Question

The order of covalent character of $KF,KI,KCl$ is :
A) $KCl < KF < KI$
B) $KI < KCl < KF$
C) $KF < KI < KCl$
D) $KF < KCl < KI$

Answer 1

Recall the postulates of Fajan’s rule. One of its postulates says that larger the size of the anion, more will be the covalent character. Since, the cation (${K^ + }$ ) is the same in $KF,KI,KCl$ compounds, compare the size of anions. $F,Cl{\text{ and }}I$ belongs to the same group number 17 and as we move down the group, atomic size increases.

Complete step by step answer:
According to Fajan’s rule, smaller the size of the cation and larger the size of the anion, greater is the covalent character of the ionic bond. When size of the anion is larger, lesser is the effective nuclear charge that holds the valence electrons of the ion. Valence or the last electron in a large anion is loosely bound, so it can easily be polarised by the cation and thus makes the compound more covalent.
In the compounds, $KF,KI,KCl$, cation is the same, that is, potassium ($K$ ) and anion is different in each compound. $F,{\text{ }}Cl{\text{ }} and {\text{ }}I$ belong to the same group number 17 and as we move down the group, atomic size increases. Consequently, order of size is: $I > Cl > F$.
And as we know, larger is the size of anion, more is the covalent character or covalency of the compound. Iodine has the largest size, therefore, $KI$ will have the most covalent character and fluorine has the smallest size, so $KF$ will be least covalent.
Therefore, the order of covalent character of $KF,KI,KCl$ is: $KF < KCl < KI$. So, the correct answer is “Option D”.

Note: We can also give the order of covalent character on the basis of electronegativity difference between the cation and anion. Greater is the electronegativity difference, smaller will be the covalent character of the ionic compound. Fluorine is the most and iodine is the least electronegative. Therefore, electronegativity difference will be the maximum in $KF$ and will be minimum in $KI$. Consequently, $KI$ will have the most and $KF$ will have the least covalent character. Thus, again the order of covalent characters of $KF,KI,KCl$ will come out as: $KF < KCl < KI$.