Question

Find the number of sigma bonds in ${P_4}{O_{10}}$.

Answer 1

Hint:
The bond that is formed by the axial overlapping of half-filled atomic orbitals is called a sigma bond. A sigma bond is stronger due to a larger overlapping of atomic orbitals. A pi bond is formed by the side wise or the lateral overlapping of half-filled atomic orbitals. This bond is relatively weaker due to less overlapping of atomic orbitals. In a multiple bond, one of the bonds is a sigma bond the other bonds are pi bonds.

Step-by-step explanation:
Step 1:
Phosphorous is a group 15 element. The first member of group 15 i.e. nitrogen forms oxide whose structure is different from the oxides of the other elements as it has the ability to form $p\pi - p\pi$ multiple bonds.
Step 2:
The remaining elements of Group 15 i.e. phosphorus, arsenic, antimony and bismuth are unstable to form $p\pi - p\pi$ multiple bonds. Thus, their oxides usually possess a cage like structure.
Step 3:
In ${P_4}{O_{10}}$ each phosphorus atom is surrounded by four oxygen atoms. With one oxygen atom phosphorus forms a double bond and with the other three oxygen atoms it forms a single bond.
Thus, the total number of sigma bonds for each phosphorus atom is three for the three $P - O$ single bonds and one for the one $P = O$ double bond, so total of four sigma bonds.
Step 4:
There are four phosphorus atoms present in ${P_4}{O_{10}}$. So, the total number of sigma bonds present in the compound is $4 \times 4 = 16$ and the total number of $\pi bonds$ are $4 \times 1 = 4$.

Hence, ${P_4}{O_{10}}$ contains 16 sigma bonds.

Note: Phosphorus also has a tri-oxide ${P_2}{O_6}$ where all the $P - O$ bonds are sigma bonds.