Electrons in Shells (Bohr–Bury Rules) – Long Answer Questions

Medium Level (Application & Explanation)

Q1. State the Bohr–Bury rules for electron distribution. Explain with Sodium, Neon, and Magnesium.

Answer:

The maximum electrons in a shell = 2n². Here, n is the shell number.
The outermost shell can have a maximum of 8 electrons.
Electrons fill lower energy shells first. This is step-wise filling.
Sodium (11): 2, 8, 1. One electron in the outer shell.
Neon (10): 2, 8. Outer shell complete. It is stable.
Magnesium (12): 2, 8, 2. Two electrons in the outer shell.

Q2. Use the 2n² formula to find capacities of K, L, M, N shells. Then explain why the outermost shell still has only 8 electrons.

Answer:

For K (n=1): 2n² = 2 × 1² = 2 electrons.
For L (n=2): 2n² = 2 × 2² = 8 electrons.
For M (n=3): 2n² = 2 × 3² = 18 electrons.
For N (n=4): 2n² = 2 × 4² = 32 electrons.
But the outermost shell can hold only a maximum of 8 electrons.
This makes atoms aim for an octet. It gives stability like noble gases.

Q3. Explain step-wise filling of shells. Use Carbon, Oxygen, and Fluorine as examples.

Answer:

Electrons fill the lowest energy shell first.
This is called step-wise filling of shells.
Carbon (6): 2, 4. K gets 2 first, then L gets 4.
Oxygen (8): 2, 6. K gets 2 first, then L gets 6.
Fluorine (9): 2, 7. Outer shell is almost full.
A filled or nearly filled outer shell controls reactivity.

Q4. Write electronic configurations for Z = 6, 8, 11, 12, 17. Explain the outer shell and likely behavior.

Answer:

Carbon (6): 2, 4. Four in the outer shell. Can share electrons.
Oxygen (8): 2, 6. Needs two to complete octet.
Sodium (11): 2, 8, 1. Loses one electron easily.
Magnesium (12): 2, 8, 2. Loses two electrons.
Chlorine (17): 2, 8, 7. Gains one electron.
Outer shell electrons decide valency and reactivity.

Q5. Why does a full outer shell make atoms stable? Explain with Neon, Sodium, and Magnesium.

Answer:

A full outer shell means 8 electrons (octet).
Neon (10): 2, 8. It is inert and stable.
Sodium (11): 2, 8, 1. Loses one to reach 2, 8 like Neon.
Magnesium (12): 2, 8, 2. Loses two to reach 2, 8.
Atoms change by losing, gaining, or sharing electrons for stability.
A complete octet lowers energy. So atoms become less reactive.

High Complexity (Analysis & Scenario-Based)

Q6. An atom has 15 electrons. Using Bohr–Bury rules, write its configuration and predict its behavior.

Answer:

Atomic number 15 gives electrons = 15.
Configuration: 2, 8, 5. K has 2, L has 8, M has 5.
Outermost electrons = 5. It is not a complete octet.
It tends to gain 3 or share electrons to reach 8.
So it can show valency 3. It forms stable compounds by sharing.
This matches the idea of octet and step-wise filling.

Q7. Two atoms A (Z = 20) and B (Z = 19) are compared. Who loses electrons more easily? Explain with shells.

Answer:

A (20): 2, 8, 8, 2.
B (19): 2, 8, 8, 1.
B has one electron in the outer shell. It is easier to lose.
A has two outer electrons. It needs to lose both to get stable.
So B (Z = 19) is more reactive in losing electrons.
Both aim for a stable octet in the next inner shell.

Q8. A student says, “The outermost shell can have 10 electrons.” Evaluate this using Bohr–Bury rules.

Answer:

2n² gives shell capacities like 2, 8, 18, 32.
But the Bohr–Bury rule says the outermost shell has maximum 8 electrons.
Example: Argon (18) is 2, 8, 8, not 2, 8, 10.
Even though M can hold 18, it does not exceed 8 when it is outermost.
The octet makes atoms stable.
So the statement is incorrect as per these rules.

Q9. Propose a quick method to write configurations up to Z = 20. Show it for Z = 19 and Z = 16.

Answer:

Step 1: Use 2n² to know shell limits: 2, 8, 18, 32.
Step 2: Fill K first, then L, then M, and so on.
Step 3: Keep outermost ≤ 8 electrons.
For Z = 19 (K to N): 2, 8, 8, 1.
For Z = 16 (Sulfur): 2, 8, 6.
This method follows step-wise filling and the octet rule.

Q10. Why can’t the M shell have more than 8 electrons when it is the outermost shell? What changes when N starts filling?

Answer:

M shell capacity by 2n² is 18.
But when M is the outermost shell, it holds only up to 8.
This follows the Bohr–Bury rule for stability.
When the N shell starts to fill, M can then have more than 8 overall.
Example: Before N starts, we see 2, 8, 8. After N starts, 2, 8, 8, 1, etc.
The change keeps the outermost within 8 until the next shell opens.

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