Q11.45CP - Chemistry: Molecular Nature Of Matter And Change

Question

The ionosphere lies about 100 km above Earth’s surface. This layer consists mostly of $\mathbf{NO}$ ,  ${\mathbf{O}}_2$ , and ${\mathbf{N}}_2$ , and photoionization creates ${\mathbf{NO}}^{+}$  , ${{\mathbf{O}}_2}^{+}$  , and  ${{\mathbf{N}}_2}^{+}$. 

(a) Use MO theory to compare the bond orders of the molecules and ions.

 

(b) Does the magnetic behaviour of each species change when its ion forms?

Step-by-Step Solution

Answer

1. The Bond order of  $\mathrm{NO}$ , ${\mathrm{O}}_2$  , and ${\mathrm{N}}_2$  is 3, 2, 2.5 whereas the bond order of  ,  , and  is 3.5, 2.5, 3.
2. The molecule  ${\mathrm{O}}_2$ , and ${\mathrm{N}}_2$ are diamagnetic which become paramagnetic in ${{\mathrm{O}}_2}^{+}$  , and  ${{\mathrm{N}}_2}^{+}$  whereas $\mathrm{NO}$  molecule is paramagnetic which become diamagnetic in ${\mathrm{NO}}^{+}$.  

Step 1: Bonding

Bonding may be defined as the force of attraction between two or more atoms to form a compound. The molecule can have the same atom and different atoms.

 

Hybridization may be defined as the mixing of the atomic orbital to form a hybrid orbital before bonding with another atomic orbital having a similar atomic orbital.

Step 2: Molecular Orbital

Molecular orbital theory may be used to find the bond order of the molecule by filling the electron in bonding and anti-bonding orbital.

$\mathbf{Bond}\mathbf{ }\mathbf{order}\mathbf{=}\mathbf{ }\mathbf{[}\mathbf{(}\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{e}\mathbf{-}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{BMO}\mathbf{)}\mathbf{-}\mathbf{(}\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{e}\mathbf{-}\mathbf{in}\mathbf{ }\mathbf{ABMO}\mathbf{)}\mathbf{]}\mathbf{/}\mathbf{2}$

$\begin{array}{l}\mathrm{Bond} \mathrm{order} \mathrm{of} {\mathrm{N}}_2\text{}=1/2(10-4)\text{}=6/2\text{}=3 \\ \mathrm{Bond} \mathrm{order} \mathrm{of} {{\mathrm{O}}_2}_{\text{}}=1/2(10-6)\text{}=4/2\text{}=2 \\ \mathrm{Bond} \mathrm{order} \mathrm{of} \mathrm{NO}= 1/2 [10-5]=2.5\\ \begin{array}{l}\mathrm{Bond} \mathrm{order} \mathrm{of} {{\mathrm{N}}_2}^{+}\text{}=1/2(10-3)\text{}=7/2\text{}=3 .5 \\ \mathrm{Bond} \mathrm{order} \mathrm{of} {{\mathrm{O}}_2}^{+}\text{}=1/2(10-5)\text{}=5/2\text{}=2.5 \end{array}\\ \mathrm{Bond} \mathrm{order} \mathrm{of} {\mathrm{NO}}^{+}= 1/2 [10-4]=3\end{array}$

Step 3: Magnetic Character

The magnetic character depends upon the electron- they are single or in pair.

 

In ${\mathrm{N}}_2$  molecule, the magnetic character changed from diamagnetic to paramagnetic. As in  ${\mathrm{N}}_2$ molecule all the electrons are paired up in molecular orbital configuration but in ${\mathrm{NO}}^{+}$molecule the is a single electron which made the molecule paramagnetic

In ${\mathrm{O}}_2$molecule, the magnetic character changed from the diamagnetic to paramagnetic. As in ${\mathrm{O}}_2$molecule all the electrons are paired up in molecular orbital configuration but in ${{\mathrm{O}}_2}^{+}$ molecule the is single electron which made the molecule paramagnetic.

In $\mathrm{NO}$  molecule, the magnetic character changed from the paramagnetic to diamagnetic. As in  $\mathrm{NO}$  molecule there is one unpaired electron in molecular orbital configuration but in ${{\mathrm{O}}_2}^{+}$  molecule all the electron paired up which made the molecule diamagnetic.