Explanation:
To delineate the the nature of the bonds that would be formed between the two elements, let us first write the electronic configuration of the two species;
Be = 2, 2
F = 2, 7
Beryllium is a metal with two valence electrons whereas fluorine is a halogen with seven valence electrons.
When Be loses two electrons it becomes isoelectronic with He;
Be → Be²⁺ + 2e⁻
Also, when fluorine gains an electron, it becomes isoelectronic with Ne;
F + e⁻ → F⁻
This loss and gain of electrons between the two elements creates an electrostatic attraction them and they enter into an electrovalent bond.
Hence;
Be²⁺ + 2F⁻ → BeF₂
Answer:
Mass= 2.77g
Explanation:
Applying
P=2.09atm, V= 1.13L, R= 0.082, T= 291K, Mm of N2= 28
PV=nRT
NB
Moles(n) = m/M
PV=m/M×RT
m= PVM/RT
Substitute and Simplify
m= (2.09×1.13×28)/(0.082×291)
m= 2.77g
An SI base unit for measuring length would be meters.
