Question

The hybrid state of ${\text{S}}$ in $S{O_3}$ is similar to that of:
A) ${\text{C}}$ in ${C_2}{H_2}$
B) ${\text{C}}$ in ${C_2}{H_4}$
C) ${\text{C}}$ in $C{H_4}$
D) ${\text{C}}$ in $C{O_2}$

Answer 1

The hybrid state of an atom is the hybridization state of that atom present in that molecule. First, discover the hybridization state of ${\text{S}}$ in the $S{O_3}$ molecule and then relate it with the given options for the correct choice.

Complete step by step answer:

1. First of all let's calculate the hybridization of $S{O_3}$ molecule and the hybrid state of sulfur in that molecule. The hybridization is equal to the sum of No. of valence electrons of the central atom, No. of monovalent electrons, Anionic charge or cationic charge divided by two.
2. This can be written in a formula representation as,
   $H = \dfrac{{{\text{No}}{\text{. of V}}{\text{.E}}{\text{. on central atom + No}}{\text{. of M}}{\text{.V}}{\text{. + Anionic / cationic charge}}}}{2}$
   Now, let's put the values for the molecule of $S{O_3}$ in the above formula,
   $H = \dfrac{{6 + 0 + 0}}{2} = 3$
   The sulfur is the central atom in the molecule $S{O_3}$ and has six valence electrons. The oxygen atom given is the divalent hence, the number of monovalent electrons is zero. There is no cationic charge or anionic charge present on the molecule hence the value is taken as zero.
3. Hence, the hybridization value of $S{O_3}$ molecule is three which shows the hybridization as $s{p^2}$ hybridization which means there is no sigma bond present in the structure and there is a double bond present.
4. Now while analyzing the given options where we need $s{p^2}$ hybridized carbon, we just need to find out the carbon which is $s{p^2}$ and possess a double bond. In the structure, ${C_2}{H_2}$ there is a triple bond present and hybridization as ${\text{sp}}$ which shows this option as an incorrect choice.
5. The structure ${C_2}{H_4}$ shows the presence of a double bond between the two carbon atoms which shows this option as a correct choice.
6. The structure $C{H_4}$ has only sigma bonds and no pi bond and carbon is $s{p^3}$ hybridized. And the structure $C{O_2}$ shows two oxygen atoms attached to carbon by double bonds which give the hybridization as ${\text{sp}}$. Hence, this option is also incorrect.
   Therefore, the hybrid state of ${\text{S}}$ in $S{O_3}$ is similar to that of ${\text{C}}$ in ${C_2}{H_4}$ which shows the option B as the correct choice.

Note:
While calculating the hybridization simply analyze the structure of the molecule. If the desired atom is attached to the single bond then it is $s{p^3}$ hybridized, if attached to double bond then it is $s{p^2}$ hybridized and if attached to a triple bond then it is ${\text{sp}}$ hybridized.