Bond Order Formula Importance Solved Examples

Solution: (1) Before finding the bond order, we need to find the electronic configuration of H₂.

Electronic Configuration of (Hydrogen gas) H₂:

H₂ has a total of 2 molecules in its molecule. Hence, its electronic configuration will be 1s¹.

(2) After determining the electronic configuration one can find the bond order using the bond order formula.

Bond Order (b.o.) formula = ½ (Nb - Na) (Here, Na = antibonding electrons, Nb = bonding electrons)

Now applying this formula,

b.o. = ½ (2-0) = 1

(3) And hence, we get;

BOND ORDER OF H₂ = 1

Solution: (1) Before finding the bond order we need to determine the electronic configuration of C₂. The electronic configuration of C₂ is as follows:

C₂ has a total of 12 electrons, therefore its electronic configuration will be = (σ2s)² (σ*2s) n(2px)² n(2py)²

(2) Now we can calculate the bond order from the electronic configuration thus obtained will be

BOND ORDER (B. O.)= ½ (Na - Nb)

This gives us,

BOND ORDER (B. O.) = 2

Solution: (1) First step will be finding the electronic configuration of Bromine (Br₂), its electronic configuration [Ar] 4s² 3d¹⁰ 4p⁵

(2) The next step will be to calculate the bond order of Bromine (Br₂), using the bond order formula

B.O. = ½ (Na - Nb)

This gives us,

B.O. = 1

Solution: (1) First step in determining the bond order will be finding the electronic configuration of BN. The number of electrons in BN are 12 (5+7). Therefore, the electronic configuration of BN will be (σ1s² σ*1s²) (σ2s σ*2s²) ( 2px² = 2py²)

(2) Once we have determined the electronic configuration we can determine the bond order using the bond order formula, that is,

BOND ORDER (B.O.) = ½ (Na - Nb)

(3) Therefore, (B.O.) = 2

Solution: (1) Firstly, find the electronic configuration of O₂.

Electronic configuration of O₂ = (σ1s² σ* 1s² σ2s² σ2p²z σ2p²x Π2p²y Π * 2p¹x Π 2p¹y)

(2) The next step will be to find the bond order using the formula

Bond order = ½ (Na - Nb)

This gives us the bond order as,

BOND ORDER = 2

Ques. Explain (i) bond strength (ii) bond length (iii) bond angle. (3 Marks)

Solution: (i) Bond strength refers to quite literally the strength or how strongly the atoms present are bonded. When the number of electrons connected increases the strength of the bond between the atoms also increases.

(ii) Bond Length refers to the distance in average between the atoms that are bonded.

(iii) Bond Angle refers to the average angle between the two bonded electrons around the atom.

Solution: Bond order refers to the number of chemical bonds that exist between the two atoms present. An increase in bond order means there is a higher level of attraction among the atoms. It also means that there is a higher level of stability in the atoms.

Solution: Bond order has an important relation with bond strength because bond order impacts the bond strength of the component as well. Both are directly proportional, that is, with an increase in bond order, there is an increase in bond strength as well. In terms of bond length, the relation is rather inverse because with an increase in bond order the bond length decreases. Bond order also impacts the stability of the chemical bond as well.

Solution: (1) First step here will be to determine the electronic configuration of carbon monoxide(co). Carbon Monoxide has 14 electrons and its electronic configuration is:

(σ1s) 2(σ*1s)2(σ2s)2(σ*2s)2(Π)4(2pz)2

(2) Once the electronic configuration is obtained, the bond order can be calculated using the formula:

BOND ORDER = ½ (Na - Nb)

This gives us the answer as,

Bond order of Carbon Monoxide (CO) = 3

Solution: (1) The first step will be to find out the electronic configuration. The electronic configuration of N₂ is: ( σ1s σ* ² σ2s² σ* 2s² Π2px² Π2py² Π2p²z)

(2) The next step will be to determine the bond order using the formula, that is,

BOND ORDER = ½ (Na - Nb)

Therefore, the bond order of N₂ is,

BOND ORDER of N₂ = 3