part 1: introduction to polyatomic ions (table partner - 7 min. group pairs - 3 min. class - 3 min.)
1. stop and consider: imagine for a moment that you were about to make a molecule, but the only atoms you had were one oxygen atom and one hydrogen atom. complete the table below based on this combination (even if it means that a stable electron configuration is not attained).
| oxygen (o) | | hydrogen (h) |
| ---- | ---- | ---- | ---- |
| # of p⁺ | atom e⁻ config. | # of p⁺ | atom e⁻ config. |
| ionization energy | electronegativity | ionization energy | electronegativity |
| δ ionization e. between o & h = | | | |
| δ electronegativity between o & h = | | | |
| atomic lewis e⁻ dot (o) | atomic lewis e⁻ dot (h) |
| (o with a circle) | h |
| combined lewis e⁻ dot | |
2. do all of the atoms drawn in the lewis electron - dot diagram above have a stable electron configuration? explain your answer in terms of electrons.

3. how could the atom picture above achieve a stable electron configuration?

4. stop and review: using your periodic table, complete the table to the right.
| sodium (na) | | | |
| ---- | ---- | ---- | ---- |
| # of p⁺ | atom e⁻ config. | ion e⁻ config. | ionic charge |
| | | | |
| atomic lewis dot | ionic lewis dot | | |

Question

part 1: introduction to polyatomic ions (table partner - 7 min. group pairs - 3 min. class - 3 min.)
1. stop and consider: imagine for a moment that you were about to make a molecule, but the only atoms you had were one oxygen atom and one hydrogen atom. complete the table below based on this combination (even if it means that a stable electron configuration is not attained).
| oxygen (o) | | hydrogen (h) |
| ---- | ---- | ---- | ---- |
| # of p⁺ | atom e⁻ config. | # of p⁺ | atom e⁻ config. |
| ionization energy | electronegativity | ionization energy | electronegativity |
| δ ionization e. between o & h = | | | |
| δ electronegativity between o & h = | | | |
| atomic lewis e⁻ dot (o) | atomic lewis e⁻ dot (h) |
| (o with a circle) | h |
| combined lewis e⁻ dot | |
2. do all of the atoms drawn in the lewis electron - dot diagram above have a stable electron configuration? explain your answer in terms of electrons.

3. how could the atom picture above achieve a stable electron configuration?

4. stop and review: using your periodic table, complete the table to the right.
| sodium (na) | | | |
| ---- | ---- | ---- | ---- |
| # of p⁺ | atom e⁻ config. | ion e⁻ config. | ionic charge |
| | | | |
| atomic lewis dot | ionic lewis dot | | |

Accepted Answer

### Question 1: Completing the Table for Oxygen (O) and Hydrogen (H)
#### Step 1: Oxygen (O) Details
- **# of p⁺ (protons)**: Oxygen has an atomic number of 8, so $\#\text{ of } p^+ = 8$.
- **Atom e⁻ Config. (Electron Configuration)**: The electron configuration of O is $1s^2 2s^2 2p^4$.
- **Ionization Energy**: Oxygen has a relatively high ionization energy (approx. 1314 kJ/mol, but for the table, we can note it's higher than H).
- **Electronegativity**: Oxygen has an electronegativity of 3.44 (Pauling scale).
- **Atomic Lewis e⁻ dot**: Oxygen has 6 valence electrons, so the Lewis dot is $\cdot \underset{\cdot \cdot}{\overset{\cdot \cdot}{O}} \cdot$ (but the given diagram has a circle, we can represent the valence electrons as 6 dots around O).

#### Step 2: Hydrogen (H) Details
- **# of p⁺ (protons)**: Hydrogen has an atomic number of 1, so $\#\text{ of } p^+ = 1$.
- **Atom e⁻ Config. (Electron Configuration)**: The electron configuration of H is $1s^1$.
- **Ionization Energy**: Hydrogen has an ionization energy of approx. 1312 kJ/mol.
- **Electronegativity**: Hydrogen has an electronegativity of 2.20 (Pauling scale).
- **Atomic Lewis e⁻ dot**: Hydrogen has 1 valence electron, so the Lewis dot is $\text{H} \cdot$ (matches the given diagram).

#### Step 3: Δ (Difference) Calculations
- **Δ Ionization E. between O & H**: $|1314 - 1312| = 2$ kJ/mol (approximate values).
- **Δ Electronegativity between O & H**: $|3.44 - 2.20| = 1.24$.

#### Step 4: Combined Lewis e⁻ Dot
- Oxygen has 6 valence electrons, Hydrogen has 1. When combined (as OH, though unstable), the Lewis dot would show O with its 6 dots and H with its 1 dot, e.g., $\cdot \underset{\cdot \cdot}{\overset{\cdot \cdot}{O}} - \text{H}$ (a single bond, sharing one electron from H with O).

### Question 2: Stable Electron Configuration?
#### Brief Explanations:
- **Oxygen**: Needs 2 more electrons to fill its 2p subshell (to get $2p^6$, a stable octet).
- **Hydrogen**: Needs 1 more electron to fill its 1s subshell (to get $1s^2$, a stable duet).
- In the combined Lewis dot (as drawn), O has 6 + 1 (from H) = 7 valence electrons (not an octet), and H has 1 + 1 (from O) = 2? Wait, no—if they share, H would have 2 (stable), but O would have 7 (not stable). So **no**, not all atoms have a stable configuration. O lacks an octet (has 7 valence electrons in the shared case), and if not shared, O has 6, H has 1—both unstable.

#### Answer:
No. Oxygen needs an octet (8 valence electrons) and has 6 (or 7 in the shared case), while Hydrogen needs a duet (2 valence electrons) and has 1 (or 2 in the shared case, but O is still unstable).

### Question 3: Achieving Stable Electron Configuration
#### Brief Explanations:
- **Oxygen**: Can gain 2 electrons (forming $O^{2-}$) to get a stable octet ($1s^2 2s^2 2p^6$).
- **Hydrogen**: Can gain 1 electron (forming $H^-$) to get a stable duet ($1s^2$), or lose its 1 electron (forming $H^+$) to have no electrons (but $H^+$ is stable as a proton).
- Alternatively, they can form a covalent bond and share electrons, but O would need another H (to form $H_2O$) to get 2 shared electrons (total 8 valence electrons: 6 + 2 from 2 H atoms).

#### Answer:
Oxygen can gain 2 electrons (form $O^{2-}$), Hydrogen can gain 1 electron (form $H^-$) or lose 1 (form $H^+$). Or, they can form a molecule with another H (like $H_2O$) where O shares electrons with two H atoms to achieve an octet, and each H achieves a duet.

### Question 4: Completing the Table for Sodium (Na)
#### Step 1: Sodium (Na) Details
- **# of p⁺ (protons)**: Sodium has an atomic number of 11, so $\#\text{ of } p^+ = 11$.
- **Atom e⁻ Config. (Electron Configuration)**: The electron configuration of Na is $1s^2 2s^2 2p^6 3s^1$.
- **Ion e⁻ Config. (Ion Electron Configuration)**: Na loses its 3s¹ electron, forming $Na^+$ with electron configuration $1s^2 2s^2 2p^6$ (same as Ne).
- **Ionic Charge**: Na loses 1 electron, so ionic charge is $+1$ (i.e., $Na^+$).
- **Atomic Lewis Dot**: Na has 1 valence electron, so the Lewis dot is $\text{Na} \cdot$.
- **Ionic Lewis Dot**: $Na^+$ has no valence electrons (lost its 1), so the Lewis dot is $Na^+$ (no dots, as it's a cation with a stable octet in the 2nd shell).

### Final Answers (Summarized)
#### Question 1 Table:
| Oxygen (O)       |          | Hydrogen (H)     |          |
|------------------|----------|------------------|----------|
| $\#\text{ of } p^+ = 8$ | $1s^2 2s^2 2p^4$ | $\#\text{ of } p^+ = 1$ | $1s^1$ |
| Ionization Energy ≈ 1314 kJ/mol | Electronegativity = 3.44 | Ionization Energy ≈ 1312 kJ/mol | Electronegativity = 2.20 |
| $\Delta$ Ionization E. = 2 kJ/mol | $\Delta$ Electronegativity = 1.24 |          |          |
| Atomic Lewis e⁻ dot: $\cdot \underset{\cdot \cdot}{\overset{\cdot \cdot}{O}} \cdot$ |          | Atomic Lewis e⁻ dot: $\text{H} \cdot$ |          |
| Combined Lewis e⁻ Dot: $\cdot \underset{\cdot \cdot}{\overset{\cdot \cdot}{O}} - \text{H}$ |          |          |          |

#### Question 2 Answer:
No, because Oxygen lacks an octet (has 6 or 7 valence electrons) and Hydrogen lacks a duet (has 1 or 2, but O is still unstable).

#### Question 3 Answer:
Oxygen can gain 2 electrons (form $O^{2-}$), Hydrogen can gain 1 (form $H^-$) or lose 1 (form $H^+$). Or form $H_2O$ with another H to share electrons for stability.

#### Question 4 Table:
| Sodium (Na)       |          |          |          |
|-------------------|----------|----------|----------|
| $\#\text{ of } p^+ = 11$ | $1s^2 2s^2 2p^6 3s^1$ | $1s^2 2s^2 2p^6$ | $+1$ (or $Na^+$) |
| Atomic Lewis Dot: $\text{Na} \cdot$ |          | Ionic Lewis Dot: $Na^+$ |          |

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QUESTION IMAGE

Question

Explanation:

Question 1: Completing the Table for Oxygen (O) and Hydrogen (H)

Step 1: Oxygen (O) Details

Step 2: Hydrogen (H) Details

Step 3: Δ (Difference) Calculations

Step 4: Combined Lewis e⁻ Dot

Question 2: Stable Electron Configuration?

Answer:

Question 3: Achieving Stable Electron Configuration

oxygen (o)		hydrogen (h)
ionization energy	electronegativity	ionization energy	electronegativity
δ ionization e. between o & h =
δ electronegativity between o & h =
atomic lewis e⁻ dot (o)	atomic lewis e⁻ dot (h)
(o with a circle)	h
combined lewis e⁻ dot