In the following which of the two are paramagnetic?
a.) \[{N_2}\]
b.) \[CO\]
c.) \[{B_2}\]
d.) \[N{O_2}\]
Correct answer is:
a.) a and c
b.) b and c
c.) c and d
d.) b and d
Answer
670.5k+ views
Hint: To solve this question, we must have a clear concept of molecular orbital theory. We can easily predict if a molecule is paramagnetic or diamagnetic by simply looking at its electronic configuration.
Complete step by step answer:
Perhaps the easiest way to differentiate between paramagnetic and diamagnetic substances is by saying that Paramagnetic molecules are attracted to magnetic field while diamagnetic molecules are repelled by the magnetic field and while paramagnetic compounds have unpaired electrons, diamagnetic compounds the electrons all have paired spins.
While an odd number of electrons can clearly produce a paramagnetic ion, radical or molecule, molecules with an even number of electrons may also be paramagnetic, the most famous example of this is the oxygen molecule.
Now, we should write the electronic configuration in accordance with Molecular Orbital Theory and find out which of these molecules have unpaired electrons.
We see that
\[{N_2} = {(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{(\pi 2{p_x})^2}{(\pi 2{p_y})^2}{(\sigma 2{p_z})^2}\]
\[CO = {(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{(\pi 2p)^4}{(\sigma 2p)^2}\]
\[{B_2} = {(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{(\pi 2{p_y})^1}{(\pi 2{p_z})^1}\]
From the periodic table, we know that the atomic number of nitrogen(N) is 7 and that of oxygen(O) is 8.
Hence, the total number of electrons for nitrogen dioxide is:
\[N{O_2} = 7 + 16 = 23\] this is odd, so it is definitely paramagnetic. We can say this without even finding the electronic configuration.
Hence, we can conclude that boron molecules and nitrogen dioxide molecules are paramagnetic.
Therefore, the correct answer is Option (C) c and d.
Additional information:
Electron spin is very important in determining the magnetic properties of an atom. If all of the electrons in an atom are paired up and share their orbital with another electron, then the total spin in each orbital is zero and the atom is diamagnetic.
Note: If we forget the Molecular Orbital Theory there is a way to experimentally determine if a substance is paramagnetic or diamagnetic by using a Gouy balance. It measures the apparent change in the mass of the sample as it is repelled or attracted by the region of high magnetic field between the poles.
Complete step by step answer:
Perhaps the easiest way to differentiate between paramagnetic and diamagnetic substances is by saying that Paramagnetic molecules are attracted to magnetic field while diamagnetic molecules are repelled by the magnetic field and while paramagnetic compounds have unpaired electrons, diamagnetic compounds the electrons all have paired spins.
While an odd number of electrons can clearly produce a paramagnetic ion, radical or molecule, molecules with an even number of electrons may also be paramagnetic, the most famous example of this is the oxygen molecule.
Now, we should write the electronic configuration in accordance with Molecular Orbital Theory and find out which of these molecules have unpaired electrons.
We see that
\[{N_2} = {(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{(\pi 2{p_x})^2}{(\pi 2{p_y})^2}{(\sigma 2{p_z})^2}\]
\[CO = {(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{(\pi 2p)^4}{(\sigma 2p)^2}\]
\[{B_2} = {(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{(\pi 2{p_y})^1}{(\pi 2{p_z})^1}\]
From the periodic table, we know that the atomic number of nitrogen(N) is 7 and that of oxygen(O) is 8.
Hence, the total number of electrons for nitrogen dioxide is:
\[N{O_2} = 7 + 16 = 23\] this is odd, so it is definitely paramagnetic. We can say this without even finding the electronic configuration.
Hence, we can conclude that boron molecules and nitrogen dioxide molecules are paramagnetic.
Therefore, the correct answer is Option (C) c and d.
Additional information:
Electron spin is very important in determining the magnetic properties of an atom. If all of the electrons in an atom are paired up and share their orbital with another electron, then the total spin in each orbital is zero and the atom is diamagnetic.
Note: If we forget the Molecular Orbital Theory there is a way to experimentally determine if a substance is paramagnetic or diamagnetic by using a Gouy balance. It measures the apparent change in the mass of the sample as it is repelled or attracted by the region of high magnetic field between the poles.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

