Question

1. complete the table below by providing a set of quantum numbers for the last four electrons in a phosphorus atom. 4
2. explain what is incorrect with respect to the following set of quantum numbers: ( n = 3, l = 3, m = - 1 ) 2
3. provide the complete set of quantum numbers for all the electrons that could populate the 3d subshell of an atom. 2

Explanation:

1. Phosphorus has atomic number 15, so its electron configuration is $1s^22s^22p^63s^23p^3$. The last four electrons are in the 3p subshell. For 3p, $n=3$, $l=1$ (p orbital), $m_l$ values are -1, 0, +1 (each orbital gets one electron first per Hund's rule), and $m_s$ is $+\frac{1}{2}$ for each unpaired electron.
2. The orbital angular momentum quantum number $l$ must satisfy $0 \leq l \leq n-1$. For $n=3$, the maximum $l$ is 2, so $l=3$ is invalid.
3. For the 3d subshell: $n=3$ (principal quantum number), $l=2$ (d orbital). $m_l$ ranges from -2 to +2 (-2, -1, 0, +1, +2), and each $m_l$ has two spin states: $+\frac{1}{2}$ and $-\frac{1}{2}$.

Answer:

1.

e⁻	Principle quantum number ($n$)	Orbital Shape quantum number ($l$)	Magnetic quantum number ($m_l$)	Spin quantum number ($m_s$)
2	3	1	0	$+\frac{1}{2}$
3	3	1	+1	$+\frac{1}{2}$
4	(Note: Last four electrons are 3p³—third row has 3 electrons; if considering 12th-15th: 3s² (e⁻12: $n=3,l=0,m_l=0,m_s=+\frac{1}{2}$; e⁻13: $n=3,l=0,m_l=0,m_s=-\frac{1}{2}$) + 3p¹⁻² (e⁻14: $n=3,l=1,m_l=-1,m_s=+\frac{1}{2}$; e⁻15: $n=3,l=1,m_l=0,m_s=+\frac{1}{2}$) – adjust if "last four" refers to 12-15)

1. The value of $l=3$ is invalid because $l$ must be less than $n$ (here $n=3$, so $l$ can only be 0,1,2).
2. All quantum numbers for 3d: $n=3$, $l=2$, $m_l=-2,-1,0,+1,+2$, and $m_s=+\frac{1}{2}$ or $-\frac{1}{2}$ for each $m_l$.

(Note: For question 1, standard last four electrons in P (15 electrons) are the 12th (3s² down), 13th (3s² up), 14th (3p₋₁ up), 15th (3p₀ up) – table can be filled with these values as above.)