Question

AgCl and NaCl are colourless. NaBr and NaI are also colourless but the AgBr and AgI are coloured.
This is due to:
A. ${\text{A}}{{\text{g}}^{\text{ + }}}$ polarizes ${\text{B}}{{\text{r}}^ - }$ and ${{\text{I}}^ - }$
B. ${\text{A}}{{\text{g}}^{\text{ + }}}$ has unpaired d-orbital
C. ${\text{A}}{{\text{g}}^{\text{ + }}}$ polarizes depolarizes ${\text{B}}{{\text{r}}^ - }$ and ${{\text{I}}^ - }$
D. none of these

Answer 1

Hint: To answer this question we should know the colour is produced by electrons transfer. The compound in which electrons can transfer easily will produce colour. Electrons transfer from anion to cation. So, we will check the power of cation to accept the electron and the power of anion to donate the electron.


Complete solution:
The cation having high positive charge or smallest size can abstract the electron easily from the anion having high negative charge or largest size.
The power of action to abstract the electrons from anion is known as polarizing power and the power of anion to lose the electrons is known as polarizability.
The cation having high polarizing power can abstract the electron from the anion. The transfer of electrons produces colour. Due The transfer of electrons from anion to cation it is known as charge transfer.
So, colour depends upon the polarizing power of the cation which in turn depends upon the size and charge of cation and colour also depends upon the polarizability of the anion which in turn depends upon the size and charge of anion.
The polarizing power of the cation is high if the size is small and the charge is maximum.
Polarizing power of the cation$ \propto $ Charge on cation
Polarizing power of the cation$ \propto $ $1$ / size of cation
The polarizability of the anion will be high if the size is large and the charge is maximum.
Polarizability of the anion$ \propto $ size of anion
Polarizability of the anion$ \propto $ Charge on anion
The cation present in the given compounds are,${\text{A}}{{\text{g}}^ + }$ and${\text{N}}{{\text{a}}^ + }$. Both have the same charge so, order of size is, ${\text{N}}{{\text{a}}^ + }\,{\text{ > }}\,\,{\text{A}}{{\text{g}}^ + }$.
Cation size is inversely proportional to the polarizing power so, the polarizing power is, ${\text{N}}{{\text{a}}^ + }\,{\text{ < }}\,\,{\text{A}}{{\text{g}}^ + }$.
The ions present in the given compounds are,${\text{C}}{{\text{l}}^ - },{\text{B}}{{\text{r}}^ - }$ and ${{\text{I}}^ - }$. All have the same charge so, order of size is, or we can say the polarizability is, ${\text{C}}{{\text{l}}^ - }\,{\text{ < }}\,\,{\text{B}}{{\text{r}}^ - }\,{\text{ < }}\,\,{{\text{l}}^ - }$.
So, the electron easily transfers from ,${\text{B}}{{\text{r}}^ - }$ and ${{\text{I}}^ - }$ to ${\text{A}}{{\text{g}}^{\text{ + }}}$ and so, AgBr and AgI produce colour.
So, the AgBr and AgI are coloured because${\text{A}}{{\text{g}}^{\text{ + }}}$ polarizes ${\text{B}}{{\text{r}}^ - }$ and ${{\text{I}}^ - }$.

Therefore, option (A) is correct.


Note:The covalent character in ionic compounds is determined by using Fajan’s rule. According to Fajan’s rule, the covalent character in ionic compounds is directly proportional to the polarizing power of the cation and inversely proportional to the polarizability of the anion. Covalent character is directly proportional to the colour intensity of compounds, acidic nature of oxides and solubility of compounds in non-polar solvents.