Solution:

Given: Sulfur (S), $Z = 16$ Find: Full electron configuration

Step 1: Determine number of electrons

Neutral sulfur has 16 electrons (same as protons).

Step 2: Fill orbitals in order

Order: 1s, 2s, 2p, 3s, 3p

• 1s: 2 electrons → $1s^2$ (total: 2)
• 2s: 2 electrons → $2s^2$ (total: 4)
• 2p: 6 electrons → $2p^6$ (total: 10)
• 3s: 2 electrons → $3s^2$ (total: 12)
• 3p: 4 electrons → $3p^4$ (total: 16) ✓

Answer: $1s^2 2s^2 2p^6 3s^2 3p^4$

Noble gas notation: $[Ne] 3s^2 3p^4$

Verification:

• Total electrons: $2 + 2 + 6 + 2 + 4 = 16$ ✓
• Valence electrons: 6 (matches Group 16) ✓

Solution:

(a) Complete configuration:

• Fe (Z=26): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶

(b) Noble gas notation:

• Previous noble gas is Ar (Z=18)
• Fe: [Ar] 4s² 3d⁶

(c) Unpaired electrons:

• 4s²: 2 paired electrons
• 3d⁶: ↑↓ ↑ ↑ ↑ ↑ (following Hund's rule)
• 4 unpaired electrons in the 3d subshell

(d) Subshell being filled:

• Transition metals fill the d subshell (specifically the (n-1)d subshell)

Solution:

Given: Carbon (C), $Z = 6$ Find: Orbital diagram and unpaired electrons

Step 1: Write electron configuration

Carbon: $1s^2 2s^2 2p^2$

Step 2: Draw orbital diagram

1s: ↑↓

2s: ↑↓

2p: ↑ ↑ _ (three orbitals: px, py, pz)

Step 3: Apply Hund's Rule

For the 2 electrons in 2p:

• Place one electron in first orbital (↑)
• Place one electron in second orbital (↑)
• Leave third orbital empty

Complete diagram:

      ↑↓        ↑↓      ↑  ↑  _
      1s        2s      2p

Answer:

• Orbital diagram shown above
• Unpaired electrons: 2 (both in 2p sublevel)

Explanation:

• Hund's rule requires maximizing unpaired electrons in degenerate orbitals
• Carbon's 2 unpaired electrons make it chemically reactive
• These unpaired electrons form bonds in compounds like CO₂, CH₄

Verification:

• Total electrons: 2 + 2 + 2 = 6 ✓
• Hund's rule followed ✓

Solution:

(a) Cu²⁺ configuration:

• Cu is [Ar] 4s¹ 3d¹⁰
• Remove 2 electrons from highest energy levels
• Cu²⁺: [Ar] 3d⁹

(b) Copper's anomalous configuration:

• Expected: [Ar] 4s² 3d⁹
• Actual: [Ar] 4s¹ 3d¹⁰
• Explanation: A completely filled d¹⁰ subshell is more stable than d⁹ due to exchange energy. The extra stability from having a filled d subshell outweighs the energy cost of promoting an electron from 4s to 3d.

(c) Electron removal:

• When forming ions, electrons are removed from the highest n value first (4s before 3d)
• Even though Cu fills 3d last, the 4s¹ electron is removed first, then one 3d electron
• This is why Cu²⁺ is [Ar] 3d⁹, not [Ar] 4s¹ 3d⁸

Solution:

Given: Fe (iron, $Z = 26$) and Fe²⁺ Find: Electron configurations and explanation

Step 1: Write configuration for neutral Fe

Iron has 26 electrons.

Following the filling order through 4s and 3d:

$\text{Fe: } [Ar] 4s^2 3d^6$

Expanded: $[Ar] = 1s^2 2s^2 2p^6 3s^2 3p^6$

Step 2: Form Fe²⁺ (remove 2 electrons)

Key concept: When forming cations, remove electrons from the highest $n$ value first.

For Fe: Remove from 4s before 3d

$\text{Fe}^{2+}: [Ar] 3d^6$

NOT $[Ar] 4s^2 3d^4$ (incorrect!)

Step 3: Explain the difference

Neutral Fe: $[Ar] 4s^2 3d^6$ (26 electrons)

• 4s sublevel: 2 electrons
• 3d sublevel: 6 electrons

Fe²⁺: $[Ar] 3d^6$ (24 electrons)

• 4s sublevel: 0 electrons (both removed)
• 3d sublevel: 6 electrons (unchanged)

Why remove 4s first?

Even though 4s fills before 3d, once the 3d sublevel begins filling, it becomes lower in energy than 4s. When ionizing, electrons are removed from the highest energy orbital, which is 4s.

Answer:

• Fe: $[Ar] 4s^2 3d^6$
• Fe²⁺: $[Ar] 3d^6$
• Difference: Lost both 4s electrons

Verification:

• Fe has 26 electrons ✓
• Fe²⁺ has 24 electrons (26 - 2) ✓
• 4s electrons removed before 3d ✓