Question

1. write out the orbital diagram (boxes) for only the valence electrons in the element se. label all orbitals. (4 points)

Explanation:

Step1: Determine the valence electrons of Se

Selenium (Se) has an atomic number of 34. Its electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{4}$. The valence electrons are in the $4s$ and $4p$ orbitals. The number of valence electrons is $6$ ($2$ in $4s$ and $4$ in $4p$).

Step2: Draw the orbital diagram for valence electrons

For the $4s$ orbital (which can hold 2 electrons), we have one box. For the $4p$ orbitals (three boxes as $p$ - subshell has three orbitals, each can hold 2 electrons), we draw three boxes.

• For the $4s$ orbital:

$$\begin{matrix} \uparrow\downarrow\\ 4s \end{matrix}$$

• For the $4p$ orbitals:

$$\begin{matrix} \uparrow&\uparrow&\uparrow\downarrow\\ 4p_x&4p_y&4p_z \end{matrix}$$

Answer:

The orbital diagram for the valence electrons of Se:

$$\begin{matrix} \uparrow\downarrow\\ 4s \end{matrix}\quad \begin{matrix} \uparrow&\uparrow&\uparrow\downarrow\\ 4p_x&4p_y&4p_z \end{matrix}$$