Bond Parameters — Length, Angle, Enthalpy & Order

Bond Length & Bond Angle

Every covalent bond can be described by a set of measurable numbers — the bond parameters. Let's meet them one by one.

Bond length

Definition: Bond length is the equilibrium distance between the nuclei of two bonded atoms in a molecule. It is measured in picometres (pm) or angstroms (Å); $1\ \text{Å} = 100\ \text{pm}$.

Bond length is roughly the sum of the covalent radii of the two atoms: $d_{A-B} \approx r_A + r_B$. Key trends:

• Higher bond order → shorter bond. $C \equiv C\ (120\ pm) < C=C\ (134\ pm) < C-C\ (154\ pm)$.
• Larger atoms → longer bonds. $H-F < H-Cl < H-Br < H-I$.

Bond angle

Definition: The bond angle is the angle between two adjacent bonds (more precisely, between the orbitals containing the bonding electron pairs) at the central atom.

Examples: $H_2O$ = $104.5°$, $NH_3$ = $107°$, $CH_4$ = $109.5°$, $CO_2$ = $180°$. Bond angle is decided by hybridisation and by lone-pair repulsions (Section 8).

[NEET Important] Bond length decreases as bond order increases. Remember the trio $154/134/120$ pm for C–C/C=C/C≡C.

Bond Enthalpy (Bond Energy)

Definition: Bond enthalpy is the amount of energy required to break one mole of bonds of a particular type in the gaseous state, to give gaseous atoms. $$H_2(g) \rightarrow 2H(g) \qquad \Delta_{bond}H = +435.8\ \text{kJ mol}^{-1}$$

Key points:

• It is always positive (bond breaking is endothermic).
• Larger bond order → larger bond enthalpy. $C-C\ (348) < C=C\ (610) < C \equiv C\ (839)$ kJ/mol. Note the increase is not simply additive.
• For polyatomic molecules with several identical bonds (e.g., the four C–H bonds in $CH_4$), we use the mean (average) bond enthalpy — the total atomisation energy divided by the number of bonds.

Using bond enthalpies to estimate reaction enthalpy

$$\Delta_r H = \sum (\text{bond enthalpies of bonds broken}) - \sum (\text{bond enthalpies of bonds formed})$$

[JEE Tip] 'Bonds broken minus bonds formed' — get this direction right. A negative answer means an exothermic reaction.

Bond Order & Its Consequences

Definition (Lewis sense): Bond order is the number of bonds (shared electron pairs) between two atoms. $H_2$ has bond order 1, $O_2$ has 2, $N_2$ has 3. (In MO theory we'll get a more general definition.)

Bond order ties all the parameters together:

• Higher bond order → shorter bond length.
• Higher bond order → greater bond enthalpy (stronger bond).
• Isoelectronic species have the same bond order, e.g. $CO$, $N_2$ and $NO^+$ all have bond order 3.

Fractional bond order from resonance

When a molecule is a resonance hybrid, the bond order can be fractional. For example, in ozone each O–O bond has bond order $1.5$ (average of a single and a double bond); in the carbonate ion each C–O bond has bond order $4/3 \approx 1.33$.

Key Point: $\text{Average bond order} = \dfrac{\text{total number of bonds among the resonating positions}}{\text{number of equivalent bond positions}}$.

[JEE Tip] Bond order of $CO_3^{2-}$ = (1 double + 2 single)/3 positions = 4 bonds / 3 = 1.33.

Factors Affecting Bond Parameters

Bringing it together, here is what changes a bond's length, strength and angle:

1. Bond order — the dominant factor: ↑ order ⇒ ↓ length, ↑ enthalpy.
2. Size of bonded atoms — bigger atoms ⇒ longer, weaker bonds.
3. Hybridisation / s-character — more s-character pulls electrons closer to the nucleus, shortening the bond. $sp$ (50% s) bonds are shorter than $sp^2$ (33%) which are shorter than $sp^3$ (25%). This is why the C–H bond is shortest in ethyne.
4. Electronegativity difference — affects polarity and slightly the length.
5. Resonance — equalises and shortens bonds relative to a pure single bond.
6. Lone pairs on the central atom — compress bond angles (lone-pair repulsion > bond-pair repulsion). This is why the angle falls $109.5° \to 107° \to 104.5°$ from $CH_4 \to NH_3 \to H_2O$.

Key Point: Greater s-character ⇒ shorter, stronger bonds and larger bond angles.

[NEET Important] The bond-angle order $CH_4 (109.5°) > NH_3 (107°) > H_2O (104.5°)$ is explained by increasing lone-pair repulsion.

Solved Examples

Example 1: Comparing bond lengths by bond order

Arrange the C–C, C=C and C≡C bonds in increasing order of bond length.

Solution:

1. Bond orders: C–C (1), C=C (2), C≡C (3).
2. Rule: higher bond order → shorter bond.
3. Order (increasing length): $C \equiv C\ (120\,pm) < C=C\ (134\,pm) < C-C\ (154\,pm)$.

Takeaway: Bond length is inversely related to bond order.

Example 2: Reaction enthalpy from bond enthalpies

Estimate $\Delta_r H$ for $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$ given bond enthalpies $H-H = 435$, $Cl-Cl = 242$, $H-Cl = 431$ kJ/mol.

Solution:

1. Formula: $\Delta_r H = \sum BE_{broken} - \sum BE_{formed}$.
2. Bonds broken: $H-H + Cl-Cl = 435 + 242 = 677$.
3. Bonds formed: $2 \times (H-Cl) = 2 \times 431 = 862$.
4. Compute: $\Delta_r H = 677 - 862 = -185$ kJ/mol.

Takeaway: Negative value ⇒ exothermic reaction. Always use 'broken − formed'.

Example 3: Mean bond enthalpy of C–H in methane

The enthalpy of atomisation of $CH_4$ is $1665$ kJ/mol. Find the mean C–H bond enthalpy.

Solution:

1. Atomisation breaks all four C–H bonds: $CH_4 \rightarrow C + 4H$.
2. Mean bond enthalpy: $1665 / 4 = 416.25$ kJ/mol.
3. Conclusion: mean C–H bond enthalpy $\approx 416$ kJ/mol.

Takeaway: Mean bond enthalpy = total atomisation energy ÷ number of identical bonds.

Example 4: Bond order of the carbonate ion

Find the average C–O bond order in $CO_3^{2-}$.

Solution:

1. Resonance: one C=O double bond and two C–O single bonds, delocalised over three equivalent positions.
2. Total bonds: $2 (\text{from double}) + 1 + 1 = 4$ bonds shared over 3 positions.
3. Average bond order: $4/3 \approx 1.33$.

Takeaway: Resonance gives fractional, equal bond orders — here 1.33 for every C–O bond.

Example 5: s-character and bond length

Why is the C–H bond shorter in ethyne than in ethane?

Solution:

1. Hybridisation: ethyne carbon is $sp$ (50% s); ethane carbon is $sp^3$ (25% s).
2. More s-character pulls bonding electrons closer to the nucleus.
3. Result: the $sp$ C–H bond is shorter (and stronger) than the $sp^3$ C–H bond.

Takeaway: ↑ s-character ⇒ shorter, stronger bonds.

Example 6: Explaining the bond-angle trend

Why does the bond angle decrease in the order $CH_4 > NH_3 > H_2O$?

Solution:

1. All have 4 electron pairs around the central atom ($sp^3$).
2. Lone pairs: $CH_4$ (0), $NH_3$ (1), $H_2O$ (2).
3. Lone-pair repulsion is stronger than bond-pair repulsion, compressing the bond angle.
4. Result: $109.5° \to 107° \to 104.5°$ as lone pairs increase.

Takeaway: More lone pairs on the central atom → smaller bond angle.

Example 7: Isoelectronic species and bond order

What is the bond order in $CO$, and name two isoelectronic species with the same bond order.

Solution:

1. CO has 10 valence electrons → triple bond, bond order 3.
2. Isoelectronic species (also 10 valence electrons): $N_2$ and $NO^+$ (and $CN^-$).
3. All have bond order 3 and similar bond lengths.

Takeaway: Isoelectronic species share the same bond order and similar bond parameters.

Example 8: Bond enthalpy and reaction direction

For $N_2 + 3H_2 \rightarrow 2NH_3$, given $N \equiv N = 945$, $H-H = 436$, $N-H = 391$ kJ/mol, estimate $\Delta_r H$.

Solution:

1. Bonds broken: $945 + 3(436) = 946 + 1308 = 2254$.
2. Bonds formed: $6 \times (N-H) = 6 \times 391 = 2346$ (each $NH_3$ has 3 N–H bonds, ×2).
3. Compute: $\Delta_r H = 2254 - 2346 = -92$ kJ/mol.

Takeaway: The negative value confirms ammonia synthesis is exothermic.

Example 9: Bond length order in hydrogen halides

Arrange $HF, HCl, HBr, HI$ in increasing order of bond length.

Solution:

1. Halogen size increases down the group: $F < Cl < Br < I$.
2. Bond length $\approx r_H + r_X$, so it increases with halogen size.
3. Order: $HF < HCl < HBr < HI$.

Takeaway: Larger atoms → longer bonds; bond length follows atomic size down a group.

Example 10: Bond order from resonance in nitrate

Find the average N–O bond order in $NO_3^-.$

Solution:

1. Resonance: one N=O double + two N–O single bonds over 3 equivalent positions.
2. Total bonds: $2 + 1 + 1 = 4$ over 3 positions.
3. Average bond order: $4/3 \approx 1.33$.

Takeaway: Like carbonate, nitrate has equal N–O bonds of order 1.33.

Example 11: Strongest bond by enthalpy

Given bond enthalpies $C-C = 348$, $C=C = 610$, $C \equiv C = 839$ kJ/mol, which bond is the strongest and is the rise additive?

Solution:

1. Strongest: $C \equiv C$ (839 kJ/mol) — largest enthalpy.
2. Additivity check: if additive, double would be $2 \times 348 = 696$ (observed 610) and triple $3 \times 348 = 1044$ (observed 839).
3. Conclusion: the increase is NOT simply additive — π bonds are weaker than σ bonds.

Takeaway: Higher bond order = stronger bond, but each successive π bond adds less than a full σ bond.

Example 12: Predicting bond angle from s-character

Why is the bond angle in $C_2H_2$ ($180°$) larger than in $C_2H_4$ ($120°$)?

Solution:

1. Hybridisation: ethyne C is $sp$ (50% s), ethene C is $sp^2$ (33% s).
2. More s-character spreads orbitals further apart → larger bond angle.
3. Result: $sp$ → $180°$ (linear), $sp^2$ → $120°$ (trigonal planar).

Takeaway: Bond angle increases with s-character: $sp\ (180°) > sp^2\ (120°) > sp^3\ (109.5°)$.