Paramagnetic coordination compounds contain ........ electrons
A No
B Both paired and unpaired
C Paired
D Unpaired

Hint: The paramagnetic nature of a compound depends on the number of electrons in the outermost shell of the metal. Let if the cation part of the coordinate compound is complex and the anionic part is not complex (and can be vice versa) but simple such as \[\left[ Ti{{\left( {{H}_{2}}O \right)}_{6}} \right]C{{l}_{3}}\], this compound is paramagnetic but not diamagnetic.

Complete answer:As an example taken in the hint, titanium is the central metal that is surrounded by ammonia, a neutral ligand. The atomic number of titanium is 22 so its electronic configuration is \[3{{d}^{2}},\text{ }4{{s}^{2}}\]. But to check the magnetic nature of titanium, we need to consider the oxidation state of titanium or the whole compound. As there is one cation (in complex form) and three anions (chlorine), the oxidation state of the complex cation is +3 (three chlorine anions).
As ligands formed coordinate bonds with titanium are neutral thus the oxidation state of the whole complex is the oxidation state of titanium. So the oxidation state of titanium is +3 and thus, the electronic configuration of \[T{{i}^{3+}}\]is \[3{{d}^{1}}\]. As there is one unpaired electron so it is paramagnetic.
Thus, the correct option is D

Note: Coordinate compounds are those compounds in which atoms or molecules are bonded with coordinates bonds. In a normal compound, the first part is a cation and the second part is an anion but in a coordinate compound one of the anion or cation parts of the compound is present in complex form (ligands form a coordinate bond with central metal) and also both anion and cation can be in complex form. The compound is said to be paramagnetic in nature when the central metal has an unpaired electron in the outermost electron whereas when there is no electron in the outermost shell of the electron so the compound is said to be diamagnetic in nature.