Question

The types of bonds present in the bonds of hydrogen sulphide molecule is:
A. $\sigma s - p$
B. $\sigma s - s$
C. $\sigma p - p$
D. $\sigma s{p^3} - s$

Answer 1

To solve this question, we must first find the orbitals in which the valence electrons of the given constituent atoms lie. On the basis of this information, we can determine the type of bond formed.Bonds formed in compounds can either be ionic or covalent in nature.

Complete step by step answer:
Before we move forward with the solution of the given question, let us first understand some important basic concepts.
In ionic bonds, there is a complete transfer of electrons to form ions, which then attract each other and then form bonds on the basis of their charge. On the other hand, covalent bonds are formed by sharing of electrons between the constituent atoms. Bond formation takes place to complete the octet state of atoms.
Now, in covalent bonds, the valence electrons present in the constituent atoms could be in any of the orbitals. So, when a single bond is formed, we denote that the given $\sigma$ bond is formed between the two orbitals. For example, if the valence electrons of the constituent atoms are present in the s orbital of both the atoms and these electrons are participating in bond formation, then we can say that the given single bond is $\sigma s - s$ .
In the case of hydrogen sulphide, the combining atoms are hydrogen and sulphur. The atomic numbers of hydrogen and sulphur are 1 and 16 respectively. Hence their electronic configurations are given as $1{s^1}$ and $1{s^2}2{s^2}2{p^6}3{s^2}3{p^4}$ respectively. We can see that the valence electron of hydrogen is in the s orbital and that of sulphur is in p orbital.
Hence the type of bond formed in hydrogen sulphide is $\sigma s - p$

Hence, Option A is the correct option.

Note:
Just like in the case of a single bond, as we consider it to be a $\sigma$ bond, in the case of a double or triple bond, we consider those bonds to be $\pi$ bonds. Hence, the naming of $\pi$ bonds is also done in a similar fashion to that of $\sigma$ bonds.