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QUESTION IMAGE

complete the table below by writing the symbols for the cation and anio…

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 compoundcationanion
$ce{mncl2}$$ce{mn\square}$$square$
$ce{mn2s3}$$square$$square$
$ce{vf2}$$square$$square$
$ce{nio}$$square$$square$

Explanation:

Step1: Analyze \( \text{MnCl}_2 \)

In \( \text{MnCl}_2 \), the anion is \( \text{Cl}^- \) (since chloride ion has a charge of \(-1\)) and to balance the charge (compound is neutral), the cation \( \text{Mn} \) must have a charge of \( +2 \) (because there are 2 \( \text{Cl}^- \) ions, so \( \text{Mn}^{2+} \) to balance \( 2\times(-1) = -2 \)). So cation: \( \text{Mn}^{2+} \), anion: \( \text{Cl}^- \).

Step2: Analyze \( \text{Mn}_2\text{S}_3 \)

For \( \text{Mn}_2\text{S}_3 \), the anion is sulfide ion \( \text{S}^{2-} \). Let the charge of \( \text{Mn} \) be \( x \). There are 2 \( \text{Mn} \) ions and 3 \( \text{S}^{2-} \) ions. So \( 2x + 3\times(-2) = 0 \Rightarrow 2x - 6 = 0 \Rightarrow x = +3 \). So cation: \( \text{Mn}^{3+} \), anion: \( \text{S}^{2-} \).

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

In \( \text{VF}_2 \), the anion is fluoride ion \( \text{F}^- \). Let the charge of \( \text{V} \) be \( y \). There is 1 \( \text{V} \) ion and 2 \( \text{F}^- \) ions. So \( y + 2\times(-1) = 0 \Rightarrow y = +2 \). So cation: \( \text{V}^{2+} \), anion: \( \text{F}^- \).

Step4: Analyze \( \text{NiO} \)

For \( \text{NiO} \), the anion is oxide ion \( \text{O}^{2-} \). Let the charge of \( \text{Ni} \) be \( z \). There is 1 \( \text{Ni} \) ion and 1 \( \text{O}^{2-} \) ion. So \( z + (-2) = 0 \Rightarrow z = +2 \). So cation: \( \text{Ni}^{2+} \), anion: \( \text{O}^{2-} \).

Ionic CompoundCationAnion
\( \text{MnCl}_2 \)\( \text{Mn}^{2+} \)\( \text{Cl}^- \)
\( \text{Mn}_2\text{S}_3 \)\( \text{Mn}^{3+} \)\( \text{S}^{2-} \)
\( \text{VF}_2 \)\( \text{V}^{2+} \)\( \text{F}^- \)
\( \text{NiO} \)\( \text{Ni}^{2+} \)\( \text{O}^{2-} \)

Answer:

The completed table is as shown above with cations and anions:

  • \( \text{MnCl}_2 \): Cation \( \boldsymbol{\text{Mn}^{2+}} \), Anion \( \boldsymbol{\text{Cl}^-} \)
  • \( \text{Mn}_2\text{S}_3 \): Cation \( \boldsymbol{\text{Mn}^{3+}} \), Anion \( \boldsymbol{\text{S}^{2-}} \)
  • \( \text{VF}_2 \): Cation \( \boldsymbol{\text{V}^{2+}} \), Anion \( \boldsymbol{\text{F}^-} \)
  • \( \text{NiO} \): Cation \( \boldsymbol{\text{Ni}^{2+}} \), Anion \( \boldsymbol{\text{O}^{2-}} \)