\(P_4O_{10}\) is the empirical formula for phosphorus pentoxide (tetraphosphorus decaoxide). Its structure consists of a tetrahedral arrangement of four phosphorus atoms, with each phosphorus atom bonded to one terminal oxygen atom (\(P=O\)) and three bridging oxygen atoms. These bridging oxygen atoms connect the phosphorus atoms, forming \(P-O-P\) linkages.
To determine the number of \(P-O-P\) linkages, consider the following:
* Each phosphorus atom is bonded to three bridging oxygen atoms. * Each bridging oxygen atom connects two phosphorus atoms.
Therefore, there are a total of 6 \(P-O-P\) linkages in the \(P_4O_{10}\) molecule. Imagine the four Phosphorus atoms forming a tetrahedron. Each edge of the tetrahedron is bridged by an Oxygen atom, hence the six P-O-P linkages, corresponding to the six edges of the tetrahedron.
Visualizing the structure:
Each P atom is tetrahedrally coordinated. There are four terminal oxygen atoms double bonded to the four P atoms. The remaining six oxygen atoms form bridges between the P atoms.
Therefore, the number of \(P-O-P\) linkages in \(P_4O_{10}\) is six.
Correct Answer:
Option 1: Six
Explanation:
P4O10 has a cage-like structure with six P-O-P linkages. Each of the four phosphorus atoms is connected to three other phosphorus atoms via oxygen bridges, resulting in six P-O-P linkages in total.