Neon has filled its outer shells, therefore it is very stable and does not need to react with other elements and doesn’t form compounds.
<span>By definition, the first ionization energy is the energy required to remove the most loosely held electron from one mole of gaseous atoms to produce 1 mole of gaseous ions each with a charge of 1+. </span><span />
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![\dfrac{[H^{+}] [A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5BH%5E%7B%2B%7D%5D%20%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
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Explanation:
- When an aqueous solution of a certain acid is prepared it is dissociated is as follows-
⇄ 
Here HA is a protonic acid such as acetic acid, 
- The double arrow signifies that it is an equilibrium process, which means the dissociation and recombination of the acid occur simultaneously.
- The acid dissociation constant can be given by -
=
- The reaction is can also be represented by Bronsted and lowry -
⇄ ![[H_3O^+] [A^-]](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%20%5BA%5E-%5D)
- Then the dissociation constant will be
= ![\dfrac{[H_3O^{+}] [A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5BH_3O%5E%7B%2B%7D%5D%20%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
Here, is the dissociation constant of an acid.
Answer:
a)CH₄, BH₃, and CCl₄
Explanation:
<u>London dispersion forces:-
</u>
The bond for example, in the molecule is F-F, which is non-polar in nature because the two fluorine atoms have same electronegativity values.
The intermolecular force acting in the molecule are induced dipole-dipole forces or London Dispersion forces / van der Waals forces which are the weakest intermolecular force.
Out of the given options, H₂O , NH₃ exhibits hydrogen bonding which is:-
<u>Hydrogen bonding:-
</u>
Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.
Thus option B and C rules out.
<u>Hence, the correct option which represents the molecules which would exhibit only London forces is:- a)CH₄, BH₃, and CCl₄</u>
