Consider ionization of an element " E " : E ( g ) + I P 1 E + ( g ) + I P 2 E 2 + ( g ) + I P 3 E 3 + ( g ) + I P 4 E 4 + ( g ) + I P 5 E 5 + ( g ) If E +5 does not have any electr

Question

Consider ionization of an element " E " : E ( g ) + I P 1 ​ ​ E + ( g ) + I P 2 ​ ​ E 2 + ( g ) + I P 3 ​ ​ E 3 + ( g ) + I P 4 ​ ​ E 4 + ( g ) + I P 5 ​ ​ E 5 + ( g ) If E +5 does not have any electrons, find the number of electrons in the valence shell of element "E".

TestHub · Accepted Answer

E → E⁺ → E²⁺ → E³⁺ → E⁴⁺ → E⁵⁺
and E⁵⁺ has no electron

E⁵⁺ has zero electrons, so neutral atom E must have 5 electrons.
(Because 5 electrons are removed one by one)

So, atomic number of E = 5

Electronic configuration of E:
1s² 2s² 2p¹

Valence shell = outermost shell (n = 2)
Electrons in valence shell = 2s² 2p¹ = 3 electrons

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