Exam-Style Mock Test | Class 11: Chemical Bonding Test

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Quiz: Chemical Bonding and Molecular Structure – Progressive Test
Total questions: 487
Questions in this round: 50
Time allotted: 70 minutes
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Total rounds: 10
Current round: 1^st
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Quiz: Chemical Bonding and Molecular Structure – Certificate Challenge
Total questions: 487
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Time allotted: 100 minutes
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Chemical Bonding and Molecular Structure – Progressive Test

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1 / 50

1. Hydrogen bonding is best described as

a) ordinary bonding between two hydrogen atoms

b) attraction to a lone pair without a polar bond

c) transfer to a metal atom in an ionic lattice

d) polar bond plus a lone pair

2 / 50

2. The comparison , , and is especially useful for bond-length trends because

a) same atoms with changing shared-pair count

b) all three are ionic bonds with different charges

c) the number of atoms in each symbol is different

d) carbon atoms have no valence electrons

3 / 50

3. The carbon-carbon double bond in is formed by

a) - -overlap and - -overlap

b) overlap of two -orbitals only

c) overlap of two orbitals to form two -bonds

d) complete transfer of one electron from one carbon atom to the other

4 / 50

4. The species that best illustrates an expanded octet rather than an incomplete octet or odd-electron structure is

5 / 50

5. The shape of is square planar because xenon has

a) four bond pairs and two lone pairs in an octahedral set

b) three bond pairs and two lone pairs in a trigonal bipyramidal set

c) two bond pairs and three lone pairs in a linear set

d) four bond pairs and no lone pair in a tetrahedral set

6 / 50

6. Use the table below to compare molecular polarity.

Molecule	Shape feature	Expected net dipole moment
P.	linear and symmetrical
Q.	trigonal planar and symmetrical
R.	bent	non-zero
S.	trigonal pyramidal

The row that needs correction is

a) P

b) R

c) Q

d) S

7 / 50

7. The internally consistent description of is

a) bent shape, -hybridized carbon, non-zero net dipole moment

b) trigonal planar shape, -hybridized carbon, one lone pair on carbon

c) linear shape, -hybridized carbon, zero net dipole moment

d) tetrahedral shape, -hybridized carbon, four single bonds

8 / 50

8. A carbon atom in a Lewis structure has four single bonds and no lone pair. Its formal charge is

9 / 50

9. Lone pairs reduce bond angles more strongly than bond pairs because lone pairs

a) are always outside the molecule and unrelated to shape

b) have no electrons and cannot repel anything

c) occupy more effective space around the central atom

d) turn every molecule into a linear structure

10 / 50

10. The features are three equivalent bonds around the central atom, resonance delocalization, trigonal planar geometry, and non-zero overall charge. The best matching species is

11 / 50

11. A proposed Lewis structure for has the skeleton with only single bonds and full octets on both oxygens. What is the main problem with this structure before multiple bonds are formed?

a) carbon has only electrons around it

b) the total valence electron count becomes

c) each oxygen has more than electrons

d) hydrogen is incorrectly placed at the centre

12 / 50

12. A proposed Lewis structure has formal charges , , and on three atoms in a neutral molecule. Another possible structure has all formal charges . If both satisfy octets, the second structure is usually preferred because it

a) makes the molecule ionic by force

b) uses more electrons than available

c) has less formal charge separation

d) gives every atom the same electronegativity

13 / 50

13. Use the passage below.

A learner draws a Lewis structure for . Oxygen is placed at the centre and connected to two hydrogen atoms with single bonds. The learner then tries to place the remaining electrons on hydrogen to complete hydrogen octets.

The best correction is that the remaining electrons should be placed on

a) hydrogen, because hydrogen must complete an octet

b) both hydrogens as three lone pairs each

c) oxygen, because hydrogen follows the duplet rule

d) no atom, because lone pairs are not allowed in water

14 / 50

14. With three bond pairs and two lone pairs around the central atom, VSEPR theory predicts the molecular shape as

a) trigonal planar

b) T-shaped

c) seesaw

d) trigonal pyramidal

15 / 50

15. The total number of valence electrons used to draw is

16 / 50

16. A solid ionic compound usually does not conduct electricity, but its molten form conducts. The reason is that

a) melting removes all charges from the compound

b) ions are absent in the solid but created only after melting

c) electrons become the only charge carriers in the molten state

d) fixed ions in solid, mobile ions in melt

17 / 50

17. In , the coordinate bond is formed when

a) donates all bonding electrons to form

b) oxygen donates a lone pair to

c) donates a lone pair to oxygen

d) donates an electron pair to oxygen

18 / 50

18. A -bond is formed by

a) sidewise overlap above and below the internuclear axis only

b) repulsion between two filled orbitals only

c) complete transfer of electrons to form ions

d) end-to-end overlap along the internuclear axis

19 / 50

19. In ethyne, , each carbon atom has

a) two -directions and two unhybridized -orbitals

b) five bonding directions using hybridization

c) four -bonding directions and keeps no unhybridized -orbital

d) three -bonding directions and keeps one unhybridized -orbital

20 / 50

20. The symbols and in molecular orbital theory represent

a) ionic and metallic bonds, respectively

b) lone pair and bond pair on the same atom

c) formal charge and oxidation state

d) bonding and antibonding orbitals

21 / 50

21. Use the comparison below.

Species	Central atom	Electrons around central atom in usual Lewis description
P.
Q.		odd-electron arrangement
R.
S.

The entries that represent expanded-octet cases are

a) Q and S only

b) P and Q only

c) P and R only

d) R and S only

22 / 50

22. Consider these statements about hydrogen bonding.
I. It is usually stronger than ordinary dipole-dipole attraction.
II. It is usually weaker than a normal covalent bond.
III. It requires hydrogen bonded directly to a highly electronegative atom such as , , or in the usual strong cases.
The valid set is

a) I and II only

b) II and III only

c) I and III only

d) I, II, and III

23 / 50

23. The factor that does not favour formation of a strongly ionic compound is

a) low ionization enthalpy of the metal

b) large electronegativity difference

c) highly negative electron gain enthalpy of the non-metal

d) very small lattice enthalpy magnitude

24 / 50

24. For a linear molecule with no lone pair on the central atom and two electron domains around it, the bond angle at is expected to be

25 / 50

25. A comparison of two possible ionic compounds is given below.

Compound	Metal ionization tendency	Non-metal electron gain tendency	Lattice stabilization
P	low ionization enthalpy	favourable electron gain	high
Q	high ionization enthalpy	weak electron gain	low

The better prediction is that

a) neither can form ions because lattice stabilization is listed

b) P is more likely to form an ionic solid favourably

c) Q is more likely to form an ionic solid favourably

d) both are equally favoured because both contain a metal

26 / 50

26. A graph is described below.

The x-axis lists the number of electron domains around a central atom as , , , , and . The y-axis lists the usual hybridization labels as , , , , and , respectively.

The graph is best used to decide hybridization by

a) counting only the number of atoms in the whole molecule

b) counting electron domains around the central atom

c) arranging elements by increasing atomic mass

d) counting only -bonds and ignoring -bonds

27 / 50

27. A note describes a particle as follows: it contains one carbon atom and two oxygen atoms chemically bonded together. The notation that best represents this particle is

28 / 50

28. The most common unit used for reporting ordinary covalent bond lengths is

29 / 50

29. A comparison of two ionic solids gives the following information:

Solid	Cation formation	Anion formation	Lattice formation
P	moderately easy	favourable	large energy release
Q	difficult	favourable	small energy release

The more reasonable prediction is that

a) Q is more favoured because difficult cation formation always increases stability

b) neither can form any ions because lattice formation is listed

c) P is favoured by easier ion formation and stronger lattice stabilization

d) P and Q must be equally favoured because both have anion formation

30 / 50

30. Polarization in an ionic bond means

a) anion electron-cloud distortion by a cation

b) conversion of all bonds into metallic bonds

c) complete neutralization of ionic charges

d) electron-cloud movement from cation to anion only

31 / 50

31. One common Lewis contributor of shows nitrogen with one bond and two single bonds. More than one contributor is possible because

a) oxygen atoms exchange positions in

b) the negative charge vanishes from

c) nitrogen changes into oxygen in

d) shifts among equivalent oxygen atoms

32 / 50

32. A salt is more likely to form readily when the metal has low ionization enthalpy, the non-metal has favourable electron gain enthalpy, and the ionic solid has high lattice stabilization. This combination works because

a) electron transfer is discouraged and no ions form

b) ion formation is easier and lattice formation releases energy

c) the compound becomes stable without any electrostatic attraction

d) all three changes require large energy input

33 / 50

33. Equal and opposite bond dipoles arranged in a straight line give a net dipole moment expected to be

a) zero because the bond dipoles cancel vectorially

b) independent of molecular shape

c) equal to the sum of bond lengths only

d) maximum because every polar bond adds as a scalar

34 / 50

34. For , the usual expanded-octet Lewis description places around sulfur

a) electrons

b) electrons

c) electrons

d) electrons

35 / 50

35. A compound has a central atom with only three single bonds and no lone pair in its usual Lewis structure. If is from Group , the central atom has

a) electrons around it

b) electrons around it

c) electrons around it

d) electrons around it

36 / 50

36. A symbolic record for forming magnesium oxide is shown:

The formula of the ionic compound formed from these ions is

37 / 50

37. While drawing the Lewis structure of a negative ion, the total valence electron count is adjusted by

a) subtracting electrons equal to the charge magnitude twice

b) adding electrons equal to the charge magnitude

c) ignoring the charge until after the structure is complete

d) subtracting electrons equal to the charge magnitude

38 / 50

38. Use the table below to identify the inconsistent molecular-structure record.

Species	Shape	Polarity statement
P.	linear	non-polar due to cancellation
Q.	trigonal planar	non-polar if ideal and symmetric
R.	trigonal pyramidal	polar
S.	linear	non-polar

The row that needs correction is

a) P

b) R

c) S

d) Q

39 / 50

39. The bonding in contains

a) three -bonds only

b) two -bonds and one -bond

c) three -bonds only

d) one -bond and two -bonds

40 / 50

40. For an oxygen atom with two lone pairs and one single bond in a Lewis structure, the formal charge is

41 / 50

41. Intramolecular hydrogen bonding means that the hydrogen bond is formed

a) within the same molecule

b) by complete transfer of hydrogen as

c) only between two separate molecules

d) only between two ions in a crystal lattice

42 / 50

42. The Lewis symbol of neutral sodium, , contains one dot because sodium

a) has one valence electron

b) forms only one compound in nature

c) has one electron in its first shell

d) has one completely filled shell

43 / 50

43. A central atom has three single bonds and one lone pair. The most consistent set is

a) hybridization, trigonal pyramidal shape, non-zero dipole

b) hybridization, seesaw shape, no lone pair

c) hybridization, linear shape, zero dipole moment

d) hybridization, trigonal planar shape, zero dipole moment

44 / 50

44. A -bond is produced by

a) end-to-end overlap of two -orbitals only

b) electron transfer from a metal to a non-metal

c) formation of a lattice from gaseous ions

d) sidewise overlap of parallel orbitals

45 / 50

45. A sample of pure water contains molecules. This means that each molecule contains

a) separated hydrogen and oxygen atoms with no fixed ratio

b) bonded hydrogen and oxygen in a fixed ratio

c) only oxygen atoms arranged in pairs

d) sodium and chloride ions arranged alternately

46 / 50

46. The VSEPR type is non-polar when all atoms are identical because

a) all lone pairs become bond pairs

b) every square planar molecule has zero bond polarity

c) square planar shape allows bond-dipole cancellation

d) it has no bonds

47 / 50

47. Consider the following order proposed for lattice enthalpy magnitude:

The best reason for being placed above is that

a) and have identical ion sizes

b) has no oxide ion

c) is smaller at the same charge

d) contains singly charged ions

48 / 50

48. The percentage ionic character of a covalent bond can be estimated by comparing

a) observed versus fully ionic dipole

b) bond angle with bond enthalpy only

c) formal charge with atomic mass only

d) lattice enthalpy with boiling point only

49 / 50

49. In bonding terminology, the valence shell means

a) the innermost occupied electron shell of an atom

b) the space between two bonded atoms only

c) the outermost occupied electron shell of an atom

d) the shell that must always contain electrons

50 / 50

50. The formation of a stable ionic compound depends on several energy changes. The set most directly involved is

a) surface tension, viscosity, and optical density only

b) bond angle, molecular mass, and vapour pressure only

c) bond polarity, dipole moment, and bond angle only

d) ionization, electron gain, and lattice enthalpies

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Chemical Bonding and Molecular Structure Mock Test – Class 11 Chemistry

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