Q33 / 180 · COMEDK 2022 · Chemistry
The $\mathrm{P}-\mathrm{O}$ bond order in $\mathrm{PO}_4^{3-}$ is
- A1
- B1.5
- C1.45
- Correct answerD1.25
Solution
The phosphate ion $\mathrm{PO}_4^{3-}$ has a central phosphorus atom bonded to four oxygen atoms. The total number of valence electrons is $5 + 4 \times 6 + 3 = 32$.
The Lewis structure consists of a central phosphorus atom with four single bonds to oxygen atoms, where each oxygen atom carries a formal charge of $-1$ except for one double bond to one oxygen atom to satisfy the octet rule for phosphorus. However, resonance structures exist where the double bond is delocalized over all four $\mathrm{P}-\mathrm{O}$ bonds.
The total number of bonds in the resonance hybrid is $4$ single bonds plus $1$ double bond, which equals $5$ bonds distributed over $4$ positions.
The bond order is calculated as the total number of bonds divided by the number of resonance structures or the number of bond positions. Thus, Bond Order = $\dfrac{\text{Total number of bonds}}{\text{Number of P-O linkages}} = \dfrac{5}{4} = 1.25$.
Answer: 1.25
The Lewis structure consists of a central phosphorus atom with four single bonds to oxygen atoms, where each oxygen atom carries a formal charge of $-1$ except for one double bond to one oxygen atom to satisfy the octet rule for phosphorus. However, resonance structures exist where the double bond is delocalized over all four $\mathrm{P}-\mathrm{O}$ bonds.
The total number of bonds in the resonance hybrid is $4$ single bonds plus $1$ double bond, which equals $5$ bonds distributed over $4$ positions.
The bond order is calculated as the total number of bonds divided by the number of resonance structures or the number of bond positions. Thus, Bond Order = $\dfrac{\text{Total number of bonds}}{\text{Number of P-O linkages}} = \dfrac{5}{4} = 1.25$.
Answer: 1.25
Hence, the correct answer is option: 'D'.