Question

The total number of sigma bonds in P$_4$O$_{10}$ is:

• A. 6
• B. 7
• C. 16
• D. 17

Answer 1

The Correct Option is C

The correct answer is: Option 3: 16

To determine the total number of sigma bonds in P₄O₁₀, we need to understand its structure.

The structure of P₄O₁₀ consists of a cage-like structure with four phosphorus atoms and ten oxygen atoms. Each phosphorus atom is bonded to four oxygen atoms. Six oxygen atoms bridge the phosphorus atoms, and four oxygen atoms are terminal, bonded to individual phosphorus atoms via a double bond.

Here's a breakdown of the bonds:

• Each of the 4 phosphorus atoms is bonded to 4 oxygen atoms.
• Each P atom is bonded to 3 bridging O atoms and 1 terminal O atom.
• The 6 bridging oxygen atoms form P-O-P bonds.
• The 4 terminal oxygen atoms form P=O bonds (one sigma and one pi bond).

Counting the sigma bonds:

• 6 P-O-P bonds (6 sigma bonds)
• 4 P=O bonds (4 sigma bonds)
• Each of the 4 P atoms also has 3 other P-O sigma bonds, so 4 * 3 = 12 P-O bonds. However, each of these bonds is shared with another P atom, so there are 6 bridging P-O bonds.
• 4 P=O double bonds, which have 4 sigma bonds.

Therefore, the total number of sigma bonds is 6 (bridging P-O) + 4 (terminal P=O) + 6 (from the other P-O bonds) = 16 sigma bonds.

The total number of sigma bonds is:

6 (P-O-P) + 4 (P=O) + 6(P-O) = 16