QUESTION IMAGE
Question
complete the table below. for example, in the first row decide whether $\ce{sc^{3+}}$ is a cation or anion. in the second row, write the symbol for the ion that an atom of chlorine is mostly likely to form and then decide what type of ion it is.
| most likely ion | element | scandium | chlorine | oxygen | calcium | sodium | |
|---|---|---|---|---|---|---|---|
| type of ion | $\circ$ cation $\circ$ anion | $\circ$ cation $\circ$ anion | $\circ$ cation $\circ$ anion | $\circ$ cation $\circ$ anion | $\circ$ cation $\circ$ anion |
To solve this, we analyze each element:
1. Chlorine (Cl)
- Chlorine is a non - metal. Non - metals generally gain electrons to achieve a stable electron configuration. When an atom gains electrons, it forms an anion (negative ion).
- Chlorine has 7 valence electrons. To achieve the stable octet configuration (like the noble gas argon), it gains 1 electron. So the ion symbol is $\ce{Cl^-}$.
2. Oxygen (O)
- Oxygen is a non - metal. It has 6 valence electrons. To achieve the stable octet configuration (like neon), it gains 2 electrons.
- When an atom gains electrons, it forms an anion. The ion symbol is $\ce{O^{2-}}$.
3. Calcium (Ca)
- Calcium is a metal. Metals generally lose electrons to achieve a stable electron configuration. When an atom loses electrons, it forms a cation (positive ion).
- Calcium has 2 valence electrons. It loses these 2 electrons to achieve the stable electron configuration of argon. The ion symbol is $\ce{Ca^{2+}}$.
4. Sodium (Na)
- Sodium is a metal. It has 1 valence electron. It loses this 1 electron to achieve the stable electron configuration of neon.
- When a metal loses electrons, it forms a cation. The ion symbol is $\ce{Na^+}$.
For the "type of ion" column:
- Cations are formed when atoms (usually metals) lose electrons (positive charge). So Scandium ($\ce{Sc^{3+}}$), Calcium ($\ce{Ca^{2+}}$), and Sodium ($\ce{Na^+}$) form cations.
- Anions are formed when atoms (usually non - metals) gain electrons (negative charge). So Chlorine ($\ce{Cl^-}$) and Oxygen ($\ce{O^{2-}}$) form anions.
Filling the table:
| Element | Symbol of ion | Type of ion |
|---|---|---|
| Chlorine | $\ce{Cl^-}$ | Anion |
| Oxygen | $\ce{O^{2-}}$ | Anion |
| Calcium | $\ce{Ca^{2+}}$ | Cation |
| Sodium | $\ce{Na^+}$ | Cation |
(If we were to just summarize the ion symbols and types as per the table's requirements for each element):
- Chlorine: Ion symbol $\boldsymbol{\ce{Cl^-}}$, Type: Anion
- Oxygen: Ion symbol $\boldsymbol{\ce{O^{2-}}}$, Type: Anion
- Calcium: Ion symbol $\boldsymbol{\ce{Ca^{2+}}}$, Type: Cation
- Sodium: Ion symbol $\boldsymbol{\ce{Na^+}}$, Type: Cation
Final Answers (for each element's ion symbol and type):
- Chlorine: Ion symbol $\ce{Cl^-}$, Type: Anion
- Oxygen: Ion symbol $\ce{O^{2-}}$, Type: Anion
- Calcium: Ion symbol $\ce{Ca^{2+}}$, Type: Cation
- Sodium: Ion symbol $\ce{Na^+}$, Type: Cation
(If you need to fill the table cells: For "symbol of ion" column - Chlorine: $\ce{Cl^-}$, Oxygen: $\ce{O^{2-}}$, Calcium: $\ce{Ca^{2+}}$, Sodium: $\ce{Na^+}$; For "type of ion" column - Chlorine: Anion, Oxygen: Anion, Calcium: Cation, Sodium: Cation)
Snap & solve any problem in the app
Get step-by-step solutions on Sovi AI
Photo-based solutions with guided steps
Explore more problems and detailed explanations
To solve this, we analyze each element:
1. Chlorine (Cl)
- Chlorine is a non - metal. Non - metals generally gain electrons to achieve a stable electron configuration. When an atom gains electrons, it forms an anion (negative ion).
- Chlorine has 7 valence electrons. To achieve the stable octet configuration (like the noble gas argon), it gains 1 electron. So the ion symbol is $\ce{Cl^-}$.
2. Oxygen (O)
- Oxygen is a non - metal. It has 6 valence electrons. To achieve the stable octet configuration (like neon), it gains 2 electrons.
- When an atom gains electrons, it forms an anion. The ion symbol is $\ce{O^{2-}}$.
3. Calcium (Ca)
- Calcium is a metal. Metals generally lose electrons to achieve a stable electron configuration. When an atom loses electrons, it forms a cation (positive ion).
- Calcium has 2 valence electrons. It loses these 2 electrons to achieve the stable electron configuration of argon. The ion symbol is $\ce{Ca^{2+}}$.
4. Sodium (Na)
- Sodium is a metal. It has 1 valence electron. It loses this 1 electron to achieve the stable electron configuration of neon.
- When a metal loses electrons, it forms a cation. The ion symbol is $\ce{Na^+}$.
For the "type of ion" column:
- Cations are formed when atoms (usually metals) lose electrons (positive charge). So Scandium ($\ce{Sc^{3+}}$), Calcium ($\ce{Ca^{2+}}$), and Sodium ($\ce{Na^+}$) form cations.
- Anions are formed when atoms (usually non - metals) gain electrons (negative charge). So Chlorine ($\ce{Cl^-}$) and Oxygen ($\ce{O^{2-}}$) form anions.
Filling the table:
| Element | Symbol of ion | Type of ion |
|---|---|---|
| Chlorine | $\ce{Cl^-}$ | Anion |
| Oxygen | $\ce{O^{2-}}$ | Anion |
| Calcium | $\ce{Ca^{2+}}$ | Cation |
| Sodium | $\ce{Na^+}$ | Cation |
(If we were to just summarize the ion symbols and types as per the table's requirements for each element):
- Chlorine: Ion symbol $\boldsymbol{\ce{Cl^-}}$, Type: Anion
- Oxygen: Ion symbol $\boldsymbol{\ce{O^{2-}}}$, Type: Anion
- Calcium: Ion symbol $\boldsymbol{\ce{Ca^{2+}}}$, Type: Cation
- Sodium: Ion symbol $\boldsymbol{\ce{Na^+}}$, Type: Cation
Final Answers (for each element's ion symbol and type):
- Chlorine: Ion symbol $\ce{Cl^-}$, Type: Anion
- Oxygen: Ion symbol $\ce{O^{2-}}$, Type: Anion
- Calcium: Ion symbol $\ce{Ca^{2+}}$, Type: Cation
- Sodium: Ion symbol $\ce{Na^+}$, Type: Cation
(If you need to fill the table cells: For "symbol of ion" column - Chlorine: $\ce{Cl^-}$, Oxygen: $\ce{O^{2-}}$, Calcium: $\ce{Ca^{2+}}$, Sodium: $\ce{Na^+}$; For "type of ion" column - Chlorine: Anion, Oxygen: Anion, Calcium: Cation, Sodium: Cation)