Chemical Bonding and Molecular Structure Mock Test – Class 11 Chemistry

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

2. For a hypothetical ionic solid , the following energy data are given for one mole of formation through gaseous atoms and ions: ionization enthalpy of , electron gain enthalpy of , and lattice formation enthalpy . The approximate net enthalpy change is

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

4. The lattice enthalpy magnitude of is much larger than that of mainly because

5. 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

6. A learner says that the compound with the largest electronegativity difference will always be the most stable ionic solid. The best evaluation is that the statement is

7. In the formation of , sodium and chlorine change as follows:

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

9. A graph compares boiling points of noble gases from to . The curve rises as atomic size and number of electrons increase. The best explanation is that

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

11. The molecular shapes of and are different because

12. A graph description is given below.

The most suitable interpretation is that

13. Assertion: is paramagnetic according to molecular orbital theory.
Reason: Its molecular orbital configuration contains unpaired electrons in antibonding orbitals.

14. A resonance-aware Lewis description of is needed because

15. A qualitative energy expression for the formation of an ionic solid may be written as

For favourable formation, the lattice-formation term is especially important because it is usually

16. Resonance in explains why the two bond lengths are

17. The molecule has three equivalent bonding regions around sulfur in a trigonal planar arrangement. The hybridization used in this simple description is

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

19. For a fixed cation , the covalent character among , , , and is expected to increase in the order

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

21. The central sulfur atom in is commonly described using

22. The observed dipole moment of a bond is , while the dipole moment calculated for complete ionic character is . The percentage ionic character is

23. A student tries to predict structure by saying, "First count total valence electrons, then draw a reasonable Lewis structure, then use electron domains for shape and hybridization, and finally use shape to decide dipole cancellation." This method is

24. The hybridization of the central atom in is usually taken as

25. 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

26. A structure builder has valence electrons for . After connecting carbon to three oxygen atoms with single bonds, the number of electrons used in the skeleton bonds is

27. The SI unit of dipole moment is

28. Use the graph description below.

The x-coordinate of the minimum point gives

29. The approximate bond angle order for , , and is

30. A row in a Lewis-structure table is inconsistent. Identify it.

The inconsistent row is

31. Fluorine has valence electrons. After it forms , its Lewis symbol should show

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

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

34. 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

35. In VSEPR theory, a double bond around the central atom is counted as

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

37. The octet rule fails for incomplete-octet molecules because the central atom may

38. Six electron domains around a central atom with no lone pair give the ideal molecular shape

39. The molecule expected to have zero dipole moment among the following is

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

41. In a single covalent bond between two atoms, the bond is usually

42. The three major resonance contributors of are called equivalent because they

43. Match each bond description with the number of shared electron pairs.

The suitable matching is

44. Resonance is needed when

45. VSEPR theory predicts molecular shape mainly by considering

46. A compound is formed from and . The simplest formula obtained by charge balance is

47. The main reason a complete octet cannot be assigned to every atom in an odd-electron species is that

48. Dipole moment is a measure of

49. A species has the Lewis structure with three lone pairs on oxygen. This structure best represents

50. A noble-gas configuration is often used as a stability reference because noble gases generally have