Question

In $P{O}_4^{3-}$ ion the formal change on each oxygen atom and $P-O$ bond order respectively are:

• A. $-0.75,1.25$
• B. $-0.75,1.0$
• C. $-0.75,0.6$
• D. $-3,1.25$

Answer 1

Answer: (A)

$-0.75,1.25$

In $P{O}_4^{-3}$, the formal charge on each oxygen atom and the $P-O$ bond order are −0.75, 1.25 respectively.

Bond order = $\frac{\text{Number of bonds}}{\text{Number of Resonating structures}}$ = $\frac{5}{4}$ = 1.25

In a given resonance structure, the O atom that forms double bond has formal charge of 0 and the remaining 3 O atoms have formal charge of -1 each.

In the resonance hybrid, a total of -3 charge is distributed over 4 O atoms. Thus the formal charge of each O atom is

$\frac{-3}{4}$ = −0.75.

Note:

To calculate the formal charge, the following formula is used.

Formal Charge = [Number of valence electrons on atom] – [non-bonded electrons + number of bonds]

For O atom that forms double bond with P atom,

Formal Charge = 6] – [4 +2] = 0.

For O atom that forms single bond with P atom,

Formal Charge = 6] – [6 +1] =− 1.