Question

The value of electronegativity of the atoms A and B are $ 1.2 $ and $ 0.4 $ respectively. The percent ionic character of the A-B bond is ?
(A) $ 50\% $
(B) $ 72.24\% $
(C) $ 55.3\% $
(D) $ 43\% $

Answer 1

Hint : The electronegativity of the atom can be defined as the tendency of that atom to attract the shared electrons or the electron density towards itself. The electronegativity of the atom is generally symbolised by the symbol $ '\chi ' $ .

Complete Step By Step Answer:
To proceed with this question, first we need to understand the relationship between the electronegativity and the ionic character of the bond.
The electronegativity of the element affects the ionic character of that particular bond.
The more the atom is electronegative, the more the ionic character it has and vice-versa i.e. the less the electronegative atom, the less the ionic character it has.
So, there is a formula that relates the ionic character and the electronegativities of the atoms.
Let’s have a look at the formula, so the formula is:
 $ \% $ Ionic character $ = 16[{E_A} - {E_B}] + 3.5{[{E_A} - {E_B}]^2} $
Where, $ {E_A}{\text{ and }}{{\text{E}}_B} $ are the electronegativities of the atom?
So, in the question we are given,
 $ {E_A} $ $ = $ $ 1.2 $
 $ {{\text{E}}_B} $ $ = $ $ 0.4 $
So, putting the values in the above formula, we get:
 $ \% $ Ionic character $ = 16[4 - 1.2] + 3.5{[4 - 1.2]^2} $
 $ \% $ Ionic character $ = 72.24\% $
Thus, the percent ionic character of the A-B bond is $ = 72.24\% $
Therefore, on the basis of the above discussion we can say that the correct answer is option B.

Note :
So ionic bonds are formed when one of the atoms completely donates its electrons and the other atom accepts it. So the decision that which atom will donate the electron and which atom will accept the electron depends on the electronegativities of those particular atoms.
So, the atom with high electronegativity, generally a non-metal, accepts the electron whereas on the other hand the less electronegative element, generally a metal, donates its valence electron.