Question

4s^2 is the confrigution of the outermost orbit of an element its atomic number would be


I think it's 20 but 20 is not given in the answers ...

Answer 1

Yeah it should be 20.

The electronic configuration goes like this..

$1 {s}^{2} \: \: \: \: 2 {s}^{2} \: \: \: \: 2 {p}^{6} \: \: \: \: 3 {s}^{2} \: \: \: \: 3 {p}^{6} \: \: \: \: 4 {s}^{2} \: \: \: \: 3 {d}^{0}$
Thus it can be clearly seen that the atomic number should be 20.

The reason is that the energy level of 4s is higher than 3d and thus is filled first before 3d.

<<But>>

Alternatively
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It can be also seen that after filling the 4s ² orbital 3 d is filled....

But if it is noticed than we 3d10 and 4s² is totally filled than the outermost orbital will be 4s²..

Since electrons first leave the 4s² orbital.

1s2 2s2 2p6 3s2 3p6 3d10 4s2.

Since after filling the configuration comes in a state like this.

So answer will be 30.

The atomic number required is 30 and

Thanks for such a tricky question.

Hope this is what you need.

Answer 2

the answer is 30
its electronic configuration is
1s2 2s2 2p6 3s2 3p6 3d10 4s2