Chemistry - Atomic Structure Question with Solution | TestHub

(I) $n=4,l=2,m=-1$: This describes a specific orbital within the 4d subshell. A single orbital can hold a maximum of 2 electrons (one with $s=+1/2$ and one with $s=-1/2$). Since $m=-1$ specifies one orbital, it can hold 2 electrons. So, (I) $\to$ (R).

(II) $n=3,l=1$: This describes the 3p subshell. For $l=1$, $m$ can be $-1,0,+1$. There are 3 orbitals in the 3p subshell. Each orbital can hold 2 electrons. So, the 3p subshell can hold $3\times 2=6$ electrons. So, (II) $\to$ (Q).

(III) $n=5,l=0,s=-1/2$: This describes a specific electron in the 5s subshell with a spin of $-1/2$. For $l=0$, $m$ can only be $0$. This specifies a single orbital (5s orbital) and a specific spin. Therefore, only 1 electron fits this description. So, (III) $\to$ (P).

(IV) $n=2,l=2$: This combination is NOT possible because the value of $l$ must be less than $n$. Here, $l=2$ is not less than $n=2$. Therefore, no electrons can exist with these quantum numbers. So, (IV) $\to$ (T).

Matching:

(I) $\to$ (R)

(II) $\to$ (Q)

(III) $\to$ (P)

(IV) $\to$ (T)

The incorrect combination is (III) $\to$ (R) because (III) corresponds to (P).

The final answer is $\boxed{C}$