Question

given the ionic chemical formulas below, determine oxidation state of the cation and anion. enter your answer as direction followed by magnitude (i.e. -1, +2, etc) chemical formula | cation oxidation state | anion oxidation state nibr₂ | | pbo₂ | | alf₃ | | rb₃n | | ki | |

Explanation:

Step1: Analyze \( \boldsymbol{NiBr_2} \)

• Cation: \( Ni \), Anion: \( Br^- \) (bromide). In ionic compounds, bromide usually has oxidation state \( -1 \). Let \( Ni \) oxidation state be \( x \). For \( NiBr_2 \), charge balance: \( x + 2(-1) = 0 \) → \( x = +2 \).
• Cation Oxidation State: \( +2 \), Anion Oxidation State: \( -1 \)

Step2: Analyze \( \boldsymbol{PbO_2} \)

• Cation: \( Pb \), Anion: \( O^{2-} \) (oxide, oxidation state \( -2 \) usually). Let \( Pb \) oxidation state be \( y \). Charge balance: \( y + 2(-2) = 0 \) → \( y = +4 \).
• Cation Oxidation State: \( +4 \), Anion Oxidation State: \( -2 \)

Step3: Analyze \( \boldsymbol{AlF_3} \)

• Cation: \( Al \), Anion: \( F^- \) (fluoride, oxidation state \( -1 \) as it's a halogen). Let \( Al \) oxidation state be \( z \). Charge balance: \( z + 3(-1) = 0 \) → \( z = +3 \).
• Cation Oxidation State: \( +3 \), Anion Oxidation State: \( -1 \)

Step4: Analyze \( \boldsymbol{Rb_3N} \)

• Cation: \( Rb^+ \) (alkali metal, oxidation state \( +1 \)). Let \( N \) (anion) oxidation state be \( w \). Charge balance: \( 3(+1) + w = 0 \) → \( w = -3 \).
• Cation Oxidation State: \( +1 \), Anion Oxidation State: \( -3 \)

Step5: Analyze \( \boldsymbol{KI} \)

• Cation: \( K^+ \) (alkali metal, oxidation state \( +1 \)), Anion: \( I^- \) (iodide, oxidation state \( -1 \)). Charge balance: \( +1 + (-1) = 0 \), so oxidation states match.
• Cation Oxidation State: \( +1 \), Anion Oxidation State: \( -1 \)

Answer:

Chemical Formula	Cation Oxidation State	Anion Oxidation State
\( PbO_2 \)	\( +4 \)	\( -2 \)
\( AlF_3 \)	\( +3 \)	\( -1 \)
\( Rb_3N \)	\( +1 \)	\( -3 \)
\( KI \)	\( +1 \)	\( -1 \)