Beryllium is element number 4 on the periodic table atomic 4

Beryllium is element number 4 on the periodic table (atomic #4). Fluorine is element number 9 (atomic #9). Based on what you can determine about the valence electrons of each element, predict the chemical formula for the compound that forms between beryllium and fluorine. (Chemical formulas are the simplest representation of a molecule. For example, butane is C_4H_10, as is isobutene, indicating they both contain 4 carbon atoms and 10 hydrogen atoms).

Solution

Ans. Beryllium.

                        Atomic number = 4

                        Electronic configuration = 1s² 2s²

                        Number of electrons in outermost shell (n= 2) = 2

                        Valency = 2

It’s easier to release 2 electrons to get the nearest inert gas configuration (of He); than gaining 6 electrons to get the distant inert gas configuration (of Neon).

Being group 2 metal, it readily gets ionized by donating 2 electrons from outermost shell.

            Be [1s² 2s²] --------donates 2e ----> Be²⁺ [1s²]         – Inert gas [He] configuration

Fluorine:

                        Atomic number = 9

                        Electronic configuration = 1s² 2s² 2p⁵

                        Number of electrons in outermost shell (n= 2) = 7

                        Valency = 1

Being a electronegative non-metal, it accepts an electron attain nearest inert gas configuration (of Ne).

            F [1s² 2s² 2p⁵] -----gains 1e -----> F^1- [1s² 2s² 2p⁶] – Inert gas [Ne] configuration  

Compound formation:

We have-

            1. Be loses 2 electrons

            2. Fe accepts 1 electron

So,1 Be atom can share its 2 electrons with two F- atoms (1e with each F-atom) to form the compound BeF₂.

So, the chemical formula of a compound resulting from Be and F atom is BeF₂.