Question

How many of the species are paramagnetic?
 $N_{2}O,NO,N{O}_2,O_2,N{O}_2^{+},N{O}^{+},C{O}^{+}.$
(A) $2$
(B) $3$
(C) $5$
(D) $4$

Answer 1

Hint: We know that to know whether the species are paramagnetic or not, molecular orbital theory is used to identify the electronic configuration. As we know the electron may remain unpaired during the formation of the bond between two atoms, such species are called paramagnetic species.

Complete answer:
We know that the magnetic character is categorized into three parts. One of them is diamagnetic, second is paramagnetic and third is ferromagnetic. Generally, the magnetic property is affected by the existence of magnetic fields on either one side of the magnetic bar. As we all know, the compound which is said to be paramagnetic commonly has one or more unpaired electrons. In paramagnetic form, the substance is weakly attached by the given magnetic bar.
Whenever two electrons are paired together in an orbital, their total spin is zero, they are diamagnetic electrons. Atoms with all diamagnetic electrons are called diamagnetic atoms. A paramagnetic electron is an unpaired electron. Nitrogen monoxide has eleven valence electrons; it is paramagnetic, with a single electron occupying the pair of orbitals. Paramagnetic materials are weakly attracted to a magnetic field. Nitrogen monoxide has odd no. of electrons and due to the presence of unpaired electrons; it is paramagnetic in the gaseous state. However, in the liquid and solid states, the unpaired electrons are involved in the formation of the loose dimer. In the absence of an unpaired electron, it is diamagnetic in nature.
The number of valence electrons in $N_{2}O=5+6+5=16.$ Thus, all electrons can be paired up and there are no free electrons. $N_{2}O$ molecule is diamagnetic.
The number of valence electrons in $NO=5+6=11.$ Thus, all electrons cannot be paired up and there is an odd electron responsible for magnetism. $NO$ molecule is paramagnetic.
The number of valence electrons in $N{O}_2=5+6+6=17.$ Thus, all electrons cannot be paired up and there is an odd electron responsible for magnetism. $N{O}_2$ molecule is paramagnetic.
The number of valence electrons in $O_2=6+6=12.$ Thus, all electrons can be paired up and there are no free electrons. $O_2$ molecule is an exception and is paramagnetic.
The number of valence electrons in $N{O}_2^{+}=5+6+6-1=16.$ Thus, all electrons can be paired up and there are no free electrons. $N{O}_2^{+}$ molecule is diamagnetic.
The number of valence electrons in $N{O}^{+}=5+6-1=10.$ Thus, all electrons can be paired up and there are no free electrons. $N{O}^{+}$ molecule is diamagnetic.
The number of valence electrons in $C{O}^{+}=4+6-1=9.$ Thus, all electrons cannot be paired up and there is an odd electron. $C{O}^{+}$ molecule is paramagnetic.
Therefore, the correct answer is option D, i.e. the species are paramagnetic are $4.$

Note:
Remember that the paramagnetic molecules always have the colour because they have unpaired electrons and they require less amount of energy for electron transition. Also the magnetic properties of a substance can be determined by examining its electronic configuration. A substance having ten paired electrons, but as long as it also has only one unpaired electron, it is still considered a paramagnetic.