Generally transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in solid state?
A. $A{g_2}S{O_4}$
B. $Cu{F_2}$
C. $Zn{F_2}$
D. $C{u_2}C{l_2}$

Hint: In $A{g_2}S{O_4}$, silver is present as $A{g^ + }$, in $Cu{F_2}$, copper is present as $C{u^{2 + }}$, in $Zn{F_2}$, zinc is present as $Z{n^{2 + }}$, in $C{u_2}C{l_2}$, the copper is present as $C{u^{ + 1}}$. As they are metal cations, the electrons are removed from their actual configuration to form ions.

Complete step by step answer:
In the complex $A{g_2}S{O_4}$, the metal cation is silver which is present as $A{g^ + }$ and anion is$SO_4^{2 - }$. The atomic number of silver is 47. The electronic configuration of silver is $[Kr]4{d^{10}}5{s^1}$. From Ag atom one electron is removed to form $A{g^ + }$, the resulting electronic configuration will be $[Kr]4{d^{10}}5{s^0}$. The $A{g^ + }$, have completely filled the d-orbital therefore it will not form a colored complex.
In the complex $Cu{F_2}$, the metal cation is copper which is present as $C{u^{2 + }}$ and anion is${F^ - }$. The atomic number of copper is 29. The electronic configuration of copper is $[Ar]3{d^{10}}4{s^1}$. From Cu atom two electrons are removed to form $C{u^{2 + }}$, the resulting electronic configuration will be $[Ar]3{d^9}$. The $C{u^{2 + }}$, have incomplete d-orbital therefore it will form a colored complex.
In the complex $Zn{F_2}$, the metal cation is zinc which is present as $Z{n^{2 + }}$ and anion is${F^ - }$. The atomic number of copper is 30. The electronic configuration of zinc is $[Ar]3{d^{10}}4{s^2}$. From the Zn atom two electrons are removed to form $Z{n^{2 + }}$, the resulting electronic configuration will be $[Ar]3{d^{10}}$. The $Z{n^{2 + }}$, have completely filled the d-orbital therefore it will not form a colored complex.
In the complex $C{u_2}C{l_2}$, the metal cation is copper which is present as $C{u^{ + 1}}$ and anion is$C{l^ - }$. The atomic number of copper is 29. The electronic configuration of copper is $[Ar]3{d^{10}}4{s^1}$. From Cu atom one electron is removed to form $C{u^{ + 1}}$, the resulting electronic configuration will be $[Ar]3{d^{10}}$. The $C{u^{ + 1}}$, have completely filled d-orbital therefore it will not form a colored complex.
Thus, the compound which is coloured at solid state is $Cu{F_2}$.
So, the correct answer is “Option B ”.

Note: The transition metal elements form coloured salt due to the presence of unpaired electrons as they show d-d transition. In d-d transition, the electrons present in the d- orbital of lower energy of the metal get excited by absorbing photons to the higher energy of d-orbital.