Question

Among the following compounds both coloured and paramagnetic one is:
A. ${K_2}C{r_2}{O_7}$
B. $VOS{O_4}$
C. ${(N{H_4})_2}[TiC{l_2}]$
D. ${K_2}[Cu{(CN)_4}]$

Answer 1

When the free electron in the outermost orbital of an atom is excited, it jumps to the next orbital, and while returning to the initial state (ground state), it emits energy, this energy when falling under the wavelength of the visible spectrum (VIBGYOR), we see colour. And paramagnetic behavior of a compound allows it to get weekly attracted by an external magnetic field and formed an induced internal magnetic field in the direction of the applied magnetic field.

Complete step by step answer:
We get the idea that the electron in the outermost orbital should be free, which means it should be unpaired, to absorb energy and jump to the next energy state.
Now to look for unpaired electrons we have to calculate the oxidation number of the central metal atom and see its electronic configuration.
The oxidation number and electronic configuration of each metal ion in the compound are written below.
${K_2}C{r_2}{O_7}$ , Cr is the central metal atom, its oxidation number is +6 and the electronic configuration is;
${}^{24}C{r^{ + 6}} \to \left[ {Ag} \right]4{s^0}3{d^0}$
$VOS{O_4}$, V is the central metal atom, its oxidation number is +4 and the electronic configuration is;
${}^{23}{V^{ + 4}} \to [Ag]4{s^0}3{d^1}$
${(N{H_4})_2}[TiC{l_2}]$, Ti is the central metal atom, its oxidation number is +4 and the electronic configuration is:
${}^{22}T{i^{ + 4}} \to \left[ {Ag} \right]4{s^0}3{d^0}$
${K_2}[Cu{(CN)_4}]$, Cu is the central metal atom, its oxidation number is +1, and the electronic configuration is:
${}^{29}C{u^ + } \to [Ag]4{s^0}3{d^{10}}$
By observing each of the electronic configurations of metal ion, we can come to the conclusion that $VOS{O_4}$ will show colour and also be paramagnetic because of one unpaired electron in the central metal ion .i.e., Vanadium .
Hence, the correct answer is option (B)i.e., $VOS{O_4}$

Additional information: To calculate the oxidation number of the central atom, assume it be as ‘x’ and calculate using the rest known oxidation states of atoms present in the compound. e.g., to calculate the oxidation state of Cr in ${K_2}C{r_2}{O_7}$ assume Cr has the oxidation state to be ‘x’, oxidation of K and O is known to us as +1 and -2 respectively.
$\Rightarrow 2(1) + 2x + 7( - 2) = 0$
$\Rightarrow 2x + 2 - 14 = 0$
$\Rightarrow 2x - 12 = 0$
$\Rightarrow x = \dfrac{{12}}{2}$
$\Rightarrow x = + 6$
Calculate the oxidation number of the rest of the central metal atom by yourself, using this knowledge.

Note: Although according to the rules the electronic configuration of copper is supposed to be ${}^{29}Cu \to [Ag]4{s^2}3{d^9}$, however, this is not the correct configuration for copper, one electron from the $4{s^2}$ will pair with one electron in the $3{d^9}$ subshell and form a more stable $3{d^{10}}$configuration. Hence, the accepted electronic configuration of Cu is; ${}^{29}Cu \to [Ag]4{s^1}3{d^{10}}$ .