Write the electronic configurations and calculate the bond orders of H_(2)^(+) , H_2 and He_2 Why the bond length in H_(2)^(+) is longer than that in H_2 ?

Write the electronic configurations and calculate the bond orders of H

1.	Write the electronic configurations and calculate the bond orders of H_(2)^(+) , H_2 and He_2 Why the bond length in H_(2)^(+) is longer than that in H_2 ?
Answer» Solution :`{: ("Species""""Electronic configuration"""Nb""Na"""Bond order"),(H_(2)^(+)""(SIGMA1S)^(1)""1""0""0.5),(H_(2)""(sigma1s)^(2)""2""0""1),(He_(2)""(sigma1s)^(2)(sigma**1s)^(2)""2""2""0):}` Then bond length in `H_(2)^(+)` is longer than in `H_(2)` because in `H_(2)^(+)` only ONE ELECTRON is present to SHIELD the two nuclei from mutual repulsion. In `H_(2)` there are two electrons to hold the two nuclei, thus the nuclear repulsion is less than that in `H_(2)^(+)`. Hence nuclear separation in `H_(2)^(+)` is more thanin `H_(2)`.