Chemistry - GENERAL CHEMISTRY Question with Solution | TestHub

To determine the value of $\ell \times m$ for each element, we consider the maximum possible values of $\ell$ and $m$ for their filled orbitals.

For a given principal quantum number $n$, the azimuthal quantum number $\ell$ can range from $0$ to $n-1$. The magnetic quantum number $m$ can range from $-\ell$ to $+\ell$. The maximum possible value of $m$ for a given $\ell$ is $\ell$.

Let's analyze each element:

A. Na (Sodium, Z=11): Electron configuration is $1s^{2}2s^{2}2p^{6}3s^1$.

The highest filled orbital is $2p$. For $2p$, $n=2$, $\ell =1$. The maximum $m$ for $\ell =1$ is $1$.

So, $\ell \times m=1\times 1=1$.

B. Ca (Calcium, Z=20): Electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^2$.

The highest filled orbital is $3p$. For $3p$, $n=3$, $\ell =1$. The maximum $m$ for $\ell =1$ is $1$.

So, $\ell \times m=1\times 1=1$.

C. Al (Aluminum, Z=13): Electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^1$.

The highest filled orbital is $2p$. For $2p$, $n=2$, $\ell =1$. The maximum $m$ for $\ell =1$ is $1$.

So, $\ell \times m=1\times 1=1$.

D. Ga (Gallium, Z=31): Electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^1$.

The highest filled orbital is $3d$. For $3d$, $n=3$, $\ell =2$. The maximum $m$ for $\ell =2$ is $2$.

So, $\ell \times m=2\times 2=4$.

Elements Na, Ca, and Al all have $\ell \times m=1$.

The final answer is $\boxed{A,B,C}$.