Question

write the electron configuration of the IONS in Mg, K, P, S, I

Answer 1

Mg ion = Mg+2 then

Shorthand notation of Mg = [Ne] 3s2

if Mg loses 2 electrons then [Ne] will be the ionic electronic concfiguration

K ion = K+ then

[Ar] 4s1

when losing 1 electron then

[Ar]

P has the next configuration

[Ne] 3s2 3p3

when gaining 3 electrons

then

[Ne] 3s2 3p6

For S

[Ne] 3s2 3p4

when gaining two electrons, i.e. S-2

then

[Ne] 3s2 3p6

For I, iodine then

[Kr] 4d10 5s2 5p5

it needs only one electron to transform into an anion , so

[Kr] 4d10 5s2 5p6