Question

AgCl and NaCl are colourless. NaBr and NaI are colourless but AgBr and AgI are coloured. This is because:
a.) \[A{g^ + }\] polarises \[B{r^ - }\] and \[{I^ - }\]
b.) \[A{g^ + }\] has unpaired d – orbital.
c.) \[A{g^ + }\] depolarises \[B{r^ - }\] and \[{I^ - }\]
d.) None of the above is correct

Answer 1

Hint: Polarization is an effect in which a small charge dense cation attracts the loosely bound electron cloud of the penultimate shell of a large anion and distorts its anionic electron cloud shape. This is known as ion polarization.

Complete step by step answer:
The ability of a cation to distort an anion is known as its polarization power and the tendency of the anion to become polarized by the cation is known as its polarizability. The polarizing power and the polarizability enhances the formation of covalent bonds and is favoured by small size of cation, large size of anion and large charges of cation and anion.
Small size of cation: The smaller the cation is, more is its charge density, and more is the capability of the cation to attract the electrons. Positive charge on a small area increases the charge density of the cation.
Large size of anion: The larger size of anion enhances the polarizability as the outer electrons are more loosely held and can be easily distorted by the cation.
Large charges: The increasing charges increase the electrostatic attractions of the cation for the outer electrons of the anion. Thus, it enhances polarizability.

Ion polarization can change the colour of a compound. This is explained by the Fajan’s rule.
Hence, \[A{g^ + }\] polarises \[B{r^ - }\] and \[{I^ - }\].
Therefore, the correct answer is (A).

Note: Remember that the small cations are found on the top left of the periodic table and large anions are found on the bottom right. Better polarizability requires smaller cation and larger anion.