Question

show the number of unpaired electrons of silicon (z=14) in box diagram​

Answer 1

Explanation:

1s^2,2S^2,2p^6,3s^2,3p^6

Answer 2

To show:

The number of unpaired electrons in Silicon using box diagram.

Solution:

Silicon has atomic number : 14.

So, its electronic configuration will be:

$\boxed{ \bold{Si = 1 {s}^{2} \: 2 {s}^{2} \: 2 {p}^{6} \: 3 {s}^{2} \: 3 {p}^{2} }}$

So, let's construct the box diagram:

$\bold{ Si : \boxed{ \uparrow \downarrow} \: \: \boxed{\uparrow \downarrow } \: \: \boxed{\uparrow \downarrow | \uparrow \downarrow | \uparrow \downarrow } \: \: \boxed{\uparrow \downarrow} \: \: \boxed{\uparrow | \uparrow | \_ | \_ | \_}} \\ \implies 1 {s}^{2} \: 2 {s}^{2} \: \: \: \: \: \: \: \: \: \: \: 2 {p}^{6} \: \: \: \: \: \: \: \: \: \: \: 3 {s}^{2} \: \: \: \: \: \: \: \: \: \: \: 3 {p}^{2}$

From the box diagram, we can conclude that there are 2 unpaired electrons in 3p orbital.

So, final answer is:

$\boxed{ \bold{no. \: of \: unpaired \: electrons = 2}}$