Answer:
I think option A is right answer
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:
Check the explanation
Explanation:
When,
pH = -log[H+] = 3.30
[H+] = 
![alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6](https://tex.z-dn.net/?f=alpha%5BY%5E-4%5D%20%3D%20%5BH%2B%5D%5E6%20%2B%20Ka1%5BH%2B%5D%5E5%20%2B%20Ka1Ka2%5BH%2B%5D%5E4%20%2B%20Ka1Ka2Ka3%5BH%2B%5D%5E3%20%2B%20Ka1Ka2Ka3Ka4%5BH%2B%5D%5E2%20%2B%20Ka1Ka2Ka3Ka4Ka5%5BH%2B%5D%20%2B%20Ka1Ka2Ka3Ka4Ka5Ka6)
= 
= 
When,
pH = -log[H+] = 10.15
[H+] = 
Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 
; Ka6 = 
![alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6](https://tex.z-dn.net/?f=alpha%5BY%5E%7B-4%7D%5D%20%3D%20%5BH%2B%5D%5E6%20%2B%20Ka1%5BH%2B%5D%5E5%20%2B%20Ka1Ka2%5BH%2B%5D%5E4%20%2B%20Ka1Ka2Ka3%5BH%2B%5D%5E3%20%2B%20Ka1Ka2Ka3Ka4%5BH%2B%5D%5E2%20%2B%20Ka1Ka2Ka3Ka4Ka5%5BH%2B%5D%20%2B%20Ka1Ka2Ka3Ka4Ka5Ka6)
= 
= 
