writing electron configurations
1. electrons occupy the lowest energy orbital first, then move to the next one and so on. (the \aufbau\ principle)
2. orbitals are considered to be in the same shell if they have the same first number (no matter in what order filling is done).
3. an atom will gain or lose electrons in order to have eight electrons in its outer shell. (the \octet\ rule)
4. the outer shell is the highest numbered shell which has electrons in it. only s and p orbitals are part of the outer shell.
an atom has the tendency to lose electrons (to another atom) or to gain electrons (from another atom) in order to make the outer shell complete with eight electrons. atoms with a complete outer shell (eight electrons) are considered stable. some atoms naturally have eight electrons in their outer shell and are very stable. (helium is the exception being stable with two electrons in its outer shell.) complete the following chart:

Question

Accepted Answer

To solve this, we'll use the Aufbau Principle, Octet Rule, and electron shell rules. Let's take **Oxygen (O)** as an example (we can do others similarly):

### Step 1: Atomic Number of O
Oxygen has an atomic number of 8 (from periodic table).

### Step 2: Electrons per Energy Level
Using Aufbau (fill lowest energy first: 1s, 2s, 2p...).
- Energy Level 1 (n=1): Holds up to 2 electrons (1s²).
- Energy Level 2 (n=2): Holds up to 8, but O has 8 total electrons. After filling n=1 (2e⁻), remaining: $ 8 - 2 = 6 $ (so 2s², 2p⁴ → total 6e⁻ in n=2).
Thus, electrons per level: 2 (n=1), 6 (n=2).

### Step 3: Electron Configuration
Write orbitals in order: $ 1s^2 2s^2 2p^4 $.

### Step 4: Electrons Lost/Gained (Octet Rule)
O has 6 valence electrons (outer shell: n=2, 2s²2p⁴). To reach 8 (octet), it gains $ 8 - 6 = 2 $ electrons.

### Step 5: Electrons Left After Gain
Total electrons: 8 + 2 = 10 (since it gains 2).

### Step 6: Charge on Ion
Gaining 2 electrons (negative charge) → charge = $ -2 $.

### For Other Elements (e.g., Na, S, K, etc.), repeat:
1. **Atomic Number**: From periodic table (e.g., Na=11, S=16, K=19...).
2. **Electrons per Level**: Fill 1s, 2s, 2p, 3s, 3p, 4s... (e.g., Na: 1s²2s²2p⁶3s¹ → levels: 2, 8, 1).
3. **Electron Configuration**: Write orbitals (e.g., Na: $ 1s^2 2s^2 2p^6 3s^1 $).
4. **Electrons Lost/Gained**: Valence electrons (outer shell). Metals (Na, K, Al, Sr, Ca) lose electrons; nonmetals (O, S, Cl, F, Br, N, I) gain electrons.
   - Na: 1 valence e⁻ → loses 1.
   - S: 6 valence e⁻ → gains 2.
   - K: 1 valence e⁻ → loses 1.
   - Al: 3 valence e⁻ → loses 3.
   - Cl: 7 valence e⁻ → gains 1.
   - Sr: 2 valence e⁻ → loses 2.
   - Ca: 2 valence e⁻ → loses 2.
   - F: 7 valence e⁻ → gains 1.
   - Br: 7 valence e⁻ → gains 1.
   - N: 5 valence e⁻ → gains 3.
   - I: 7 valence e⁻ → gains 1.
5. **Electrons Left**:
   - If loses: total e⁻ - lost (e.g., Na: 11 - 1 = 10).
   - If gains: total e⁻ + gained (e.g., S: 16 + 2 = 18).
6. **Charge**:
   - Loses e⁻ → positive charge (e.g., Na: $ +1 $, K: $ +1 $, Al: $ +3 $, Sr: $ +2 $, Ca: $ +2 $).
   - Gains e⁻ → negative charge (e.g., S: $ -2 $, Cl: $ -1 $, F: $ -1 $, Br: $ -1 $, N: $ -3 $, I: $ -1 $).

### Example Table for O:
| Element | Atomic Number | e⁻ per E Level | Electron Configuration | e⁻ Lost/Gained | e⁻ Left | Charge |
|---------|---------------|----------------|------------------------|----------------|---------|--------|
| O       | 8             | 2, 6           | $ 1s^2 2s^2 2p^4 $   | Gain 2         | 10      | $ -2 $ |

Repeat this process for all elements (Na, S, K, Al, Cl, Sr, Ca, F, Br, N, I) using their atomic numbers and valence electrons.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step 1: Atomic Number of O

Step 2: Electrons per Energy Level

Step 3: Electron Configuration

Step 4: Electrons Lost/Gained (Octet Rule)

Step 5: Electrons Left After Gain

Step 6: Charge on Ion

For Other Elements (e.g., Na, S, K, etc.), repeat:

Example Table for O:

Answer: