QUESTION IMAGE
Question
practice table - valence electrons & atoms vs. ions
fill in the chart below
name\tsymbol\t# of valence electrons\tp⁺\te⁻\tatomic #\tcharge\tgained or lost e⁻\tneutral atom, cation, anion
\t\t\t\t\t\t\t\t
boron atom\t\t\t\t\t\t\t\t
\t\t\t20\t\t\t+2\t\t\t
\t\t\t\t18\t\t-3\t\t\t
\t\t\t\t36\t\t0\t\t\t
\t\t\t3\t\t\t+1\t\t\t
\t\t\t17\t\t\t-1\t\t\t
Response
To solve the table, we use the following concepts:
- Atomic number ($Z$) = number of protons ($p^+$) = number of electrons ($e^-$) in a neutral atom.
- For ions: Charge = $p^+ - e^-$, so $e^- = p^+ - \text{Charge}$ (for cations) or $e^- = p^+ - \text{Charge}$ (for anions, since charge is negative, it becomes $e^- = p^+ + |\text{Charge}|$).
- Valence electrons: For main - group elements, valence electrons = group number (for A - group elements).
1. Boron atom
- Boron has atomic number $Z = 5$, so $p^+ = 5$, $e^- = 5$ (neutral atom), charge = $0$.
- Boron is in group 13, so valence electrons = $3$.
- Symbol: $\text{B}$; Name: Boron atom; $p^+ = 5$; $e^- = 5$; Atomic # = $5$; Charge = $0$; Gained/Lost $e^-$: Neutral; Atom/ion: Atom.
2. Ion with $p^+ = 20$ and charge = $+2$
- Atomic number $Z = 20$ (calcium, $\text{Ca}$).
- For a cation with charge $+2$, $e^- = p^+ - \text{Charge} = 20 - 2 = 18$.
- Calcium is in group 2, so valence electrons (in atom) = $2$, but this is a $\text{Ca}^{2+}$ ion.
- Name: Calcium ion; Symbol: $\text{Ca}^{2+}$; $p^+ = 20$; $e^- = 18$; Atomic # = $20$; Charge = $+2$; Gained/Lost $e^-$: Lost 2; Atom/ion: Ion.
3. Ion with $e^- = 18$ and charge = $-3$
- For an anion with charge $-3$, $p^+ = e^- - \text{Charge} = 18 - (-3)=15$ (phosphorus, $\text{P}$).
- Atomic number $Z = 15$.
- Phosphorus is in group 15, so valence electrons (in atom) = $5$, and as $\text{P}^{3 - }$ (gained 3 electrons).
- Name: Phosphide ion; Symbol: $\text{P}^{3 - }$; $p^+ = 15$; $e^- = 18$; Atomic # = $15$; Charge = $-3$; Gained/Lost $e^-$: Gained 3; Atom/ion: Ion.
4. Neutral species with $e^- = 36$
- Neutral, so $p^+ = e^- = 36$ (krypton, $\text{Kr}$), atomic number $Z = 36$, charge = $0$.
- Krypton is a noble gas, valence electrons = $8$.
- Name: Krypton atom; Symbol: $\text{Kr}$; $p^+ = 36$; $e^- = 36$; Atomic # = $36$; Charge = $0$; Gained/Lost $e^-$: Neutral; Atom/ion: Atom.
5. Ion with $p^+ = 3$ and charge = $+1$
- Atomic number $Z = 3$ (lithium, $\text{Li}$).
- For $\text{Li}^+$, $e^- = p^+ - \text{Charge} = 3 - 1 = 2$.
- Lithium is in group 1, valence electrons (in atom) = $1$, lost 1 electron.
- Name: Lithium ion; Symbol: $\text{Li}^+$; $p^+ = 3$; $e^- = 2$; Atomic # = $3$; Charge = $+1$; Gained/Lost $e^-$: Lost 1; Atom/ion: Ion.
6. Ion with $p^+ = 17$ and charge = $-1$
- Atomic number $Z = 17$ (chlorine, $\text{Cl}$).
- For $\text{Cl}^-$, $e^- = p^+ - \text{Charge} = 17 - (-1)=18$.
- Chlorine is in group 17, valence electrons (in atom) = $7$, gained 1 electron.
- Name: Chloride ion; Symbol: $\text{Cl}^-$; $p^+ = 17$; $e^- = 18$; Atomic # = $17$; Charge = $-1$; Gained/Lost $e^-$: Gained 1; Atom/ion: Ion.
Filled Table (Summary)
| Name | Symbol | # of Valence electrons | $p^+$ | $e^-$ | Atomic # | Charge | Gained/Lost $e^-$ | Atom, cation, anion |
|---|---|---|---|---|---|---|---|---|
| Calcium ion | $\text{Ca}^{2+}$ | $2$ (atom) | $20$ | $18$ | $20$ | $+2$ | Lost 2 | Cation |
| Phosphide ion | $\text{P}^{3 - }$ | $5$ (atom) | $15$ | $18$ | $15$ | $-3$ | Gained 3 | Anion |
| Krypton atom | $\text{Kr}$ | $8$ | $36$ | $36$ | $36$ | $0$ | Neutral | Atom |
| Lithium ion | $\text{Li}^+$ | $1$ (atom) | $3$ | $2$ | $3$ | $+1$ | Lost 1… |
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To solve the table, we use the following concepts:
- Atomic number ($Z$) = number of protons ($p^+$) = number of electrons ($e^-$) in a neutral atom.
- For ions: Charge = $p^+ - e^-$, so $e^- = p^+ - \text{Charge}$ (for cations) or $e^- = p^+ - \text{Charge}$ (for anions, since charge is negative, it becomes $e^- = p^+ + |\text{Charge}|$).
- Valence electrons: For main - group elements, valence electrons = group number (for A - group elements).
1. Boron atom
- Boron has atomic number $Z = 5$, so $p^+ = 5$, $e^- = 5$ (neutral atom), charge = $0$.
- Boron is in group 13, so valence electrons = $3$.
- Symbol: $\text{B}$; Name: Boron atom; $p^+ = 5$; $e^- = 5$; Atomic # = $5$; Charge = $0$; Gained/Lost $e^-$: Neutral; Atom/ion: Atom.
2. Ion with $p^+ = 20$ and charge = $+2$
- Atomic number $Z = 20$ (calcium, $\text{Ca}$).
- For a cation with charge $+2$, $e^- = p^+ - \text{Charge} = 20 - 2 = 18$.
- Calcium is in group 2, so valence electrons (in atom) = $2$, but this is a $\text{Ca}^{2+}$ ion.
- Name: Calcium ion; Symbol: $\text{Ca}^{2+}$; $p^+ = 20$; $e^- = 18$; Atomic # = $20$; Charge = $+2$; Gained/Lost $e^-$: Lost 2; Atom/ion: Ion.
3. Ion with $e^- = 18$ and charge = $-3$
- For an anion with charge $-3$, $p^+ = e^- - \text{Charge} = 18 - (-3)=15$ (phosphorus, $\text{P}$).
- Atomic number $Z = 15$.
- Phosphorus is in group 15, so valence electrons (in atom) = $5$, and as $\text{P}^{3 - }$ (gained 3 electrons).
- Name: Phosphide ion; Symbol: $\text{P}^{3 - }$; $p^+ = 15$; $e^- = 18$; Atomic # = $15$; Charge = $-3$; Gained/Lost $e^-$: Gained 3; Atom/ion: Ion.
4. Neutral species with $e^- = 36$
- Neutral, so $p^+ = e^- = 36$ (krypton, $\text{Kr}$), atomic number $Z = 36$, charge = $0$.
- Krypton is a noble gas, valence electrons = $8$.
- Name: Krypton atom; Symbol: $\text{Kr}$; $p^+ = 36$; $e^- = 36$; Atomic # = $36$; Charge = $0$; Gained/Lost $e^-$: Neutral; Atom/ion: Atom.
5. Ion with $p^+ = 3$ and charge = $+1$
- Atomic number $Z = 3$ (lithium, $\text{Li}$).
- For $\text{Li}^+$, $e^- = p^+ - \text{Charge} = 3 - 1 = 2$.
- Lithium is in group 1, valence electrons (in atom) = $1$, lost 1 electron.
- Name: Lithium ion; Symbol: $\text{Li}^+$; $p^+ = 3$; $e^- = 2$; Atomic # = $3$; Charge = $+1$; Gained/Lost $e^-$: Lost 1; Atom/ion: Ion.
6. Ion with $p^+ = 17$ and charge = $-1$
- Atomic number $Z = 17$ (chlorine, $\text{Cl}$).
- For $\text{Cl}^-$, $e^- = p^+ - \text{Charge} = 17 - (-1)=18$.
- Chlorine is in group 17, valence electrons (in atom) = $7$, gained 1 electron.
- Name: Chloride ion; Symbol: $\text{Cl}^-$; $p^+ = 17$; $e^- = 18$; Atomic # = $17$; Charge = $-1$; Gained/Lost $e^-$: Gained 1; Atom/ion: Ion.
Filled Table (Summary)
| Name | Symbol | # of Valence electrons | $p^+$ | $e^-$ | Atomic # | Charge | Gained/Lost $e^-$ | Atom, cation, anion |
|---|---|---|---|---|---|---|---|---|
| Calcium ion | $\text{Ca}^{2+}$ | $2$ (atom) | $20$ | $18$ | $20$ | $+2$ | Lost 2 | Cation |
| Phosphide ion | $\text{P}^{3 - }$ | $5$ (atom) | $15$ | $18$ | $15$ | $-3$ | Gained 3 | Anion |
| Krypton atom | $\text{Kr}$ | $8$ | $36$ | $36$ | $36$ | $0$ | Neutral | Atom |
| Lithium ion | $\text{Li}^+$ | $1$ (atom) | $3$ | $2$ | $3$ | $+1$ | Lost 1 | Cation |
| Chloride ion | $\text{Cl}^-$ | $7$ (atom) | $17$ | $18$ | $17$ | $-1$ | Gained 1 | Anion |
(Note: For valence electrons of ions, we usually refer to the neutral atom's valence electrons as the ion's behavior is related to the atom's valence shell. If we consider the ion's valence shell, for example, $\text{Ca}^{2+}$ has 8 valence electrons in its outer shell, but the table might expect the atom's valence electrons for reference.)