Bond Order: Nature, Important Equations and Importance

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Bond order is simply defined as the number of covalent bonds present between a pair of atoms. On the basis of bond order, the stability of molecules can be determined. This concept was first introduced by Linus Pauling to find out the molecular dynamics which is now one of the most important scientific determinants. In this article, we will look at the nature of bond order and how it works with various molecules with necessary examples and equations.


What is Bond Order?

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Basically, bond order (B.O.) is a number of bonds present between two atoms which is measured via the electrons present in the bond formation through both covalent and ionic bonds. For example, referring to the Lewis structure and homonuclear diatomic molecules, H2 (with a single bond) has a bond order of one, O2 (with a double bond) has a bond order of two, N2 (with a triple bond) has a bond pair of three, and so forth. 

In order to calculate the bond order the below given formula is used:

b.o. = ½ (Nb-Na)

where Nb is number of electrons in the bonding orbitals

and Na is the number of electrons in the anti-bonding orbitals.

Chemical Bonds

Chemical Bonds


Nature of Bond Order

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When Nb is greater than Na (Nb > Na) it is said to be a positive order which means it is a stable bond order and when Nb is less than Na (Nb < Na) or equal to Na (Nb = Na) it is said to be negative or zero order which is termed as an unstable bond order.

A distance between two bonded atoms is called a bond length, it helps in determining the measure between the two atoms. The bond length decreases as bond order increases. The length of the bond between the atoms is usually equivalent to the sum of the covalent radii of the two atoms.

Read More: Diatomic


Important Equations

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In order to find the bond order one should know the electronic configuration of the molecules which is basically the distribution of electrons of an atom or molecule in atomic or molecular orbit. Some of the examples are given as follows:

Beryllium Molecule (Be2

The electronic configuration of beryllium is 1s2, 2s2 and has 8 electrons

Therefore, the electronic configuration of Be2 is σ1s2< σ1s2< σ2s2< σ2s2

so, Nb=4 and Na=4

Bond order =½ (Nb–Na)

= ½ (4–4) 

= 0

Hence, as Nb is equal to Na the bond order is negative.

Read Also: 

Hydrides Degree of Unsaturation Formula
Activation Energy Formula Enthalpy Formula

Hydrogen Molecular Positive Ion (H2+)

The electronic configuration of H2+ is σ(1s)2

so, Nb=1 and Na=0

Bond order = ½ (Nb–Na)

= ½ (1 – 0)

= 12

Read More: Specific Heat of Water

Oxygen Molecule (O2)

The electronic configuration of O is 1s2,2s2,2px2,2py12pz1 ,

where O2 has 16 electrons, therefore the molecular orbital structure is 

σ1s2< σ∗1s2< σ2s2< σ∗2s2< σ2p2z< (π2p2x2p2y) < (π∗2p1x=π∗2p1y)

so, Nb=10 and Na=6

Bond order = ½ (Nb–Na)

= ½ (10–6)

= 2

Read More: Colorimeter

Superoxide Ion (O-2)

This ion is formed when it gains one electron by O2 molecule, which is added in the π∗2px or π∗2py molecular orbital structure. 

Hence, the electronic configuration of O-2 is 

σ1s2< σ1s2< σ2s2< σ2s2< σ2pz2< (π2px22py2) <(π2px22py1)

so, Nb=10 and Nb=7

Bond order = ½ (Nb–Na)

= ½ (10 – 7)

= 32

= 1.5

Molecular Orbital Theory, Bond Order, Bond Strength, Magnetic Properties

Molecular Orbital Theory, Bond Order, Bond Strength, Magnetic Properties


Importance of Bond Order

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  • Bond order helps determine the number of participating electrons in the formation of bonds.
  • Bond order determines the stability of the bond. Higher bond order confers more stability.
  • Bond order helps us to understand bond strength. Higher bond order implies more energy is required to break the bond.
  • Bond order helps us to understand the bond length.
  • Bond order gives us an indication of the hybridization of the molecule.
  • A fractional bond order value implies that no bond is formed.

Read More: Oxidizing Agent


Things to Remember

  • Bond order is determined with the number of bonding and antibonding electrons. Bond order formula is used to calculate bond. Several properties of bond can be known from bond order values.
  • Stable bonds have a positive bond order.
  • Bond order is an index of bond strength and is used extensively in valence bond theory.
  • Key Terms:
    • acetyleneethyne; the simplest alkyne, a hydrocarbon of formula HC≡CH; a colorless gas, with a peculiar, unpleasant odor, formerly used as an illuminating gas but now used in welding and metallurgy
    • sigma bonda covalent atomic bond that is rotationally symmetric about its axis
    • bond orderthe number of overlapping electron pairs between a pair of atoms
    • antibondingan atomic or molecular orbital whose energy increases as its constituent atoms converge, generating a repulsive force that hinders bonding

Read More: Difference between Cations and Anions


Sample Questions

Ques. How do you express the bond strength in terms of bond order? (1 Mark)

Ans. Bond strength is directly proportional to the bond order. Greater the bond order, more is the bond strength.

Ques. Define the bond-length. (1 Mark)

Ans. Bond-length is the equilibrium distance between the nuclei of two bonded atoms in a molecule. Bond-lengths are measured by spectroscopic methods.

Ques. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3. (1 Mark)

Ans. N2 < SO2 < ClF3 < K2O < LiF

Ques. Arrange O2, O2, O22-, O2+in increasing order of bond energy. (1 Mark)

Ans. O22-< O2– < O22-< O2+

Ques. How is bond order related to the stability of a molecule? (1 Mark)

Ans. Higher the bond order, greater is the stability.

Ques. Give reason why H2+ ions are more stable than H2- though they have the same bond order. (1 Mark)

Ans. In H2- ion, one electron is present in anti-bonding orbital due to which destabilizing effect is more and thus the stability is less than that of H2+ ion.

Ques. Explain the formation of a chemical bond. (2 Marks)

Ans. According to Kossel and Lewis, atoms combine together in order to complete their respective octets so as to acquire the stable inert gas configuration. This can occur in two ways; by transfer of one or more electrons from one atom to another or by sharing of electrons between two or more atoms.

Ques. What is the Bond Order for the Electronic Configuration of a Compound = (σ2s)2 (σ*2s)2 n(2px)2 n(2py)2? (3 Marks)

Ans. The electronic configuration of the compound = (σ2s)2 (σ*2s)2 n(2px)2 n(2py)2

Number of bonding electrons = 8

Number of antibonding electrons = 4

Bond order = (Nb – Na) / 2

=( 8 – 4 )/ 2

= 2

Therefore, two bonds are formed between the atoms in the molecule.

Ques. What is meant by the term bond order? Calculate the bond order of N2, O2, O2+, O2. (5 Marks)

Ans. Bond order is defined as the half of the difference between the number of electrons present in bonding and antibonding molecular orbitals.

Bond order = ½ (Nb – Na)

E.C of N2 = 1s2 2s2 2px12py12pz1

(i) M.O. configuration of N2 = [σ1s]2 [σ*1s]2 [σ2s]2 [σ*2s]2 [π2px]2 [π2py]2 [σ2pz]2

Bond order (b.o.) = ½(Nb-Na)

= ½[8-2] = 3

(ii) M.O. configuration of O2[ σ1s]2 [σ*1s]2 [σ2s]2 [σ*2s]2 [σ2pz]2

b.o = ½[Nb-Na]

= ½[8 – 4] = 2

(iii) M.O.configuration of O2+  = KK[σ2s]2 [σ*2s]2 [σ2pz]2 [π2px]2 [π2py]2 [π2px]1

= ½[8 - 3] = 2.5

(iv) M.O.configuration of O2-  = KK[σ2s]2 [σ*2s]2 [σ2pz]2 [π2px]2 [π2py]2[π*2px]2[π*2py]1

= ½[8 - 3] = 1.5

Also Read:

Electrophilic Aromatic Substitution Copolymers Cis Trans Isomerism
Polytetrafluoroethene (Teflon) High Density Polythene VSEPR Theory
Order of Reaction Mendeleev Periodic Table Rate of Reaction

CBSE CLASS XII Related Questions

1.
Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible: 
(i) Fe3+ (aq) and I- (aq) 
(ii) Ag+ (aq) and Cu(s) 
(iii) Fe3+(aq) and Br-(aq) 
(iv) Ag(s) and Fe3+(aq) 
(v) Br2 (aq) and Fe2+(aq).

      2.
      Write the Nernst equation and emf of the following cells at 298 K : 
      (i) Mg(s) | Mg2+ (0.001M) || Cu2+(0.0001 M) | Cu(s) 
      (ii) Fe(s) | Fe2+ (0.001M) || H+ (1M)|H2(g)(1bar) | Pt(s) 
      (iii) Sn(s) | Sn2+(0.050 M) || H+ (0.020 M) | H2(g) (1 bar) | Pt(s) 
      (iv) Pt(s) | Br2(l) | Br-  (0.010 M) || H+ (0.030 M) | H2(g) (1 bar) | Pt(s).

          3.
          Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

              4.

              Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.

                  5.
                  In the button cells widely used in watches and other devices the following reaction takes place:
                  Zn(s) + Ag2O(s) + H2O(l) \(\rightarrow\) Zn2+(aq) + 2Ag(s) + 2OH-  (aq) 
                  Determine \(\triangle _rG^\ominus\) and \(E^\ominus\) for the reaction.

                      6.

                      Draw the structures of optical isomers of: 
                      (i) \([Cr(C_2O_4)_3]^{3–}\)
                      (ii) \([PtCl_2(en)_2]^{2+}\)
                      (iii) \([Cr(NH_3)2Cl_2(en)]^{+}\)

                          CBSE CLASS XII Previous Year Papers

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