Question

$B{F_3}$,$BC{l_3}$,$BB{r_3}$,$B{I_3}$ which have a more acidic nature.

Answer 1

Hint:We know that species that is more electron deficient would attract electrons more easily and hence, would be more acidic. All the given compounds are boron halides. In boron halides, the number of electrons present in the outermost orbitals is six and it could accept electrons to attain its octet.

Complete step by step answer:
We must remember that back bonding is seen between atoms in a compound in which one atom has a lone pair of electrons and another shows vacant orbital adjacent to each other.
We can call back bonding as a $\pi $-coordinate bond. Two types of compounds generally show back bonding.
-Compounds in which the side atoms donate a pair of electrons to central atom
-Compounds in which the central atoms donate a pair of electron to side atoms
The conditions for back bonding are,
-Any one atom in the pair must contain one lone pair of electrons
-At Least one atom should belong to the 2nd period and other atoms should belong to the 3rd period.
We know that all halides contain lone pairs of electrons and they share their one lone pair of electrons to boron by overlapping with p orbital with the atom of boron. Effective overlap between the orbital reduces the large size of p orbital of halides. Therefore, the electron deficiency of boron in its halides is seen in the order,
$B{I_3} > BB{r_3} > BC{l_3} > B{F_3}$
From that trend, we observe that $B{I_3}$ is more acidic because of pi back bonding.
Therefore, the option (A) is correct.

Note:
We have to remember that the back bonding permits the molecule to be stable as it satisfies its octet. Back bonding results in decrease in bond length and increase in bond order. In a pi-back bonding electron, it moves from an atomic orbital of one atom to anti-bonding orbital of another atom (or) ligand. Nickel carbonyl and Zeise’s salt are examples of compounds that show pi-back bonding.