Question

Which of the following overlaps gives σ− bond along x-axis as internuclear axis?
A.$\mathop P\nolimits_Z$and $\mathop P\nolimits_Z$
B.S and $\mathop P\nolimits_Z$
C.S and $\mathop P\nolimits_Y$
D.$d\mathop x\nolimits_2 - \mathop y\nolimits_2$and $d\mathop x\nolimits_2 - \mathop y\nolimits_2$

Answer 1

The sigma bond is formed by the end to end (head -on) overlap of bonding orbitals along the internuclear axis. This is called head on overlap or axial overlap. This can be formed by s-s overlapping that is when overlap of two half filled s-orbitals along the internuclear axis. It can also be formed between s-p orbital and also between p-p orbital but the overlapping should occur headways.

Complete step by step answer:
A.$\mathop P\nolimits_Z$ and $\mathop P\nolimits_Z$:- When we take x axis as internuclear axis these two orbitals can’t do sideways overlapping because these two lie in z axis. Thus they both do head on overlapping so form sigma bonds . Thus this option is correct.
B. S and $\mathop P\nolimits_Z$ :- As we take x axis as internuclear axis s can exist in any of the axis so do not form a bond in x direction. Hence this option is not correct.
C.S and $\mathop P\nolimits_Y$ :- s can again lie in any axis and not only in y. Thus it can not give sigma bonds only in x direction . So this option is also not correct.
D.$d\mathop x\nolimits_2 - \mathop y\nolimits_2$and $d\mathop x\nolimits_2 - \mathop y\nolimits_2$ :- This option is not correct as d orbitals do not take part in sigma bond formation.
Our required answer is A that is $\mathop P\nolimits_Z$and $\mathop P\nolimits_Z$.

Note:
Pi-bond is formed due to sideways overlapping of the orbital. When both orbital belongs to the same axis as $\mathop P\nolimits_Z$ $\mathop P\nolimits_Z$ they give a bond in the direction of the internuclear axis. But when both the orbitals do not belong to the same axis they do not give bond in the internuclear axis.