Question

The black coloured compound may be:
A.${\text{NiS}}$
B.${\text{CdS}}$
C.${\text{PbS}}$
D.${\text{CoS}}$

Answer 1

Paramagnetic compounds are almost always coloured. Paramagnetic substances are those substances which have unpaired electrons.

Complete step by step answer:
Based on the presence or absence of unpaired electrons, two types of compounds are there namely paramagnetic and diamagnetic. Compounds having unpaired electrons are known as paramagnetic compounds while compounds with zero unpaired electrons are called diamagnetic. The unpaired electrons in the paramagnetic compounds show electron transitions even with a small amount of energy provided by visible light. The colour corresponds to the complementary colour of light absorbed. That's why paramagnetic compounds are coloured. Also they are attracted by a magnetic field due to unpaired electrons. On the other hand diamagnetic compounds are mostly colourless.
According to our question, given compounds as ${\text{NiS, CdS,PbS,CoS}}$ . The cations present in the compound are ${\text{N}}{{\text{i}}^{2 + }}{\text{, C}}{{\text{d}}^{2 + }}{\text{, P}}{{\text{b}}^{2 + }}{\text{, C}}{{\text{o}}^{2 + }}$ . Electronic configuration of ${\text{N}}{{\text{i}}^{2 + }}$ is $3{{\text{d}}^8}$ so it is having 2 unpaired electrons or it is paramagnetic. Electronic configuration of ${\text{C}}{{\text{d}}^{2 + }}$ is ${\text{4}}{{\text{d}}^{10}}$ so it does not having any unpaired electron or it is diamagnetic. Electronic configuration of ${\text{P}}{{\text{b}}^{2 + }}$ is ${\text{6}}{{\text{p}}^2}$ so it is having 2 unpaired electron or it is paramagnetic. Electronic configuration of ${\text{C}}{{\text{o}}^{2 + }}$ is $3{{\text{d}}^7}$ so it is having 3 unpaired electrons or it is paramagnetic.
These electron configurations are due to loss of 2 electrons from the valence shell of their neutral state.

Thus, ${\text{NiS,PbS, CoS}}$ may be black coloured. Hereby correct options are A, C and D.

Note:
${\text{KMn}}{{\text{O}}_4}$ is not having any unpaired electron still it is coloured due to an electronic transition, which is not a d-d transition. It arises from a charge transfer within a molecule where an electron from the molecular orbital of Manganese - oxygen is transferred to the empty d orbital of Manganese. This results in a purple colour of the compound.