Q.In PO4^3- ion the formal charge on the oxygen atom of P–O bond is

Concept and Intuition

Resonance structures are hypothetical representations of a molecule's electron distribution, where electrons are delocalized over multiple atoms. The true structure, called the resonance hybrid, is a weighted average of these contributing structures. The more stable a resonance structure is, the more it contributes to the overall resonance hybrid, and the more stable the hybrid itself becomes.

To determine the relative stability of different resonance structures, we follow a set of rules, listed in decreasing order of importance:

1. Complete Octets: Structures where all atoms (especially carbon, nitrogen, oxygen, and fluorine) have a complete octet (8 valence electrons) are significantly more stable than those with incomplete octets. This is the most important rule.
2. Maximum Covalent Bonds: Among structures that satisfy the octet rule, those with a greater number of covalent bonds are more stable. More bonds generally mean stronger overall attraction and greater stability.
3. Minimum Charge Separation: Structures with less charge separation (fewer formal charges, or charges closer together) are more stable. Creating and separating charges requires energy, making such structures less favorable.
4. Favorable Charge Placement: Among structures with similar charge separation, those where negative charges reside on more electronegative atoms (like oxygen or nitrogen) and positive charges reside on less electronegative atoms (like carbon) are more stable.
5. Like Charges: Structures with like charges on adjacent atoms are highly unstable and contribute very little to the resonance hybrid.

Step-by-Step Analysis

Let's consider three common types of resonance structures that illustrate these rules. We will assume the given structures (I, II, III) correspond to these types, as this is a standard pattern for such problems.

Hypothetical Structures:

Let's consider a molecule that can form the following resonance structures:

Structure I:

• Octets:
  • Carbon: 2 C-H bonds + 2 C=N bonds = 4 bonds (8 electrons). Octet complete.
  • Nitrogen: 2 C=N bonds + 2 N=O bonds = 4 bonds (8 electrons). Octet complete.
  • Oxygen: 2 N=O bonds + 2 lone pairs = 4 bonds + 2 lone pairs (8 electrons). Octet complete.
• Formal Charges: All atoms have a formal charge of 0.
• Covalent Bonds: 5 covalent bonds (2 C-H, 1 C=N, 1 N=O).
• Summary: All atoms have complete octets, and there is no charge separation. This is generally the most stable type of resonance structure.

Structure II:

• Octets:
  • Carbon: 2 C-H bonds + 1 C-N bond = 3 bonds (6 electrons). Incomplete octet.
  • Nitrogen: 1 C-N bond + 2 N=O bonds = 3 bonds + 1 lone pair (8 electrons). Octet complete.
  • Oxygen: 1 N=O bond + 3 lone pairs = 1 bond + 3 lone pairs (8 electrons). Octet complete.
• Formal Charges: Carbon has a +1 charge, Oxygen has a -1 charge.
• Covalent Bonds: 4 covalent bonds (2 C-H, 1 C-N, 1 N=O).
• Summary: Carbon has an incomplete octet. There is charge separation. This type of structure is highly unstable.

Structure III:

• Octets: …

Method. For a xenon compound the lone pairs on Xe follow from its steric number: count the $\sigma$-bonds Xe forms (a Xe=O double bond occupies one $\sigma$ position), then

Representative values: $\text{XeF}_2$ has 3 lone pairs and $\text{XeF}_4$ has 2, while $\text{XeF}_6$, $\text{XeOF}_4$, $\text{XeO}_3$ and $\text{XeO}_2\text{F}_2$ each have 1 lone pair on Xe. The molecules are compared on this count, and the correct group is the one whose three members share the same value. …

Concept and Intuition

Lone pairs are pairs of valence electrons that are not involved in bonding. To count them, we need the correct Lewis structure. For $\mathrm{HClO_4}$, a common pitfall is forgetting that chlorine can expand its octet (it is in period 3) and that the molecule has a specific connectivity: H is bonded to O, and that O is bonded to Cl, with the other three O atoms double-bonded to Cl. Once we draw the structure properly, counting lone pairs becomes straightforward.

Step-by-Step Solution

1. Determine total valence electrons

   • H: 1 valence electron
   • Cl: 7 valence electrons
   • O (×4): 4 × 6 = 24 valence electrons
   • Total = 1 + 7 + 24 = 32 valence electrons.

2. Establish connectivity

   In perchloric acid, the hydrogen is attached to one oxygen (forming an –OH group), and that oxygen is single-bonded to chlorine. The other three oxygens are double-bonded to chlorine. So the skeleton is:

3. Place single bonds and count used electrons

   • H–O: 1 bond (2 electrons)
   • O–Cl: 1 bond (2 electrons)
   • Three Cl=O double bonds: each double bond uses 4 electrons → 3 × 4 = 12 electrons
   • Total used in bonds so far: 2 + 2 + 12 = 16 electrons.

4. Distribute remaining electrons as lone pairs

   Remaining electrons: 32 − 16 = 16 electrons = 8 lone pairs.

   • Each of the three double-bonded oxygens already has 2 bonds (4 electrons), so they need 4 more electrons each to complete an octet → 3 × 2 lone pairs = 6 lone pairs.
   • The oxygen in the –OH group has 2 bonds (one to H, one to Cl), so it needs 4 more electrons → 2 lone pairs.
   • Chlorine already has 4 bonds (one single, three double) = 8 electrons around it, so it has no lone pairs. …

Concept & Intuition

Bond length is inversely related to bond order — the more electron pairs shared between two atoms, the stronger and shorter the bond. For molecules with resonance, the bond order is the average over contributing structures. So to compare C–O bond lengths, we first determine the C–O bond order in each species.

1. Carbon monoxide (CO)

   • Lewis structure: C≡O (triple bond) with a lone pair on each atom.
   • Bond order = 3.
   • This is the shortest C–O bond among common carbon oxides.

2. Carbon dioxide (CO₂)

   • Lewis structure: O=C=O (two double bonds).
   • Each C–O bond is a double bond, so bond order = 2.
   • Longer than a triple bond, shorter than a single bond.

3. Carbonate ion (CO₃²⁻)

   • Three equivalent resonance structures: one C=O double bond and two C–O⁻ single bonds.
   • The actual structure is an average: each C–O bond has bond order = (1 + 1 + 2)/3 = 4/3 ≈ 1.33.
   • This is the longest C–O bond among the three. …