Worksheet focused on electron configurations and quantum numbers, including exercises on writing configurations, drawing orbital diagrams, and identifying elements based on electron arrangements.

Let's solve each question step-by-step with clear explanations.

---

1. Give the name and symbol for each of the following elements:

We determine the total number of electrons from the electron configuration and match it to the atomic number.

a. $ 1s^2 2s^2 2p^6 3s^1 $

- Total electrons = 2 + 2 + 6 + 1 = 11
- Element with atomic number 11 is Sodium (Na)

✔ Answer: Sodium, Na

---

b. $ 1s^2 2s^2 2p^3 $

- Total electrons = 2 + 2 + 3 = 7
- Atomic number 7 → Nitrogen (N)

✔ Answer: Nitrogen, N

---

c. $ 1s^2 2s^2 2p^6 3s^2 3p^6 3d^3 4s^2 $

- Total electrons = 2+2+6+2+6+3+2 = 29
- Atomic number 29 → Copper (Cu)

⚠️ Wait! Let's double-check:
Standard order: after 3p⁶ comes 4s before 3d.
But here: 3d³ 4s² → this is correct for Copper, but Cu is [Ar] 3d¹⁰ 4s¹ — exception!

Wait! This configuration has only 3d³, not 3d¹⁰. So let's count again:

- 1s² → 2
- 2s² → 2 (total 4)
- 2p⁶ → 6 (total 10)
- 3s² → 2 (12)
- 3p⁶ → 6 (18)
- 3d³ → 3 (21)
- 4s² → 2 (23)

So total electrons = 23

Atomic number 23 → Vanadium (V)

✔ Answer: Vanadium, V

---

d. $ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 $

- Total electrons = 2+2+6+2+6+1 = 19
- Atomic number 19 → Potassium (K)

✔ Answer: Potassium, K

---

✔ Answers:

a. Sodium, Na
b. Nitrogen, N
c. Vanadium, V
d. Potassium, K

---

2. What is the maximum number of electrons in each subshell?

Use formula: Maximum electrons = 2 × (number of orbitals)
Orbitals per subshell:
- s → 1 orbital → 2e⁻
- p → 3 orbitals → 6e⁻
- d → 5 orbitals → 10e⁻
- f → 7 orbitals → 14e⁻
- g → 9 orbitals → 18e⁻

a) 2s → 2 electrons
b) 4p → 6 electrons
c) 3d → 10 electrons
d) 1p → Invalid! No 1p subshell (n=1 only has s). But if we assume it exists, it would be 6e⁻, but actually, 1p doesn't exist → 0 electrons

But since it’s listed, perhaps just treat as theoretical: p subshell holds 6e⁻ regardless of n, but 1p is not allowed. So 0

But in many contexts, they expect the max based on type.

So:
e) 1p → 6 electrons (if allowed), but in reality, no 1p orbital → 0
But likely they want: 6 (based on p subshell capacity)

Similarly, g) 6g → g subshell has l=4 → 2(2l+1) = 2(9)=18 electrons

So:

| Subshell | Max Electrons |
|---------|---------------|
| a) 2s | 2 |
| b) 4p | 6 |
| c) 3d | 10 |
| d) 1p | 0 (invalid) or 6? → 0 (since n=1 can’t have p) |
| e) 1p | 0 |
| f) 2p | 6 |
| g) 2d | 0 (n=2, l≤1 → no d) → 0 |
| h) 6g | 18 (g subshell: l=4 → 2(2×4+1)=18) |

✔ Answers:
a) 2
b) 6
c) 10
d) 0
e) 0
f) 6
g) 0
h) 18

---

3. Number of orbitals in each subshell

Use: number of orbitals = 2l + 1

- s → l=0 → 1 orbital
- p → l=1 → 3 orbitals
- d → l=2 → 5 orbitals
- f → l=3 → 7 orbitals
- g → l=4 → 9 orbitals

Also, subshells like 1p, 2d don’t exist, so number of orbitals = 0

a) 2s → s → 1 orbital
b) 4p → p → 3 orbitals
c) 3d → d → 5 orbitals
d) 1p → invalid → 0
e) 4f → f → 7 orbitals
f) 2p → p → 3 orbitals
g) 2d → invalid → 0
h) 6g → g → 9 orbitals

✔ Answers:
a) 1
b) 3
c) 5
d) 0
e) 7
f) 3
g) 0
h) 9

---

4. Write electron configurations for elements

a) P (Phosphorus, Z=15)
→ $ 1s^2 2s^2 2p^6 3s^2 3p^3 $

b) Y (Yttrium, Z=39)
→ $ [Kr] 5s^2 4d^1 $
Full: $ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^1 $

c) Cs (Cesium, Z=55)
→ $ [Xe] 6s^1 $

d) Re (Rhenium, Z=43)
→ $ [Kr] 5s^2 4d^5 5p^6 $ → wait, Re is actually $ [Kr] 5s^1 4d^5 5p^6 $? No.

Actually: Re (Z=43):
After Kr (36), add 7 more electrons:
→ 5s² 4d⁵ → but Re is [Kr] 5s¹ 4d⁵ 5p⁶? No.

Correct: Re is [Kr] 5s¹ 4d⁵ 5p⁶? No.

Standard: Re → [Kr] 5s¹ 4d⁵ 5p⁶? Let's check:

Kr = 36
Re = 43 → 7 electrons after Kr

Order: 5s, then 4d, then 5p?

But 4d fills before 5p.

So: 5s² → 2
Then 4d⁵ → 5 → total 7 → yes.

But Re is an exception: it's [Kr] 5s¹ 4d⁶? Wait, no.

Actually, Re is [Kr] 5s¹ 4d⁵ 5p⁶? No.

Correct configuration: [Kr] 5s¹ 4d⁵ 5p⁶? That would be 36 + 1 + 5 + 6 = 48 → too much.

Wait: After Kr, Re adds 7 electrons.

The correct configuration is: [Kr] 5s¹ 4d⁵ → 36 + 1 + 5 = 42 → still short.

Wait: Re is [Kr] 5s¹ 4d⁵? No.

Correct: Re is [Kr] 5s¹ 4d⁵ 5p⁶? No.

Actually, Re (Z=43):
→ $ [Kr] 5s^1 4d^5 $ → 36 + 1 + 5 = 42 → need one more.

No: Re is [Kr] 5s¹ 4d⁵? Still 42.

Wait: I think I'm missing something.

Correct: Re → [Kr] 5s¹ 4d⁵ → that’s 42 electrons.

But Re is 43 → so must be [Kr] 5s¹ 4d⁵ 5p⁶? No, that’s 36+1+5+6=48.

No — after Kr, the order is: 5s, 4d, 5p, 6s...

So for Re (Z=43):

- Kr = 36
- Then: 5s² → 2 → 38
- 4d⁵ → 5 → 43 → done

So Re → [Kr] 5s² 4d⁵

But actual configuration is [Kr] 5s¹ 4d⁶ due to half-filled stability?

Yes! Re is [Kr] 5s¹ 4d⁶ → 36 + 1 + 6 = 43

So: [Kr] 5s¹ 4d⁶

✔ Re: $ [Kr] 5s^1 4d^6 $

e) Cm (Curium, Z=96)
→ Actinide series. Cm is [Rn] 7s² 5f⁷

But let’s confirm: Rn = 86, Cm = 96 → 10 electrons after Rn.

Filling: 7s², then 5f⁷ → 2+7=9 → need one more? 86+9=95 → one more.

So: 7s² 5f⁷ 6d¹? Or 5f⁸?

Actually, Cm is [Rn] 5f⁷ 7s² → 86 + 7 + 2 = 95 → no, 96.

Wait: Cm is [Rn] 5f⁷ 6d¹ 7s²? That's 86+7+1+2=96.

But standard: Cm is [Rn] 5f⁷ 7s²? No.

Actually, Curium (Cm, Z=96):
→ [Rn] 5f⁷ 6d¹ 7s²? No.

Correct: [Rn] 5f⁷ 7s² → 86 + 7 + 2 = 95 → wrong.

Wait: Rn is 86, Cm is 96 → 10 electrons added.

After Rn: 7s² (2), 5f⁸ (8) → 10 → so [Rn] 5f⁸ 7s²

But actually, Cm is [Rn] 5f⁷ 6d¹ 7s²? No.

Standard: Cm is [Rn] 5f⁷ 7s² → 86+7+2=95 → no.

Wait: Cm is [Rn] 5f⁷ 6d¹ 7s²? That's 86+7+1+2=96.

But according to periodic table, Cm is [Rn] 5f⁷ 7s²? No.

Actually, Cm is [Rn] 5f⁷ 6d¹ 7s²? Let me recall:

- Lanthanides: after La, fill 4f
- Actinides: after Ac, fill 5f

So:
- Ac (89): [Rn] 6d¹ 7s²
- Th (90): [Rn] 6d² 7s²
- Pa (91): [Rn] 5f² 6d¹ 7s²
- U (92): [Rn] 5f³ 6d¹ 7s²
- Np (93): [Rn] 5f⁴ 6d¹ 7s²
- Pu (94): [Rn] 5f⁶ 7s²
- Am (95): [Rn] 5f⁷ 7s²
- Cm (96): [Rn] 5f⁷ 6d¹ 7s²

Yes! So Cm: [Rn] 5f⁷ 6d¹ 7s²

✔ Cm: $ [Rn] 5f^7 6d^1 7s^2 $

f) He (Helium, Z=2)
→ $ 1s^2 $

g) Ni (Nickel, Z=28)
→ $ [Ar] 4s^2 3d^8 $

✔ So final answers:

a) P: $ 1s^2 2s^2 2p^6 3s^2 3p^3 $
b) Y: $ [Kr] 5s^2 4d^1 $
c) Cs: $ [Xe] 6s^1 $
d) Re: $ [Kr] 5s^1 4d^6 $
e) Cm: $ [Rn] 5f^7 6d^1 7s^2 $
f) He: $ 1s^2 $
g) Ni: $ [Ar] 4s^2 3d^8 $

---

5. Draw electron box diagrams (orbital notation)

We'll describe how to draw them.

a) P (Z=15)
- 1s: ↑↓
- 2s: ↑↓
- 2p: ↑↓ ↑↓ ↑↓
- 3s: ↑↓
- 3p: ↑ ↑ ↑ (three unpaired)

b) Y (Z=39)
- [Kr] → full up to 36
- Then: 5s² ↑↓
- 4d¹ ↑ (one electron)

So: 5s: ↑↓, 4d: ↑ in first box

c) Cs (Z=55)
- [Xe] → full up to 54
- 6s¹: ↑

d) Re (Z=43)
- [Kr] → 36
- 5s¹: ↑
- 4d⁶: ↑↓ ↑↓ ↑ ↑ ↑ ↑ (6 electrons in 5 orbitals: 2 paired, 4 unpaired)

e) Cm (Z=96)
- [Rn] → 86
- 5f⁷: ↑↓ ↑↓ ↑↓ ↑ ↑ ↑ ↑ (7 electrons: 3 pairs, 1 unpaired? No: 7 electrons in 7 orbitals → all unpaired? No, 7 orbitals → 7 electrons → all ↑)

Hund's rule: all spins same → 7 unpaired electrons in 5f

- 6d¹: ↑
- 7s²: ↑↓

f) He (Z=2)
- 1s: ↑↓

g) Ni (Z=28)
- [Ar] → 18
- 4s²: ↑↓
- 3d⁸: ↑↓ ↑↓ ↑↓ ↑ ↑ (two unpaired)

You draw boxes for each orbital, arrows for electrons.

---

6. Electron configurations for ions

a) P³⁻
P (Z=15) gains 3 electrons → 18 electrons
→ $ 1s^2 2s^2 2p^6 3s^2 3p^6 $ → same as Ar

b) Mg²⁺
Mg (Z=12) loses 2 electrons → 10 electrons
→ $ 1s^2 2s^2 2p^6 $ → Ne

c) Cs⁺
Cs (Z=55) loses 1 → 54 electrons → Xe
→ $ [Xe] $

d) O²⁻
O (Z=8) gains 2 → 10 electrons → Ne
→ $ 1s^2 2s^2 2p^6 $

e) U³⁺
U (Z=92) → loses 3 electrons
U: [Rn] 5f³ 6d¹ 7s²
U³⁺: remove 7s² and one 6d → [Rn] 5f³

So: $ [Rn] 5f^3 $

✔ Answers:
a) $ 1s^2 2s^2 2p^6 3s^2 3p^6 $
b) $ 1s^2 2s^2 2p^6 $
c) $ [Xe] $
d) $ 1s^2 2s^2 2p^6 $
e) $ [Rn] 5f^3 $

---

7. Maximum number of electrons for quantum numbers

Quantum numbers: (n, l, m_l, m_s)

Each set defines one electron unless unspecified.

a) (2,1,,) → n=2, l=1 → p subshell
m_l can be -1,0,+1 → 3 values
m_s = ±½ → 2 possibilities
But only m_l and m_s unspecified → so 3 orbitals × 2 spins = 6 electrons

b) (5,,) → n=5, l unspecified → l can be 0 to 4 → s,p,d,f,g
For each l, multiple m_l → total orbitals = 1+3+5+7+9 = 25 → 50 electrons

But if only n given, max electrons = 2n² = 2×25 = 50

So 50 electrons

c) (4,4,,) → n=4, l=4 → g subshell
l=4 → m_l = -4 to +4 → 9 values
Each can have 2 spins → 18 electrons

d) (3,2,-1,-½) → specific set: all four quantum numbers given → only 1 electron

e) (3,0,1,) → n=3, l=0 → s orbital
m_l = 0 → only possible value
m_s = ±½ → two electrons possible
But m_l is given as 1 → but for l=0, m_l must be 0 → so no such electron possible → 0 electrons

f) (4,3,,) → n=4, l=3 → f subshell
m_l = -3 to +3 → 7 values
each with 2 spins → 14 electrons

g) (3,3,-3,) → n=3, l=3 → f subshell → l ≤ n-1 → l=3 → n=3 → l=3 is allowed (since l ≤ n-1 → 3 ≤ 2? No!)
n=3 → l max = 2 → so l=3 is impossible → 0 electrons

h) (3,,1,) → n=3, m_l=1, m_s unspecified
l can be 1 or 2 (since m_l=1 → l ≥ 1)
- If l=1 (p): m_l=1 → possible → m_s = ±½ → 2 electrons
- If l=2 (d): m_l=1 → possible → m_s = ±½ → 2 electrons
Total: 4 electrons

But m_l=1 → for l=1: one orbital; l=2: one orbital → total 2 orbitals → 4 electrons

✔ Answers:
a) 6
b) 50
c) 18
d) 1
e) 0
f) 14
g) 0
h) 4

---

8. Quantum numbers from electron box diagram

Given diagram:

```
1s: ↑ ↓
2s: [↑] ↓ ← boxed
2p: ↑↓ ← diamond
↑
↑
2p: ↑ ↓
3s: ↑ ↓
3p: ↑ ↓ ← circled
↑
↑
```

Wait, the diagram shows:

- 1s: ↑↓
- 2s: ↑↓ → but one arrow is boxed → the boxed is the up arrow in 2s
- 2p: three orbitals: first has ↑↓ (diamond), second has ↑, third has ↑
- 3s: ↑↓
- 3p: three orbitals: first ↑↓ (circled), second ↑↓, third ↑

Now:

Circled → the first 3p orbital, which has ↑↓ → the down arrow is circled? Or the whole orbital?

It says "Circled =" and points to the first 3p orbital which has both arrows.

But it says "the circled" → probably the electron in the circled box.

Assuming the box is circled → the last electron placed in that box.

But the circled box has ↑↓ → so two electrons.

But likely, the last electron placed is the down arrow.

But the problem says: "Circled =", "Boxed =", etc.

From diagram:

- Circled: the first 3p orbital → contains ↑↓ → the down arrow is the last placed.
- Boxed: the up arrow in 2s
- Diamond: the first 2p orbital, which has ↑↓ → the down arrow is diamond?
- Last one placed: the last electron added → probably the rightmost 3p orbital, which has ↑ (only one electron)

Let’s list electrons in order:

1. 1s: ↑↓ → 2 electrons
2. 2s: ↑↓ → 2 electrons (boxed is ↑)
3. 2p: ↑↓ (diamond), ↑, ↑ → 5 electrons
4. 3s: ↑↓ → 2 electrons
5. 3p: ↑↓ (circled), ↑↓, ↑ → 5 electrons

Total electrons: 2+2+5+2+5 = 16 → Sulfur (S)

Now:

- Circled: the first 3p orbital → the down arrow in it → this is the second electron in 3p₁
- n=3, l=1 (p), m_l = -1 (for first p orbital), m_s = -½ → (3,1,-1,-½)

- Boxed: the up arrow in 2s → first electron in 2s
- n=2, l=0, m_l=0, m_s=+½ → (2,0,0,+½)

- Diamond: the down arrow in first 2p orbital → 2pₓ → m_l=-1 → (2,1,-1,-½)

- Last one placed: the single ↑ in the last 3p orbital → 3p_z → m_l=+1 → (3,1,+1,+½)

✔ Answers:

- Circled = (3,1,-1,-½)
- Boxed = (2,0,0,+½)
- Diamond = (2,1,-1,-½)
- Last one placed = (3,1,+1,+½)

---

9. What element is the diagram showing?

Total electrons = 16 → Sulfur (S)

✔ Answer: Sulfur

---

10. Draw electron box diagram for chlorine (Cl, Z=17)

Chlorine: $ 1s^2 2s^2 2p^6 3s^2 3p^5 $

Orbitals:

- 1s: ↑↓
- 2s: ↑↓
- 2p: ↑↓ ↑↓ ↑↓
- 3s: ↑↓
- 3p: ↑↓ ↑↓ ↑ (three orbitals: first two full, third has one)

Label orbitals:

```
1s: [↑↓]
2s: [↑↓]
2p: [↑↓] [↑↓] [↑↓]
3s: [↑↓]
3p: [↑↓] [↑↓] [↑]
```

Only use needed boxes.

So:

```
□ □ □ □ □ □ □ □ □
```

Label above:

1s 2s 2p 2p 2p 3s 3p 3p 3p

Fill:

- 1s: ↑↓
- 2s: ↑↓
- 2p: ↑↓ ↑↓ ↑↓
- 3s: ↑↓
- 3p: ↑↓ ↑↓ ↑

So:

```
[↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑]
```

But only 9 boxes → label:

1s 2s 2px 2py 2pz 3s 3px 3py 3pz

Fill:

- 1s: ↑↓
- 2s: ↑↓
- 2px: ↑↓
- 2py: ↑↓
- 2pz: ↑↓
- 3s: ↑↓
- 3px: ↑↓
- 3py: ↑↓
- 3pz: ↑

---

11. Quantum numbers of last electron in Cl

Last electron is in 3p_z → n=3, l=1, m_l=+1, m_s=+½

Because 3p has three orbitals: m_l = -1, 0, +1

We fill: 3px (-1), 3py (0), 3pz (+1)

Last electron goes into 3pz → ↑ → spin up → m_s = +½

✔ (3,1,+1,+½)

---

✔ Final Answers Summary:

1.
a) Sodium, Na
b) Nitrogen, N
c) Vanadium, V
d) Potassium, K

2.
a) 2
b) 6
c) 10
d) 0
e) 0
f) 6
g) 0
h) 18

3.
a) 1
b) 3
c) 5
d) 0
e) 7
f) 3
g) 0
h) 9

4.
a) $ 1s^2 2s^2 2p^6 3s^2 3p^3 $
b) $ [Kr] 5s^2 4d^1 $
c) $ [Xe] 6s^1 $
d) $ [Kr] 5s^1 4d^6 $
e) $ [Rn] 5f^7 6d^1 7s^2 $
f) $ 1s^2 $
g) $ [Ar] 4s^2 3d^8 $

5. Draw orbital diagrams (not shown here)

6.
a) $ 1s^2 2s^2 2p^6 3s^2 3p^6 $
b) $ 1s^2 2s^2 2p^6 $
c) $ [Xe] $
d) $ 1s^2 2s^2 2p^6 $
e) $ [Rn] 5f^3 $

7.
a) 6
b) 50
c) 18
d) 1
e) 0
f) 14
g) 0
h) 4

8.
- Circled = (3,1,-1,-½)
- Boxed = (2,0,0,+½)
- Diamond = (2,1,-1,-½)
- Last one placed = (3,1,+1,+½)

9. Sulfur (S)

10. Diagram:
```
1s: [↑↓] 2s: [↑↓] 2p: [↑↓][↑↓][↑↓] 3s: [↑↓] 3p: [↑↓][↑↓][↑]
```

11. (3,1,+1,+½)

---

Let me know if you'd like a visual version of any diagram!