Question

Choose the correct order of bond strength by overlapping of atomic orbitals
 (A) $1s-1s>1s-2s>1s-2p$
 (B) $2s-2s>2s-2p>2p-2p$
 (C) $2s-2p>2s-2s>2p-2p$
 (D) $1s-1s>1s-2p>1s-2s$

Answer 1

Hint: When the principal quantum number n is less, then overlapping is stronger. The extent of overlap is higher for s orbital than for p orbital. Higher is the extent of overlap, stronger is the bond formed.

Complete step by step answer:
- As we know, the atoms combine by colliding with each other. In this process, two atoms come very close to each other and they penetrate each other's orbital and form a new hybridized orbital. The bonding pair of electrons reside in this new hybridized orbital. This orbital has lower energy than the atomic orbital and hence it is stable. This entire process is known as orbital overlap.
 - The extent of overlap depends on several parameters. In general, we can say that, the stronger is the bond formed between the two atoms, greater the overlap.
- The bond strength mainly depends upon two factors. They are the effectiveness of overlapping (Angular concentration) and extent of overlapping (size of lobe). Bond strength is directly proportional to the percentage s character. As the s character increases, bond strength of orbitals will also increase.
-Thus, $s-s$ orbital overlapping will have a higher bond strength. $s-s$ overlap is the mutual overlap between two half filled s orbitals of two atoms. The spherical shape of s orbital and overlapping taking place in all directions to some extent makes the newly formed $s-s$ bond non-directional.
-Another important thing that matters in the bond strength by overlapping orbitals is the principal quantum number. In a 1s orbital 1 is the principal quantum number. As the principal quantum number increases the extent of overlapping would decrease.
Hence by taking this factor into account, we can eliminate options B and C. Because 1s orbital would have a higher bond strength. The remaining options are A and D. When one set is totally non-directional and n is also different, then the bond with directional nature is stronger and this is option (D).
So, the correct answer is “Option D”.

Note: It should be noted that bond strength is maximum in $s-s$ overlap because here the orbital is closest to the nucleus and thus are the electrons close to it making it more stable. The extent of bond strength follows the order $s-s>s-p>p-p$.