Chapter 5, Problem 5.111QA

Interpretation Introduction

To check

If 1+ cations of homonuclear diatomic molecules of the second-row elements always have shorter bond lengths than the corresponding neutral molecules.

Answer to Problem 5.111QA

Solution:

The cations of homonuclear diatomic molecules of the second row elements do not have shorter bond lengths than 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 and vice versa. We will find out the bond order of neutral diatomic molecules and their 1+ cations. From that we will check if the bond length of the cation is shorter or not.

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 molecule with the loss of two electrons forming the corresponding 1+ cation is as shown below. With the help of this diagram we will calculate bond order of each molecule and ion of the second row element.

Bond order of Li₂ and Li₂¹⁺.

$Bond order of {Li}_2= \frac{1}{2}\left[\left(2\right)-\left(0\right)\right]=1$

$Bond order of {Li}_2^{1+}= \frac{1}{2}\left[\left(1\right)-\left(0\right)\right]=0.5$

Bond order of Be₂ and Be₂¹⁺.

$Bond order of {Be}_2= \frac{1}{2}\left[\left(2\right)-\left(2\right)\right]=0$

$Bond order of {Be}_2^{1+}= \frac{1}{2}\left[\left(2\right)-\left(1\right)\right]=0.5$

Bond order of B₂ and B₂¹⁺.

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

$Bond order of B_2^{1+}= \frac{1}{2}\left[\left(3\right)-\left(2\right)\right]=0.5$

Bond order of C₂ and C₂¹⁺.

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

$Bond order of C_2^{1+}= \frac{1}{2}\left[\left(5\right)-\left(2\right)\right]=1.5$

Bond order of N₂ and N₂¹⁺.

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

$Bond order of N_2^{1+}= \frac{1}{2}\left[\left(7\right)-\left(2\right)\right]=2.5$

Bond order of O₂ and O₂¹⁺.

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

$Bond order of O_2^{1+}= \frac{1}{2}\left[\left(8\right)-\left(3\right)\right]=2.5$

Bond order of F₂ and F₂¹⁺.

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

$Bond order of F_2^{1+}= \frac{1}{2}\left[\left(8\right)-\left(5\right)\right]=1.5$

Bond order of Ne₂ and Ne₂¹⁺.

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

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

Neutral diatomic molecule	Bond order of neutral molecule	Molecular cation	Bond order of cation
Li2	1	Li21+	0.5
Be2	0	Be21+	0.5
B2	1	B21+	0.5
C2	2	C21+	1.5
N2	3	N21+	2.5
O2	2	O21+	2.5
F2	1	F21+	1.5
He2	0	He21+	0.5

From the bond order of neutral diatomic molecule and bond order of corresponding +1 cation we can say the bond order is not always increasing. Therefore, 1+ cations of homonuclear diatomic molecules of the second-row elements do not always have shorter bond lengths than the corresponding neutral molecules.

Conclusion:

Bond order is inversely proportional to bond length. From bond order of the neutral diatomic molecule and cations we cannot say the bond length is always shorter than neutral atom.