# Question #728e3

Jan 2, 2017

Alkali earth metals have a valance configuration of ${s}^{2}$
Halogens have a valance configuration of ${s}^{2} {p}^{5}$
Both families of elements are chemically active.

#### Explanation:

Alkali Earth metals with a valance configuration of ${s}^{2}$ can most easily become stable by sharing these valance electrons with an element that is electrophilic. This leaves the Alkali Earth metals with the outside valance electron structure of a noble gas of the next lower electron shell.

Halogens with a valance configuration of ${s}^{2} {p}^{5}$ can most easily become stable by acquiring an electron from another element. This leaves the Halogens with the outside valance electron structure of a noble gas in the same period.

Both of these families of elements are very active chemically.

Jan 25, 2017

Here's what I get.

#### Explanation:

$\boldsymbol{\text{Alkaline Earth"color(white)(ml)"Halogen}}$
${\text{Be"color(white)(m)["He"] "2s"^2color(white)(mmmm)"F"color(white)(m)["He"] "2s"^2 "2p}}^{5}$
${\text{Mg"color(white)(ll)["Ne"] "3s"^2color(white)(m)color(white)(mmm)"Cl"color(white)(ll)["Ne"] "3s"^2 "3p}}^{5}$

${\text{Ca"color(white)(m)["Ar"] "4s"^2color(white)(m)color(white)(mmm)"Br"color(white)(ll)["Ar"]"4s"^2 "3d"^10 "4p}}^{5}$
${\text{Sr"color(white)(ml)["Kr"] "5s"^2color(white)(m)color(white)(mmm)"I"color(white)(ml)["Kr"] "5s"^2 "4d"^10 "5p}}^{5}$

${\text{Ba"color(white)(m)["Xe"] "6s"^2color(white)(m)color(white)(mmm)"At"color(white)(ll)["Xe"]"6s"^2 "4f"^14 "5d"^10 "6p}}^{5}$
${\text{Ra"color(white)(m)["Rn"] "7s"^2color(white)(m)color(white)(mmll)"Ts"color(white)(m)["Rn"] "7s"^2 "5f"^14 "6d"^10 "7p}}^{5}$

The alkaline earth metals have a electron configuration ending in $n {\text{s}}^{2}$.

They can get a noble gas configuration by losing their two $\text{s}$ electrons.

The halogens have an electron configuration ending in $n {\text{s"^2 n"p}}^{5}$.

They can get a noble gas configuration by gaining one electron to complete their octet.

Thus, an alkali metal $\text{M}$ will react by donating an electron to each of two halogen $\text{X}$ atoms.

The metal will become a metal cation $\text{M"^"2+}$ and the halogen will become an anion $\text{X"^"-}$.

$\text{M + 2X" → "M"^"2+", "2X"^"-}$