Question

AB is predominantly ionic as ${{A}^{+}}{{B}^{-}}$, if (IE stand for ionization energy, EA for electron affinity, and EN for electronegativity).
(a)- $I{{P}_{A} < }I{{P}_{B}}$
(b)- $E{{A}_{A} < }E{{A}_{B}}$
(c)- $E{{N}_{A} < }E{{N}_{B}}$
(d)- $I{{P}_{A}} > I{{P}_{B}}$

Answer 1

Hint: When a compound splits into ions, they form positive and negative charges. The higher the electronegativity, electron affinity, and ionization potential of the element, the more negative will be the charge.

Complete answer:
Ionization energy is the energy that is required to remove the last electron from an isolated gaseous atom so that it can be converted into a gaseous cation.
As we move down the group the ionization energy decreases and as we move along the period the ionization energy increases. As the ionization energy increases the negative charge increases.
Hence in AB, B will have greater ionization energy.
$I{{P}_{A} < }I{{P}_{B}}$
Electron affinity is the tendency of an element to accept an extra electron and acquire a negative charge.
As we move down the group the electron affinity decreases and as we move along the period the electron affinity increases. As the electron affinity increases the negative charge increases.
Hence B will have more electron affinity.
$E{{A}_{A} < }E{{A}_{B}}$
Electronegativity is the ability of an atom to attract a shared pair electron from a bond towards itself.
In a bond, the atom which has more electronegativity will split into an anion and the atom with less electronegativity will split as a cation.
Hence, B will have more electronegativity.
$E{{N}_{A} < }E{{N}_{B}}$
Since all the factors ionization energy, electron affinity, and electronegativity are more for B than A,
So, the correct answer is “OptionA, B and C”.

Note: The trend of all factors should be kept in mind. If A and B are replaced with any atoms we can easily find the atom which will form a cation and which will form an anion.