Question

complete the table below by writing the symbols for the cation and anion that make up each ionic compound. the first row has been completed for you.
ionic compound | cation | anion
--- | --- | ---
nacl | na⁺ | cl⁻
cucl₂ | cu²⁺ | cl⁻
nh₄f | nh₄⁺ | f⁻
vi₃ | v³⁺ | i⁻
cr(po₄)₂ | cr⁴⁺ | po₄³⁻

Explanation:

Step1: Analyze \( \text{VI}_3 \)

The ionic compound is \( \text{VI}_3 \). To find the cation and anion, we look at the formula. The cation is the positive ion, and the anion is the negative ion. For \( \text{VI}_3 \), the cation should be \( \text{V}^{3+} \) (as given in the table, and the subscript in the formula relates to the charge balance). The anion is \( \text{I}^- \) (iodide ion). Let's verify the charge balance: \( \text{V}^{3+} \) and \( \text{I}^- \). The formula \( \text{VI}_3 \) means 1 \( \text{V}^{3+} \) and 3 \( \text{I}^- \), so \( 3+ + 3\times(1 -)= 0 \), which balances.

Step2: Analyze \( \text{NH}_4\text{F} \)

The ionic compound is \( \text{NH}_4\text{F} \). The cation is \( \text{NH}_4^+ \) (ammonium ion) and the anion is \( \text{F}^- \) (fluoride ion). Charge balance: \( \text{NH}_4^+ \) (1+) and \( \text{F}^- \) (1-), so \( 1+ + (1 -)= 0 \), which balances.

Step3: Analyze \( \text{CuCl}_2 \)

The ionic compound is \( \text{CuCl}_2 \). The cation is \( \text{Cu}^{2+} \) (copper(II) ion) and the anion is \( \text{Cl}^- \) (chloride ion). Charge balance: \( \text{Cu}^{2+} \) (2+) and 2 \( \text{Cl}^- \) (2×1 -), so \( 2+ + 2\times(1 -)= 0 \), which balances.

Step4: Analyze \( \text{Cr(PO}_4\text{)}_2 \)

The ionic compound is \( \text{Cr(PO}_4\text{)}_2 \). Let's find the cation and anion. The anion is \( \text{PO}_4^{3-} \) (phosphate ion). Let the charge of \( \text{Cr} \) be \( x \). The formula has 1 \( \text{Cr} \) and 2 \( \text{PO}_4^{3-} \). So charge balance: \( x + 2\times(3 -)= 0 \) → \( x - 6 = 0 \) → \( x = 6+ \)? Wait, but the table has \( \text{Cr}^{4+} \), which was incorrect. Wait, no, let's recalculate. Wait, the formula is \( \text{Cr(PO}_4\text{)}_2 \). So \( \text{Cr} \) cation and \( \text{PO}_4^{3-} \) anion. Let the charge of \( \text{Cr} \) be \( y \). Then \( y + 2\times(-3)= 0 \) → \( y = 6+ \)? But the table had \( \text{Cr}^{4+} \), which was wrong. Wait, maybe a mistake in the table, but according to the problem, we need to write the correct cation and anion. Wait, no, the problem says "the first row has been completed for you" (NaCl, CuCl₂, NH₄F, VI₃, Cr(PO₄)₂). Wait, for Cr(PO₄)₂, the anion is \( \text{PO}_4^{3-} \). Let's find the cation charge. Let cation be \( \text{Cr}^n+ \), anion \( \text{PO}_4^{3-} \). Formula: \( \text{Cr(PO}_4\text{)}_2 \), so 1 Cr and 2 PO₄³⁻. Charge balance: \( n + 2\times(-3)= 0 \) → \( n = 6+ \). But the table had \( \text{Cr}^{4+} \), which is incorrect. Wait, maybe the compound is \( \text{Cr}_3(\text{PO}_4)_2 \)? No, the given compound is \( \text{Cr(PO}_4\text{)}_2 \). Wait, perhaps a typo, but according to the problem's table, we need to use the given cation and anion as a reference? Wait, no, the problem is to write the symbols for the cation and anion that make up each ionic compound. Let's correct the Cr(PO₄)₂ part. The anion is \( \text{PO}_4^{3-} \). Let the cation be \( \text{Cr}^{6+} \)? But that's unusual. Wait, maybe the compound is \( \text{Cr}_2(\text{PO}_4)_3 \), but no, the given is \( \text{Cr(PO}_4\text{)}_2 \). Wait, maybe the charge of Cr is 4+, then let's check: \( 4+ + 2\times(-3)= 4 - 6 = -2 \), which is not zero. So that's wrong. So the correct cation should be \( \text{Cr}^{6+} \), but maybe the problem has a mistake. However, following the table's pattern, maybe we proceed with the given (even if incorrect) or correct it. Wait, the first row (NaCl) is correct: \( \text{Na}^+ \) and \( \text{Cl}^- \). CuCl₂: \( \text{Cu}^{2+} \) and \( \text{Cl}^- \) (correct, since 2 Cl⁻ balance 1 Cu²⁺). NH₄F: \( \text{NH}_4^+ \) and \( \text{F}^- \) (correct). VI₃: \( \text{V}^{3+} \) and \( \text{I}^- \) (correct, 3 I⁻ balance 1 V³⁺). For Cr(PO₄)₂, the anion is \( \text{PO}_4^{3-} \). Let's find the cation. Let's assume the table's \( \text{Cr}^{4+} \) is wrong, but maybe the problem expects us to use the given (even if incorrect) or correct it. Wait, no, the problem is to write the cation and anion for each compound. So for Cr(PO₄)₂:

Anion: \( \text{PO}_4^{3-} \) (phosphate ion)

Cation: Let's calculate the charge. Let cation be \( \text{Cr}^x+ \). Then \( x + 2\times(-3)= 0 \) → \( x = 6 \). So \( \text{Cr}^{6+} \). But the table has \( \text{Cr}^{4+} \), which is incorrect. Maybe a typo in the problem, but we'll go with the correct charge balance.

But let's summarize the correct cations and anions for each compound:

- \( \text{VI}_3 \): Cation \( \text{V}^{3+} \), Anion \( \text{I}^- \)
- \( \text{NH}_4\text{F} \): Cation \( \text{NH}_4^+ \), Anion \( \text{F}^- \)
- \( \text{CuCl}_2 \): Cation \( \text{Cu}^{2+} \), Anion \( \text{Cl}^- \)
- \( \text{Cr(PO}_4\text{)}_2 \): Cation \( \text{Cr}^{6+} \), Anion \( \text{PO}_4^{3-} \) (but the table has \( \text{Cr}^{4+} \), which is wrong)

Wait, maybe the compound is \( \text{Cr}_2(\text{PO}_4)_3 \), then cation \( \text{Cr}^{3+} \), anion \( \text{PO}_4^{3-} \), formula \( \text{Cr}_2(\text{PO}_4)_3 \), but the given is \( \text{Cr(PO}_4\text{)}_2 \). So there's a mistake in the problem's compound formula, but we'll proceed with the given table's pattern.

But the problem is to complete the table, so for each compound:

1. \( \text{VI}_3 \): Cation \( \text{V}^{3+} \), Anion \( \text{I}^- \) (matches the table)
2. \( \text{NH}_4\text{F} \): Cation \( \text{NH}_4^+ \), Anion \( \text{F}^- \) (matches the table)
3. \( \text{CuCl}_2 \): Cation \( \text{Cu}^{2+} \), Anion \( \text{Cl}^- \) (matches the table)
4. \( \text{Cr(PO}_4\text{)}_2 \): Anion \( \text{PO}_4^{3-} \), Cation should be \( \text{Cr}^{6+} \) (but table has \( \text{Cr}^{4+} \), which is incorrect)

But maybe the problem has a typo, and the compound is \( \text{Cr}_3(\text{PO}_4)_2 \), then cation \( \text{Cr}^{2+} \), but no. Alternatively, maybe the phosphate ion is \( \text{PO}_4^{2-} \), but no, phosphate is \( \text{PO}_4^{3-} \).

Alternatively, maybe the table's \( \text{Cr}^{4+} \) is a mistake, and the correct cation is \( \text{Cr}^{6+} \) for \( \text{Cr(PO}_4\text{)}_2 \).

But let's list the correct cations and anions for each compound as per charge balance:

- \( \text{VI}_3 \): \( \text{V}^{3+} \) (cation), \( \text{I}^- \) (anion)
- \( \text{NH}_4\text{F} \): \( \text{NH}_4^+ \) (cation), \( \text{F}^- \) (anion)
- \( \text{CuCl}_2 \): \( \text{Cu}^{2+} \) (cation), \( \text{Cl}^- \) (anion)
- \( \text{Cr(PO}_4\text{)}_2 \): \( \text{Cr}^{6+} \) (cation), \( \text{PO}_4^{3-} \) (anion)

But since the table has \( \text{Cr}^{4+} \), maybe the problem expects us to use that (even if incorrect), but we'll go with the correct charge balance.

Answer:

For each compound, the cation and anion are:

• \( \text{VI}_3 \): Cation \( \boldsymbol{\text{V}^{3+}} \), Anion \( \boldsymbol{\text{I}^-} \)
• \( \text{NH}_4\text{F} \): Cation \( \boldsymbol{\text{NH}_4^+} \), Anion \( \boldsymbol{\text{F}^-} \)
• \( \text{CuCl}_2 \): Cation \( \boldsymbol{\text{Cu}^{2+}} \), Anion \( \boldsymbol{\text{Cl}^-} \)
• \( \text{Cr(PO}_4\text{)}_2 \): Cation \( \boldsymbol{\text{Cr}^{6+}} \) (correct charge balance), Anion \( \boldsymbol{\text{PO}_4^{3-}} \)

(Note: If the table's \( \text{Cr}^{4+} \) is to be used despite the charge imbalance, then cation \( \text{Cr}^{4+} \), but that would be incorrect as \( 4+ + 2\times(-3)= -2
eq 0 \).)