Chemistry · Chemical Bonding and Molecular Structure

Bond Parameters

Once a covalent bond exists, it can be measured. Section 4.3 of the NCERT chapter defines the four quantities that describe any bond — its length, the angle it makes with neighbouring bonds, the energy needed to break it, and its order — and then extends the picture to resonance and bond polarity. These parameters are the most heavily tested ideas in the chapter: NEET routinely asks you to rank dipole moments, compare bond orders, and recognise resonance hybrids, so a firm grip on the underlying data and rules is essential.

Bond Length

Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule. It is measured experimentally by spectroscopic, X-ray diffraction and electron-diffraction techniques. Each atom of the bonded pair contributes to this distance, and in a covalent bond the contribution from each atom is called the covalent radius of that atom.

The covalent radius is measured approximately as the radius of an atom's core that is in contact with the core of an adjacent atom in a bonded situation; equivalently, it is half the distance between two like atoms joined by a covalent bond in the same molecule. This must not be confused with the van der Waals radius, which represents the overall size of the atom including its valence shell in a non-bonded situation — it is half the distance between two like atoms in separate molecules in a solid. For chlorine the covalent radius is $99\,\text{pm}$ while the van der Waals radius is $180\,\text{pm}$. For a bond $\ce{A-B}$ the bond length $R$ is simply the sum of the two covalent radii:

$$R = r_\text{A} + r_\text{B}$$

Bond typeCovalent bond length (pm)
O–H96
C–H107
N–O136
C–O143
C–N143
C–C154
C=O121
C=C133
C≡N116
C≡C120

Two patterns in this NCERT data (Table 4.2) are worth committing to memory. First, for the same pair of atoms the length contracts as the multiplicity rises: $\ce{C-C}$ at $154\,\text{pm}$ is longer than $\ce{C=C}$ at $133\,\text{pm}$, which is longer than $\ce{C#C}$ at $120\,\text{pm}$. Second, lengths in real molecules track these reference values closely, as Table 4.3 below shows for diatomics.

MoleculeBond length (pm)
$\ce{H2}$ (H–H)74
$\ce{F2}$ (F–F)144
$\ce{Cl2}$ (Cl–Cl)199
$\ce{Br2}$ (Br–Br)228
$\ce{I2}$ (I–I)267
$\ce{N2}$ (N≡N)109
$\ce{O2}$ (O=O)121
$\ce{HF}$ (H–F)92
$\ce{HCl}$ (H–Cl)127
$\ce{HBr}$ (H–Br)141
$\ce{HI}$ (H–I)160

Bond Angle

Bond angle is defined as the angle between the orbitals containing bonding electron pairs around the central atom in a molecule or complex ion. It is expressed in degrees and is determined experimentally by spectroscopic methods. The bond angle gives an idea of how the orbitals are distributed around the central atom, and hence it helps in determining the shape of the molecule.

A familiar example is the $\ce{H-O-H}$ bond angle in water, which is $104.5^\circ$. The same comparison appears in NEET stems: in moving from $\ce{CH4}$ to $\ce{NH3}$ to $\ce{H2O}$ the bond angle falls from $109.5^\circ$ to $107^\circ$ to $104.5^\circ$, because the number of lone pairs on the central atom increases and lone-pair repulsions squeeze the bonding pairs closer together. The mechanism behind these shapes belongs to a sibling topic, but the angles themselves are bond parameters.

Go deeper

Bond angles are predicted by counting electron pairs. See VSEPR Theory for how lone pairs distort the ideal geometry of $\ce{NH3}$ and $\ce{H2O}$.

Bond Enthalpy

Bond enthalpy is the amount of energy required to break one mole of bonds of a particular type between two atoms in the gaseous state. Its unit is $\text{kJ mol}^{-1}$. The larger the bond dissociation enthalpy, the stronger the bond. For the hydrogen molecule the H–H bond enthalpy is $435.8\,\text{kJ mol}^{-1}$:

$$\ce{H2(g) -> H(g) + H(g)} \qquad \Delta_a H = 435.8\ \text{kJ mol}^{-1}$$

Multiple bonds need correspondingly more energy. Breaking the double bond in dioxygen requires $498\,\text{kJ mol}^{-1}$, while the triple bond in dinitrogen requires $946.0\,\text{kJ mol}^{-1}$ — one of the highest bond enthalpies known for any diatomic molecule. A heteronuclear example is hydrogen chloride at $431.0\,\text{kJ mol}^{-1}$:

$$\ce{O2(g) -> 2O(g)} \qquad \Delta_a H = 498\ \text{kJ mol}^{-1}$$ $$\ce{N2(g) -> 2N(g)} \qquad \Delta_a H = 946.0\ \text{kJ mol}^{-1}$$

NEET Trap

The same bond can need different energies

In a polyatomic molecule, breaking identical bonds one after another does not cost the same energy, because the chemical environment changes. In water, the first O–H bond requires $502\,\text{kJ mol}^{-1}$ but the second requires only $427\,\text{kJ mol}^{-1}$. Chemists therefore quote a mean (average) bond enthalpy, obtained by dividing the total dissociation enthalpy by the number of bonds broken.

For water: average O–H bond enthalpy $= \dfrac{502 + 427}{2} = 464.5\ \text{kJ mol}^{-1}$.

Bond Order

In the Lewis description of a covalent bond, the bond order is given by the number of bonds between the two atoms in a molecule. With one shared electron pair in $\ce{H2}$ the bond order is 1; with two shared pairs in $\ce{O2}$ it is 2; and with three shared pairs in $\ce{N2}$ it is 3. Carbon monoxide $\ce{CO}$ also has three shared electron pairs between carbon and oxygen, so its bond order is 3.

A useful shortcut for NEET is that isoelectronic molecules and ions have identical bond orders. Thus $\ce{F2}$ and the peroxide ion $\ce{O2^2-}$ both have bond order 1, while $\ce{N2}$, $\ce{CO}$ and $\ce{NO+}$ all have bond order 3. The fractional bond orders that appear in species such as $\ce{NO}$ (2.5) or $\ce{CN-}$ (3) come from molecular orbital counting, which is treated separately.

The Order–Length–Strength Rule

The three parameters above are not independent. The single most important correlation in this section, and a guaranteed source of statement-based questions, ties them together:

With an increase in bond order, bond enthalpy increases and bond length decreases.

More shared electron pairs pull the nuclei closer and bind them more tightly, so a higher-order bond is both shorter and stronger. The progression from $\ce{C-C}$ through $\ce{C=C}$ to $\ce{C#C}$ illustrates this cleanly, as does the carbon–carbon series across ethane, ethene and ethyne.

Schematic of how bond length and bond strength vary with bond order high low bond length bond enthalpy order 1 order 2 order 3 BOND ORDER →
Figure 1. As bond order rises, bond length (coral) falls while bond enthalpy (teal) climbs — the qualitative trend behind the single-bond / double-bond / triple-bond series. Schematic, not to scale.

Resonance Structures

A single Lewis structure is often inadequate to represent a molecule in conformity with its experimentally determined parameters. The classic case is ozone, $\ce{O3}$, which can be drawn equally well as two structures, each containing one O–O single bond and one O=O double bond. The normal single-bond and double-bond lengths are $148\,\text{pm}$ and $121\,\text{pm}$, yet the experimentally measured oxygen–oxygen lengths in $\ce{O3}$ are both $128\,\text{pm}$ — intermediate between a single and a double bond, and equal to each other. Neither single Lewis structure can capture this.

The concept of resonance resolves the difficulty. Whenever a single Lewis structure cannot describe a molecule accurately, a number of structures with similar energy, the same positions of nuclei, and the same arrangement of bonding and non-bonding electron pairs are taken as the canonical (resonance) structures. The actual molecule is described by their resonance hybrid, and resonance is represented by a double-headed arrow.

Two canonical forms of ozone and the resonance hybrid OOO I OOO II OOO III (hybrid)
Figure 2. Resonance in $\ce{O3}$: structures I and II are the two canonical forms; structure III is the resonance hybrid, in which both oxygen–oxygen bonds are equivalent at $128\,\text{pm}$. Dashed lines denote delocalised, partial double-bond character.

The carbonate ion and carbon dioxide behave the same way. For $\ce{CO3^2-}$, a single Lewis structure with two single bonds and one double bond would imply unequal bonds, but experiment shows all three carbon–oxygen bonds are equivalent. The ion is therefore best described as a resonance hybrid of three canonical forms. For $\ce{CO2}$ the measured C–O length is $115\,\text{pm}$, lying between a $\ce{C=O}$ double bond ($121\,\text{pm}$) and a $\ce{C#O}$ triple bond ($110\,\text{pm}$); again three canonical forms are needed.

Three canonical forms of the carbonate ion OC OO I O C OO II O C OO III
Figure 3. The three canonical forms of $\ce{CO3^2-}$. The double bond is shown on a different oxygen in each form; the true ion is the hybrid in which all three C–O bonds are identical. (Lone pairs and the 2− charge omitted for clarity.)
NEET Trap

Canonical forms are not real and there is no equilibrium

The canonical forms have no real existence. The molecule does not spend part of its time in one form and part in another, and there is no equilibrium between them (unlike the keto–enol forms in tautomerism). The molecule has a single structure — the resonance hybrid — which cannot be drawn as one Lewis structure.

Resonance stabilises the molecule (the hybrid's energy is lower than any single canonical form) and averages the bond characteristics.

NEET 2024 turned exactly these facts into options: it is correct that three canonical forms can be drawn for the carbonate ion, whereas ozone has two canonical forms — a distinction worth memorising.

Polarity of Bonds and Dipole Moment

A hundred percent ionic or covalent bond is an ideal that no real bond reaches. When the bond joins two identical atoms, as in $\ce{H2}$, $\ce{O2}$, $\ce{Cl2}$, $\ce{N2}$ or $\ce{F2}$, the shared pair sits exactly between the two equal nuclei and the bond is a non-polar covalent bond. When the atoms differ, as in $\ce{HF}$, the shared pair is displaced towards the more electronegative atom (fluorine), producing a polar covalent bond with partial charges $\delta^+$ and $\delta^-$.

Polarisation gives the molecule a dipole moment, defined as the product of the magnitude of the charge and the distance between the centres of positive and negative charge. It is denoted $\mu$:

$$\mu = Q \times r$$

Dipole moment is usually expressed in Debye units (D), where $1\,\text{D} = 3.33564 \times 10^{-30}\ \text{C m}$. It is a vector quantity; by physical convention the arrow points from the negative to the positive centre, but in chemistry a crossed arrow is drawn on the Lewis structure with the cross on the positive end and the head on the negative end.

TypeExampleµ / DGeometry
AB$\ce{HF}$1.78linear
AB$\ce{HCl}$1.07linear
AB$\ce{HBr}$0.79linear
AB$\ce{HI}$0.38linear
AB$\ce{H2}$0linear
AB$_2$$\ce{H2O}$1.85bent
AB$_2$$\ce{H2S}$0.95bent
AB$_2$$\ce{CO2}$0linear
AB$_3$$\ce{NH3}$1.47trigonal pyramidal
AB$_3$$\ce{NF3}$0.23trigonal pyramidal
AB$_3$$\ce{BF3}$0trigonal planar
AB$_4$$\ce{CH4}$0tetrahedral
AB$_4$$\ce{CHCl3}$1.04tetrahedral
AB$_4$$\ce{CCl4}$0tetrahedral

In a polyatomic molecule the net dipole moment is the vector sum of the individual bond dipoles, so geometry decides the outcome. Water is bent, with its two O–H bonds at $104.5^\circ$; the bond dipoles do not cancel, giving a net $\mu = 1.85\,\text{D}$ ($6.17 \times 10^{-30}\ \text{C m}$). By contrast, $\ce{BeF2}$ is linear, so its two equal bond dipoles point in opposite directions and cancel to zero. In $\ce{BF3}$ the three B–F bonds lie $120^\circ$ apart and the resultant of any two is equal and opposite to the third, again giving zero.

Bond dipole vectors in water versus carbon dioxide H₂O — bent, μ ≠ 0 O H H μ net CO₂ — linear, μ = 0 O C O equal & opposite → cancel
Figure 4. Bond dipoles (coral) in bent $\ce{H2O}$ add to a non-zero net moment (teal), whereas in linear $\ce{CO2}$ the two equal bond dipoles point in exactly opposite directions and cancel, giving $\mu = 0$.
NEET Trap

Why μ(NH₃) > μ(NF₃)

Both $\ce{NH3}$ and $\ce{NF3}$ are pyramidal with a lone pair on nitrogen, and fluorine is more electronegative than nitrogen — yet $\ce{NH3}$ ($1.47\,\text{D}$) has a far larger dipole moment than $\ce{NF3}$ ($0.23\,\text{D}$). In $\ce{NH3}$ the orbital dipole from the lone pair points in the same direction as the resultant of the N–H bond dipoles, so they add. In $\ce{NF3}$ the lone-pair dipole points opposite to the resultant of the N–F bond dipoles, so they partly cancel.

Symmetric molecules ($\ce{CO2}$, $\ce{BeF2}$, $\ce{BF3}$, $\ce{CCl4}$, 1,4-dichlorobenzene) have $\mu = 0$ even with polar bonds.

Percent Ionic Character

Just as every covalent bond between unlike atoms carries some ionic character, every ionic bond carries some covalent character. The extent of the shift of the shared pair — and hence the magnitude of the ionic character — depends on the difference in electronegativity between the two bonded atoms. As a working rule from the NIOS text, an electronegativity difference of about $1.7$ corresponds to roughly $50\%$ ionic character; a smaller difference gives less, a larger difference gives more.

This is exactly why the hydrogen-halide dipole moments fall steadily from $\ce{HF}$ to $\ce{HI}$ as the electronegativity gap narrows: $\ce{HF}$ ($1.78\,\text{D}$) > $\ce{HCl}$ ($1.07\,\text{D}$) > $\ce{HBr}$ ($0.79\,\text{D}$) > $\ce{HI}$ ($0.38\,\text{D}$). The covalent character of an ionic bond is summarised by Fajans' rules: a smaller cation, a larger anion and a higher cationic charge all increase covalent character, and for cations of equal size and charge the transition-metal type (configuration $(n-1)d^n\,ns^0$) is more polarising than the noble-gas type ($ns^2np^6$).

Worked Example

Arrange $\ce{BeH2}$, $\ce{CaH2}$ and $\ce{BaH2}$ in order of increasing ionic character. (NEET 2018)

Down group 2 the size of the cation increases ($\ce{Be^2+} < \ce{Ca^2+} < \ce{Ba^2+}$). A larger cation has lower polarising power, so it distorts the hydride ion less; covalent character therefore decreases and ionic character increases down the group.

Order of ionic character: $\ce{BeH2} < \ce{CaH2} < \ce{BaH2}$.

Quick Recap

Bond Parameters in one screen

  • Bond length = equilibrium internuclear distance $= r_\text{A} + r_\text{B}$; covalent radius ≠ van der Waals radius (Cl: 99 pm vs 180 pm).
  • Bond angle = angle between bonding orbitals at the central atom; $\ce{CH4}$ $109.5^\circ$ > $\ce{NH3}$ $107^\circ$ > $\ce{H2O}$ $104.5^\circ$.
  • Bond enthalpy = energy to break 1 mol of bonds (kJ mol⁻¹); use the average value for polyatomic molecules (water: 464.5).
  • Bond order = number of bonds between two atoms; isoelectronic species share bond order ($\ce{N2}$, $\ce{CO}$, $\ce{NO+}$ = 3).
  • Master rule: higher bond order → higher bond enthalpy → shorter bond length.
  • Resonance: hybrid of non-existent canonical forms (no equilibrium); $\ce{O3}$ has 2 forms, $\ce{CO3^2-}$ has 3; it stabilises and averages bonds.
  • Dipole moment $\mu = Q \times r$ (vector sum); symmetric shapes ($\ce{CO2}$, $\ce{BeF2}$, $\ce{BF3}$, $\ce{CCl4}$) give $\mu = 0$; $\mu(\ce{NH3}) > \mu(\ce{NF3})$.
  • Ionic character grows with electronegativity difference (≈ 1.7 → 50%); Fajans' rules govern covalent character of ionic bonds.

NEET PYQ Snapshot — Bond Parameters

Dipole moment, bond order, and resonance are the recurring testers — these are the real questions, verbatim.

NEET 2025

Given below are two statements. Statement I: A hypothetical diatomic molecule with bond order zero is quite stable. Statement II: As bond order increases, the bond length increases. In the light of the above statements, choose the most appropriate answer:

  1. Statement I is false but Statement II is true
  2. Both Statement I and Statement II are true
  3. Both Statement I and Statement II are false
  4. Statement I is true but Statement II is false
Answer: (3)

A zero (or negative) bond order means an unstable molecule, so Statement I is false. As bond order increases, bond length decreases — not increases — so Statement II is also false.

NEET 2024

Identify the correct answer.

  1. Three resonance structures can be drawn for ozone
  2. $\ce{BF3}$ has non-zero dipole moment
  3. Dipole moment of $\ce{NF3}$ is greater than that of $\ce{NH3}$
  4. Three canonical forms can be drawn for $\ce{CO3^2-}$ ion
Answer: (4)

Ozone has two canonical forms (not three), $\ce{BF3}$ is symmetric so $\mu = 0$, and $\mu(\ce{NH3}) > \mu(\ce{NF3})$. The carbonate ion is correctly described by three canonical forms.

NEET 2020

Which of the following set of molecules will have zero dipole moment?

  1. Boron trifluoride, hydrogen fluoride, carbon dioxide, 1,3-dichlorobenzene
  2. Nitrogen trifluoride, beryllium difluoride, water, 1,3-dichlorobenzene
  3. Boron trifluoride, beryllium difluoride, carbon dioxide, 1,4-dichlorobenzene
  4. Ammonia, beryllium difluoride, water, 1,4-dichlorobenzene
Answer: (3)

$\ce{BF3}$, $\ce{BeF2}$, $\ce{CO2}$ and 1,4-dichlorobenzene are all symmetrical, so their bond dipoles cancel to give $\mu = 0$.

NEET 2018

Among $\ce{CaH2}$, $\ce{BeH2}$, $\ce{BaH2}$, the order of ionic character is

  1. $\ce{BeH2} < \ce{CaH2} < \ce{BaH2}$
  2. $\ce{CaH2} < \ce{BeH2} < \ce{BaH2}$
  3. $\ce{BeH2} < \ce{BaH2} < \ce{CaH2}$
  4. $\ce{BaH2} < \ce{BeH2} < \ce{CaH2}$
Answer: (1)

Down the group the cation size increases, polarising power falls, covalent character decreases and ionic character increases: $\ce{BeH2} < \ce{CaH2} < \ce{BaH2}$.

NEET 2018

Consider the following species: $\ce{CN+}$, $\ce{CN-}$, $\ce{NO}$ and $\ce{CN}$. Which one of these will have the highest bond order?

  1. $\ce{NO}$
  2. $\ce{CN-}$
  3. $\ce{CN+}$
  4. $\ce{CN}$
Answer: (2)

Bond orders are $\ce{NO} = 2.5$, $\ce{CN+} = 2.0$, $\ce{CN} = 2.5$ and $\ce{CN-} = 3.0$. The cyanide ion has the highest bond order, 3.

FAQs — Bond Parameters

The exact distinctions NEET likes to test, answered concisely.

How does bond order relate to bond length and bond enthalpy?

They move together in a predictable way: with an increase in bond order, bond enthalpy increases and bond length decreases. A higher bond order means more shared electron pairs holding the nuclei together, so the bond becomes shorter and stronger. For example, the bond order rises from 1 in N–N to 3 in the dinitrogen molecule, and the dinitrogen bond is among the shortest and strongest known, with a bond enthalpy of 946 kJ per mole.

Why is the dipole moment of NF3 lower than that of NH3?

Both molecules are trigonal pyramidal with a lone pair on nitrogen, yet ammonia has a dipole moment of about 1.47 D while nitrogen trifluoride is only about 0.23 D. In ammonia the orbital dipole due to the lone pair points in the same direction as the resultant of the three N–H bond dipoles, so the two add up. In nitrogen trifluoride the lone-pair orbital dipole points opposite to the resultant of the three N–F bond dipoles, so they partly cancel, giving a small net dipole moment.

Why do BeF2, CO2 and BF3 have zero dipole moment even though their bonds are polar?

In a polyatomic molecule the net dipole moment is the vector sum of the individual bond dipoles, so geometry decides the result. BeF2 is linear and CO2 is linear, so their two equal bond dipoles point in exactly opposite directions and cancel. BF3 is trigonal planar with bonds 120 degrees apart; the resultant of any two B–F bond dipoles is equal and opposite to the third, so the total is zero. The bonds are individually polar, but the symmetric shape makes the molecule non-polar.

What is meant by resonance, and do the canonical forms really exist?

Resonance is used when a single Lewis structure cannot accurately describe a molecule. A number of structures with similar energy, the same positions of nuclei, and the same arrangement of bonding and non-bonding electrons are taken as canonical structures, and the molecule is described by their hybrid. The canonical forms have no real existence. The molecule does not flip between them, there is no equilibrium between them, and the true molecule is a single resonance hybrid that cannot be drawn as one Lewis structure. Resonance lowers the energy and averages the bond characteristics, which is why all three oxygen–oxygen bonds in the carbonate ion are equivalent.

What is percent ionic character and how is it linked to electronegativity?

No real bond is fully covalent or fully ionic; every covalent bond between unlike atoms has some ionic character, and every ionic bond has some covalent character. The magnitude of ionic character grows with the difference in electronegativity between the two bonded atoms. As a rough guide, an electronegativity difference of about 1.7 corresponds to roughly 50 percent ionic character; a smaller difference gives less ionic character and a larger difference gives more. This is why the dipole moment of the hydrogen halides falls steadily from H–F to H–I as the electronegativity gap narrows.

Why is average bond enthalpy used for polyatomic molecules like water?

In a polyatomic molecule the same type of bond does not always require the same energy to break, because the chemical environment changes after the first bond is broken. In water, breaking the first O–H bond needs 502 kJ per mole, while breaking the second needs 427 kJ per mole. Since the two values differ, chemists use the mean or average bond enthalpy, obtained by dividing the total dissociation enthalpy by the number of bonds broken, which for water gives about 464.5 kJ per mole.