Question

The colorless species is:
A.) $VC{l_3}$
B.) $VOS{O_4}$
C.) $N{a_3}V{O_4}$
D.) $[V{({H_2}O)_6}]S{O_4}.{H_2}O$

Answer 1

Hint: In this question, to find out the colorless species the concept of unpaired electrons can be used. According to which if in a compound the central metal atom has unpaired electrons then it is colored and if there are no unpaired electrons then the compound will be colorless.

Complete step by step solution:
The color that is exhibited by transition metal complexes are caused due to the excitation of an electron from a lower energy orbital to higher energy orbital. If a compound contains a transition element then the compound will be colored if the number of unpaired electrons on the transition metal is not zero. Also, the compound will be colorless if there is no unpaired electron present on the central metal atom in the compound. Therefore, no unpaired electrons mean colorless compounds.
As we know that Vanadium($V$) is a transition metal element having atomic number as $23$. Vanadium has electronic configuration as : ${[Ar]^{18}}3{d^5}$.
In $VC{l_3}$, as we know that chlorine has $ - 1$ charge on it and a total of $ - 3$ from three chlorines. Also, the total charge on the compound is zero. Therefore, the oxidation state or charge on vanadium is x:
$
  x - 3 = 0 \\
  x = + 3 \\
 $
Thus, Electronic configuration of ${V^{3 + }}:{[Ar]^{18}}3{d^2}$. As it has two unpaired electrons so it is a colored species.
In $VOS{O_4}$, as we know that sulphate ion has $ - 2$ charge on it and oxygen has $ - 2$ on it. Also, the total charge on the compound is zero. Therefore, the oxidation state or charge on vanadium is x:
$
  x - 2 - 2 = 0 \\
  x = + 4 \\
 $
Thus, Electronic configuration of ${V^{4 + }}:{[Ar]^{18}}3{d^1}$. As it has one unpaired electron so it is a colored species.
In$N{a_3}V{O_4}$, as we know that sodium ion has $ + 1$ charge on it , total of three sodium so $ + 3$ charge and oxygen has $ - 2$ on it so for four oxygen it is $ - 8$. Also, the total charge on the compound is zero. Therefore, the oxidation state or charge on vanadium is x:
$
  x + 3 - 8 = 0 \\
  x = + 5 \\
 $
Thus, Electronic configuration of ${V^{5 + }}:{[Ar]^{18}}3{d^0}$. As it has no unpaired electrons so it is a colorless species.
In $[V{({H_2}O)_6}]S{O_4}.{H_2}O$, as we know that sulphate ion has $ - 2$ charge on it so charge on $[V{({H_2}O)_6}]$ is $ + 2$ and water has zero charge on it. Also, the total charge on the compound is $ + 2$. Therefore, the oxidation state or charge on vanadium is x:
$
  x + 0 = + 2 \\
  x = + 2 \\
 $
Thus, Electronic configuration of ${V^{2 + }}:{[Ar]^{18}}3{d^3}$. As it has three unpaired electrons so it is a colored species.

Hence, option C.) is the correct answer.

Note: Always remember that the transition metal elements exhibit colors due to the excitation of electrons from lower energy state to higher energy state which in turn depend on unpaired electrons.