Chemical Bonding and Molecular Structure Mock Test – Class 11 Chemistry

1. Assertion: A highly negative electron gain enthalpy of a non-metal can favour ionic compound formation.
Reason: Such an atom releases energy readily when it gains an electron to form an anion.

2. One molecule contains one bond and four bonds. The carbon atoms are -hybridized. The total number of - and -bonds is

3. 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

4. The open structure of ice is mainly produced by

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

The table mainly shows that

6. A chlorine molecule, , forms when two chlorine atoms

7. The description that best separates from is

8. When changes to , the Lewis symbol of the ion is commonly written without the original valence dot because

9. A comparison of two bonds shows that Bond P has length and bond enthalpy , while Bond Q has length and bond enthalpy . The better conclusion is that

10. 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

11. The total number of valence electrons in is

12. Consider these statements about Lewis structures of polyatomic ions.
I. The total electron count must be adjusted for charge.
II. Brackets with charge help show the whole ion as a charged unit.
III. Resonance contributors may have different positions of atoms.
The valid set is

13. Use the table below to identify the row that gives the best description of each species.

The set of descriptions that fits the basic bonding vocabulary is

14. The bond angle in a molecule is the angle between

15. For , each neon atom has electronic configuration . Using the -type MO order for the second shell, the valence molecular orbital filling may be represented as

The bond order and stability conclusion are

16. The high melting points of many ionic compounds are mainly due to

17. A molecule contains two atoms joined by one line in its Lewis structure. The line represents

18. A table compares two bonding situations.

The comparison mainly separates the two cases by

19. The statement that best summarizes hydrogen bonding is

20. In a comparison of , , and , the lattice enthalpy magnitude is expected to decrease in the order

21. A graph is described below.

The graph supports the relation

22. For type , if all six surrounding atoms are identical, the molecule is expected to be non-polar mainly because

23. Consider these statements about weak intermolecular attractions.
I. Dispersion forces can occur in non-polar molecules.
II. Hydrogen bonding is usually stronger than ordinary dispersion forces.
III. van der Waals forces are the same as normal covalent bonds.
The valid set is

24. Hydrogen bonding is best described as

25. For , the statement "one carbon-oxygen bond is always shorter than the other two" is not suitable because

26. The final Lewis structure of is commonly written as because this arrangement

27. Hybridization is introduced in valence bond theory to explain

28. The total valence electron count in and the number of shared pairs in its usual Lewis structure are respectively

29. van der Waals forces are best described as

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

31. When a chlorine atom gains one electron during ion formation, the resulting species is represented as

32. A -bond is produced by

33. A graph description is given below.

The correct interpretation is

34. A compound contains and ions. The formula is decided by charge balance as

35. In the usual Lewis description of gaseous , the central beryllium atom is surrounded by how many electrons?

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

37. In the Lewis structure of , each fluorine atom has

38. Assertion: has polar bonds but zero dipole moment.
Reason: The three equal bond dipoles cancel in the trigonal planar geometry.

39. The central carbon atom in has two electron domains and no lone pair. The molecular shape and bond angle are

40. A formal-charge calculation for an atom uses , , and . The formal charge is

41. If has bond order , the bond order of is

42. The VSEPR type is expected to have the molecular shape

43. A representative element belongs to Group . The number of dots in the Lewis symbol of its neutral atom is normally

44. A data note lists two lattice processes:
Case 1: ,
Case 2: ,
The value represents

45. A student assigns hybridization to carbon in because the molecule has more than four atoms in total. The best correction is that

46. A proposed description says, " is polar because every bond is polar." The best evaluation is that the description is

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

48. Use the energy-step table for a simple ionic compound formed from and .

The step that represents lattice formation is

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

50. Use the arrangement described below. In a water molecule, oxygen is bonded to two hydrogen atoms and still has two pairs of electrons not used in bonding. The electron pairs not used in bonding are called