Exam-Style Mock Test | Class 11: Chemical Bonding Test
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Chemical Bonding and Molecular Structure Mock Test – Class 11 Chemistry

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Chemical Bonding and Molecular Structure – Progressive Test

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1. A structural and molecular-orbital comparison is given below.

Species Key data Conclusion
P. bond order , all electrons paired strong and diamagnetic
Q. bond order , two unpaired electrons paramagnetic
R. extra electron in antibonding orbital bond order lower than
S. bond order stable strong molecule

The row that needs correction is

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2. A bonding molecular orbital is lower in energy than the combining atomic orbitals because

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3. 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

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

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5. The SI unit of dipole moment is

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6. A claim says, "In covalent bonding, each atom completely gives away its electron to the other atom." The best evaluation is that the claim

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7. A common Lewis contributor of has one bond and two single bonds. Each singly bonded oxygen has three lone pairs, the double-bonded oxygen has two lone pairs, and nitrogen has no lone pair. Using , the formal charge set on nitrogen, one singly bonded oxygen, and the double-bonded oxygen is respectively

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

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

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10. The pair of terms that describes shared and non-shared electron pairs in bonding is

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11. A completed Lewis structure should be checked against the total valence electron count because

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12. A molecule that can act as a Lewis acid because its central atom has an incomplete octet is

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13. 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.

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14. Interpret the graph description below.

A graph compares the boiling points of group hydrides. The points for , , and follow a gradual trend with increasing molar mass, but appears much higher than expected.

The unusually high position of is best explained by

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15. The electrons most directly involved in ordinary chemical bonding are

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16. The most reasonable skeleton for before placing lone pairs and multiple bonds is

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17. 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

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18. A learner says that carbonate ion rapidly flips among three separate structures. The better statement is that carbonate ion is

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19. The molecular orbital bond order is calculated using

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20. In bonding terminology, the valence shell means

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21. In a Lewis symbol, the dots placed around the element symbol represent

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22. Use the table below for main-group atoms approaching a noble-gas-like arrangement.

Atom Valence electrons Likely simple ionic change
P. loses electron
Q. loses electrons
R. gains electrons
S. loses electron

The row that does not fit the usual Kossel-Lewis tendency is

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23. Hydrogen is placed as the central atom in a proposed Lewis skeleton. The best reason to reject that skeleton is that hydrogen

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24. In the resonance hybrid of , the three bonds are expected to have

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

Species Shape Hybridization of central atom 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

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

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

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28. Water has a bent shape because the oxygen atom has

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29. For , the total valence electron count is

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30. Assertion: Bond formation is often associated with a decrease in energy of the system.
Reason: A lower-energy arrangement is generally more stable than a higher-energy arrangement.

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

If and , the percentage ionic character is

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32. van der Waals forces are best described as

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33. 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

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34. A -bond is produced by

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35. For the Lewis structure of , the valence electron total is

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

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37. In the formation of from and , the donor and acceptor are respectively

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38. 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

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39. 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

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40. In resonance contributors, the feature that must remain unchanged is

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41. The role of a small highly charged ion in ionic bonding is best understood through

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

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43. The hybridization and approximate bond angle around each carbon in are

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44. A set of proposed resonance contributors is being checked. The proposal that must be rejected as a resonance contributor is the one in which

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45. Assertion: has more covalent character than .
Reason: is larger and more polarizable than .

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46. Valence bond theory successfully explains directional bonding, but one important limitation is that it does not simply explain

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

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48. The formal charges in one contributor of are most suitably described as

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49. 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

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

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