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

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2. A molecule has five electron domains around its central atom, out of which one is a lone pair. The hybridization and molecular shape are respectively

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3. A graph is described below.

The x-axis lists the number of electron domains around a central atom as , , , , and . The y-axis lists the usual hybridization labels as , , , , and , respectively.

The graph is best used to decide hybridization by

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4. A graph description is given below.

A curve shows bond angle on the y-axis for molecules with four electron domains around the central atom. Moving from to to , the curve decreases.

The decreasing trend is mainly due to

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

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6. Equal and opposite bond dipoles arranged in a straight line give a net dipole moment expected to be

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7. 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

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

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9. In ozone, , two equivalent resonance contributors show one single bond and one double bond. The average bond order of each bond in the resonance hybrid is

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10. A Lewis structure for a molecule is rejected only because one atom has fewer than electrons around it. The safest conclusion is that the rejection is

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11. One common Lewis contributor of shows nitrogen with one bond and two single bonds. More than one contributor is possible because

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12. For carbon-carbon bonds involving the same pair of atoms, the usual order of increasing bond length is

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

Row Combination Orbital formed Energy effect
P in-phase bonding lower energy
Q out-of-phase antibonding higher energy
R in-phase antibonding higher energy
S out-of-phase bonding lower energy

The correct rows are

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14. The molecular shapes of and are different because

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15. In ethyne, , each carbon atom has

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

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17. For a diatomic molecule, increasing bond order usually corresponds to

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18. 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

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19. Use the passage below.

A neutral molecule contains two identical atoms of an element . Each atom has valence electrons. The molecule is stable when each atom shares enough electrons to complete an octet, and no complete electron transfer occurs.

The most suitable bond between the two atoms is

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20. A student writes the Lewis symbol of with dots because aluminium has electrons in more than one shell. The best correction is that Lewis symbols

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

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22. For , the usual expanded-octet Lewis description places around sulfur

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23. The internally consistent description of is

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24. Read the case below.

A neutral molecule has valence electrons. The central atom is bonded to three identical atoms . After forming three single bonds and completing octets on the terminal atoms, the central atom has one lone pair.

The most suitable VSEPR shape of the molecule is

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25. The Kossel-Lewis approach connects chemical bonding mainly with

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26. In ice, water molecules are arranged through hydrogen bonding in an open structure. This explains why ice

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

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

Row Molecule Central atom description
P has four bond pairs and no lone pair
Q has three bond pairs and one lone pair
R has two bond pairs and two lone pairs
S has two single bonds and two lone pairs

The inconsistent row is

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

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30. Consider this arrangement: a central atom has six electron domains, four bond pairs, and two lone pairs. The two lone pairs occupy opposite positions in an octahedral set. The expected hybridization and molecular shape are

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

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32. Ionic compounds often dissolve better in polar solvents such as water because

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33. For comparing the magnitude of lattice enthalpy, the most useful qualitative relation is

This relation suggests that lattice enthalpy magnitude increases when

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34. The table below includes one incorrect hybridization-geometry link.

Hybridization Number of hybrid orbitals Ideal geometry
P. linear
Q. trigonal planar
R. trigonal bipyramidal
S. octahedral

The row that needs correction is

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35. A molecular orbital comparison gives the following data.

Species Unpaired electrons Magnetic nature
P. paramagnetic
Q. paramagnetic
R. paramagnetic
S. paramagnetic

The row that needs correction is

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

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37. In choosing the central atom for many simple Lewis structures, the atom usually selected is

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38. Before any multiple-bond representation is considered, the skeleton of has how many sigma-type connections?

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39. A molecule contains an bond and another molecule has a nitrogen atom with a lone pair. The interaction between the hydrogen of the first molecule and the lone pair of the second molecule is best described as

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

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41. Each resonance contributor has one bond and two single bonds. The average bond order in the resonance hybrid is

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42. The Lewis structure for contains

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43. 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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44. In hybridization, the number of hybrid orbitals formed is

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45. A Lewis-structure attempt uses electrons for . The most likely error is

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46. Match each molecule with the simplest shared-pair description.

Molecule Shared-pair description
P. 1. one shared pair
Q. 2. two shared pairs
R. 3. three shared pairs
S. 4. one shared pair

The proper matching is

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47. The species is predicted by simple molecular orbital theory to have bond order and no unpaired electrons. Its magnetic nature is

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

Bond description Number of shared electron pairs
P. Single bond
Q. Double bond
R. Triple bond

The suitable matching is

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

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50. A molecule or ion with resonance is usually more stable than any single contributor because

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