Chapter 6, Problem 84QP

Interpretation Introduction

Interpretation:

The quantum numbers of each electron of the element with the electron configuration $1s^{2}2s^{2}2p^{4}3s^2$ are to be determined, and the name of that element is to be identified.

Concept introduction:

The principle quantum number $\left(n\right)$, the angular quantum number $\left(l\right)$, the magnetic spin quantum number $\left(m_l\right)$, and the electron spin quantum number $\left(m_s\right)$ are the four quantum numbers.

The principle quantum number designates the orbital size. The angular momentum quantum number tells the shape of the atomic orbital. The possible range of l is from $0$ to $n-1$. The magnetic quantum number tells the orientation of the atomic orbital in space. The possible range of $m_l$ is from $-l\text{ to}+l$. The electron spin quantum number tells the electron spin in an orbital in space. The possible spins are $+\frac{1}{2}\text{ or }-\frac{1}{2}$.

Answer to Problem 84QP

Solution:

The quantum numbers for the electrons in $1s$ are $\left(1,0,0,+\frac{1}{2}\right)$ and $\left(1,0,0,-\frac{1}{2}\right)$.

The quantum numbers for the electrons in $2s$ are $\left(2,0,0,+\frac{1}{2}\right)$ and $\left(2,0,0,-\frac{1}{2}\right)$.

The quantum numbers for the electrons in $2p$ are $\left(2,1,-1,+\frac{1}{2}\right)$, $\left(2,1,-1,-\frac{1}{2}\right)$, $\left(2,1,0,+\frac{1}{2}\right)$, $\left(2,1,0,-\frac{1}{2}\right)$, $\left(2,1,+1,+\frac{1}{2}\right)$, and $\left(2,1,+1,-\frac{1}{2}\right)$.

The quantum numbers for the electrons in $3s$ are $\left(3,0,0,+\frac{1}{2}\right)$ and $\left(3,0,0,-\frac{1}{2}\right)$.

The name of the element is magnesium.

Explanation of Solution

For $1s$, the principle quantum number $\left(n\right)$ is $1$. The angular momentum quantum number $\left(l\right)$ is $0$ to $n-1$; but for $n=1$, $l\text{ is }0$. The value of the magnetic quantum number $\left(m_l\right)$ is $-l\mathrm{\dots \dots .0\dots \dots .}+l$. For $l=0$, $m_l\text{ is 0}$. The spin of the electron in one orbital can either be $+\frac{1}{2}\text{ or }-\frac{1}{2}$. Therefore, the quantum numbers for the electrons in $1s$ are $\left(1,0,0,+\frac{1}{2}\right)$ and $\left(1,0,0,-\frac{1}{2}\right)$.

For $2s$, the principle quantum number $\left(n\right)$ is $2$. The angular momentum quantum number $\left(l\right)$ is $0$ to $n-1$. The value of $l\text{ is }0\text{ and 1}$, but the value of $l$ for $2s$ is $0$. The value of the magnetic quantum number $\left(m_l\right)$ is $-l\mathrm{\dots \dots .0\dots \dots .}+l$; so, for $l=0$, $m_l\text{ is 0}$. The spin of the electron in one orbital can either be $+\frac{1}{2}\text{ or }-\frac{1}{2}$. Therefore, the quantum numbers for the electrons in $2s$ are $\left(2,0,0,+\frac{1}{2}\right)$ and $\left(2,0,0,-\frac{1}{2}\right)$.

For $2p$, the principle quantum number $\left(n\right)$ is $2$. The angular momentum quantum number $\left(l\right)$ is $0$ to $n-1$. So, the value of $l\text{ is }0\text{ and 1}$, but the value of $l$ for $2p$ is $1$. The value of the magnetic quantum number $\left(m_l\right)$ is $-l\mathrm{\dots \dots .0\dots \dots .}+l$; so, for $l=1$, $m_l\text{ is }-1,\text{0,+1}$. The spin of the electron in one orbital can either be $+\frac{1}{2}\text{ or }-\frac{1}{2}$. Therefore, the quantum numbers for the electrons in $2p$ are $\left(2,1,-1,+\frac{1}{2}\right)$, $\left(2,1,-1,-\frac{1}{2}\right)$, $\left(2,1,0,+\frac{1}{2}\right)$, $\left(2,1,0,-\frac{1}{2}\right)$, $\left(2,1,+1,+\frac{1}{2}\right)$, and $\left(2,1,+1,-\frac{1}{2}\right)$.

For $3s$, the principle quantum number $\left(n\right)$ is $3$. The angular momentum quantum number $\left(l\right)$ is $0$ to $n-1$. So, the values of $l\text{ are }0,1\text{ and 2}$, but the value of $l$, for $3s$, is $0$. The value of the magnetic quantum number $\left(m_l\right)$ is $-l\mathrm{\dots \dots .0\dots \dots .}+l$; so, for $l=0$, $m_l\text{ is 0}$. The spin of the electron in one orbital can either be $+\frac{1}{2}\text{ or }-\frac{1}{2}$. Therefore, the quantum numbers for the electrons in $3s$ are $\left(3,0,0,+\frac{1}{2}\right)$ and $\left(3,0,0,-\frac{1}{2}\right)$.

Since the total number of electrons in the element is 12, the element is magnesium.

Conclusion

There are 12 electrons in the element having electron configuration $1s^{2}2s^{2}2p^{4}3s^2$. Atomic number of neutral atom is 12 and its name is magnesium $\text{Mg}$.