Question

The bond length in $LiF$ will be:
A. Less than that of $NaF$
B. Equal to that of $KF$
C. More than that of $KF$
D. Equal to that of $NaF$

Answer 1

Hint: Lithium is an element of the alkali metal series or the group one elements. It has a very small size and is electropositive in nature. Sodium is also an element of the group one series and has a size comparatively greater than the size of the lithium atom and is slightly more electropositive than the lithium.

Complete step by step answer:
The bond lengths are dependent on the covalent nature of the elements. The covalent character depends on the polarizing potential of a cation. It can be represented mathematically as:
$\phi = \dfrac{{charge}}{{radius}}$
This means that the greater the charge and the smaller the charge of the cation or anion, the greater will be the polarizing power and thus, the greater is the covalent character and hence, the shorter the bond length is. Based on this, we can see that in the case of both $LiF$ and $NaF$ , the charge is the same on the cations (+1) but the size of the sodium atom is greater than the size of the lithium atom. Thus, the covalent character of the lithium fluoride bond will be more than the sodium fluoride bond and the bond length will be reverse of this phenomenon i.e. the bond length in $LiF$ will be less than that of $NaF$ .
Thus, the correct option is A. less than that of $NaF$ .

Note:
Bond lengths are typically in the range of \[100 - 200{\text{ }}pm{\text{ }}\left( {1 - 2{\text{ }}{A^o}} \right)\] . As a general trend, bond length decreases across a row in the periodic table and increases down a group. Atoms with multiple bonds between them have shorter bond lengths than singly bonded ones.