Class 11 Chemistry Worksheet on Chapter 2 Structure of Atom Set - 4 - Free Printable

Let's solve each question from the Chemistry Worksheet step by step and explain the reasoning behind each answer.

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Q1. The orbital nearest to the nucleus is:

Options:
a.) 4f
b.) 5d
c.) 4s
d.) 7p

Answer: c.) 4s

Explanation:
The distance of an orbital from the nucleus depends on its principal quantum number (n) and azimuthal quantum number (l). Although 4s has a higher n than 3s, 3p, etc., among the given options, we compare:

- 4f → n = 4, l = 3
- 5d → n = 5, l = 2
- 4s → n = 4, l = 0
- 7p → n = 7, l = 1

Even though all have different n values, the orbital with the lowest n is closest, but here 4s and 4f both have n=4, so we must consider penetration effect.

Among orbitals with same n, s-orbitals penetrate closer to the nucleus than p, d, or f due to their spherical shape and higher probability near the nucleus.

So even though 4f and 4s are in the same shell, 4s is closer to the nucleus than 4f.

Also, comparing across shells:
- 4s (n=4) is much closer than 5d (n=5) or 7p (n=7)

Thus, 4s is the nearest among the options.

✔ Correct Answer: c.) 4s

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Q2. The electronic configuration of Cu²⁺ ion is:

Options:
a.) [Ar] 3d⁸4s¹
b.) [Ar] 3d⁹4s⁰
c.) [Ar] 3d⁷4s²
d.) [Ar] 3d⁸4s⁰

Answer: b.) [Ar] 3d⁹4s⁰

Explanation:
Copper (Cu) has atomic number 29.

- Ground state electron configuration of Cu:
[Ar] 3d¹⁰ 4s¹
(Note: Exception to Aufbau principle – half-filled or fully filled d-subshell is more stable.)

Now, for Cu²⁺, we remove two electrons.

Electrons are removed first from the outermost shell, i.e., 4s before 3d.

So:
- First electron removed: from 4s → now [Ar] 3d¹⁰ 4s⁰
- Second electron removed: from 3d → now [Ar] 3d⁹ 4s⁰

So, Cu²⁺ = [Ar] 3d⁹ 4s⁰

✔ Correct Answer: b.) [Ar] 3d⁹4s⁰

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Q3. What is the maximum number of emission lines obtained when the excited electrons of a hydrogen atom in n = 5 drop to the ground state?

Options:
a.) 10
b.) 5
c.) 12
d.) 15

Answer: d.) 15

Explanation:
When an electron in a hydrogen atom drops from a higher energy level (n = 5) to lower levels, it can emit photons corresponding to various transitions.

To find the maximum number of spectral lines, we use the formula:

> Number of spectral lines = n(n - 1)/2

This is because the electron can transition from n=5 to any lower level (n=4, 3, 2, 1), and every possible transition between these levels contributes a line.

But since we want all possible transitions from n=5 down to n=1, including intermediate steps, the total number of unique transitions is:

> Number of ways to choose 2 different levels from 5 levels = C(5, 2) = 5×4/2 = 10

Wait! But that’s not right — actually, this formula gives number of possible transitions from n=5 to lower states, but only if we consider direct transitions.

Actually, the correct way is:

An electron starting at n=5 can go directly to n=4, n=3, n=2, or n=1 → 4 transitions.

Then from n=4, it can go to n=3, n=2, n=1 → 3 transitions.

From n=3 → n=2, n=1 → 2 transitions.

From n=2 → n=1 → 1 transition.

But since each transition emits one spectral line, and we are interested in how many different wavelengths (i.e., distinct energy differences), we count all possible downward transitions.

Total number of transitions =
From n=5: to n=4, 3, 2, 1 → 4
From n=4: to n=3, 2, 1 → 3
From n=3: to n=2, 1 → 2
From n=2: to n=1 → 1

Total = 4 + 3 + 2 + 1 = 10

Wait! That gives 10.

But the correct formula for maximum number of spectral lines when an electron falls from level n to ground state is:

> Number of lines = n(n−1)/2

For n = 5:
→ 5×4 / 2 = 10

So why is option d.) 15?

Wait — maybe there's confusion.

But let's double-check.

Actually, if the electron starts at n=5, it can fall to any lower level, and every possible path leads to a unique energy difference.

But the total number of spectral lines observed is equal to the number of possible transitions between different energy levels.

So, for n = 5, the number of possible transitions is:

All pairs (i, j) where i > j and i ≤ 5, j ≥ 1

That is:
Transitions:
5→4, 5→3, 5→2, 5→1
4→3, 4→2, 4→1
3→2, 3→1
2→1

Total = 4 + 3 + 2 + 1 = 10

So, 10 spectral lines.

But wait — is there a possibility of multiple paths leading to same wavelength? No, each transition has a unique ΔE.

So maximum number of distinct emission lines is 10.

But let's see the options again:
a.) 10 ✔
b.) 5
c.) 12
d.) 15

So a.) 10 should be correct.

But wait — perhaps the question means: "when excited electrons in n=5 drop to ground state", meaning all possible ways the electron can de-excite, and each transition produces a line.

Yes, and that’s exactly what we calculated.

But sometimes people get confused and think of combinations.

Alternatively, some might think of total number of possible transitions from n=5 to all lower levels, which is indeed 10.

But let’s verify with standard result:

- From n=2 → 1 line (2→1)
- From n=3 → 3 lines (3→2, 3→1, 2→1)
- From n=4 → 6 lines (4→3, 4→2, 4→1, 3→2, 3→1, 2→1)
- From n=5 → 10 lines

Yes, 10 lines.

So why is option d.) 15?

Wait — could the question mean something else?

No. Standard formula: n(n−1)/2 = 5×4/2 = 10

✔ So Answer: a.) 10

But hold on — let's recheck the options:

a.) 10
b.) 5
c.) 12
d.) 15

So a.) 10 is correct.

But I recall that some sources say that if the electron is in n=5, it can make multiple transitions, and the total number of possible spectral lines is 10.

Yes.

However, wait — there's a common mistake: sometimes people confuse with the number of spectral series or lines in a series, but here it's asking for maximum number of emission lines.

So yes, 10.

But let me check online or recall:
For hydrogen, when electron goes from n=5 to n=1, number of possible spectral lines = 10

✔ So Correct Answer: a.) 10

But the worksheet says option d.) 15 — that would be for n=6: 6×5/2 = 15.

So probably a typo in the question or options?

No — wait, perhaps the student thinks of all possible transitions between all levels up to n=5, including those not starting from n=5.

But no — the question says: "excited electrons of a hydrogen atom in n = 5 drop to the ground state"

So only transitions from n=5 down, including via intermediate levels.

Each time the electron jumps, it emits a photon.

But the total number of distinct wavelengths is still 10.

For example:
- 5→1
- 5→2
- 5→3
- 5→4
- 4→1
- 4→2
- 4→3
- 3→1
- 3→2
- 2→1

→ 10 distinct transitions.

Yes.

So Answer: a.) 10

But wait — let's see if the formula is correctly applied.

Another way: number of spectral lines = n(n−1)/2 = 5×4/2 = 10

Yes.

So ✔ Answer: a.) 10

But earlier I thought d.) 15 was tempting — but it's wrong.

So Final Answer: a.) 10

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Q4. The maximum number of electrons in a subshell is given by the expression:

Options:
a.) 4l − 2
b.) 4l + 2
c.) 2l + 1
d.) 2n²

Answer: None of the above? Wait...

Let’s analyze.

A subshell is defined by the azimuthal quantum number l.

For a given l, the magnetic quantum number mₗ ranges from −l to +l → total of 2l + 1 orbitals.

Each orbital holds 2 electrons → max electrons = 2 × (2l + 1) = 4l + 2

So, maximum number of electrons in a subshell = 4l + 2

Let’s test:

- s subshell: l = 0 → 4(0)+2 = 2 → correct
- p subshell: l = 1 → 4(1)+2 = 6 → correct
- d subshell: l = 2 → 4(2)+2 = 10 → correct
- f subshell: l = 3 → 4(3)+2 = 14 → correct

Perfect!

So the correct expression is 4l + 2

Now look at options:

a.) 4l − 2 → wrong
b.) 4l + 2 → ✔ correct
c.) 2l + 1 → number of orbitals, not electrons
d.) 2n² → total electrons in a shell

So ✔ Answer: b.) 4l + 2

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Q5. In Bohr’s theory, the radius r of the orbit is proportional to _____.

Options:
a.) n
b.) n²
c.) n⁻¹
d.) n⁻²

Answer: b.) n²

Explanation:
In Bohr’s model, the radius of the nth orbit is given by:

> r ∝ n²

More precisely:
r = (0.529 Å) × n² / Z

Where n is the principal quantum number, Z is atomic number.

So radius is proportional to n²

✔ Answer: b.) n²

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✔ Final Answers:

| Q | Answer |
|---|--------|
| Q1 | c.) 4s |
| Q2 | b.) [Ar] 3d⁹4s⁰ |
| Q3 | a.) 10 |
| Q4 | b.) 4l + 2 |
| Q5 | b.) n² |

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Summary of Explanations:

1. 4s is closest due to better penetration than other orbitals.
2. Cu²⁺ loses two electrons: first from 4s, then from 3d → 3d⁹4s⁰.
3. From n=5 to ground state, number of spectral lines = 5×4/2 = 10.
4. Max electrons in subshell = 2 × (2l+1) = 4l+2.
5. Bohr’s radius ∝ n².

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