Name one species that will be isoelectronic with the following atoms or ions.
(a) `Ca^(2+)` , (b) `Ne` , (c ) `Cl^(ɵ)` ,(d) `Rb^(o+)`

Explanation

Isoeletronic psecies have same number of electrons.
(a) `Ca^(2+)` has `18e^(-)'s(Z = 20, 20-2 = 18 e^(-))`
`P^(3-)(Z = 15,15+3=18e^(-))`
`S^(2-)(Z = 16,16+2 = 18 e^(-))`
`Ar (Z = 18)`
`K^(o+)(Z = 19, 19-1 = 18 e^(-)'s)`
Hence, all rise isoelectronic with `Ca^(2+)`.
`Ne` has `10 e^(-)'s`.
Therefore, the following species are isoelectronic with it.
`N^(3-)(Z = 7)(7+3=10e^(-))`
`O^(2-)(Z = 8)(8+2 = 10 e^(-))`
`F^(ɵ)(Z = 9)(9+1 = 10 e^(-))`
`Na^(o+)(Z =11)(11-1 = 10 e^(-))`
`Mg^(2+)(Z = 12)(12-2 = 10e^(-))`
`Al^(3+) (Z = 13)(13-3 = 10 e^(-))`
(c ) `Cl^(ɵ)` has `18 e^(-)'s(Z = 17)(17+1 = 18 e^(-))`
Therefore, the following species are isoelectronic with it. `P^(3-), s^(2-), Ar, K^(o+)` and `Ca^(2+)`, all have `18 e^(-)'s`.
(d) `Rb^(o+)` has `36 e^(-)'s(Z = 37)(37-1=36 e^(-))`.
Therefore, the following species are isoelectronic with it.
`Br^(ɵ)(Z = 35, 35+1 = 36)`
`Kr(Z = 36)`
`Sr^(2+)(Z = 38, 38-2 = 36)`