Question

The types of bonds present in sulphuric anhydride are
A. ${{3\sigma }}$ and ${{3}}{{{p}}_{{\pi }}} - {{{d}}_{{\pi }}}$
B. ${{3\sigma }}$, ${\text{1}}\,{{{p}}_{{\pi }}} - {{\text{p}}_{{\pi }}}$ and ${\text{2}}{{\text{p}}_{{\pi }}} - {{\text{d}}_{{\pi }}}$
C. ${{2\sigma }}$ and ${{3}}{{{p}}_{{\pi }}} - {{\text{d}}_{{\pi }}}$
D. ${{2\sigma }}$ and $2p_{\pi}- d_{\pi}$

Answer 1

Hint: To determine the answer we should know about hybridization. First we will draw the Lewis structure and by using VSEPR theory we will determine the hybridization to determine the number of sigma and pi bonds. We will also draw the hybridization to determine the orbitals forming sigma and pi bonds.

Complete step-by-step answer:
The sulphur trioxide is known as sulphuric anhydride. The chemical formula of sulphuric anhydride is ${\text{S}}{{\text{O}}_{\text{3}}}$.

We will write the Lewis structure for this we will write the basic structure. Then we will arrange all the valance electrons around each atom to complete the octet. According to valence shell electron pair repulsion theory, the electron pairs get arranged around the central to minimize the repulsion giving a specific geometry. The hybridization and geometry is decided based on the number of sigma and lone pairs around the central atom.
The Lewis structure of ${\text{S}}{{\text{O}}_{\text{3}}}$is as follows:
Total valence electrons are as follows:
$ = \,\left( {6 \times 1} \right) + \left( {6 \times 3} \right)$
$ = \,24$

The sigma electron pairs around the sulphur atom is three. So, the hybridization and geometry of ${\text{S}}{{\text{O}}_{\text{3}}}$is ${\text{s}}{{\text{p}}^{\text{2}}}$ and trigonal planar.

So, sulphuric anhydride has three sigma and three pi bonds.
The hybridization in ${\text{S}}{{\text{O}}_{\text{3}}}$ is shown as follows:

The valence configuration of sulphur is, ${\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{4}}}{\text{3d}}$ .
The valence configuration of oxygen anhydride is, ${\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{4}}}$ .
One s and two p-orbitals of sulphur combine to form three ${\text{s}}{{\text{p}}^{\text{2}}}$ hybridised orbitals. Two electrons one form s and one from p-orbital, transfer into d-orbital. Three ${\text{s}}{{\text{p}}^{\text{2}}}$ hybridised orbitals forms three sigma bonds with oxygen atoms. The next p-orbital forms one pi bond. The two d-orbital form two remaining two pi bonds.
So, the types of bonds present in sulphuric anhydride are ${{3\sigma }}$, ${\text{1}}\,{{\text{p}}_{{\pi }}} - {{{p}}_{{\pi }}}$ and ${\text{2}}{{{p}}_{{\pi }}} - {{\text{d}}_{{\pi }}}$.

Therefore, option (B) ${{3\sigma }}$, ${\text{1}}\,{{{p}}_{{\pi }}} - {{\text{p}}_{{\pi }}}$ and ${\text{2}}{{\text{p}}_{{\pi }}} - {{\text{d}}_{{\pi }}}$ is correct.

Note: Hybridised orbitals form sigma bonds only. So, the number of hybridised orbitals are equal to the number of sigma bonds. s-orbital always forms a sigma bond only. The pi bonds are formed by p and d-orbitals. Each atom donates an electron for both sigma and pi bond. Hydride orbits have the same energy.