Question

Which of the following has $\text{p}\pi -\text{d}\pi$ bonding?
A. $\text{NO}_{3}^{-}$
B. $\text{SO}_{3}^{2-}$
C. $\text{BO}_{3}^{3-}$
D. $\text{CO}_{3}^{2-}$

Answer 1

The main concept behind this is not to check the hybridization and orbitals overlapping in all the options but to simply check if the elements contain d-orbitals or not. Because if the elements in the ionic species will not have d-orbitals then how come they form $\text{p}\pi -\text{d}\pi$ with one another. If the elements belong to ${{3}^{\text{rd}}}$ period, then it contains d-orbitals.

Complete step by step answer:
Let us check if the elements have d-orbitals or not, which can be easily found by writing the electronic configuration of the elements. If their electronic configuration has a third shell present then, they are likely to form $\text{p}\pi -\text{d}\pi$ bonding.

A. $\text{NO}_{3}^{-}$ : This compound is nitrate ion. It contains one nitrogen atom $\left( \text{N} \right)$ and three oxygen atoms $\left( \text{O} \right)$ in it. The atomic number of nitrogen is 7 then, its electronic configuration is $1{{\text{s}}^{2}}\text{2}{{\text{s}}^{2}}\text{2}{{\text{p}}^{3}}$. The atomic number of oxygen is 8 then, its electronic configuration is $1{{\text{s}}^{2}}\text{2}{{\text{s}}^{2}}\text{2}{{\text{p}}^{4}}$. The electrons in both the elements are filled in 1 and 2 shells only. So, it will not form $\text{p}\pi -\text{d}\pi$ bonding.

B. $\text{SO}_{3}^{2-}$: This compound is sulphite ion. It contains one sulphur atom $\left( \text{S} \right)$ and three oxygen atoms $\left( \text{O} \right)$ in it. The atomic number of sulphur is 16 then, its electronic configuration is $1{{\text{s}}^{2}}\text{2}{{\text{s}}^{2}}\text{2}{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{4}}$. The atomic number of oxygen is 8 then, its electronic configuration is $1{{\text{s}}^{2}}\text{2}{{\text{s}}^{2}}\text{2}{{\text{p}}^{4}}$. The electrons in sulphur are filled in 1, 2 and 3 shells. So, it will form $\text{p}\pi -\text{d}\pi$ bonding.

C. $\text{BO}_{3}^{3-}$: This compound is borate ion. It contains one boron atom $\left( \text{B} \right)$ and three oxygen atoms $\left( \text{O} \right)$ in it. The atomic number of boron is 5. The electronic configuration of boron is $1{{\text{s}}^{2}}\text{2}{{\text{s}}^{2}}\text{2}{{\text{p}}^{1}}$ . The atomic number of oxygen is 8 then, its electronic configuration is $1{{\text{s}}^{2}}\text{2}{{\text{s}}^{2}}\text{2}{{\text{p}}^{4}}$. The electrons in both the elements are filled in 1 and 2 shells only. So, it will not form $\text{p}\pi -\text{d}\pi$ bonding.

D. $\text{CO}_{3}^{2-}$: This compound is carbonate ion. It contains one carbon atom $\left( \text{C} \right)$ and three oxygen atoms $\left( \text{O} \right)$ in it. The atomic number of carbon is 6 then, its electronic configuration is $1{{\text{s}}^{2}}\text{2}{{\text{s}}^{2}}\text{2}{{\text{p}}^{2}}$. The atomic number of oxygen is 8 then, its electronic configuration is $1{{\text{s}}^{2}}\text{2}{{\text{s}}^{2}}\text{2}{{\text{p}}^{4}}$. The electrons in both the elements are filled in 1 and 2 shells only. So, it will not form $\text{p}\pi -\text{d}\pi$ bonding.
$\text{SO}_{3}^{2-}$ has $\text{p}\pi -\text{d}\pi$ bonding
So, the correct answer is “Option B”.

Note: The elements present in the third shell are said to have d-orbital, although they do not have electrons present in d-orbitals but, this in ground state only. At excited state, when compounds undergo hybridization through d-orbitals present in it and extend their octets, to form bonds. So, they tend to be called as third period elements that have d-orbitals.