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NEET CHEMISTRYEasy

In which of the following molecule, the internuclear distance will be maximum:

A

CsI

B

CsF

C

LiF

D

LiI

Step-by-Step Solution

The internuclear distance is the sum of the ionic radii of the cation and the anion (d=r++rd = r_+ + r_-). To maximize this distance, we need the largest cation and the largest anion.

  1. Cations: Cs is in Group 1 (Alkali metals), Period 6. Li is in Group 1, Period 2. Atomic/ionic size increases down a group. Therefore, r(Cs+)>r(Li+)r(Cs^+) > r(Li^+).
  2. Anions: I is in Group 17 (Halogens), Period 5. F is in Group 17, Period 2. Atomic/ionic size increases down a group. Therefore, r(I)>r(F)r(I^-) > r(F^-).

Comparing the combinations:

  • CsI: Largest cation (Cs+Cs^+) + Largest anion (II^-) \rightarrow Maximum internuclear distance.
  • LiF: Smallest cation (Li+Li^+) + Smallest anion (FF^-) \rightarrow Minimum internuclear distance.

Therefore, CsI has the maximum internuclear distance.

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