 on Cl)
• $\ce{Ca(ClO3)2}$: The $\ce{ClO3^-}$ ion has 1 Cl per ion, and there are 2 ions (because of the subscript 2 outside the parentheses), so 2*1 = 2 Cl? Wait, no, wait $\ce{Cr(ClO3)3}$: 3 $\ce{ClO3^-}$ ions, so 3 Cl atoms. $\ce{Ca(ClO3)2}$: 2 $\ce{ClO3^-}$ ions, so 2 Cl atoms? Wait, no, I made a mistake earlier. Let's do each compound:
• $\ce{HClO2}$: Chemical formula is $\ce{HClO2}$, so Cl atoms: 1.
• $\ce{LiCl}$: $\ce{LiCl}$, Cl atoms: 1.
• $\ce{BaCl2}$: $\ce{BaCl2}$, Cl atoms: 2 (subscript 2).
• $\ce{CCl2F2}$: $\ce{CCl2F2}$, Cl atoms: 2 (subscript 2 on Cl).
• $\ce{Ca(ClO3)2}$: The formula is $\ce{Ca(ClO3)2}$, so the number of $\ce{ClO3^-}$ groups is 2, and each $\ce{ClO3^-}$ has 1 Cl, so total Cl: 21 = 2? Wait, no, wait $\ce{Cr(ClO3)3}$: $\ce{Cr(ClO3)3}$, number of $\ce{ClO3^-}$ groups is 3, so Cl atoms: 31 = 3.
• $\ce{CCl4}$: $\ce{CCl4}$, Cl atoms: 4 (subscript 4).
• $\ce{PCl5}$: $\ce{PCl5}$, Cl atoms: 5 (subscript 5).
• $\ce{Cr(ClO3)3}$: $\ce{Cr(ClO3)3}$, number of $\ce{ClO3^-}$ groups is 3, so Cl atoms: 3*1 = 3.

Step2: Classify each compound

• 1 Cl atom: $\ce{HClO2}$, $\ce{LiCl}$
• 2 Cl atoms: $\ce{BaCl2}$, $\ce{CCl2F2}$, $\ce{Ca(ClO3)2}$
• 3 Cl atoms: $\ce{Cr(ClO3)3}$
• 4 Cl atoms: $\ce{CCl4}$
• 5 Cl atoms: $\ce{PCl5}$
• 6 Cl atoms: None of the given compounds have 6 Cl atoms.

Answer:

• 1 Cl atom: $\ce{HClO2}$, $\ce{LiCl}$
• 2 Cl atoms: $\ce{BaCl2}$, $\ce{CCl2F2}$, $\ce{Ca(ClO3)2}$
• 3 Cl atoms: $\ce{Cr(ClO3)3}$
• 4 Cl atoms: $\ce{CCl4}$
• 5 Cl atoms: $\ce{PCl5}$
• 6 Cl atoms: (No compounds)