Question

Statement - 1: Nucleophilicity order in polar-protic solvent is $I^{-}<B{r}^{-}<C{l}^{-}<F^{-}$.
Statement - 2: Due to the bigger size of $I^{-}$ it is less solvated in polar-protic solvent.
A: Statement - 1 is true, statement - 2 is true and statement -2 is correct explanation of statement - 1.
B: Statement -1 true, statement - 2 is true and statement - 2 is not the correct explanation for Statement -1
C: Statement -1 is true, statement - 2 is false.
D: Statement -1 is false, statement - 2 is true.

Answer 1

Hint: Nucleophilicity reflects the capability of any molecule or anion to attack an electrophile. There are certain factors that can impact the nucleophilicity which includes size, solvent and basicity.

Complete answer:
A protic solvent consists of an H atom that is bounded to O or N. It can utilize its H atom to take part in H-bonding along-with a nucleophile. Thus, this creates a "shell" of solvent molecules around the nucleophile. Now, the nucleophile pushes this shell of solvent molecules to attack the carbon possessing the leaving group.
Halogens are not nucleophile in nature when they exist in diatomic form (for instance, $I_2$ is not nucleophilic in nature), while their anionic form (for instance, $I^{-}$) is a good nucleophile.
$F^{-}$ is a small anion having a high charge density. Thus, it is tightly solvated and proves to be the weakest nucleophile in case of polar-protic solvents.
While, $I^{-}$ is a large ion having a low charge density. Thus, it is loosely or less solvated. Only a few solvent molecules are there to push out of the way.
In polar-protic solvents, the order of nucleophilicity is:
$I^{-}>B{r}^{-}>C{l}^{-}>F^{-}$

As a result, Statement – 1 is false but Statement – 2 is true. Thus, option D is correct.

Note: In a polar-aprotic solvent, hydrogen atom is not present to participate in hydrogen bonding. Here, the nucleophile has less molecules in its solvent shell. Thus, nucleophiles can easily attack the substrate. In polar-aprotic solvents, the order of nucleophilicity is reversed as shown below:
$F^{-}>C{l}^{-}>B{r}^{-}>I^{-}$