Question

Find the number of chemical species which are paramagnetic in nature ${{O}_{3}},N{{O}_{2}},K{{O}_{2}},{{\left[ PtC{{l}_{4}} \right]}^{2-}},{{N}_{2}},N{{a}_{2}}\left[ Fe{{\left( CN \right)}_{5}}NO \right]$

Answer 1

Hint: The electron has an electron magnetic dipole moment, which is generated by the electron's intrinsic spin property, making it an electric charge in motion. There are various magnetic behaviors such as paramagnetism, diamagnetism, ferromagnetism etc.

Complete answer:
- We are asked to find out the molecules which are paramagnetic in nature. The term paramagnetism refers to the magnetic state of an atom with one or more unpaired electrons.
- According to Hund's Rule, the electrons must occupy every orbital singly before any orbital is doubly occupied and this may leave the atom with many unpaired electrons. Since unpaired electrons can spin in either direction, they can display magnetic moments in any direction and this capability allows paramagnetic atoms to be attracted to magnetic fields.
(i) ${{O}_{3}}$ - ${{O}_{3}}$ is an ozone molecule which is made up of three oxygen atoms. When we look into the MO diagram of ozone, there are 5 non-bonding orbitals which do not contribute to the bonding structure of the molecule and essentially merely hold electrons and there are three pi orbitals, the two lowest are occupied with two electrons each. Hence, all the electrons are paired up and this makes the molecule diamagnetic in its ground state.
(ii) $N{{O}_{2}}$- $N{{O}_{2}}$ is nitrogen dioxide and is a highly reactive oxide of nitrogen. We know that nitrogen has 7 electrons and two oxygen atoms will contribute 16 electrons. Thus the total electrons in $N{{O}_{2}}$ is 23 and it’s an odd number. Therefore definitely there will be unpaired electrons and $N{{O}_{2}}$ is paramagnetic.
(iii) $K{{O}_{2}}$- $K{{O}_{2}}$ is a superoxide of potassium in which, only one electron is released from the dioxygen atom and the superoxide ion is represented as ${{O}^{2-}}$ . Therefore in $K{{O}_{2}}$the oxygen atoms behave as a free radical species, having an unpaired electron. Due to the presence of unpaired electrons, $K{{O}_{2}}$ is also paramagnetic.
(iv) ${{\left[ PtC{{l}_{4}} \right]}^{2-}}$ - In ${{\left[ PtC{{l}_{4}} \right]}^{2-}}$ the platinum is in +2 oxidation state and hence the electronic configuration of $P{{t}^{2+}}$becomes ${{d}^{8}}$ and 8 electrons will fill in the 5d orbital evenly leaving one orbital vacant. Since all the electrons are paired, ${{\left[ PtC{{l}_{4}} \right]}^{2-}}$ is diamagnetic.
(v) ${{N}_{2}}$- ${{N}_{2}}$ is molecular nitrogen and it will have 14 electrons in total and all the electrons are paired up. Therefore ${{N}_{2}}$ is diamagnetic in nature.
(vi) $N{{a}_{2}}\left[ Fe{{\left( CN \right)}_{5}}NO \right]$ - The molecule $N{{a}_{2}}\left[ Fe{{\left( CN \right)}_{5}}NO \right]$ is sodium nitroprusside and in this molecule Fe is in +2 oxidation state and $N{{O}^{+}}$ acts as the ligand. Due to the +2 oxidation state Fe has ${{d}^{6}}$ configuration and these electrons will be paired up due to the influence ligands. Since there is no unpaired electrons, the complex $N{{a}_{2}}\left[ Fe{{\left( CN \right)}_{5}}NO \right]$ will be diamagnetic.

The number of chemical species which are paramagnetic in nature are two and they are $N{{O}_{2}}$ and$K{{O}_{2}}$.

Note: It should be noted that if an electron is single in an orbital, the orbital has a net spin, since the spin of the lone electron does not get canceled out and if even one orbital has a net spin, the entire atom will have a net spin. Or in other words, an atom can have 10 paired (diamagnetic) electrons, but as long as it also has one unpaired (paramagnetic) electron, it is still considered as a paramagnetic atom.