Question

According to octet rule, $S{{O}_{3}}$ contains_______ dative bonds
A.1
B. 2
C.3
D.4

Answer 1

In terms of electron-counting method, sulphur atom has an oxidation state of +6 and a formal charge of 0. According to the Lewis structure $S{{O}_{3}}$ consists of S=O double bond and two S–O dative bonds without utilizing d-orbitals.

Complete answer:
Let us know about octet rule, to answer this question. According to the Octet rule, all the elements tend to form their bonds in such a way that they have 8 electrons in their outermost shell. The outermost shell can contain a maximum of 8 electrons and if an element is able to achieve it, then the element attains maximum stability and it is called noble gas configuration. So, during formation of bonds, sharing of electrons occurs so that both the participating elements fill up their outermost or valence shell.

This is the structure of $S{{O}_{3}}$ where each element achieves octet stability
Valence electrons in Sulphur(S)=6
Total no of valence electrons=$6+3\times 6=24$
According to the octet rule we can say that the Lewis structure consist of an double bond (S=O) and two dative bonds to complete its octet as:
Two dative bonds , 2 bond$\times$2electrons = 4
One double bond , 2$\times$2electrons = 4
together containing 8 electrons.
Structure of the gas state $S{{O}_{3}}$ is a trigonal planar molecule of ${{D}_{3}}H$ symmetry, as predicted by the famous VSEPR theory. The $S{{O}_{3}}$ belongs to ${{D}_{3}}H$ point group.
The electrical dipole moment of the gaseous sulphur trioxide is zero. This is a consequence of ${{120}^{0}}$ angle between the S-O bonds.
Sulphur trioxide can be found as colorless to white crystalline solid which will fume in air as colorless liquid and gas. It has a molecular weight of about $8.066g\,mo{{l}^{-1}}$. It has a density of $1.92g\,c{{c}^{-1}}$in liquid form.
Melting point of Sulphur trioxide is ${{16.9}^{0}}C$(${{62.4}^{0}}F$; 290.0 K) and boiling point is ${{45}^{0}}C$ (${{113}^{0}}F$; 318 K)

Note:
In $S{{O}_{3}}$ there are a total of 6 covalent bonds in which there are $3-\sigma$bonds and $3-\pi$ bonds. There are no co-ordinate covalent bonds present because no bond in $S{{O}_{3}}$ is formed by donating an electron pair. So, the total number of coordinate bonds present in $S{{O}_{3}}$ molecule is zero.