Question

Electronic Configuration Of O=8 F=9 Ar=18 Ne=10

Answer 1

Answer:

O = 8

It is Oxygen and it's electronic configuration is ===> 1s2 , 2s2 , 2p4 .

Fl = 9

It is flourine and it's electronic configuration is ===> 1s2 , 2s2 , 2p5

Ar = 18

It is Argon and it's electronic configuration is ===> 1s2 , 2s2 , 2p6 , 3s2 , 3p6

Ne = 10

It is neon and it's electronic configuration is ===> 1s2 , 2s2 , 2p6

Explanation:

Hope this will help you buddy .

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Answer 2

Answer :-

Electronic configuration :- The arrangement of electrons in different shells of an atom of an elements is called electronic configuration.

Every atoms want to be inert or stable.

If they have complete octet or duplet then they are inert or stable in nature.

O = 8

Now, arranged in different shell by bhor's law.

The arrangement of electrons in different shell is given by :-

$\boxed{\sf{ 2n^2 ; n = shell no. }}$

Take K = 1

k = 2 n²

k = 2 ×1²

k = 2 e

L = 2

L = 2 n²

L = 2 × 2²

L = 2 × 4

L = 8 e

M = 3

M = 2n²

M = 2 × 3²

M = 2 × 9

M = 18 e

Now,

Arranged it according to octet rule,

$O = 2 , 6$

$F = 2 , 7$

$Ar = 2 , 8 , 8$

$Ne = 2 , 8$