Question

The silver salts like AgBr and AgI are colored because of:
A. d-d transition
B. charge transfer
C. polarization of halid $A{g^ + }$ by
D. both (b) and (c)

Answer 1

Hint: As we know that on moving down in the group, the size of the anions increases. In the given compounds, halogens ions are present. Fluorine is the smallest halogen ion and iodine is the largest halogen ion.

Complete step by step answer:
The atomic number of silver is 47. The electronic configuration of silver is $[Kr]4{d^{10}}5{s^1}$. In silver ions, one electron is removed from the outermost s-orbital. So the electronic configuration of $A{g^ + }$is$[Kr]4{d^{10}}$. In the d orbital of$A{g^ + }$, all the electrons are paired. As no unpaired electrons are present d-d transition cannot take place.The color of the compound AgCl is white but silver salt of AgBr and AgI are colored.
On moving down in the group the atomic size of the increases, the order is given as shown below.
$I > Br > Cl > F$
With the increase in the size of anions, the polarization of anions increases.
The property of anion to get polarized by the deformation of the electron cloud formed by the
cation is known as polarization. With the increase in polarization, the intensity of compound color increases. Thus, the silver salts like AgBr and AgI are colored. The order of the color intensity in the covalent compound is shown below.
$AgI > AgBr > AgCl > AgF$
Thus, the silver salts like AgBr and AgI are colored because of the polarization of halogen anion
by the silver cation.

So, the correct answer is Option C.

Note:
The covalent character of the compound increases with the increase in polarization of anion. Therefore, AgI shows a more covalent character than AgF. The order for increasing the covalent character is shown below.
$AgI > AgBr > AgCl > AgF$