Question

In which of the following diatomic molecules would the bond strength be expected to be weaken as an electron is removed?

a. ${\mathit{H}}_{\mathbf{2}}$

b. ${\mathit{B}}_{\mathbf{2}}$

c. ${{\mathbf{C}}_{\mathbf{2}}}^{\mathbf{2}\mathbf{-}}$

d. OF

Short answer

The weakest bond strength is for option b

The bond strength depends upon the bond order.

The greater the bond order more is the bond strength

The bond order is$\frac{1}{2}(n_b-n_a)$

Where

role="math" localid="1663770311020" $$\begin{gathered} n_b=bondingelectrons \\ {n}_a=noofantibondingelectron \end{gathered}$$

Step-by-step solution

Step 1

On removing an electron, ${H^{+}}_2\left(1electron\right)$ is formed

So the bond filling is ${{\sigma }^1}_{1s}$

And the bond order is $B.O=\frac{1}{2}$

On removing an electron is formed ${B^{+}}_2\left(9e^{-}\right)$

So bond filling is ${{\sigma }^2}_{1s}({{\sigma }^{*}}_{1s}{)}^2({\sigma }_{2s}{)}^2({{\sigma }^{*}}_{2s}{)}^2{{\pi }^1}_{2px}$

And the bond order is

localid="1663770859533" $$\begin{gathered} B.O=\frac{1}{2}(5-4) \\ B.O=0.5 \end{gathered}$$

For c) on removing an electron ${C_2}^{-}\left(13e^{-}\right)$ is formed

Bond filling order is ${{\sigma }^2}_{1s}({{\sigma }^{*}}_{1s}{)}^2({\sigma }_{2s}{)}^2({{\sigma }^{*}}_{2s}{)}^2{{\pi }^2}_{2px}{{\pi }^2}_{2py}{{\sigma }^1}_{2pz}$

Bond order is

localid="1663772662156" $$\begin{gathered} B.O=\frac{1}{2}(9-4) \\ B.O=2.5 \end{gathered}$$

For d) on removing an electron $O{F}^{+}\left(16e^{-}\right)$

The bond filling is as follows ${{\sigma }^2}_{1s}({{\sigma }^{*}}_{1s}{)}^2({\sigma }_{2s}{)}^2({{\sigma }^{*}}_{2s}{)}^2{{\sigma }^2}_{2pz}{{\pi }^2}_{2px}{{\pi }^2}_{2py}{{\pi }^{*}}_{2pz}{{\pi }_{2py}}^{*}$

Bond order is

$$\begin{gathered} B.O=\frac{1}{2}(10-6) \\ B.O=2 \end{gathered}$$

Step 2

The bond order of $B_2$ on removing an electron is the lowest so it will have the weakest bond strength.

Final answer:

The weakest bond strength is for option b