Question

How would the bond length vary in dicarbon species${\mathrm{C}}_2,{{\mathrm{C}}_2}^{+},{{\mathrm{C}}_2}^{-}$

Answer 1

Bond length and Bond order:

• The distance between the centres of two covalently bound atoms is denoted as bond length.
• The number of bound electrons determines the length of the bond which is called the bond order.
• The stronger the attraction between the two atoms and the shorter the bond length, the higher the bond order.
• Bond order can be calculated as:

$\mathrm{Bond}\mathrm{order}=\frac{1}{2}(\mathrm{number}\mathrm{of}\mathrm{electrons}\mathrm{in}\mathrm{bonding}\mathrm{orbitals}-\mathrm{number}\mathrm{of}\mathrm{electrons}\mathrm{in}\mathrm{antibonding}\mathrm{orbitals})$

• The bond order of the dicarbon species ${\mathrm{C}}_2,{{\mathrm{C}}_2}^{+},{{\mathrm{C}}_2}^{-}$ can be calculated using the same formula.
  • Energy level diagram of ${\mathrm{C}}_2$ molecule:

• According to molecular orbital theory, ${\mathrm{C}}_2$​ molecule has one $\mathrm{\sigma }$ and one $\mathrm{\pi }$ bond.
• The M.O. electronic configuration is $\mathrm{\sigma }2{\mathrm{s}}^2{\mathrm{\sigma }}^{\ast }2{\mathrm{s}}^2\mathrm{\pi }2{\mathrm{px}}^2\mathrm{\pi }2{\mathrm{py}}^2$​
  $$\begin{gathered} \mathrm{Bond}\mathrm{order}=\frac{1}{2}(\mathrm{number}\mathrm{of}\mathrm{electrons}\mathrm{in}\mathrm{bonding}\mathrm{orbitals}-\mathrm{number}\mathrm{of}\mathrm{electrons}\mathrm{in}\mathrm{antibonding}\mathrm{orbitals}) \\ \mathrm{Bond}\mathrm{order}=\frac{6-2}{2}=2 \end{gathered}$$
• The bond order results show that the ${{\mathrm{C}}_2}^{-}$ has the highest number of valence electrons and the highest bond energy. Therefore, ${{\mathrm{C}}_2}^{-}$ supposedly has the shortest bond length.
• ${{\mathrm{C}}_2}^{+}$ has lesser bond energy than ${\mathrm{C}}_2$. Therefore, ${{\mathrm{C}}_2}^{+}$ has a longer bond length than ${\mathrm{C}}_2$.

So, the order starting with the highest bond length is ${{\mathrm{C}}_2}^{-}<{\mathrm{C}}_2<{{\mathrm{C}}_2}^{+}$