Question

The compound which does not show paramagnetism is
A \[\left[ Cu{{\left( N{{H}_{3}} \right)}_{4}} \right]C{{l}_{2}}\]
B \[\left[ Ag{{\left( N{{H}_{3}} \right)}_{2}} \right]C{{l}_{1}}\]
C NO
D \[N{{O}_{2}}\]

Answer 1

Hint: The magnetic nature of any compound depends on the arrangement of electrons that are present in the outermost shell of the metal. If the number of electrons in the outermost subshell (considering the oxidation state of the metal atom) is not paired (following Hund’s and Pauli's exclusion principle) then the compound is said to be paramagnetic otherwise diamagnetic. In this question, we need to find which compound does not have paired electrons (which means diamagnetic).

Complete answer:In the first compound, \[\left[ Cu{{\left( N{{H}_{3}} \right)}_{4}} \right]C{{l}_{2}}\]the central metal atom is copper (Cu) whose atomic number is 29 and its electronic configuration is \[[Ar]3{{d}^{10}},\text{ }4{{s}^{2}}\]. In the given compound primary valency is +2 (because of two chlorine atoms). In the anionic part of complex compound ligands, ammonia is the neutral one so the primary valency is the oxidation state of copper such as \[C{{u}^{2+}}\]so, its electronic configuration is given as \[[Ar]3{{d}^{9}}\]. As in the d orbital maximum of 10 electrons can be filled (two electrons in each five orbital with opposite spin), but in the above compound, copper with its oxidation state has 9 electrons in the 3d orbital so one electron is unpaired so it is paramagnetic in nature.

Thus, the correct option is B

Note: In options, 3rd and 4th compounds are not complex so to determine the magnetic nature of such compounds just add the total number of electrons present in the compound and if it is odd then it is said to be paramagnetic. The compounds NO and NO2 both contain odd numbers of electrons (15 and 23) so both are paramagnetic in nature. As per Hund’s rule, we need to fill a single electron in all orbital of the subshell before filling a double, and Pauli’s exclusion principle tells us that two electrons in one orbital always lie with opposite spin.