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Samacheer Kalvi solutions for Chemistry - Volume 1 and 2 [English] Class 11 TN Board chapter 10 - Chemical bonding [Latest edition]

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Samacheer Kalvi solutions for Chemistry - Volume 1 and 2 [English] Class 11 TN Board chapter 10 - Chemical bonding - Shaalaa.com
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Solutions for Chapter 10: Chemical bonding

Below listed, you can find solutions for Chapter 10 of Tamil Nadu Board of Secondary Education Samacheer Kalvi for Chemistry - Volume 1 and 2 [English] Class 11 TN Board.


Evaluation
Evaluation [Pages 104 - 107]

Samacheer Kalvi solutions for Chemistry - Volume 1 and 2 [English] Class 11 TN Board 10 Chemical bonding Evaluation [Pages 104 - 107]

Choose the best answer

Evaluation | Q I. 1. | Page 104

In which of the following Compounds does the central atom obey the octet rule?

  • XeF4

  • AlCl3

  • SF6

  • SCl2

Evaluation | Q I. 2. | Page 104

In the molecule OA = C = OB, the formal charge on OA, C and OB are respectively.

  • -1, 0, +1

  • +1, 0, -1

  • -2, 0, +2

  • 0, 0, 0

Evaluation | Q I. 3. | Page 104

Which of the following is electron deficient?

  • PH3

  • (CH3)2

  • BH3

  • NH3

Evaluation | Q I. 4. | Page 104

Which of the following molecule contain no л bond?

  • SO2

  • SO2

  • CO2

  • H2O

Evaluation | Q I. 5. | Page 104

The ratio of number of sigma (σ) and pi (л) bonds in 2- butynal is ______.

  • `8/3`

  • `5/3`

  • `8/2`

  • `9/2`

Evaluation | Q I. 6. | Page 104

Which one of the following is the likely bond angles of sulphur tetrafluoride molecule?

  • 120°, 80°

  • 109°28’

  • 90°

  • 89°, 117°

Evaluation | Q I. 7. | Page 104

Assertion: Oxygen molecule is paramagnetic.

Reason: It has two unpaired electron in its bonding molecular orbital.

  • Both assertion and reason are true and reason is the correct explanation of assertion

  • Both assertion and reason are true but reason is not the correct explanation of assertion

  • Assertion is true but reason is false

  • Both assertion and reason are false

Evaluation | Q I. 8. | Page 104

According to Valence bond theory, a bond between two atoms is formed when ______.

  • fully filled atomic orbitals overlap

  • half-filled atomic orbitals overlap

  • non – bonding atomic orbitals overlap

  • empty atomic orbitals overlap

Evaluation | Q I. 9. | Page 104

In ClF3, NF3 and BF3 molecules the chlorine, nitrogen and boron atoms are ______.

  • sp3 hybridised

  • sp3, sp3 and sp2 respectively

  • sp3 hybridised

  • sp3d, sp3 and sp hybridised respectively

Evaluation | Q I. 10. | Page 104

When ones and three p orbitals hybridise,

  • four equivalent orbitals at 90° to each other will be formed

  • four equivalent orbitals at 109°28’ to each other will be formed

  • four equivalent orbitals, that are lying the same plane will be formed

  • none of these

Evaluation | Q I. 11. | Page 105

Which of these represents the correct order of their increasing bond order.

  • C2+ < C22- < O22- < O2

  • C22- < C2+ < O2 < O22-

  • O22- < O2 < C22- < C2+

  • O22- < C2+ < O2 < C22-

Evaluation | Q I. 12. | Page 105

Hybridisation of central atom in PCl5 involves the mixing of orbitals.

  • s, Px, Py, dx2, `"d"_("x"^2 – "y"^2)`

  • s, px, py, pxy, `"d"_("x"^2 – "y"^2)`

  • s, px, py, pz, `"d"_("x"^2 – "y"^2)`

  • s, px, Py, dxy, `"d"_("x"^2 – "y"^2)`

Evaluation | Q I. 13. | Page 105

The correct order of O – O bond length in hydrogen peroxide, ozone and oxygen is

  • H2O2 > O3 > O2

  • O2 > O3 > H2O2

  • O2 > H2O2 > O3

  • O3 > O2 > H2O2

Evaluation | Q I. 14. | Page 105

Which one of the following is diamagnetic?

  • O2

  • O22-

  • O2+

  • None of these

Evaluation | Q I. 15. | Page 105

Bond order of a species is 2.5 and the number of electons in its bonding molecular orbital is formd to be 8 The no. of electons in its antibonding molecular orbital is

  • three

  • four

  • Zero

  • can not be calculated from the given information.

Evaluation | Q I. 16. | Page 105

Shape and hybridisation of IF5 are ______.

  • Trigonal bipyramidal, sp3d2

  • Trigonal bipyramidal, sp3d

  • Square pyramidal, sp3d2

  • Octahedral, sp3d2

Evaluation | Q I. 17. | Page 105

Pick out the incorrect statement from the following:

  • sp3 hybrid orbitals are equivalent and are at an angle of 109°28’ with each other

  • dsp2 hybrid orbitals are equivalent and bond angle between any two of them is 90°

  • All five sp3d hybrid orbitals are not equivalent out of these five sp3d hybrid orbitals, three are at an angle of 120° remaining two are perpendicular to the plane containing the other three

  • none of these

Evaluation | Q I. 18. | Page 105

The molecules having same hybridisation, shape and number of lone pairs of electons are ______.

  • SeF4, XeO2F2

  • SF4, XeF2

  • XeOF4, TeF4

  • SeCl4, XeF4

Evaluation | Q I. 19. | Page 105

In which of the following molecules / ions BF3, NO2 H20 the central atom is sp2 hybridised?

  • NH2 and H2O

  • NO2 and H2O

  • BF3 and NO2

  • BF3 and NH2

Evaluation | Q I. 20. | Page 106

Some of the following properties of two species, NO3 and H3O+ are described below. Which one of them is correct?

  • dissimilar in hybridization for the central atom with different structure.

  • isostructural with same hybridisation for the central atom.

  • different hybridisation for the central atom with same structure.

  • none of these

Evaluation | Q I. 21. | Page 106

The types of hybridiration on the five carbon atom from right to left in the, 2,3 pentadiene.

  • sp3, sp2, sp, sp2, sp3

  • sp3, sp, sp, sp, sp3

  • sp2, sp, sp2, sp2, sp3

  • sp3, sp3, sp2, sp3, sp3

Evaluation | Q I. 22. | Page 106

XeF2 is isostructural with ______.

  • SbCl2

  • BaCl2

  • TeF2

  • ICl2

Evaluation | Q I. 23. | Page 106

The percentage of s-character of the hybrid orbitals in methane, ethane, ethene and ethyne are respectively.

  • 25, 25, 33.3, 50

  • 50, 50, 33.3, 25

  • 50, 25, 33.3, 50

  • 50, 25, 25, 50

Evaluation | Q I. 24. | Page 106

Of the following molecules, which have shape similar to carbon dioxide?

  • SnCl2

  • NO2

  • C2H2

  • All of these

Evaluation | Q I. 25. | Page 106

According to VSEPR theory, the repulsion between different parts of electrons obey the order.

  • l.p – l.p > b.p–b.p> l.p–b.p

  • b.p–b.p> b.p–l.p> l.p–b.p

  • l.p–l.p> b.p–l.p > b.p–b.p

  • b.p–b.p> l.p–l.p> b.p–l.p

Evaluation | Q I. 26. | Page 106

Shape of ClF3 is ______.

  • Planar triangular

  • Pyramidal

  • "T” Shaped

  • none of these

Evaluation | Q I. 27. | Page 106

Non – Zero dipole moment is shown by ______.

  • CO2

  • p – dichlorobenzene

  • carbon tetrachloride

  • water

Evaluation | Q I. 28. | Page 106

Which of the following conditions is not correct for resonating structures?

  • the contributing structure must have the same number of unpaired electrons

  • the contributing structures should have similar energies

  • the resonance hybrid should have higher energy than any of the contributing structure.

  • none of these

Evaluation | Q I. 29. | Page 106

Among the following, the compound that contains, ionic, covalent and Coordinate linkage is ______.

  • NH4Cl

  • NH3

  • NaCl

  • none of these

Evaluation | Q I. 30. | Page 107

CaO and NaCl have the same crystal structure and approximately the same radii. If U is the lattice energy of NaCl, the approximate lattice energy of CaO is ______.

  • U

  • 2U

  • `"U"/2`

  • 4U

Write brief answer to the following questions.

Evaluation | Q II. 31. i) | Page 106

Define bond order.

Evaluation | Q II. 31. ii) | Page 107

Define Hybridisation.

Evaluation | Q II. 31. iii) | Page 107

Define σ – bond.

Evaluation | Q II. 32. | Page 107

What is a pi - bond?

Evaluation | Q II. 33. | Page 107

In CH4, NH3, and H2O, the central atom undergoes sp3 hybridization – yet their bond angles are different. Why?

Evaluation | Q II. 34. | Page 107

Explain Sp2 hybridization in BF3.

Evaluation | Q II. 35. | Page 107

Draw the M.O diagram for oxygen molecule calculate its bond order and show that O2 is paramagnetic.

Evaluation | Q II. 36. | Page 107

Draw MO diagram of CO and calculate its bond order.

Evaluation | Q II. 37. | Page 107

What do you understand by Linear combination of atomic orbitals in MO theory?

Evaluation | Q II. 38. | Page 107

Discuss the formation of N2 molecule using MO Theory.

Evaluation | Q II. 39. | Page 107

What is dipole moment?

Evaluation | Q I. 40. | Page 107

Linear form of carbondioxide molecule has two polar bonds. Yet the molecule has Zero dipole moment. Why?

Evaluation | Q II. 41. i) | Page 107

Draw the Lewis structure for the following species.

\[\ce{NO^-3}\]

Evaluation | Q II. 41. ii) | Page 107

Draw the Lewis structure for the following species.

\[\ce{SO^{2-}4}\]

Evaluation | Q II. 41. iii) | Page 107

Draw the Lewis structure for the following species.

HNO3

Evaluation | Q II. 41. iv) | Page 107

Draw the Lewis structure for the following species.

O3

Evaluation | Q II. 42. | Page 107

Explain the bond formation if BeCl2 and MgCl2.

Evaluation | Q II. 43. | Page 107

Which bond is stronger σ or π? Why?

Evaluation | Q II. 44. | Page 107

Define bond energy.

Evaluation | Q II. 45. | Page 107

Hydrogen gas is diatomic whereas inert gases are monoatomic – Explain on the basis of MO theory.

Evaluation | Q II. 46. a. | Page 107

What is the Polar Covalent bond?

Evaluation | Q II. 46. b. | Page 107

Explain an example of a Polar Covalent bond.

Evaluation | Q II. 47. i) | Page 107

Considering x-axis as the molecular axis which out of the following will form a sigma bond.

1s and 2py

Evaluation | Q II. 47. ii) | Page 107

Considering x-axis as the molecular axis which out of the following will form a sigma bond.

2px and 2py

Evaluation | Q II. 47. iii) | Page 107

Considering x-axis as the molecular axis which out of the following will form a sigma bond.

2px and 2pz

Evaluation | Q II. 47. iv) | Page 107

Considering x-axis as the molecular axis which out of the following will form a sigma bond.

1s and 2pz

Evaluation | Q II. 48. | Page 107

Explain resonance with reference to a carbonate ion.

Evaluation | Q II. 49. a. | Page 107

Explain the bond formation in ethylene.

Evaluation | Q II. 49. b. | Page 107

Explain the bond formation in acetylene.

Evaluation | Q II. 50. a) | Page 107

What type of hybridisation is possible in the following geometeries?

octahedral

Evaluation | Q II. 50. b) | Page 107

What type of hybridisation is possible in the following geometeries?

tetrahedral

Evaluation | Q II. 50. c) | Page 107

What type of hybridisation is possible in the following geometeries?

square planer

Evaluation | Q II. 51. | Page 107

Explain VSEPR theory. Applying this theory to predict the shapes of IF7 and SF6.

Evaluation | Q II. 52. | Page 107

CO2 and H2O both are triatomic molecule but their dipole moment values are different. Why?

Evaluation | Q II. 53. | Page 107

Which one of the following has the highest bond order?

  1. N2
  2. N2+
  3. N2
Evaluation | Q II. 54. | Page 107

Explain the covalent character in ionic bond.

Evaluation | Q II. 55. | Page 107

Describe Fajan’s rule.

Solutions for 10: Chemical bonding

Evaluation
Samacheer Kalvi solutions for Chemistry - Volume 1 and 2 [English] Class 11 TN Board chapter 10 - Chemical bonding - Shaalaa.com

Samacheer Kalvi solutions for Chemistry - Volume 1 and 2 [English] Class 11 TN Board chapter 10 - Chemical bonding

Shaalaa.com has the Tamil Nadu Board of Secondary Education Mathematics Chemistry - Volume 1 and 2 [English] Class 11 TN Board Tamil Nadu Board of Secondary Education solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. Samacheer Kalvi solutions for Mathematics Chemistry - Volume 1 and 2 [English] Class 11 TN Board Tamil Nadu Board of Secondary Education 10 (Chemical bonding) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.

Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. Samacheer Kalvi textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.

Concepts covered in Chemistry - Volume 1 and 2 [English] Class 11 TN Board chapter 10 Chemical bonding are Introduction of Chemical Bonding, Types of Chemical Bonds, Ionic or Electrovalent Bond, Coordinate Covalent Bond, Bond Parameters, Valence Shell Electron Pair Repulsion Theory (VSEPR), Valence Bond Theory, Valence Bond Theory - Orbital Overlap Concept, Hybridisation - Introduction, Molecular Orbital Theory.

Using Samacheer Kalvi Chemistry - Volume 1 and 2 [English] Class 11 TN Board solutions Chemical bonding exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in Samacheer Kalvi Solutions are essential questions that can be asked in the final exam. Maximum Tamil Nadu Board of Secondary Education Chemistry - Volume 1 and 2 [English] Class 11 TN Board students prefer Samacheer Kalvi Textbook Solutions to score more in exams.

Get the free view of Chapter 10, Chemical bonding Chemistry - Volume 1 and 2 [English] Class 11 TN Board additional questions for Mathematics Chemistry - Volume 1 and 2 [English] Class 11 TN Board Tamil Nadu Board of Secondary Education, and you can use Shaalaa.com to keep it handy for your exam preparation.

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