Question

The total number of electrons with magnetic quantum number \(m_l = 0\) in Fe\(^{3+}\) and Cr\(^+\) is respectively

• A. 14, 13
• B. 13, 14
• C. 11, 11
• D. 13, 13

Hint:

Electrons in orbitals with \(m_l = 0\): count 1 from each s orbital, 2 from full p, and 1 from singly filled d.

Answer 1

The Correct Option is C

We need to count electrons with magnetic quantum number \(m_l = 0\), which corresponds to:
- s-orbitals (\(l = 0\)) → \(m_l = 0\)
- p-orbitals (\(l = 1\)) → \(m_l = -1, 0, +1\)
- d-orbitals (\(l = 2\)) → \(m_l = -2, -1, 0, +1, +2\)
⇒ Each orbital has one \(m_l = 0\) orbital
Now determine electron configurations:
For Fe\(^{3+}\) (Atomic number 26 → loses 3e⁻ → 23e⁻)
Electronic configuration:
\[ 1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 3d^5 \] Electrons in \(m_l = 0\) orbitals:
- 1s, 2s, 3s → 3 orbitals × 1 = 3 electrons
- 2p, 3p → each has 1 orbital with \(m_l = 0\) → 2
- 3d: \(d^5\) fills each orbital singly → includes 1 in \(m_l = 0\) → 1
Total = \(3 + 2 + 1 = 6\)
(But this is incorrect with our original assumption—let's reanalyze with all orbitals that have \(m_l = 0\))
Actually, for each full orbital set:
- s: 1 orbital → 2 electrons (all with \(m_l = 0\))
- p: 3 orbitals → 1 has \(m_l = 0\) → max 2 electrons
- d: 5 orbitals → 1 with \(m_l = 0\)
Let’s sum for each shell:
Fe\(^{3+}\): 23 electrons
- 1s² → 2 (m_l = 0) → 2
- 2s² → 2 (m_l = 0) → 2
- 2p⁶ → 2 (in \(m_l = 0\))
- 3s² → 2 (m_l = 0) → 2
- 3p⁶ → 2 (in \(m_l = 0\))
- 3d⁵ → 1 (each orbital singly filled) → 1 in \(m_l = 0\)
Total = \(2 + 2 + 2 + 2 + 2 + 1 = 11\)
Cr\(^+\): Z = 24, loses 1e⁻ → 23e⁻ Expected config: \[ 1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 3d^5 \] (Same as Fe\(^{3+}\))
→ So total is also 11

Answer 2

Step 1: Determine the electronic configuration of Fe\(^{3+}\)
Neutral Fe has atomic number 26, so its electron configuration is [Ar] 3d\(^{6}\) 4s\(^{2}\).
Fe\(^{3+}\) means 3 electrons are removed, usually first from 4s and then 3d orbitals.
So, Fe\(^{3+}\) configuration: [Ar] 3d\(^{5}\).

Step 2: Determine the electronic configuration of Cr\(^{+}\)
Neutral Cr has atomic number 24, configuration [Ar] 3d\(^{5}\) 4s\(^{1}\).
Cr\(^{+}\) means 1 electron is removed, usually from 4s.
So, Cr\(^{+}\) configuration: [Ar] 3d\(^{5}\).

Step 3: Count electrons with magnetic quantum number \(m_l = 0\)
For 3d orbitals, the magnetic quantum number \(m_l\) can be –2, –1, 0, +1, +2.
Each orbital can hold 2 electrons (with opposite spins).
Electrons fill each orbital singly before pairing (Hund's rule).

For 3d\(^{5}\) configuration (Fe\(^{3+}\) and Cr\(^{+}\)):
- Five electrons occupy five different d orbitals singly.
- Each d orbital has 1 electron, so the orbital with \(m_l = 0\) has 1 electron.

For inner shells ([Ar]) electrons, count electrons with \(m_l=0\) in filled orbitals:
- 1s, 2s, 3s orbitals each have 2 electrons with \(m_l=0\) (s orbitals have \(l=0\), so \(m_l=0\) only).
- 1s, 2s, 3s total 6 electrons with \(m_l=0\).

For 3p orbitals (3p\(^{6}\)):
- 3p orbitals have \(l=1\), so \(m_l = -1,0,+1\).
- Each p orbital has 2 electrons, total 6 electrons.
- Electrons with \(m_l=0\) in 3p orbitals = 2 (since one orbital has \(m_l=0\) with 2 electrons).

Total electrons with \(m_l=0\) in [Ar]:
6 (from s orbitals) + 2 (from 3p) = 8 electrons.

Add 1 electron from 3d orbital with \(m_l=0\) (since 3d\(^{5}\) fills each orbital singly):
8 + 1 = 9 electrons.

Check 4s and 4p orbitals:
- Fe\(^{3+}\) and Cr\(^{+}\) do not have 4s electrons.
- So, no additional electrons here.

Therefore, total electrons with \(m_l=0\) = 9 (from [Ar]) + 1 (from 3d) = 10.

However, the correct given answer is 11, which indicates we should consider paired spins or review inner shell orbitals:
Actually, each s orbital has 2 electrons with \(m_l=0\), so:
- 1s: 2 electrons
- 2s: 2 electrons
- 3s: 2 electrons
Total s electrons with \(m_l=0\) = 6

In 3p orbitals (3p\(^{6}\)), 2 electrons have \(m_l=0\)

In 3d\(^{5}\) orbitals, 1 electron has \(m_l=0\)

Sum = 6 + 2 + 1 = 9 electrons.
To get 11, consider also 2 electrons in 2p with \(m_l=0\) and 2 electrons in 1p (no 1p), so correction:
- 1s: 2 (m_l=0)
- 2s: 2 (m_l=0)
- 2p: 6 electrons but only 2 with \(m_l=0\)
- 3s: 2 (m_l=0)
- 3p: 6 electrons, 2 with \(m_l=0\)

So total electrons with \(m_l=0\):
2 (1s) + 2 (2s) + 2 (2p) + 2 (3s) + 2 (3p) + 1 (3d) = 11 electrons.

Step 4: Final answer
Therefore, both Fe\(^{3+}\) and Cr\(^{+}\) have 11 electrons with magnetic quantum number \(m_l = 0\).