Chemical Bonding and Molecular Structure Mock Test – Class 11 Chemistry

1. If , a dipole moment of equals

2. From the table below, identify the row that correctly describes molecular orbital formation.

The correct rows are

3. The anion that is most easily polarized is usually

4. In hybridization, the number of hybrid orbitals formed is

5. 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

6. A consistency check for an unknown neutral molecule gives these facts: total valence electrons , skeleton , and both oxygen atoms must complete octets without formal charge separation. The best final structure and carbon hybridization are

7. A statement about lattice enthalpy says, "The energy released in lattice formation is unimportant because the ions are already formed." The best response is that

8. A central atom has the VSEPR type . The molecular shape is

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

10. 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

11. Bent shape, polar bonds, and strong intermolecular hydrogen bonding together point most strongly to

12. The enthalpy change for the reaction

is to be estimated from average bond enthalpies. Use:


The approximate enthalpy change is

13. The set most likely to show strong hydrogen bonding is

14. Match the hydrogen-bonding description with the suitable example.

The proper matching is

15. Match the property of an ionic compound with its microscopic reason.

The proper matching is

16. A substance contains only particles. The most suitable classification is

17. The Lewis structure of contains

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

19. The following data refer to two possible ionic formation paths with the same non-metal: Path I has , , and . Path II has , , and . The better comparison is

20. The percentage ionic character of a bond is sometimes estimated by

If and , the percentage ionic character is

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

22. Dispersion forces arise mainly because

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

24. Use the arrangement described below. A central atom has five electron domains: three bond pairs and two lone pairs. The two lone pairs occupy equatorial positions in a trigonal bipyramidal arrangement. The molecular shape is

25. For VSEPR type with no lone pair on the central atom, the ideal shape is

26. The open structure of ice is mainly produced by

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

28. An atom becomes an ion when it

29. Use the graph description below.

The x-coordinate of the minimum point gives

30. 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

31. In a trigonal bipyramidal electron-domain arrangement, a lone pair prefers an equatorial position because it has

32. In the Lewis structure of , brackets and the positive charge are used because

33. A molecule contains hydrogen, but the hydrogen is bonded to carbon rather than to , , or . The safest conclusion is that the molecule

34. A graph description is given below.

The decreasing trend is mainly due to

35. The central atom in is commonly assigned

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

37. In , the electron-pair geometry and molecular shape are both tetrahedral because the central carbon has

38. Use the table below to identify the incorrect shape prediction.

The row that needs correction is

39. A formula is formed from and . This formula shows that

40. In bonding terminology, the valence shell means

41. In the molecular orbital description of , two electrons occupy degenerate orbitals singly. The molecule is therefore

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

43. A polar covalent bond has effective charge separation and bond length . If the dipole moment for complete ionic separation at the same distance would use , the observed dipole moment and percentage ionic character are respectively

44. A molecule has four identical polar bonds arranged tetrahedrally and no lone pair on the central atom. Its net dipole moment is expected to be

45. Use the table below to identify the hybridization and geometry link that needs correction.

The row that needs correction is

46. Four surrounding atoms around a central atom lead a student to conclude that the shape must be tetrahedral. The best evaluation is that the conclusion is

47. In an ionic solid such as , the formula unit represents

48. In gaseous , the central beryllium atom is linear in a simple valence-bond description. The hybridization assigned to beryllium is

49. Two hypothetical salts have the same electron gain enthalpy term for the non-metal. Salt P uses a metal with ionization enthalpy , while Salt Q uses a metal with ionization enthalpy . If their lattice formation enthalpies are similar, the more favourable salt formation is expected for

50. A sample is tested in three forms: solid, molten, and aqueous solution. It conducts only in the molten and aqueous forms. This result supports the presence of