Chapter 5, Problem 5.112QA

Interpretation Introduction

To check

If any of the anions of the homonuclear diatomic molecules formed by B, C, N, O and F have shorter bond lengths than those of the corresponding neutral molecules.

Answer to Problem 5.112QA

Solution:

The bond lengths of the anions of the homonuclear diatomic molecules formed by B and C have shorter bond lengths than those of the corresponding neutral molecules.

Explanation of Solution

Bond order and bond length are inversely proportional to each other. When bond order increases the bond length decreases. When bond order decreases the bond length increases. We will find out the bond order of neutral diatomic molecules and their 1- and 2- anions. From the molecular orbital diagram we can find out number of bonding and antibonding electrons. The formula to calculate bond order is as below.

$Bond order= \frac{1}{2}[\left(number of bonding electrons\right)-\left(number of antibonding electrons\right)]$

The molecular orbital diagram for the given molecules with the gain of one and two electrons forming the corresponding 1- and 2- anion is as shown below. With the help of this diagram we will calculate bond order of each molecule and ion.

Bond order of B₂ and B₂^1- and B₂^2-.

$Bond order of B_2= \frac{1}{2}\left[\left(4\right)-\left(2\right)\right]=1$

$Bond order of B_2^{1-}= \frac{1}{2}\left[\left(5\right)-\left(2\right)\right]=1.5$

$Bond order of B_2^{2-}= \frac{1}{2}\left[\left(6\right)-\left(2\right)\right]=2$

Bond order of C₂, C₂^1- and C₂^2-.

$Bond order of C_2= \frac{1}{2}\left[\left(6\right)-\left(2\right)\right]=2$

$Bond order of C_2^{1-}= \frac{1}{2}\left[\left(7\right)-\left(2\right)\right]=2.5$

$Bond order of C_2^{2-}= \frac{1}{2}\left[\left(8\right)-\left(2\right)\right]=3$

Bond order of N₂, N₂^1- and N₂^2-.

$Bond order of N_2= \frac{1}{2}\left[\left(8\right)-\left(2\right)\right]=3$

$Bond order of N_2^{1-}= \frac{1}{2}\left[\left(8\right)-\left(3\right)\right]=2.5$

$Bond order of N_2^{2-}= \frac{1}{2}\left[\left(8\right)-\left(4\right)\right]=2$

Bond order of O₂, O₂^1- and O₂^2-.

$Bond order of O_2= \frac{1}{2}\left[\left(8\right)-\left(4\right)\right]=2$

$Bond order of O_2^{1-}= \frac{1}{2}\left[\left(8\right)-\left(5\right)\right]=1.5$

$Bond order of O_2^{2-}= \frac{1}{2}\left[\left(8\right)-\left(6\right)\right]=1$

Bond order of F₂, F₂^1- and F₂^2-.

$Bond order of F_2= \frac{1}{2}\left[\left(8\right)-\left(6\right)\right]=1$

$Bond order of F_2^{1-}= \frac{1}{2}\left[\left(8\right)-\left(7\right)\right]=0.5$

$Bond order of F_2^{2-}= \frac{1}{2}\left[\left(8\right)-\left(8\right)\right]=0$

Neutral diatomic molecule	Bond order of neutral molecule	Molecular anion	Bond order of anion	Molecular anion	Bond order of anion
B2	1	B21-	1.5	B22-	2
C2	2	C21-	2.5	C22-	3
N2	3	N21-	2.5	N22-	2
O2	2	O21-	1.5	O22-	1
F2	1	F21-	0.5	F22-	0

From the data in the table we can say that bond order of B₂ is smaller than B₂^1- and bond order of B₂^1- is smaller than B₂^2-. Therefore bond length of B₂ is larger than B₂^1-. Also bond length of B₂^1- is larger than B₂^2-. Similarly bond length of C₂ is larger than C₂^1-. Also bond length of C₂^1- is larger than C₂^2-. Therefore the bond lengths of the anions of the homonuclear diatomic molecules formed by B and C have shorter bond lengths than those of the corresponding neutral molecules.

Conclusion:

Bond order is inversely proportional to bond length. From bond order of each molecule and corresponding anion, the bond lengths of the anions of the homonuclear diatomic molecules formed by B and C have shorter bond lengths than those of the corresponding neutral molecules.