How many electrons in an atom can have the following designations?

[tex]$5d_{z^2}$[/tex] : [tex]$\square$[/tex] electron(s)

[tex]$1d$[/tex] : [tex]$\square$[/tex] electron(s)

[tex]$5d$[/tex] : [tex]$\square$[/tex] electron(s)

[tex]$7p$[/tex] : [tex]$\square$[/tex] electron(s)

[tex]$6d$[/tex] : [tex]$\square$[/tex] electron(s)

[tex]$n=3$[/tex] : [tex]$\square$[/tex] electron(s)



Answer :

Certainly! Let's determine the number of electrons for the given designations one by one:

1. [tex]\( 5 d_{z^2} \)[/tex] :

This refers to a specific orbital in the [tex]\( d \)[/tex] subshell of the 5th energy level. Each orbital can hold 2 electrons. Therefore, [tex]\( 5 d_{z^2} \)[/tex] can have:
[tex]\[ 5 d_{z^2} : 2 \text{ electron(s)} \][/tex]

2. [tex]\( 1 d \)[/tex] :

The [tex]\( d \)[/tex] subshell does not exist in the 1st principal energy level ([tex]\( n=1 \)[/tex]). Therefore, [tex]\( 1 d \)[/tex] can have:
[tex]\[ 1 d : 0 \text{ electron(s)} \][/tex]

3. [tex]\( 5 d \)[/tex] :

The [tex]\( d \)[/tex] subshell in the 5th energy level can have 5 orbitals, and each orbital can hold 2 electrons. Therefore, [tex]\( 5 d \)[/tex] can have:
[tex]\[ 5 d : 5 \text{ orbitals} \times 2 \text{ electrons/orbital} = 10 \text{ electron(s)} \][/tex]

4. [tex]\( 7 p \)[/tex] :

The [tex]\( p \)[/tex] subshell does not exist in the 7th principal energy level ([tex]\( n=7 \)[/tex]). Therefore, [tex]\( 7 p \)[/tex] can have:
[tex]\[ 7 p : 0 \text{ electron(s)} \][/tex]

5. [tex]\( 6 d \)[/tex] :

The [tex]\( d \)[/tex] subshell in the 6th energy level can have 5 orbitals, and each orbital can hold 2 electrons. Therefore, [tex]\( 6 d \)[/tex] can have:
[tex]\[ 6 d : 5 \text{ orbitals} \times 2 \text{ electrons/orbital} = 10 \text{ electron(s)} \][/tex]

6. [tex]\( n=3 \)[/tex] :

The total number of electrons in the 3rd energy level ([tex]\( n=3 \)[/tex]) can be calculated by summing the electrons in the 3s, 3p, and 3d subshells:
[tex]\[ \begin{align*} 3s & : 1 \text{ orbital} \times 2 \text{ electrons/orbital} = 2 \text{ electrons} \\ 3p & : 3 \text{ orbitals} \times 2 \text{ electrons/orbital} = 6 \text{ electrons} \\ 3d & : 5 \text{ orbitals} \times 2 \text{ electrons/orbital} = 10 \text{ electrons} \\ \end{align*} \][/tex]
Therefore, the total number of electrons for [tex]\( n=3 \)[/tex] is:
[tex]\[ n=3 : 2 + 6 + 10 = 18 \text{ electron(s)} \][/tex]

In summary:
[tex]\[ \begin{align*} 5 d_{z^2} & : 2 \text{ electron(s)} \\ 1 d & : 0 \text{ electron(s)} \\ 5 d & : 10 \text{ electron(s)} \\ 7 p & : 0 \text{ electron(s)} \\ 6 d & : 10 \text{ electron(s)} \\ n=3 & : 18 \text{ electron(s)} \end{align*} \][/tex]