Question

question 11 (2 points)
state whether this compound is ionic or covalent and then explain how you would determine the name:
\\(\text{nh}_4\text{br}\\)

question 12 (3 points)
state whether this compound is ionic or covalent and then explain how you would determine the name:
\\(\text{cocro}_4\\)

Explanation:

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<reasoning>

Analyze NH4Br bonding

Using the Ionic Bonding and Polyatomic Ion Compounds knowledge points:

• It contains the polyatomic ammonium cation \(\text{NH}_4^+\).
• It contains the monatomic bromide anion \(\text{Br}^-\).
• The electrostatic attraction makes it an ionic compound.

Determine NH4Br name

Using the Chemical Nomenclature and Polyatomic Ion Compounds knowledge points:

• Identify the cation \(\text{NH}_4^+\) as ammonium.
• Identify the anion \(\text{Br}^-\) as bromide.
• Combine them to get ammonium bromide.

Analyze CoCrO4 bonding

Using the Ionic Bonding and Polyatomic Ion Compounds knowledge points:

• It contains the transition metal cobalt (\(\text{Co}\)).
• It contains the polyatomic chromate anion \(\text{CrO}_4^{2-}\).
• The metal-nonmetal combination forms an ionic compound.

Determine CoCrO4 name

Using the Transition Metal Nomenclature and Polyatomic Ion Compounds knowledge points:

• Identify the anion as chromate, \(\text{CrO}_4^{2-}\).
• Balance the charge: cobalt must be \(\text{Co}^{2+}\).
• Name it cobalt(II) chromate.

</reasoning>

<answer>

Question 11

• Classification: Ionic
• Explanation: This compound is ionic because it is composed of the polyatomic ammonium cation (\(\text{NH}_4^+\)) and the bromide anion (\(\text{Br}^-\)) held together by ionic bonds. To determine the name, identify the polyatomic cation as "ammonium" and the monatomic anion as "bromide," then combine them to get ammonium bromide.

Question 12

• Classification: Ionic
• Explanation: This compound is ionic because it consists of a transition metal cation (cobalt) and a polyatomic anion (chromate, \(\text{CrO}_4^{2-}\)). To determine the name, identify the anion as "chromate" which has a \(2-\) charge. Since the compound is neutral, the cobalt ion must have a \(2+\) charge, written as cobalt(II). Combine these to get cobalt(II) chromate.

</answer>

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Answer:

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<reasoning>

Analyze NH4Br bonding

Using the Ionic Bonding and Polyatomic Ion Compounds knowledge points:

• It contains the polyatomic ammonium cation \(\text{NH}_4^+\).
• It contains the monatomic bromide anion \(\text{Br}^-\).
• The electrostatic attraction makes it an ionic compound.

Determine NH4Br name

Using the Chemical Nomenclature and Polyatomic Ion Compounds knowledge points:

• Identify the cation \(\text{NH}_4^+\) as ammonium.
• Identify the anion \(\text{Br}^-\) as bromide.
• Combine them to get ammonium bromide.

Analyze CoCrO4 bonding

Using the Ionic Bonding and Polyatomic Ion Compounds knowledge points:

• It contains the transition metal cobalt (\(\text{Co}\)).
• It contains the polyatomic chromate anion \(\text{CrO}_4^{2-}\).
• The metal-nonmetal combination forms an ionic compound.

Determine CoCrO4 name

Using the Transition Metal Nomenclature and Polyatomic Ion Compounds knowledge points:

• Identify the anion as chromate, \(\text{CrO}_4^{2-}\).
• Balance the charge: cobalt must be \(\text{Co}^{2+}\).
• Name it cobalt(II) chromate.

</reasoning>

<answer>

Question 11

• Classification: Ionic
• Explanation: This compound is ionic because it is composed of the polyatomic ammonium cation (\(\text{NH}_4^+\)) and the bromide anion (\(\text{Br}^-\)) held together by ionic bonds. To determine the name, identify the polyatomic cation as "ammonium" and the monatomic anion as "bromide," then combine them to get ammonium bromide.

Question 12

• Classification: Ionic
• Explanation: This compound is ionic because it consists of a transition metal cation (cobalt) and a polyatomic anion (chromate, \(\text{CrO}_4^{2-}\)). To determine the name, identify the anion as "chromate" which has a \(2-\) charge. Since the compound is neutral, the cobalt ion must have a \(2+\) charge, written as cobalt(II). Combine these to get cobalt(II) chromate.

</answer>

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