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Chapter 4 Chemical Bonding And Molecular Structure
Multiple Choice Questions (Type - I)
Question 1. Isostructural species are those which have the same shape and hybridisation. Among the given species identify the isostructural pairs.
(i) [$NF_3$ and $BF_3$]
(ii) [$BF_4^-$ and $NH_4^+$]
(iii) [$BCl_3$ and $BrCl_3$]
(iv) [$NH_3$ and $NO_3^-$]
Answer:
Question 2. Polarity in a molecule and hence the dipole moment depends primarily on electronegativity of the constituent atoms and shape of a molecule. Which of the following has the highest dipole moment?
(i) $CO_2$
(ii) $HI$
(iii) $H_2O$
(iv) $SO_2$
Answer:
Question 3. The types of hybrid orbitals of nitrogen in $NO_2^+$, $NO_3^-$ and $NH_4^+$ respectively are expected to be
(i) sp, $sp^3$ and $sp^2$
(ii) sp, $sp^2$ and $sp^3$
(iii) $sp^2$, sp and $sp^3$
(iv) $sp^2$, $sp^3$ and sp
Answer:
Question 4. Hydrogen bonds are formed in many compounds e.g., $H_2O$, $HF$, $NH_3$. The boiling point of such compounds depends to a large extent on the strength of hydrogen bond and the number of hydrogen bonds. The correct decreasing order of the boiling points of above compounds is :
(i) $HF > H_2O > NH_3$
(ii) $H_2O > HF > NH_3$
(iii) $NH_3 > HF > H_2O$
(iv) $NH_3 > H_2O > HF$
Answer:
Question 5. In $PO_4^{3-}$ ion the formal charge on the oxygen atom of P–O bond is
(i) + 1
(ii) – 1
(iii) – 0.75
(iv) + 0.75
Answer:
Question 6. In $NO_3^-$ ion, the number of bond pairs and lone pairs of electrons on nitrogen atom are
(i) 2, 2
(ii) 3, 1
(iii) 1, 3
(iv) 4, 0
Answer:
Question 7. Which of the following species has tetrahedral geometry?
(i) $BH_4^-$
(ii) $NH_2^-$
(iii) $CO_3^{2-}$
(iv) $H_3O^+$
Answer:
Question 8. Number of $\pi$ bonds and $\sigma$ bonds in the following structure is–
(i) 6, 19
(ii) 4, 20
(iii) 5, 19
(iv) 5, 20
Answer:
Question 9. Which molecule/ion out of the following does not contain unpaired electrons?
(i) $N_2^+$
(ii) $O_2$
(iii) $O_2^{2-}$
(iv) $B_2$
Answer:
Question 10. In which of the following molecule/ion all the bonds are not equal?
(i) $XeF_4$
(ii) $BF_4^-$
(iii) $C_2H_4$
(iv) $SiF_4$
Answer:
Question 11. In which of the following substances will hydrogen bond be strongest?
(i) HCl
(ii) $H_2O$
(iii) HI
(iv) $H_2S$
Answer:
Question 12. If the electronic configuration of an element is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2$, the four electrons involved in chemical bond formation will be_____.
(i) $3p^6$
(ii) $3p^6, 4s^2$
(iii) $3p^6, 3d^2$
(iv) $3d^2, 4s^2$
Answer:
Question 13. Which of the following angle corresponds to $sp^2$ hybridisation?
(i) 90°
(ii) 120°
(iii) 180°
(iv) 109°
Answer:
The electronic configurations of three elements, A, B and C are given below. Answer the questions 14 to 17 on the basis of these configurations.
A $1s^2 2s^2 2p^6$
B $1s^2 2s^2 2p^6 3s^2 3p^3$
C $1s^2 2s^2 2p^6 3s^2 3p^5$
Question 14. Stable form of A may be represented by the formula :
(i) A
(ii) $A_2$
(iii) $A_3$
(iv) $A_4$
Answer:
Question 15. Stable form of C may be represented by the formula :
(i) C
(ii) $C_2$
(iii) $C_3$
(iv) $C_4$
Answer:
Question 16. The molecular formula of the compound formed from B and C will be
(i) BC
(ii) $B_2C$
(iii) $BC_2$
(iv) $BC_3$
Answer:
Question 17. The bond between B and C will be
(i) Ionic
(ii) Covalent
(iii) Hydrogen
(iv) Coordinate
Answer:
Question 18. Which of the following order of energies of molecular orbitals of $N_2$ is correct?
(i) $(\pi 2p_y) < (\sigma 2p_z) < (\pi^*2p_x) \approx (\pi^*2p_y)$
(ii) $(\pi 2p_y) > (\sigma 2p_z) > (\pi^*2p_x) \approx (\pi^*2p_y )$
(iii) $(\pi 2p_y) < (\sigma 2p_z) > (\pi^*2p_x) \approx (\pi^*2p_y )$
(iv) $(\pi 2p_y) > (\sigma 2p_z) < (\pi^*2p_x) \approx (\pi^*2p_y )$
Answer:
Question 19. Which of the following statement is not correct from the view point of molecular orbital theory?
(i) $Be_2$ is not a stable molecule.
(ii) $He_2$ is not stable but $He_2^+$ is expected to exist.
(iii) Bond strength of $N_2$ is maximum amongst the homonuclear diatomic molecules belonging to the second period.
(iv) The order of energies of molecular orbitals in $N_2$ molecule is
$\sigma 2s < \sigma^* 2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y ) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^*2p_z$
Answer:
Question 20. Which of the following options represents the correct bond order :
(i) $O_2^– > O_2 > O_2^+$
(ii) $O_2^– < O_2 < O_2^+$
(iii) $O_2^– > O_2 < O_2^+$
(iv) $O_2^– < O_2 > O_2^+$
Answer:
Question 21. The electronic configuration of the outer most shell of the most electronegative element is
(i) $2s^22p^5$
(ii) $3s^23p^5$
(iii) $4s^24p^5$
(iv) $5s^25p^5$
Answer:
Question 22. Amongst the following elements whose electronic configurations are given below, the one having the highest ionisation enthalpy is
(i) $[Ne]3s^23p^1$
(ii) $[Ne]3s^23p^3$
(iii) $[Ne]3s^23p^2$
(iv) $[Ar]3d^{10}4s^24p^3$
Answer:
Multiple Choice Questions (Type - II)
In the following questions two or more options may be correct.
Question 23. Which of the following have identical bond order?
(i) $CN^–$
(ii) $NO^+$
(iii) $O_2^–$
(iv) $O_2^{2–}$
Answer:
Question 24. Which of the following attain the linear structure:
(i) $BeCl_2$
(ii) $NCO^+$
(iii) $NO_2$
(iv) $CS_2$
Answer:
Question 25. CO is isoelectronic with
(i) $NO^+$
(ii) $N_2$
(iii) $SnCl_2$
(iv) $NO_2^–$
Answer:
Question 26. Which of the following species have the same shape?
(i) $CO_2$
(ii) $CCl_4$
(iii) $O_3$
(iv) $NO_2^–$
Answer:
Question 27. Which of the following statements are correct about $CO_3^{2–}$ ?
(i) The hybridisation of central atom is $sp^3$.
(ii) Its resonance structure has one C–O single bond and two C=O double bonds.
(iii) The average formal charge on each oxygen atom is 0.67 units.
(iv) All C–O bond lengths are equal.
Answer:
Question 28. Dimagnetic species are those which contain no unpaired electrons. Which among the following are dimagnetic?
(i) $N_2$
(ii) $N_2^{2–}$
(iii) $O_2$
(iv) $O_2^{2–}$
Answer:
Question 29. Species having same bond order are :
(i) $N_2$
(ii) $N_2^–$
(iii) $F_2^+$
(iv) $O_2^–$
Answer:
Question 30. Which of the following statements are not correct?
(i) NaCl being an ionic compound is a good conductor of electricity in the solid state.
(ii) In canonical structures there is a difference in the arrangement of atoms.
(iii) Hybrid orbitals form stronger bonds than pure orbitals.
(iv) VSEPR Theory can explain the square planar geometry of $XeF_4$.
Answer:
Short Answer Questions
Question 31. Explain the non linear shape of $H_2S$ and non planar shape of $PCl_3$ using valence shell electron pair repulsion theory.
Answer:
Question 32. Using molecular orbital theory, compare the bond energy and magnetic character of $O_2^+$ and $O_2^–$ species.
Answer:
Question 33. Explain the shape of $BrF_5$.
Answer:
Question 34. Structures of molecules of two compounds are given below :
(a) Which of the two compounds will have intermolecular hydrogen bonding and which compound is expected to show intramolecular hydrogen bonding.
(b) The melting point of a compound depends on, among other things, the extent of hydrogen bonding. On this basis explain which of the above two compounds will show higher melting point.
(c) Solubility of compounds in water depends on power to form hydrogen bonds with water. Which of the above compounds will form hydrogen bond with water easily and be more soluble in it.
Answer:
Question 35. Why does type of overlap given in the following figure not result in bond formation?
Answer:
Question 36. Explain why $PCl_5$ is trigonal bipyramidal whereas $IF_5$ is square pyramidal.
Answer:
Question 37. In both water and dimethyl ether ($CH_3–O–CH_3$), oxygen atom is central atom, and has the same hybridisation, yet they have different bond angles. Which one has greater bond angle? Give reason.
Answer:
Question 38. Write Lewis structure of the following compounds and show formal charge on each atom.
$HNO_3, NO_2, H_2SO_4$
Answer:
Question 39. The energy of $\sigma 2p_z$ molecular orbital is greater than $\pi 2p_x$ and $\pi 2p_y$ molecular orbitals in nitrogen molecule. Write the complete sequence of energy levels in the increasing order of energy in the molecule. Compare the relative stability and the magnetic behaviour of the following species :
$N_2, N_2^+, N_2^–, N_2^{2+}$
Answer:
Question 40. What is the effect of the following processes on the bond order in $N_2$ and $O_2$?
(i) $N_2 \rightarrow N_2^+ + e^–$
(ii) $O_2 \rightarrow O_2^+ + e^–$
Answer:
Question 41. Give reasons for the following :
(i) Covalent bonds are directional bonds while ionic bonds are non-directional.
(ii) Water molecule has bent structure whereas carbon dioxide molecule is linear.
(iii) Ethyne molecule is linear.
Answer:
Question 42. What is an ionic bond? With two suitable examples explain the difference between an ionic and a covalent bond?
Answer:
Question 43. Arrange the following bonds in order of increasing ionic character giving reason.
N—H, F—H, C—H and O—H
Answer:
Question 44. Explain why $CO_3^{2-}$ ion cannot be represented by a single Lewis structure. How can it be best represented?
Answer:
Question 45. Predict the hybridisation of each carbon in the molecule of organic compound given below. Also indicate the total number of sigma and pi bonds in this molecule.
Answer:
Question 46. Group the following as linear and non-linear molecules :
$H_2O, HOCl, BeCl_2, Cl_2O$
Answer:
Question 47. Elements X, Y and Z have 4, 5 and 7 valence electrons respectively. (i) Write the molecular formula of the compounds formed by these elements individually with hydrogen. (ii) Which of these compounds will have the highest dipole moment?
Answer:
Question 48. Draw the resonating structure of
(i) Ozone molecule
(ii) Nitrate ion
Answer:
Question 49. Predict the shapes of the following molecules on the basis of hybridisation.
$BCl_3, CH_4, CO_2, NH_3$
Answer:
Question 50. All the C—O bonds in carbonate ion ($CO_3^{2-}$) are equal in length. Explain.
Answer:
Question 51. What is meant by the term average bond enthalpy? Why is there difference in bond enthalpy of O—H bond in ethanol ($C_2H_5OH$) and water?
Answer:
Matching Type Questions
Question 52. Match the species in Column I with the type of hybrid orbitals in Column II.
Column I
(i) $SF_4$
(ii) $IF_5$
(iii) $NO_2^+$
(iv) $NH_4^+$
Column II
(a) $sp^3d^2$
(b) $d^2sp^3$
(c) $sp^3d$
(d) $sp^3$
(e) sp
Answer:
Question 53. Match the species in Column I with the geometry/shape in Column II.
Column I
(i) $H_3O^+$
(ii) $HC \equiv CH$
(iii) $ClO_2^–$
(iv) $NH_4^+$
Column II
(a) Linear
(b) Angular
(c) Tetrahedral
(d) Trigonal bipyramidal
(e) Pyramidal
Answer:
Question 54. Match the species in Column I with the bond order in Column II.
Column I
(i) NO
(ii) CO
(iii) $O_2^–$
(iv) $O_2$
Column II
(a) 1.5
(b) 2.0
(c) 2.5
(d) 3.0
Answer:
Question 55. Match the items given in Column I with examples given in Column II.
Column I
(i) Hydrogen bond
(ii) Resonance
(iii) Ionic solid
(iv) Covalent solid
Column II
(a) C
(b) LiF
(c) $H_2$
(d) HF
(e) $O_3$
Answer:
Question 56. Match the shape of molecules in Column I with the type of hybridisation in Column II.
Column I
(i) Tetrahedral
(ii) Trigonal
(iii) Linear
Column II
(a) $sp^2$
(b) sp
(c) $sp^3$
Answer:
Assertion and Reasong Type Questions
In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Question 57. Assertion (A) : Sodium chloride formed by the action of chlorine gas on sodium metal is a stable compound.
Reason (R) : This is because sodium and chloride ions acquire octet in sodium chloride formation.
(i) A and R both are correct, and R is the correct explanation of A.
(ii) A and R both are correct, but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A and R both are false.
Answer:
Question 58. Assertion (A) : Though the central atom of both $NH_3$ and $H_2O$ molecules are $sp^3$ hybridised, yet H–N–H bond angle is greater than that of H–O–H.
Reason (R) : This is because nitrogen atom has one lone pair and oxygen atom has two lone pairs.
(i) A and R both are correct, and R is the correct explanation of A.
(ii) A and R both are correct, but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A and R both are false.
Answer:
Question 59. Assertion (A): Among the two O–H bonds in $H_2O$ molecule, the energy required to break the first O–H bond and the other O–H bond is the same.
Reason (R) : This is because the electronic environment around oxygen is the same even after breakage of one O–H bond.
(i) A and R both are correct, and R is correct explanation of A.
(ii) A and R both are correct, but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A and R both are false.
Answer:
Long Answer Questions
Question 60. (i) Discuss the significance/ applications of dipole moment.
(ii) Represent diagrammatically the bond moments and the resultant dipole moment in $CO_2$, $NF_3$ and $CHCl_3$.
Answer:
Question 61. Use the molecular orbital energy level diagram to show that $N_2$ would be expected to have a triple bond, $F_2$, a single bond and $Ne_2$, no bond.
Answer:
Question 62. Briefly describe the valence bond theory of covalent bond formation by taking an example of hydrogen. How can you interpret energy changes taking place in the formation of dihydrogen?
Answer:
Question 63. Describe hybridisation in the case of $PCl_5$ and $SF_6$. The axial bonds are longer as compared to equatorial bonds in $PCl_5$ whereas in $SF_6$ both axial bonds and equatorial bonds have the same bond length. Explain.
Answer:
Question 64. (i) Discuss the concept of hybridisation. What are its different types in a carbon atom.
(ii) What is the type of hybridisation of carbon atoms marked with star.
(a)
(b)
(c)
(d)
(e)
Answer:
Comprehension given below is followed by some multiple choice questions. Each question has one correct option. Choose the correct option.
Molecular orbitals are formed by the overlap of atomic orbitals. Two atomic orbitals combine to form two molecular orbitals called bonding molecular orbital (BMO) and anti bonding molecular orbital (ABMO). Energy of anti bonding orbital is raised above the parent atomic orbitals that have combined and the energy of the bonding orbital is lowered than the parent atomic orbitals. Energies of various molecular orbitals for elements hydrogen to nitrogen increase in the order :
$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$
and for oxygen and fluorine order of energy of molecular orbitals is given below :
$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < (\pi 2p_x \approx \pi 2p_y) < (\pi^* 2p_x \approx \pi^* 2p_y) < \sigma^* 2p_z$
Different atomic orbitals of one atom combine with those atomic orbitals of the second atom which have comparable energies and proper orientation. Further, if the overlapping is head on, the molecular orbital is called ‘Sigma’, ($\sigma$) and if the overlap is lateral, the molecular orbital is called ‘pi’, ($\pi$). The molecular orbitals are filled with electrons according to the same rules as followed for filling of atomic orbitals. However, the order for filling is not the same for all molecules or their ions. Bond order is one of the most important parameters to compare the strength of bonds.
Question 65. Which of the following statements is correct?
(i) In the formation of dioxygen from oxygen atoms 10 molecular orbitals will be formed.
(ii) All the molecular orbitals in the dioxygen will be completely filled.
(iii) Total number of bonding molecular orbitals will not be same as total number of anti bonding orbitals in dioxygen.
(iv) Number of filled bonding orbitals will be same as number of filled anti bonding orbitals.
Answer:
Question 66. Which of the following molecular orbitals has maximum number of nodal planes?
(i) $\sigma^*1s$
(ii) $\sigma^*2p_z$
(iii) $\pi 2p_x$
(iv) $\pi^*2p_y$
Answer:
Question 67. Which of the following pair is expected to have the same bond order?
(i) $O_2, N_2$
(ii) $O_2^+, N_2^–$
(iii) $O_2^–, N_2^+$
(iv) $O_2^–, N_2^–$
Answer:
Question 68. In which of the following molecules, $\sigma 2p_z$ molecular orbital is filled after $\pi 2p_x$ and $\pi 2p_y$ molecular orbitals?
(i) $O_2$
(ii) $Ne_2$
(iii) $N_2$
(iv) $F_2$
Answer: