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# Bond Parameters Questions

In order to become stable, various atoms must combine. This combination occurs through the formation of bonds. Bonds are classified into three types: ionic or electrovalent bonds, covalent bonds, and coordinate bonds. This, in turn, demonstrates that every bond has some characteristic associated with it.

Here are some of the various bond features or characteristics that can be referred to as bond parameters. Certain parameters define a particular covalent bond:

• Bond Length
• Bond Angle
• The Bond Enthalpy
• Bond Order
 Definition: Covalent bonds can be characterized on the basis of several bond parameters such as bond length, bond angle, bond order, and bond energy (also known as bond enthalpy). These bond parameters offer insight into the stability of a chemical compound and the strength of the chemical bonds holding its atoms together.

## Bond Parameters Chemistry Questions with Solutions

Q1. The angle between hydrogen through oxygen is-

a.) 104.5°

b.) 104°

c.) 105.4°

d.) 105°

Q2. The bond enthalpy of H2O and OH are 502 kJ mol–1 and 427 kJ mol–1. Then what will be the average bond enthalpy?

a.) 502 kJ mol–1

b.) 464.5 kJ mol–1

c.) 427 kJ mol–1

d.) 75 kJ mol–1

Correct Answer– (b.) 464.5 kJ mol–1

Q3. The bond order of CO is-

a.) 1

b.) 2

c.) 3

d.) 4

Q4. Out of the given options, choose the one which cannot be used to measure bond lengths?

a.) Spectroscopy

b.) X-ray diffraction

c.) Electron diffraction

d.) Young’s Double-slit method

Correct Answer– (d.) Young’s Double-slit method

Q5. Which of the following pairs has dipole-induced dipole interactions?

a.) HCl and He atoms

b.) SiF4 and He atoms

c.) H2O and alcohol

d.) Cl2 and CCl4

Correct Answer– (a.) HCl and He atoms.

Q6. State True or False.

Strength of the bond between the two atoms can be known from bond dissociation enthalpy.

Bond dissociation enthalpy is the energy required to break a molecule into atoms. The stronger the bond between the atoms, the higher the bond dissociation energy. Bond dissociation enthalpy can thus be used to calculate the strength of the bond between two atoms.

Q7. Fill in the blank.

All the ___ species have the same bond order.

Answer. All the isoelectronic species have the same bond order.

Isoelectronic species are the molecules and ions that contain the same number of electrons, and they have the same bond order.

Q8. Give the increasing order of O2, O2+, O22– and O2

Answer. According to Molecular Orbital Theory,

$$\begin{array}{l}Bond order = \frac{\left ( No. of electrons in bonding orbital \right )-\left ( No. of electrons in anti-bonding orbital \right )}{2}\end{array}$$

Bond Order for O2 is 2

Bond Order for O2+ is 2.5

Bond Order for O22–is 1

Bond Order for O2 is 1.5

Therefore, the increasing order will be O22 < O2 < O2 < O2+.

Q9. Out of sigma and pi bonds, which one is stronger bond and why?

Answer. A Sigma (σ) bond is stronger than a pi(𝜋) bond. This is because the sigma(σ) bond is formed by head-on overlapping of orbitals and therefore, overlapping is large. On the other hand, the pi(𝜋) bond is formed by the sidewise overlapping which is small.

Q10. State the factors affecting bond length?

Answer. The factors affecting bond length are as follows-

• Size of the atoms- The bond length increases with an increase in the size of the atoms.
• Multiplicity of bond- The bond length decreases with the multiplicity of the bond. For example- the bond length of t=carbon – carbon bonds are in order Alkynes (C ☰ C) < Alkenes (C = C) < Alkanes (C – C).
• TYpe of hybridization- As an s orbital is smaller in size, shorter is the hybrid orbital and hence shorter is the bond length.

Q11. Calculate the bond order of: N2, O2, O2+, and O2.

Answer. According to Molecular Orbital Theory,

$$\begin{array}{l}Bond order = \frac{\left ( No. of electrons in bonding orbital \right )-\left ( No. of electrons in anti-bonding orbital \right )}{2}\end{array}$$

In N2, Bond order will be (10 – 4) / 2 = 3.

In O2, Bond order will be (10 – 6) / 2 = 2

In O2+ Bond order will be (10 – 5) / 2 = 5/2 = 2.5

In O2 Bond order will be (10 – 7) / 2 = 3/2 = 1.5

Q12. Determine the bond order for hydrogen gas, H2 and Cyanide.

Answer. The lewis structure of H2 is-

There is only 1 pair of shared electrons and no lone pairs. This indicates that there is a single bond present. Therefore, the bond order of H2 is 1.

The Lewis structure of cyanide CN is

There are three pairs of shared electrons. This indicates that there is a triple bond between the atoms. Therefore, the bond order of CN is 3.

Q13. Predict the dipole moment of:

(i) a molecule of the type AX2 having a linear geometry.

(ii) a molecule of type AX4 having tetrahedral geometry.

(iii) a molecule of the type AX2 having angular geometry.

(iv) a molecule of the type AX4 having square planar geometry.

(i) The dipole moment will be zero in a molecule of the type AX2 having a linear geometry.

(ii) The dipole moment will be zero in a molecule of type AX4 having tetrahedral geometry.

(iii) There will be some appreciable dipole moment in a molecule of the type AX2 having angular geometry.

(iv) The dipole moment will be zero in a molecule of the type AX4 having square planar geometry.

Q14. Out of NH3 and NF3, which one has a higher dipole moment and why?

Answer. The dipole moment of ammonia (1.47D) is greater than that of NF3 (0.24D). Both molecules have pyramidal molecular geometry. N atoms in each molecule have one lone pair. F has a higher electronegative potential than H, and the NF bond is more polar than the NH bond. As a result, NF3 is expected to have a much higher dipole moment than NH3. However, in the case of ammonia, the direction of the lone pair dipole moment and the bond pair dipole moment are the same, whereas, in the case of NF3, they are opposite. Individual dipole moment vectors in ammonia molecules add whereas they cancel in NF3.

Q15. CO2 is non-polar while H2O is polar. What conclusions can be drawn about their structures?

Answer. Co2 is a linear molecule, therefore the resultant dipole moment of two C = O bonds gets cancelled, giving zero dipole moment. On the other hand, the water molecule is not linear, it has an angular shape and the bond moments of two O–H bonds give the resultant dipole moment.

## Practise Questions on Bond Parameters

Q1. Which of the following molecules have a dipole moment?

a.) N2

b.) CH4

c.) BeF2

d.) H2O

Q2. What can be the covalent radius and Van der Waal’s radius between chlorine molecules?

a.) 99 pm,198 pm

b.) 198 pm, 99 pm

c.) 198 pm, 198 pm

d.) 99 pm, 99 pm

Q3. Does Resonance stabilize the molecule?

Q4. Select the molecule or ion having larger property mentioned in each of the following pairs:

(i) NF3, NH3 : Dipole moment

(ii) NH3, PH3 :Bond angle

(iii) CO2, BF3 : Bond angle

Q5. Define Bond enthalpy and state the factors on which it depends?

Click the PDF to check the answers for Practice Questions.