Given hypothetical rule

Normally, for a shell with principal quantum number $n$, the allowed values of azimuthal quantum number are $l=0,1,2,\dots,n-1.$

But here, hypothetically, $l=0,1,2,\dots,n+1.$

So for each $n$, there are two extra subshells compared to the real world.

Capacity of each subshell

For a given $l$, number of orbitals $=2l+1$, so maximum electrons in that subshell: $2(2l+1)=4l+2.$

Thus:

• $l=0 \Rightarrow s$ holds $2$
• $l=1 \Rightarrow p$ holds $6$
• $l=2 \Rightarrow d$ holds $10$
• $l=3 \Rightarrow f$ holds $14$

Subshells available for small $n$ under this rule

• For $n=1$: $l=0,1,2$ i.e. $1s,1p,1d$
• For $n=2$: $l=0,1,2,3$ i.e. $2s,2p,2d,2f$

Order of filling (Aufbau principle)

We use the usual $n+l$ rule. If two subshells have same $n+l$, lower $n$ fills first.

Compute for early subshells:

• $1s$: $n+l=1$
• $1p$: $n+l=2$
• $2s$: $n+l=2$
• $1d$: $n+l=3$
• $2p$: $n+l=3$
• $3s$: $n+l=3$

So filling order begins as: $1s < 1p < 2s < 1d < 2p < 3s < \dots$

Write first few electronic configurations

Fill electrons in this order:

• $Z=1,2$: $1s^1,1s^2$

• Next $1p$ can hold 6 electrons, so $Z=3\text{ to }8: 1s^2 1p^{1\text{ to }6}$

• Thus at $Z=8$: $1s^2 1p^6$ This is a completely filled outer subshell, so this would be the first noble gas.

• Next $2s$ starts, so $Z=9: 1s^2 1p^6 2s^1$ This has one electron in the outermost subshell, so this is the first alkali metal.

• Continue filling $2s^2$ at $Z=10$.

• Then $1d$ starts at $Z=11$: $Z=11: 1s^2 1p^6 2s^2 1d^1$

• Since $1d$ can hold 10 electrons, half-filled means $1d^5$. That occurs at $Z=11+4=15.$ So $Z=13$ is $1s^2 1p^6 2s^2 1d^3,$ not half-filled.

• For $Z=6$: $1s^2 1p^4$ so valence subshell is $1p$, not $2p$.

Check each option

A: 13 has a half-filled valence subshell

At $Z=13$: $1s^2 1p^6 2s^2 1d^3$. Valence subshell is $1d$, but it is not half-filled ($d^5$ would be half-filled).

A is false.

B: 9 is the first alkali metal

At $Z=8$: $1s^2 1p^6$ is noble gas. Next element $Z=9$: $1s^2 1p^6 2s^1$. This is the first alkali metal.

B is true.

C: 8 is the first noble gas

Yes, $Z=8$ gives $1s^2 1p^6$, a closed-shell configuration.

C is true.

D: 6 has a 2p-valence subshell

At $Z=6$: $1s^2 1p^4$. Valence subshell is $1p$, not $2p$.

D is false.

Conclusion

Correct options are: $\boxed{B,\ C}$

Comparison with stored answer

Stored correct answer is A, but from the filling order under the given hypothetical quantum-number rule, A is false while B and C are true. So the stored answer appears incorrect.