Q11.52CP - Chemistry: Molecular Nature Of Matter And Change

Question

Molecular nitrogen, carbon monoxide, and cyanide ion are isoelectronic.

a) Draw an MO diagram for each.

b)  $\mathbf{CO}$and${\mathbf{CN}}^{-}$ are toxic. What property may explain why${\mathbf{N}}_2$   isn’t? 

 

Step-by-Step Solution

Answer

Molecular nitrogen, carbon monoxide, and cyanide ion are isoelectronic because all are having a total of 14 electrons.

The answer for

(a).

 $\begin{array}{l}{\text{N}}_2:\text{KK}\text{'}{\left(\sigma 2s\right)}^{2 }{(\sigma *2s)}^2 {\left(\pi 2P_x\right)}^2 {\left(\pi 2p_y\right)}^{2 }{\left(\sigma 2p_z\right)}^2\\ \text{CO}:\text{KK}\text{'}{\left(\sigma 2s\right)}^{2 }{(\sigma *2s)}^2 {\left(\pi 2P_x\right)}^2 {\left(\pi 2p_y\right)}^{2 }{\left(\sigma 2p_z\right)}^2\\ {\text{CN}}^{\text{-}}:\text{KK}\text{'}{\left(\sigma 2s\right)}^{2 }{(\sigma *2s)}^2 {\left(\pi 2P_x\right)}^2 {\left(\pi 2p_y\right)}^{2 }{\left(\sigma 2p_z\right)}^2\end{array}$

(b). Both are polar

Step 1:Definition

A molecular orbital (MO) diagram is a qualitative descriptive tool that shows the relative energy and number of electrons in each MO, as well as the AOs from which they formed.

Those species which have the same number of electronsare called isoelectronic species.

 

Step2:MO of N 2 , CO and CN MO of N 2

The electronic configuration of nitrogen.$(Z=7)=1s^2 2{\text{s}}^2 2{{\text{p}}_x}^{1}2{{\text{p}}_y}^{1}2{{\text{p}}_z}^1$

The total number of electrons present in the nitrogen molecule $\left({\text{N}}_2\right)$  is 14.These electrons can be accommodated in the increasingorder of energy orbitals.

 ${\text{N}}_2:\text{KK}\text{'}{\left(\sigma 2s\right)}^{2 }{(\sigma *2s)}^2 {\left(\pi 2P_x\right)}^2 {\left(\pi 2p_y\right)}^{2 }{\left(\sigma 2p_z\right)}^2$

Step3:MO of CO

The electronic configuration of carbon .$(\text{Z}=6)=1{\text{s}}^2 2{\text{s}}^2 2{{\text{p}}_x}^{1}2{{\text{p}}_y}^1$

The electronic configuration of oxygen.$(Z=8)=1{\text{s}}^2 2{\text{s}}^2 2{{\text{p}}_x}^{2}2{{\text{p}}_y}^{1}2{{\text{p}}_z}^1$

The total number of electrons present in the carbon monoxide molecule (CO) is 14.These electrons can be accommodated in the increasingorder of energy orbitals.

 $\text{CO}:\text{KK}\text{'}{\left(\sigma 2s\right)}^{2 }{(\sigma *2s)}^2 {\left(\pi 2{\text{P}}_x\right)}^2 {\left(\pi 2{\text{p}}_y\right)}^{2 }{\left(\sigma 2{\text{p}}_z\right)}^2$

Step4:MO of CN

The electronic configuration of carbon  .$(\text{Z}=6)=1{\text{s}}^2 2{\text{s}}^2 2{p_x}^{1}2{{\text{p}}_y}^1$

The electronic configuration of nitrogen  $(Z=7)=1s^2 2s^2 2{p_x}^{1}2{p_y}^{1}2{p_z}^1$.

The total number of electrons present in the cyanide ion$\left({\text{CN}}^{\text{-}}\right)$ is. $14(6+7+1)$The one electron is from ion(-ve).These electrons can be accommodated in the increasing order of energy orbitals.

 ${\text{CN}}^{\text{-}}:\text{KK}\text{'}{\left(\sigma 2s\right)}^{2 }{(\sigma *2s)}^2 {\left(\pi 2P_x\right)}^2 {\left(\pi 2p_y\right)}^{2 }{\left(\sigma 2p_z\right)}^2$

Step 5: Toxicity of N2

$CO$  and  data-custom-editor="chemistry" $C{N}^{-}$both are polar and can bind to haemoglobin in blood while   is non-toxic because of the absence of polarity of the bond.