sch4u
structure and
properties of matter
assignment
1. complete the table below by providing a set of quantum numbers for the last four
electrons in an phosphorus atom. 4

Question

Accepted Answer

# Explanation:
## Step1: Determine the electron configuration of phosphorus
Phosphorus has an atomic number of 15. Its electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$. The last four electrons are in the $3s^{2}3p^{2}$ sub - shells.
## Step2: Recall the rules for quantum numbers
- **Principal quantum number ($n$)**: For the $3s$ and $3p$ sub - shells, $n = 3$.
- **Orbital shape quantum number ($l$)**: For $s$ sub - shell $l=0$, for $p$ sub - shell $l = 1$.
- **Magnetic quantum number ($m_{l}$)**: For $l = 0$, $m_{l}=0$; for $l = 1$, $m_{l}=- 1,0,1$.
- **Spin quantum number ($m_{s}$)**: Can be $+\frac{1}{2}$ or $-\frac{1}{2}$

## Step3: Assign quantum numbers for each electron
- **Electron 1 ($3s^{1}$)**: $n = 3$, $l = 0$, $m_{l}=0$, $m_{s}=+\frac{1}{2}$
- **Electron 2 ($3s^{2}$)**: $n = 3$, $l = 0$, $m_{l}=0$, $m_{s}=-\frac{1}{2}$
- **Electron 3 ($3p^{1}$)**: $n = 3$, $l = 1$, $m_{l}=-1$, $m_{s}=+\frac{1}{2}$
- **Electron 4 ($3p^{2}$)**: $n = 3$, $l = 1$, $m_{l}=0$, $m_{s}=+\frac{1}{2}$

# Answer:
| e- | Principle quantum number | Orbital Shape quantum numbers | Magnetic quantum number | Spin quantum number |
|---|---|---|---|---|
| 1 | 3 | 0 | 0 | $+\frac{1}{2}$ |
| 2 | 3 | 0 | 0 | $-\frac{1}{2}$ |
| 3 | 3 | 1 | - 1 | $+\frac{1}{2}$ |
| 4 | 3 | 1 | 0 | $+\frac{1}{2}$ |

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QUESTION IMAGE

Question

Explanation:

Step1: Determine the electron configuration of phosphorus

Step2: Recall the rules for quantum numbers

Step3: Assign quantum numbers for each electron

Answer:

e-	Principle quantum number	Orbital Shape quantum numbers	Magnetic quantum number	Spin quantum number
2	3	0	0	\(-\frac{1}{2}\)
3	3	1	- 1	\(+\frac{1}{2}\)
4	3	1	0	\(+\frac{1}{2}\)