1. complete the table below by providing a set of quantum numbers for the last four electrons in an phosphorus atom. 4
2. explain what is incorrect with respect to the following set of quantum numbers: n = 3, l = 3, m_l=-1
structure and properties of matter
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1. **For Phosphorus (P) with atomic number $Z = 15$, electron - configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$. The last four electrons are in the $3p$ sub - shell**:
   - **Principal Quantum Number ($n$)**:
     - **Explanation**: The principal quantum number represents the energy level of the electron. For the $3p$ electrons, $n = 3$ as they are in the third energy level.
     - $n = 3$
   - **Azimuthal Quantum Number ($l$)**:
     - **Explanation**: For $p$ orbitals, the azimuthal quantum number $l=1$. It determines the shape of the orbital ($l = 0$ for $s$, $l = 1$ for $p$, $l = 2$ for $d$, $l = 3$ for $f$).
     - $l = 1$
   - **Magnetic Quantum Number ($m_l$)**:
     - **Explanation**: The possible values of $m_l$ range from $-l$ to $+l$. For $l = 1$, $m_l=- 1,0, + 1$. For the three $3p$ electrons, we can have different $m_l$ values. Let's consider the three $3p$ electrons filling the three $3p$ orbitals ($3p_x$, $3p_y$, $3p_z$). For the fourth electron in the $3p$ sub - shell, we can choose one of the values. Let $m_l=-1$ (we could also choose $0$ or $+1$).
     - $m_l=-1$
   - **Spin Quantum Number ($m_s$)**:
     - **Explanation**: The spin quantum number can have two values, $m_s=+\frac{1}{2}$ or $m_s =-\frac{1}{2}$. Let's assume $m_s=+\frac{1}{2}$ (either value is possible for an unpaired electron in the orbital).
     - $m_s=+\frac{1}{2}$

2. **For the set $n = 3$, $l = 3$, $m_l=-1$**:
   - **Brief Explanation**: The value of $l$ cannot be equal to $n$. The allowed values of $l$ range from $0$ to $n - 1$. Here, $n = 3$, so the possible values of $l$ are $0$, $1$, and $2$ (corresponding to $3s$, $3p$, and $3d$ orbitals respectively). Since $l = 3$ is given when $n = 3$, this set of quantum numbers is incorrect because $l$ exceeds the maximum allowed value ($n-1$) for the given $n$.

# Answer:
1. For the last four electrons in a phosphorus atom:
   - $n = 3$, $l = 1$, $m_l=-1$, $m_s=+\frac{1}{2}$ (other valid combinations for $m_l$ and $m_s$ are also possible for the $3p$ electrons)
2. The set $n = 3$, $l = 3$, $m_l=-1$ is incorrect because for $n = 3$, the maximum value of $l$ is $n - 1=2$, and $l = 3$ is not allowed.

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