The total number of sigma bonds present in P₄O₁₀ are:

The correct option is 16.

To determine the total number of σ (sigma) bonds present in a molecule of tetraphosphorus decaoxide ($P_4{O}_{10}$), let us analyze its cage-like molecular structure:
1. Each of the four phosphorus (P) atoms in $P_4{O}_{10}$ is located at the vertices of a tetrahedron.
2. There are two types of oxygen (O) atoms in the molecule:
- Bridging oxygen atoms (P-O-P bonds): There are 6 bridging oxygen atoms, each forming two single covalent σ bonds between two phosphorus atoms. This contributes:

$6\times 2=12\text{ σ bonds}$

- Terminal oxygen atoms (P=O bonds): Each of the 4 phosphorus atoms is also bonded to one terminal oxygen atom via a double bond. A double bond consists of one σ bond and one π bond. Therefore, the 4 terminal P=O double bonds contribute:

$4\times 1=4\text{ σ bonds}$

Adding these together, the total number of σ bonds in $P_4{O}_{10}$ is: