Answer
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Hint: We have to remember that the internuclear distance is the distance between two nuclei in a molecule. It can be explained by electronegative and size of the atom. It is the distance from the radius of the one atom to the radius of another atom.
Complete step by step answer:
Now let us see the compounds in the options ie., There are four compound $CsI$ ,
$CsF$ , $LiF$ and $LiI$ . We will see the properties of individual compound cesium $(Cs)$, Lithium \[\left( {Li} \right)\] , Iodine \[\left( I \right)\] , Fluorine \[\left( F \right)\] . Cesium $(Cs)$ and Lithium \[\left( {Li} \right)\] are non-metal, \[Li\] is more electronegative than \[Cs\]. We have to remember that the fluoride \[\left( F \right)\] and Iodine \[\left( I \right)\] are halogens, $F$ is more electronegative than $I$. Lesser the electronegative lesser the attraction. Internuclear distance also depends upon the size of the two atoms. \[Cs\] is larger than \[Li\] and \[I\]is larger than $F$ .
Now let us consider the first option, \[CsI\] in this compound both the atoms are less electronegative and larger in size.
Now consider the second option, \[CsF\], \[Cs\] is lower in electronegative and larger in size whereas $F$ is larger in electronegative and smaller in size.
Consider third option, \[LiI\] , Li is smaller in size and higher in electronegative whereas \[I\] is smaller in size and lesser in electronegative
Consider the fourth option, \[LiF\] , \[Li\] smaller in size and higher in electronegative whereas \[F\] is larger in electronegative and smaller in size.
Therefore, \[CsI\] have the maximum internuclear distance.
Therefore, the option A is correct.
Note: As we already discussed the internuclear distance depends upon electronegative character of an atom, the atom with less electronegative will have less attraction of atom and also atom with larger radium will have maximum internuclear distance. It can also be calculated by the formula if the radius of the atom is known.
Complete step by step answer:
Now let us see the compounds in the options ie., There are four compound $CsI$ ,
$CsF$ , $LiF$ and $LiI$ . We will see the properties of individual compound cesium $(Cs)$, Lithium \[\left( {Li} \right)\] , Iodine \[\left( I \right)\] , Fluorine \[\left( F \right)\] . Cesium $(Cs)$ and Lithium \[\left( {Li} \right)\] are non-metal, \[Li\] is more electronegative than \[Cs\]. We have to remember that the fluoride \[\left( F \right)\] and Iodine \[\left( I \right)\] are halogens, $F$ is more electronegative than $I$. Lesser the electronegative lesser the attraction. Internuclear distance also depends upon the size of the two atoms. \[Cs\] is larger than \[Li\] and \[I\]is larger than $F$ .
Now let us consider the first option, \[CsI\] in this compound both the atoms are less electronegative and larger in size.
Now consider the second option, \[CsF\], \[Cs\] is lower in electronegative and larger in size whereas $F$ is larger in electronegative and smaller in size.
Consider third option, \[LiI\] , Li is smaller in size and higher in electronegative whereas \[I\] is smaller in size and lesser in electronegative
Consider the fourth option, \[LiF\] , \[Li\] smaller in size and higher in electronegative whereas \[F\] is larger in electronegative and smaller in size.
Therefore, \[CsI\] have the maximum internuclear distance.
Therefore, the option A is correct.
Note: As we already discussed the internuclear distance depends upon electronegative character of an atom, the atom with less electronegative will have less attraction of atom and also atom with larger radium will have maximum internuclear distance. It can also be calculated by the formula if the radius of the atom is known.
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