Question

 

Recalling that the electron configuration for neon (10Ne) is 1s²2s²2p⁶, explain the chemical properties of the adjacent elements (₉F and ₁₁Na).

Answer 1

1) Fluorine properties :

a) 'F' is 7th 17th group element. its electronic configuration 1s²2s²2p⁵ . it just nearest electronic conguration of Ne inert gas.

b) Fluorine readily accept electron to form anion F^- to get inert gas configuration 1s²2s²2p⁶ .

c) F is highest electronnegative atom in the periodic table . it is due to small atomic size and high electronegativity.

d) 'F' is non-metal due to high electronegativity.

e) 'F' shows only two oxidation states 0, -1.

f) F' forms two oxides OF2 and O2F2 . these are called Oxygen fluorides because of high electronegativity.

those oxides never be called as Fluorine oxides.

g) due Nearest electronic configuration of Ne , it forms many metal fluorides

   example : NaF , LiF, MgF2 , AlF3   etc

h) finally Fluorine is non- metal, high electronegative, always accept electrons , never loose electrons . these all properties are due to which is present nearset to inert gas Ne

2) Na properties:

a) 'Na' is 1st group element. its electronic configuration 1s²2s²2p⁶3s¹ . it is having one more electron than inert gas Ne

b) Na' readily loose one electron form Na+   stabel cation this cation electronic configuration is 1s²2s²2p⁶.

c) sodium is highly electropoistive element and readily form cation Na+

d) sodium is metal due to high electropositivity.

e) sodium forms metallic oxides in tis +1 oxidation states

   ex: Na2O , Na2O2

f) sodium oxides are basic in nature . it is due to high elecropositivity of Na

g) Na forms many other metallic compounds in its +1 oxidation states

ex: NaF, NaCl, Na2SO4, Na2CO3, NaHCO3

h) so fianally sodium is metal, high electropositive , always shows +1 oxidation state.