Chemical Bonding and Molecular Structure Mock Test – Class 11 Chemistry

1. The central sulfur atom in is commonly described using

2. A small highly charged cation is especially effective in producing covalent character because it has

3. A claim says, "A molecule with polar bonds must always have a non-zero dipole moment." The best correction is that

4. A claim says, "A double bond makes the central atom -hybridized because it contains two shared pairs." The best evaluation is that the claim

5. Atomic orbitals combine effectively to form molecular orbitals only when they have

6. The table below includes one incorrect hybridization assignment.

The row that needs correction is

7. A Lewis description of says oxygen forms two bonds and has two lone pairs. The number of bond pairs around oxygen is

8. With five electron domains around the central atom and no lone pair, the ideal electron-domain geometry is

9. Consider these statements about coordinate bonds.
I. The shared pair is supplied by one atom at the time of bond formation.
II. A lone-pair donor and an electron-pair acceptor are needed.
III. After formation, the coordinate bond is always weaker and visibly different from all ordinary covalent bonds in the same ion.
The valid set is

10. The molecular orbital configuration of may be written as . Its bond order is

11. Use the arrangement described below. Two fluorine atoms approach each other. Each fluorine atom has valence electrons and one unpaired electron available for sharing. The molecule formed is best represented as

12. Water has a much higher boiling point than mainly because water molecules

13. The correct comparison of carbon hybridization in , , and is

14. For formula with one lone pair on the central atom, the expected shape is not linear because

15. A species has molecular orbital data and , and its diagram shows all electrons paired. Its bond order and magnetic nature are

16. A metal with low ionization enthalpy is more likely to form a simple cation because

17. The bonding in contains

18. A central atom has six -bonding domains and no lone pair. The hybridization and ideal bond angles are best represented as

19. In a trigonal bipyramidal molecule such as , the axial bonds are often more strained than equatorial bonds because each axial bond has

20. A molecule contains two carbon atoms joined by a triple bond and one bond on each carbon. For each carbon, the hybridization is . The percentage -character of each carbon hybrid orbital and the total number of - and -bonds in the molecule are respectively

21. A formula-writing record for ionic compounds is shown below.

The row that should be simplified before accepting the formula is

22. Greater orbital overlap generally forms a stronger covalent bond because

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

24. Consider the statements about the octet rule.
I. It is useful for many main-group atoms.
II. It means every stable species must contain exactly electrons around every atom.
III. It helps explain both electron transfer and electron sharing in many simple substances.
The suitable evaluation is

25. The set of conditions that most strongly favours ionic bond formation is

26. The correct decreasing order of bond angle is

27. 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

28. For the neutral atoms , , , and , the Lewis-dot counts are respectively

29. The oxygen species table below summarizes bond order and relative bond length.

The table mainly shows that

30. Assertion: A hydrogen atom can become stable in by sharing one electron pair.
Reason: Sharing one electron pair gives each hydrogen atom access to two electrons.

31. Consider these statements about ionic compounds.
I. They generally have strong electrostatic forces in the solid state.
II. They commonly conduct electricity in solid state because ions are mobile.
III. They may conduct in molten or aqueous state because ions can move.
The valid set is

32. In the simple ionic model for , the electron transfer can be described as

33. A data table for three bonds between the same two elements is shown below.

The most suitable interpretation is that

34. Hydrogen usually follows the duplet rule rather than the octet rule because its stable outer arrangement contains

35. A diatomic species has the molecular orbital electron count shown below.

The bond order and magnetic nature of the species are

36. After drawing the single-bond skeleton of , the central carbon has only electrons around it. A common next step is to

37. A central atom uses two equivalent hybrid orbitals arranged at . The hybridization is

38. The description that best separates from is

39. A table of properties for a substance is shown below.

The observations together most strongly suggest that the substance is

40. In , nitrogen forms three bonds and keeps one lone pair. The total number of valence electron pairs around nitrogen is

41. A central atom in is bonded by one double bond to each atom and one single bond to each atom. The central atom has no lone pair. Using VSEPR domain counting, the steric number and the number of bonds around are respectively

42. The total number of valence electrons in is

43. In drawing , the total valence electron count is . After making four single bonds, the number of electrons left for lone pairs is

44. Use the arrangement described below. A molecule has two identical polar bonds. In Case 1, the molecule is linear. In Case 2, the molecule is bent. If the bond dipoles have equal magnitude in both cases, the molecule with non-zero dipole moment is

45. A qualitative ranking is needed for these salts: , , and . The expected order of decreasing lattice enthalpy magnitude is

46. Use the table below to compare molecular polarity.

The row that needs correction is

47. For , each beryllium atom has electronic configuration . Considering the valence molecular orbitals, the filling is

The best conclusion is

48. Interpret the graph description below.

The unusually high position of is best explained by

49. A central atom has six electron domains, with two lone pairs and four bond pairs. The most stable arrangement of the two lone pairs is opposite to each other because this

50. A graph is described below.

The graph is best used to decide hybridization by