Question

The $d\pi -p\pi$ bond present in:
(a)- $CO_{3}^{2-}$
(b)- $PO_{4}^{3-}$
(c)- $NO_{3}^{-}$
(d)- $NO_{2}^{-}$

Answer 1

$d\pi -p\pi$ means that the pi-bond is formed between the p-orbital of the one element and the d-orbital of the other element. Those elements that don't have vacant d-orbital, cannot form the $d\pi -p\pi$ bond.

Complete answer: $d\pi -p\pi$ means that the pi-bond is formed between the p-orbital of the one element and the d-orbital of the other element. Those elements that don't have vacant d-orbital, cannot form the $d\pi -p\pi$ bond. So, the elements that belong to period 2, don't have vacant d-orbital and only have vacant p-orbital, therefore, they can form $p\pi -p\pi$ multiple bonds. But the elements of period 3 or higher have vacant d-orbital and are able to form $d\pi -p\pi$ multiple bonds.
Let us study all the compounds in the given options above:
(a)- $CO_{3}^{2-}$
This compound contains two elements, i.e., a carbon atom and an oxygen element. Both the elements belong to period 2, therefore, it can form $p\pi -p\pi$ multiple bonds.
(b)- $PO_{4}^{3-}$
This compound contains two elements, i.e., phosphorus atoms and oxygen elements. Oxygen element belongs to period 2 and phosphorus element belongs to period 3, therefore, it can form $d\pi -p\pi$ multiple bonds.
(c)- $NO_{3}^{-}$
This compound contains two elements, i.e., a nitrogen atom and an oxygen element. Both the elements belong to period 2, therefore, it can form $p\pi -p\pi$ multiple bonds.
(d)- $NO_{2}^{-}$
This compound contains two elements, i.e., a nitrogen atom and an oxygen element. Both the elements belong to period 2, therefore, it can form $p\pi -p\pi$ multiple bonds.

Therefore, the correct answer is an option (b).

Note: Since the elements of a period don't have vacant d-orbital, therefore, the covalency of these elements cannot exceed 4, but the elements of period 3 can exceed their covalency 4.