Question

If the quantum number l could have the value 'n' also then, Sc(21) would have electronic configuration as (other rules strictly followed)

• A. 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1
• B. 1s^2 1p^6 2s^2 2p^6 3s^2 3p^3
• C. 1s^2 1p^6 2s^2 2p^6 3s^2 2d^3
• D. 1s^2 1p^6 2s^2 2p^6 3s^2 3d^3

Answer 1

Answer: (C)

1s^2 1p^6 2s^2 2p^6 3s^2 2d^3

Answer 2

The usual rules for quantum numbers are:

• Principal quantum number $n = 1, 2, 3, ...$
• Azimuthal quantum number $l = 0, 1, ..., n-1$
• Magnetic quantum number $m_l = -l, -l+1, ..., 0, ..., l-1, l$
• Spin quantum number $m_s = +1/2, -1/2$

The capacity of a subshell with quantum number $l$ is $2(2l+1)$.

The modified rule states that the azimuthal quantum number $l$ can have values from $0$ to $n$. All other rules (Aufbau principle, Hund's rule, Pauli exclusion principle) are strictly followed. This implies we fill the subshells in order of increasing energy. We assume the energy order follows the $(n+l)$ rule, with lower $n$ for ties in $(n+l)$.

Let's list the available subshells and their capacities for the first few principal quantum numbers under the modified rule:

• $n=1$: $l$ can be $0, 1$.
  • $l=0$: 1s subshell, capacity $2(2 \times 0 + 1) = 2$. $(n+l = 1+0 = 1)$
  • $l=1$: 1p subshell, capacity $2(2 \times 1 + 1) = 6$. $(n+l = 1+1 = 2)$
• $n=2$: $l$ can be $0, 1, 2$.
  • $l=0$: 2s subshell, capacity $2(2 \times 0 + 1) = 2$. $(n+l = 2+0 = 2)$
  • $l=1$: 2p subshell, capacity $2(2 \times 1 + 1) = 6$. $(n+l = 2+1 = 3)$
  • $l=2$: 2d subshell, capacity $2(2 \times 2 + 1) = 10$. $(n+l = 2+2 = 4)$
• $n=3$: $l$ can be $0, 1, 2, 3$.
  • $l=0$: 3s subshell, capacity $2(2 \times 0 + 1) = 2$. $(n+l = 3+0 = 3)$
  • $l=1$: 3p subshell, capacity $2(2 \times 1 + 1) = 6$. $(n+l = 3+1 = 4)$
  • $l=2$: 3d subshell, capacity $2(2 \times 2 + 1) = 10$. $(n+l = 3+2 = 5)$
  • $l=3$: 3f subshell, capacity $2(2 \times 3 + 1) = 14$. $(n+l = 3+3 = 6)$
• $n=4$: $l$ can be $0, 1, 2, 3, 4$.
  • $l=0$: 4s subshell, capacity $2$. $(n+l = 4+0 = 4)$

We arrange the subshells in order of increasing $(n+l)$, and for subshells with the same $(n+l)$, we order them by increasing $n$.

1. $(n+l)=1$: 1s ($n=1$). Order: 1s.
2. $(n+l)=2$: 1p ($n=1$), 2s ($n=2$). Order: 1p, 2s.
3. $(n+l)=3$: 2p ($n=2$), 3s ($n=3$). Order: 2p, 3s.
4. $(n+l)=4$: 2d ($n=2$), 3p ($n=3$), 4s ($n=4$). Order: 2d, 3p, 4s.
5. $(n+l)=5$: 3d ($n=3$), 4p ($n=4$). Order: 3d, 4p.
6. $(n+l)=6$: 3f ($n=3$), 4d ($n=4$). Order: 3f, 4d.

The filling order of subshells is: 1s, 1p, 2s, 2p, 3s, 2d, 3p, 4s, 3d, ...

Now we fill 21 electrons for Scandium (Sc, Z=21) according to this order:

1. 1s: fills with 2 electrons. Configuration: $1s^2$. Total electrons: 2. Remaining: 19.
2. 1p: fills with 6 electrons. Configuration: $1s^2 1p^6$. Total electrons: 8. Remaining: 13.
3. 2s: fills with 2 electrons. Configuration: $1s^2 1p^6 2s^2$. Total electrons: 10. Remaining: 11.
4. 2p: fills with 6 electrons. Configuration: $1s^2 1p^6 2s^2 2p^6$. Total electrons: 16. Remaining: 5.
5. 3s: fills with 2 electrons. Configuration: $1s^2 1p^6 2s^2 2p^6 3s^2$. Total electrons: 18. Remaining: 3.
6. 2d: needs 3 electrons. The 2d subshell has a capacity of 10. It fills with the remaining 3 electrons. Configuration: $1s^2 1p^6 2s^2 2p^6 3s^2 2d^3$. Total electrons: $18 + 3 = 21$. Remaining: 0.

The electronic configuration of Sc(21) under the modified rule is $1s^2 1p^6 2s^2 2p^6 3s^2 2d^3$.