Electron Configuration Quiz - Answer Key

Section A Full Electron Configurations - Answers

Problem 1

Complete electron configurations:

a. Ca: 1s²2s²2p⁶3s²3p⁶4s²
b. P: 1s²2s²2p⁶3s²3p³
c. Be: 1s²2s²
d. Cl: 1s²2s²2p⁶3s²3p⁵

Work:
a. Ca (Z=20): Fill 1s²(2) + 2s²(4) + 2p⁶(10) + 3s²(12) + 3p⁶(18) + 4s²(20) ✓
b. P (Z=15): Fill through 3s², then 3p³ (15 total)
c. Be (Z=4): Just 1s² + 2s²
d. Cl (Z=17): Fill through 3s², then 3p⁵ (17 total)

Problem 2

Fluorine electron configuration:

a. 1s²2s²2p⁵

F has Z = 9 electrons
1s²(2) + 2s²(4) + 2p⁵(9) ✓

Why others are wrong:
b. Has 13 electrons (too many)
c. Has 9 but wrong order (3s before 2p is full)
d. Has only 7 electrons

Problem 3

Cadmium (₄₈Cd) configuration:

c. 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰

Cd has 48 electrons. Check option c:
2+2+6+2+6+2+10+6+2+10 = 48 ✓

Note: 4d fills after 5s (filling order!)

Section B Noble Gas Shorthand - Answers

Problem 4

Noble gas configurations:

a. N: [He]2s²2p³
b. Na: [Ne]3s¹
c. Fe: [Ar]4s²3d⁶

a. N (Z=7): [He]=2, remaining 5 → 2s²2p³
b. Na (Z=11): [Ne]=10, remaining 1 → 3s¹
c. Fe (Z=26): [Ar]=18, remaining 8 → 4s²3d⁶

Problem 5

Vanadium (₂₃V) configuration:

a. [Ar]4s²3d³

V has 23 electrons, Ar has 18
Remaining: 23 - 18 = 5 electrons
Fill 4s² first (2), then 3d³ (3) = 5 ✓

Why others are wrong:
b. 4p fills after 3d, not before
c. 4d doesn't exist at this point
d. [Kr] has 36 electrons (more than V!)

Problem 6

Lead (Pb, Z=82):

[Xe]6s²4f¹⁴5d¹⁰6p²

Xe has 54 electrons
Remaining: 82 - 54 = 28 electrons
Fill: 6s²(2) + 4f¹⁴(16) + 5d¹⁰(26) + 6p²(28) ✓

Section C Identifying Elements - Answers

Problem 7

Identify elements:

a. Chlorine (Cl)
b. Rubidium (Rb)
c. Cobalt (Co)

a. 2+2+6+2+5 = 17 electrons → Cl
b. 2+2+6+2+6+2+10+6+1 = 37 electrons → Rb
c. [Ar]=18 + 2+7 = 27 electrons → Co

Problem 8

Element 1s²2s²2p⁶3s²3p² is in Group:

d. 14

This is Silicon (Si, Z=14)
Outer electrons: 3s²3p² = 4 valence electrons
Group 14 elements have 4 valence electrons

Alternative: Count from periodic table - Si is in Group 14

Section D Sublevel Questions - Answers

Problem 9

Orbitals in d sublevel:

b. 5

d sublevel has 5 orbitals
Each orbital holds 2 electrons
5 × 2 = 10 electrons max in d sublevel

Problem 10

Electrons to fill 3d sublevel:

10 electrons

d sublevel = 5 orbitals × 2 electrons each = 10 electrons

Problem 11

Elements in s-block:

14 elements

s-block = Groups 1 and 2 (plus H and He)
Group 1: H, Li, Na, K, Rb, Cs, Fr (7 elements)
Group 2: Be, Mg, Ca, Sr, Ba, Ra (6 elements)
He is also s-block (1s²)
Total: 7 + 6 + 1 = 14 elements

Problem 12

Electrons in n=3 level for 1s²2s²2p⁶3s²3p²:

b. 4

Third energy level (n=3) = 3s + 3p + 3d
In this configuration: 3s² + 3p² = 2 + 2 = 4 electrons

Problem 13

Orbitals with at least one electron in 1s²2s²2p⁶3s²3p²:

c. 8

Count orbitals (not electrons):
1s: 1 orbital (full)
2s: 1 orbital (full)
2p: 3 orbitals (full)
3s: 1 orbital (full)
3p: 2 orbitals (with 1 electron each, by Hund's rule)
Total: 1+1+3+1+2 = 8 orbitals

Section E Orbital Diagrams - Answers

Problem 14

Aluminum orbital configuration:

[Ne] 3s: ↑↓ 3p: ↑

a. Metal
b. Give electrons (metals lose electrons)
c. 3+ charge
d. 3s² and 3p¹ electrons (all 3 valence electrons)

Al (Z=13) = [Ne]3s²3p¹
Al is in Group 13 (metals give electrons)
Loses all 3 valence electrons to form Al³⁺

Problem 15

Errors in orbital diagrams:

Diagram A: Two electrons with the same spin in one orbital (↑↑ and ↓↓)
Violates Pauli Exclusion Principle!

Diagram B: Each d orbital has not received one electron before any pairs are made
Violates Hund's Rule!

Diagram A fix: All paired electrons must be ↑↓
Diagram B fix: With 5 electrons in 3d, should be:

3d: ↑ ↑ ↑ ↑ ↑

Section F Unpaired Electrons - Answers

Problem 16

Most unpaired electrons:

d. N (Nitrogen)

F (Z=9): [He]2s²2p⁵ → 2p has ↑↓ ↑↓ ↑ = 1 unpaired
S (Z=16): [Ne]3s²3p⁴ → 3p has ↑↓ ↑ ↑ = 2 unpaired
Cu (Z=29): [Ar]4s¹3d¹⁰ → all d paired = 1 unpaired
N (Z=7): [He]2s²2p³ → 2p has ↑ ↑ ↑ = 3 unpaired ✓

Problem 17

Unpaired in [Ar]4s¹3d⁵:

d. 6

4s: ↑ 3d: ↑ ↑ ↑ ↑ ↑

1 (in 4s) + 5 (in 3d) = 6 unpaired electrons

Problem 18

Unpaired electrons:

a. O: 2 unpaired
b. Si: 2 unpaired
c. Fe: 4 unpaired

a. O (Z=8): [He]2s²2p⁴ → 2p: ↑↓ ↑ ↑ = 2 unpaired
b. Si (Z=14): [Ne]3s²3p² → 3p: ↑ ↑ _ = 2 unpaired
c. Fe (Z=26): [Ar]4s²3d⁶ → 3d: ↑↓ ↑ ↑ ↑ ↑ = 4 unpaired

Section G Challenge Problems - Answers

Problem 19

Two s and six p electrons in outer level:

d. Ar (Argon)

Looking for ns²np⁶ in outermost level:
He: 1s² (only 2 electrons, no p)
O: 2s²2p⁴ (only 4 p electrons)
Si: 3s²3p² (only 2 p electrons)
Ar: 3s²3p⁶ (2 s + 6 p = 8 valence) ✓

Problem 20

Groups 13-18 are filling:

a. p sublevels

Groups 13-18 = the p-block
Group 13: ns²np¹
Group 14: ns²np²
Group 15: ns²np³
Group 16: ns²np⁴
Group 17: ns²np⁵
Group 18: ns²np⁶

The s electrons are from the previous sublevel; the distinguishing electrons are p.

Problem 21

Lanthanides and actinides:

d. filling in f level electrons

Lanthanides (elements 57-71): filling 4f
Actinides (elements 89-103): filling 5f

They are the f-block of the periodic table.

Problem 22

[Kr]5s²4d¹⁰5p⁴:

Tellurium (Te), 6 valence electrons

Kr = 36 electrons
Add: 2 + 10 + 4 = 16 more
Total: 36 + 16 = 52 → Tellurium (Te)

Valence electrons = outer s and p = 5s² + 5p⁴ = 6

Problem 23

Cl⁻ ion configuration:

Cl⁻: 1s²2s²2p⁶3s²3p⁶

This is identical to Argon (Ar)!

Cl (Z=17) has 17 electrons
Cl⁻ gains 1 electron → 18 electrons
Configuration: 1s²2s²2p⁶3s²3p⁶

Ar (Z=18) also has 18 electrons with the same configuration.
Cl⁻ is isoelectronic with Ar (same electron configuration).