Chemical Bonding Quizzes

Mock Tests of Chemical Bonding

Note:

Solutions of the MCQs will be shown after finishing the quiz.

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Quiz-1 (Chemical Bonding)

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1. A neutral atom will have the lowest ionization energy, when its configuration is
A. 1s² 2s² 2p⁶ 3s¹
B. 1s²
C. 1s² 2s² 2p⁵
D. 1s²2s²2p¹

Solution: A
An atom having configuration 1S², 2S²,2P⁶,3S¹, (i.e Na) has great tendency to lose an electron to get noble gas configuration. That is why these have lowest ionization energy values.

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2. Which of the following has maximum ionization potential?

A. Al

B. P

C. Mg

D. Si

Solution: B
Phosphorous has the highest ionization potential. The graph shows phosphorous high ionization potential.

Ionization energy generally increases across period 3 because the nuclear charge increases but the shielding of the outer electrons remains relatively the same. This means that the electrostatic force of attraction between the outer electrons and the nucleus is becoming greater, so more energy is needed to remove the electrons. Since the elements are in period 3, this means that all of them have an outer electron in energy level 3. However, the trend has two anomalies.

The first is between Mg and Al, because the outer electron of Mg is in the orbital 3s, whereas that of Al is in 3p. The 3p electron has more energy than the 3s electron, so the ionization energy of Al is actually less than that of Mg. This makes sense because the 3p electron requires less energy to be removed from the atom. Remember that if an electron has more energy, it needs less external energy to be removed from the atom because having more energy makes it more unstable.

The second anomaly is between P and S. Although both have outer electrons in the 3p orbital, that of P is unpaired (3p³) but that of S is paired (3p⁴). 3p orbitals can fit up to 6 electrons, but if there are 3 or less electrons, they are unpaired and take up a "space" just for themselves. Since the fourth electron in 3p⁴ is paired, it will experience a repulsive force from the third electron. This makes it easier to remove, and therefore the ionization energy of S is less than that of P.

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3. Measurement of the degree of polarity is:

A. Electron affinity

B. Ionic character

C. Ionization energy

D. Dipole moment

Solution: D
Dipole moments occur when there is a separation of charge. They can occur between two ions in an ionic bond or between atoms in a covalent bond; dipole moments arise from differences in electronegativity. The dipole moment is a measure of the polarity of the molecule.

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4. Which of the following doesn’t affect ionization energy?

A. Shielding effect

B. Nuclear charge

C. Size of atoms

D. Number of neutrons

Solution: D
Ionization energy value depends upon shielding effect, nuclear charge, size of an atom. It does not depend upon a number of neutrons.

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5. Formation of a chemical bond takes place when:

A. Energy is absorbed

B. Forces of repulsion overcome forces of attraction

C. Forces of attraction are equal to forces of repulsion

D. Forces of attraction overcome forces of repulsion

Solution: D
When an atom loses an electron, it becomes positively charged. Now, a chemical bond is formed when the forces of attraction (between the protons, and electrons of other atom) overcome the forces of repulsion (between the electrons in the outermost shell of atoms of both the elements or compounds.)

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6. The force which holds the atoms together to form a compound is known as

A. Van der Waal's

B. London force

C. Chemical bond

D. Dispersion force

Solution: C
The chemical bond holds the atoms together to form a compound.

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7. The bond angle in water suggest
A. Sp³-hybridzation
B. Bent shape
C. Tetrahedral geometry
D. All of these

Solution: D
Water has a bond angle of 104^o which shows that it has sp³ hybridization, it has nearly tetrahedral geometry & bent shape.

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8. The energy required to remove an electron from the outermost shell of its isolated gaseous atom in its ground state is

A. Electron affinity

B. Lattice energy

C. Crystal energy

D. Ionization energy

Solution: D
The amount of energy required to remove an electron from the valences shell of an isolated gaseous atom in its ground state is called ionization energy.

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9. A sudden large difference between the values of second and third ionization energies of elements would be associated with which of the following electronic configuration?
A. 1s², 2s², 2p⁶, 3s⁴
B. 1s², 2s², 2p⁶, 3s², 3p¹
C. 1s², 2s², 2p⁶, 3s²
D. 1s², 2s², 2p⁶, 3s², 3p²

Solution: C
When two electrons are removed then the atom acquire noble gas configuration as in Choice C

1s², 2s², 2p⁶, 3s²

After removing two electrons ⇒ 1s², 2s², 2p⁶

Then it is very difficult to remove the 3rd electron due to strong attraction of nucleus on the electrons.

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10. While moving down the group in the periodic table, _____________ will increase

A. Number of valence electrons

B. Electronegativity

C. Atomic radii

D. Effective nuclear charge

Solution: C
Atomic radii increase down the group and decrease along the period from left to right.

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11. In which of the following pairs, the radius of the second atom is greater than that of the first atom

A. Na, Mg

B. P, N

C. Sr, Ca

D. Cl, Br

Solution: D
Atomic radius increase down the group. As Br is below Cl in the group so Br has a bigger size.

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12. Which order of ionization energy is incorrect?
A. O₂ < O
B. N₂ > N
C. N > O
D. F₂ > F

Solution: D
The ionization energy of molecular fluorine is 1515 kJ mol^-1, which is smaller than the ionization energy of a fluorine atom, 1681 kJ mol^-1.

Because F₂ is a stable molecule, we must conclude that the lowering of energy for the electrons in the bonding orbitals is greater than the raising of energy for the electrons in the antibonding orbitals. Overall, this distribution of electrons is, net, equivalent to having two electrons paired in a single bonding orbital. This also explains why the ionization energy of F₂ is less than that of an F atom. The electron with the highest energy requires the least energy to remove from the molecule or atom.

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13. Moving down in groups, the binding energy of metals

A. Decreases

B. Remains same

C. Increases

D. Unpredictable

Solution: A
Atomic size increase down the group. Due to this the binding energy of metals decreases. It is due to the reason that the binding forces present between large-sized atoms are relatively weaker as compare to those between smaller atoms.

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14. All of the following processes do not represent ionization of sodium except.
A. Na_(g) → Na⁺_(aq) + e^-
B. Na_(s) → Na⁺_(g) + e^-
C. Na_(l) → Na⁺_(g) + e^-
D. Na_(g) → Na⁺_(g) + e^-

Solution: D
For the ionization of sodium, sodium should be in isolated gaseous form.

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15. Great shielding effect corresponds to ionization energy value:

A. Greater

B. Lesser

C. Remain same

D. No effect

Solution: B
Ionization energy ∝ 1/Shielding effect

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16. A neutral atom will have highest first ionization energy, when its configuration is
A. 1s² 2s² 2p⁶ 3s¹
B. 1s² 2s² 2p⁵
C. 1s²
D. 1s²2s² 2p¹

Solution: B
Fluorine is the most electronegative element in the periodic table. The removal of electron from its outermost shell is most difficult. So, its ionization is energy is highest. its configuration is:
1s²2s²2p⁵

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17. Consecutive elements X, Y, Z are in period 3 of the periodic table. Element Y has the highest first ionization energy and the lowest melting point. What could be the identities of X, Y and Z?

A. Sodium, Magnesium, aluminium

B. Aluminium, silicon, phosphorus

C. Silicon, phosphorus, sulphur

D. none of the above

Solution: C
3rd period contains elements: Na¹¹, Mg¹², Al¹³, Si¹⁴, P¹⁵, S¹⁶, Cl¹⁷, Ar¹⁸.

P has higher ionization energy than S & Si.

P¹⁵= 1s², 2s², 2p⁶, 3s², 3p³

It is due to the reason that P has half-filled p orbital. Half filled & complete filled orbitals are more stabilized than other orbitals. That is why it is reluctant to lose electrons. Furthermore, P has a lower melting point.

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18. The correct order of ionization energies?
A. Li > He > H
B. Mg⁺ > Mg > Ca
C. Cs > Fr > Li
D. All the above

Solution: B
Ionization energy decrease from top to bottom in a group it increases from left to right in a period. The further ionization energy of positive ions is always higher than its neutral atom.

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19. The bond angle in NF₃ is

A. 104.5

B. 106.5

C. 102

D. 120

Solution: C
NF₃ has three bond pairs and one lone pair. It has trigonal pyramidal geometry and bond angle of 102.5^o.

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20. The extent of ionization of an electrolyte does not depend on

A. Its nature

B. Its volume

C. Its dilution

D. Its temperature

Solution: B
The degree of ionization depends on nature of electrolyte, dilution of electrolyte, and temperature but does not depend on the volume of electrolyte

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21. Which of the following transitions require the maximum amount of energy?
A. M → M
B. M → M⁺
C. M⁺ → M⁺²
D. Mg⁺² → Mg⁺³

Solution: D
The transition M⁺²_(g) → M⁺³_(g) involves maximum amount of energy.
The charge on the metal cation is maximum. Due to this, Zeff is maximum. This results in maximum attraction between the nucleus and valency electrons.

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22. For non-bonded atoms, the size of an atom is described by

A. Ionic radii

B. Covalent radii

C. Van der Waal

D. None of these

Solution: C
Non-bonded atoms are described by Van der Waal’s radii, which is defined as ''half of the distance between the closest approach of two non-bonded atoms of a given element''.

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23. Which of the following pair of groups show an abnormal trend of ionization energy?

A. II, III

B. II, V

C. III, V

D. III, VI

Solution: D
Ionization energy indicates the amount of energy required to remove an electron (for the first ionization energy, this is the highest energy electron in the electron configuration), thus is an indication of the relative attraction of the negative electron to the positive nucleus of the atom. Higher ionization energy indicates a stronger attraction between them. Within a period (row), moving left-to-right, the nuclear charge increases, but added electrons remain in the same primary energy level, so the attraction to the nucleus increases as does the ionization energy.

But there are two groups (columns) in which this general trend breaks down, at least in the 2nd and 3rd energy levels. Elements in group 3A have their 3rd electron placed into a p orbital, which is slightly higher in energy than the s² electrons that come before it. This allows the s2 electrons to partially shield the p electron from the positive charge of the added proton. This reduces the attraction of the highest energy electron compared to group 2A, thus reduces the ionization energy.

The other exception occurs between groups 5A and 6A. The electron configuration of elements in 5A is p³, which has 3 unpaired electrons in each of the 3 p orbitals (Hund’s Rule). The next electron in group 6A must pair up, which increases electron-electron repulsion (essentially intra-orbital shielding), which reduces the attraction of the highest energy electron to the nucleus and reduces the ionization energy compared to group 5A.

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24. Which of the following element has highest ionization energy?
A. [Ne] 3s² 3p1
B. [Ne] 3s² 3p²
C. [Ne] 3s² 3p³
D. [Ar] 3d¹⁰ 4s²

Solution: C
(Ne) 3s²3p³ is stable configuration because the outer sub-shell is half-filled. Half-filled configurations are stable configurations.
The one half-filled and one completely filled orbital gives stability. Hence, it will be difficult to remove an electron from it and will take more energy.
Hence, [Ne]3s²3p³ has the highest ionization energy among the given elements.

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25. Which specie is non-polar although it contains polar bonds?A. HCl
B. SO3
C. H2O
D. SO2

Solution: D
SO₂ is a bent molecule, with a central sulfur atom bonded to two oxygen atoms and a lone pair of electrons on the sulfur atom. The oxygen atoms are more electronegative than the sulfur atom, so the electronegativity difference between the sulfur atom and the oxygen atoms is enough to create polar bonds. However, the molecule is non-polar because the two polar bonds are arranged symmetrically around the sulfur atom, so the dipole moments cancel each other out.

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26. Which one has the smallest radius?
A. N3
B. C4
C. O2
D. F

Solution: D
F has a smaller radius than F− because an additional electron causes greater repulsion in F− . ... F has a larger radius than F− because an additional electron causes greater repulsion in F.

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27. From left to right in a period, the atomic radii decrease due to

A. Increase in number of electrons

B. Increase in nuclear charge

C. Decrease in nuclear charge

D. Decrease in the number of shells

Solution: B
Atomic radii decrease along the period from left to right due to increase in nuclear charge.

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28. Which one shows the different structure?
A. 𝐻𝑔𝐶𝑙₂
B. 𝐵𝑒𝐶𝑙₂
C. 𝐶₂𝐻₂
D. 𝑆𝑛𝐶𝑙₂

Solution: D
SnCl₂ has a trigonal planar structure with the bond angle of 120^o.

Choice A, Choice B, and Choice C all have a linear structure with the bond angle of 180^o.

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29. The total number of electrons in NH₂^- is

A. 8

B. 12

C. 10

D. 14

Solution: C

has total 10 electrons. Nitrogen possesses seven electrons. It shares two electrons with hydrogen (7+2=9). It gains one more electron to form i.e. 9 + 1=10.

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30. The graph below shows the first six successive ionization energies, of an element. In which Group of the Periodic Table is the element found

A. Group 11

B. Group IV

C. Group V

D. Group VI

Solution: B
It contains the elements titanium (Ti), zirconium (Zr), hafnium (Hf) and rutherfordium (Rf). This group lies in the d-block of the periodic table. The group itself has not acquired a trivial name; it belongs to the broader grouping of the transition metals.

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31. According to the size of atom, which one of the following is correct order

A. Mg > Sr

B. Lu > Ce

C. Ba > Mg

D. Cl > I

Solution: C
Atomic size of an atom increase down the group. Ba is below Mg in II A group. So, Ba has large size then Mg.

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32. 3^rd ionization energy of an element can be represented as
A. M⁺²_(s) → M⁺³_(s) + e^-
B. M⁺²_(g) → M⁺³_(g) + e^-
C. M⁺¹_(s) → M⁺²_(g) + e^-
D. M_(g) → M⁺³_(g) + e^-

Solution: B
3rd ionization energy of an element can be represented as M⁺²_(g)→ M⁺³_(g) + e^-

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33. Which of the following has the largest size
A. N-3
B. F-
C. O-2
D. Na+

Solution: A
N^-3 has highest electron-electron repulsion so it has the largest size.

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34. The group that has maximum ionization energy

A. II-A

B. III-A

C. V-A

D. VII-A

Solution: D
Halogens have the highest proton number in their period thus, they have the highest ionization energy due to strong attraction of nucleus on electrons.

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35. Among the following -------- is the electron deficient from the octet rule
A. BF₃
B. SO₂
C. PCl₅
D. H₂O

Solution: A
The electronic configuration of boron is B₍₅₎ = Is²2s² 2p¹. Since it has three electrons in the valence shell, it can share only these three electrons and have three pairs (i.e., 6) electrons in the outermost Shell. Thus, BF₃ is the electron deficient from the octet rule.

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36. The 1st ionization energy of Mg is lower than the 1st ionization energy of

A. Be

B. Na

C. Ca

D. Li

Solution: A
Ionization energy decreases down the group as electrons move far away from the nucleus so it is easier to remove them hence low energy is required so in this case Magnesium has lower ionization energy than Be.

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37. The structure of SF₆molecule is

A. Square planner

B. Tetrahedral

C. Octahedral

D. Trigonal bipyramidal

Solution: C
Sulfur hexafluoride has 12 electrons around the central sulfur atom, which means that there are six electron pairs arranged in an octahedral shape with 90° F-S-F bond angles. There are no lone pairs.

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38. The pair having similar geometry is
A. NH₃, PH₃
B. CO₂, SO₂
C. H₂O, C₂H₂
D. BF₃, NH₃

Solution: A
Both NH₃and PH₃ have three Bond pairs and one lone pair. Due to lone pair, their structure is distorted and it becomes trigonal pyramidal. PH₃is isoelectronic with NH₃ as both have same molecular geometry and both N and P have s²p³ electronic configuration.

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39. The first ionization energies of Na, Mg, Al, and Si are in order

A. Na > Mg > Al > Si

B. Na Al < Si

C. Na > Mg Si

D. Na < Mg < Al < Si

Solution: B
Ionization energy increase from left to right in a period. But Mg has higher Ionization energy than Al.

Mg-12 = 1s², 2s², 2p⁶, 3s²

Due to complete s orbital, Mg show reluctant to lose an electron.

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40. Keeping in view the size of the atom which order is correct

A. Mg > Sr

B. Ba > Mg

C. Lu > Ce

D. Cl > I

Solution: B
Barium has a larger atomic radius because it has more energy levels and a larger shielding effect. Mg has a smaller atomic radius because it has a stronger nuclear charge.

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41. Due to the addition of protons in the nucleus, the regular decrease in the size of an atom is most prominent in the elements of _________

A. s block

B. d block

C. p block

D. f block

Solution: D
Due to the involvement of p-subshell, the decrease in size along a period is more prominent in f block elements.

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42. ________ is the approximate value of O-C-O bond angle in ethanoic acid
A. 109^o
B. 180^o
C. 120^o
D. 90^o

Solution: C
To determine the approximate O-C-O bond angle in ethanoic acid (CH₃COOH), consider the molecular geometry around the central carbon atom involved in the carboxylic group (-COOH):

Geometry of the -COOH group:

The carbon atom in the carboxyl group is sp² hybridized, resulting in a trigonal planar geometry.
The bond angle around a trigonal planar carbon is close to 120°.

O-C-O bond in ethanoic acid:

The two oxygen atoms attached to the central carbon form bonds in this planar arrangement.
Hence, the O-C-O bond angle is approximately 120°. So, the correct option is C.

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43. In which of the following bond pair angle is minimum

A. Water

B. Hydrogen Sulphide

C. Ammonia

D. Nitrogen trifluoride

Solution: B
Due to the repulsion of two lone pairs, the bonded pair come close to each other and show minimum bond angle in H₂S.
As we know in both structure H₂O and H₂S two bond pair and two lone pair are present and both has tetrahedral geometry. In H₂O, oxygen is more electronegative, electrons of two O-H bonds are more attracted towards oxygen atom. Consequently, more bond pair- bond pair repulsion take place, so bond angle get increase
On the other hand, in H₂S, because of less electronegative character of S, less repulsion is present so angle get decrease.
In ammonia sp³ hybridization occurs. So, its bond angle should have been 109°28'. But due to the presence of lone pair it is slightly less

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44. In the period Li to F the atomic size

A. Increases

B. Does not change

C. Decreases

D. First increases then decreases

Solution: C
Atomic radii decrease with increasing atomic number from Li to F due to the fixed number of shells but increase in electrons and protons.

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45. Elements having the smallest atomic radius of the following is

A. Sr

B. Ar

C. Rb

D. Ne

Solution: D
Atomic size increases across the period due to increase in a number of electrons as Neon is present at the end of the period that's why it has smallest atomic.

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46. The minimum angle between the bond pair-bond pair is shown by
A. 𝑆𝑛𝐶𝑙₂
B. 𝑁𝐹₃
C. 𝑆𝑂₃
D. 𝐻₂𝑂

Solution: B
Due to the repulsion of two lone pairs, the bond angle decreases from 109^o to 104.5^o.

Choice A SnCl₂has a bond angle of 120^o

Choice B The NF₃ bond angle is 102°. There is more distortion than for NH3 because the single bonds are taking up less room, close to the nitrogen. Fluorine is more electronegative than hydrogen and the electron density in the N–F bond is skewed towards the fluorine.
Choice C SO₃ has a bond angle of 120^o

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47. An atom loses or gains electrons, to

A. Gain stability

B. Form a bond

C. Complete its outermost shell

D. All are accurate justifications

Solution: D
When an atom loses or gains an electron it produces ions which react to form a bond in this way atom gains stability.

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48. Which of the following equations represent the 2nd Ionization Energy of sodium?
A. Na_(g) → Na²⁺_(g) + 2e⁻
B. Na⁺_(s) → Na²⁺_(g) + e⁻
C. Na_(s) → Na²⁺_(g) + 2e⁻
D. Na⁺_(g) → Na²⁺_(g) + e⁻

Solution: D
Choice D represents the 2nd ionization energy as an electron is removed from a cation. Na⁺_(g) → Na²⁺_(g) + e⁻

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49. Which of the following represent elements in order of increasing atomic size?

A. I > Br > Cl

B. CO

C. Na > Mg < C

D. Li K

Solution: A
To determine the correct order of increasing atomic size, we need to consider trends in the periodic table:

Atomic size decreases across a period (left to right) because the effective nuclear charge increases.
Atomic size increases down a group because new electron shells are added.

Option-by-Option Analysis:

I > Br > Cl

Incorrect: This order is reversed. Atomic size increases down a group, so Cl < Br < I.

Unclear or incomplete: This option doesn't explicitly compare sizes of elements.

Na > Mg < C

Incorrect: Atomic size decreases across a period, so Na > Mg > C, not Na > Mg < C.

Li K

Correct: Lithium (Li) is above Potassium (K) in Group 1 of the periodic table. Atomic size increases down the group, so Li < K.

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50. A factor which is not responsible for variation in I.E down the groups:

A. Shielding effect

B. Atomic size

C. Nature of orbital

D. Nuclear attraction

Solution: C
The nature of orbital does not affect the ionization energy

There are basically three:

Nuclear charge: The larger the atomic number, the higher the amount of charge at the nuclei, and therefore the strongest the attraction that it exerts into its electrons. This therefore implies a higher ionization energy. This is why along a period, ionization energy usually increases from left to right, because the nuclear charge of the atom increases.
The number of energy levels filled: Ionization will always occur from the highest in energy atomic orbital occupied, that is the most external. Therefore, the higher the number of energy cores filled, the lowest the ionization energy. This is why in a group, ionization energy decreases as we move from the top to the bottom. For Li, the released electron from ionization comes from the second energetic level, for Na it is from the third, for K it is for the fourth.
Electronic repulsion: The highest the electronic charge of an atom, the highest the electronic repulsion will present and therefore it will be more favored to release one of those electrons to reduce the electronic density. In other words, if the electronic density of a atom increases, its ionization energy will decrease and if it decreases it will increase. This is why cations have higher ionization energies than their correspondent neutral atoms. Releasing an electron decreases the electronic repulsion in the cation, making it even more difficult to release a new electron than the first one (note that another effect to take into consideration in this case is whether or not the second released electron comes from a different energetic level than the first one, as it happens in all alkaline elements)

There are other effects to take into account in specifically cases (full or semi-full valency core configurations), but in general these three are the main effects with which one can explain ionization energy differences between all elements in the periodic table.

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51. In which of the following, atomic radii of elements decrease with increasing atomic number

A. Elements from C to Si

B. Elements from Be to Ba

C. Elements from Li to Ne

D. Elements from F to At

Solution: C
Atomic radii decrease with increasing atomic number from Li to Ne due to the fixed number of shells but increase in electrons and protons.

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52. Elements having low I.E values are:

A. Metals

B. Non-metals

C. Liquids

D. Solids

Solution: A
Metal atoms tend to have low ionization energies relative to nonmetal atoms. Metal atoms lose electrons to non-metal atoms because metals typically have relatively low ionization energies. Metals at the bottom of a group lose electrons more easily than those at the top.

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53. CO₂ is isostructural with
A. 𝑆𝑂₂
B. 𝐻₂𝑆
C. 𝑂𝐹₂
D. 𝐶₂𝐻₂

Solution: D
CO₂ and C₂H₂ both are isostructural because both have a linear structure.CO₂ is a linear molecule and it is isostructural with C₂H₂. By viewing the structure of CO₂ you might understand that O=C=O. Thus, it is a liner in structure by utilizing hybridization. By checking all the compounds, we can see that C₂H₂ has got the linear structure since it is carbon hybridized.

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54. Which one shows high %age of the ionic character:
A. H₂O
B. HF
C. HCl
D. HBr

Solution: B
Electronegativity difference ∝ % ionic character

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55. Three elements P, Q and R have consecutive increasing atomic numbers. lf Q is a noble gas then which of the following is correct about P and R?

A. P and R belong to the same period of the periodic table.

B. P is a non-metal while R is a metal.

C. P and R belong to same group of the periodic table.

D. None of these

Solution: B
P, Q and R have consecutive increasing atomic numbers. As noble gases are present at the end of each period, P must be a non-metal of the same period and R must be a metal of the next period.

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56. Boron, Carbon, Nitrogen, Oxygen belong to the second period, The element with highest ionization energy is

A. O

B. C

C. N

D. B

Solution: C
Generally ionization energy increase along a period from left to right, but there are some deviations as well. N has more ionization energy than O.

N-7 = 1s², 2s², 2p³

O-8 = 1s², 2s², 2p⁴

N has half-filled p-orbital, so it is more stabilized. It is always difficult half-filled atomic orbital to lose its electrons.

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57. In a period, the atomic radii:

A. Increases

B. Decreases

C. Remain the same

D. First decreases, then increases

Solution: B
Left to right, effective nuclear charge increases. So atomic size decreases.

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58. Highest ionization energy

A. C

B. N

C. F

D. O

Solution: C
Elements on the right side of the periodic table have high ionization energy value. Fluorine has highest ionization energy in the periodic table.

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59. The atoms with high values of first ionization potential values always have

A. Metallic properties

B. Small ionic size

C. Large atomic size

D. Strongly force on valence electrons

E. both B & D

Solution: E
The ionization energy of a chemical species (i.e., an atom or molecule) is the energy required to remove electrons from gaseous atoms or ions. Large atoms or molecules have low ionization energy, while small molecules tend to have higher ionization energies
And when electrons are strongly held by nuclear force so ionization potential is high

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60. Ionization energy is affected by some factors. Which of the following factors is inversely related to the ionization energy?

I. Atomic Radius II. Nuclear charge III. The shielding effect of inner Electron

A. I, II

B. I, II, III,

C. I, III

D. II, III

Solution: C
Ionization energy depends upon atomic radius and the shielding effect of inner Electrons.

The more electrons shielding the outer electron shell from the nucleus, the less energy is required to expel an electron from an atom. The higher the shielding effect the lower the ionization energy

Your score is

Quiz-2 (Chemical Bonding)

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1. Hydrogen chloride molecules contains

A. Covalent bond

B. Double bond

C. Co-ordinate Bond

D. Electrovalent Bond

Solution: A
Chlorine has 7 electrons in the outermost shell while hydrogen has one. The hydrogen atom shares its one electron with the chlorine to form a bond. As this bond is formed due to sharing of electrons that's why it is covalent.

2 / 50

2. The correct order of electron affinity:

A. F > Cl > Br > I

B. Cl > F > Br > I

C. Br > F > Cl > I

D. I > Cl > Br > F

Solution: B
Electron affinity decreases top to bottom but in case of F, electron affinity is less than chlorine due to the thick electronic cloud and high influence of F^-

Electron Affinity (decreases down the group). Since the atomic size increases down the group, electron affinity generally decreases (At < I < Br < F < Cl). An electron will not be as attracted to the nucleus, resulting in a low electron affinity. However, fluorine has a lower electron affinity than chlorine.

3 / 50

3. Which pair of the atom has the same electronegativity?

A. F and Cl

B. I and C

C. I and H

D. I and F

Solution: B
E.N of carbon is 2.55 whereas the E.N of Iodine is 2.66 which are nearly equal to each other.

4 / 50

4. Which is not possible?

A. Pure (100%) covalent bond

B. Pure (100 %) ionic bond

C. Partial covalent bond

D. Both A and B

Solution: B
There is no 100% ionic bond because even if an element has more electronegativity than another the element with lower electronegativity would still attract the bonding pair.

Homopolar bonds are 100 percent covalent so b is the correct option

Covalent bonds (also known as homopolar bonds). A single covalent bond between two atoms is due to an electron pair shared by the atoms. These bonds are very common in organic chemistry.

5 / 50

5. Which of the following solids does not contain covalent bond?

A. Copper

B. Ice

C. Diamond

D. Graphite

Solution: A
Copper does not have any covalent bond. Metals simply do not hold on to electrons with enough strength to form much in the way of covalent bonds. For a covalent bond to form, we need two atoms that both attract electrons with high electron affinity. Hence, the great majority of covalent bonds will be formed between two non-metals.

6 / 50

6. Ionization energy depends on

A. Atomic/ionic radii

B. Electron to proton ratio

C. Shielding effect

D. All of the above

Solution: D
Ionization energy depends upon atomic/ionic radii, shielding effect and electron to proton ratio.

7 / 50

7. The electronegativity of elements in a period from left to right

A. Increases

B. Decreases

C. First increases then decrease

D. First decreases than increases

Solution: A
The positively charged protons in the nucleus attract the negatively charged electrons as the number of protons in the nucleus increases electronegativity will increase.

8 / 50

8. Which of the followings do not affect the electron affinity in a period

A. Atomic size

B. Nuclear charge

C. Nature of orbital

D. Shielding effect

Solution: D

Shielding effect does not affect the electron affinity in a period as the shell remains the e same so the shielding effect also remains the same.
Electron Affinity depends on orbital energy, which depends on the type of orbital and the effective nuclear charge. Thus, it follows predictable patterns in the periodic table. As you go down, n increases, and the energy of the orbital increases. That means the orbital is less stable, so it's easier to pull off the electron, so Electron Affinity decreases. As you go across the periodic table, usually the type of orbital is the same, and the effective nuclear charge increases, making the orbital more stable, so Electron Affinity increases.
If the size of the atom is more, the greater will be the distance between the nucleus and the incoming electron. Hence the incoming electron experiences fewer forces of attraction towards the nucleus of an atom. Therefore, the smaller will be the electron affinity value.
If the nuclear charge increases, the attraction for the incoming electron increases. Hence the electron affinity value increases.
Atoms with stable electronic configurations have less tendency to accept an electron. Hence they possess smaller electron affinity values.

9 / 50

9. Which one is not the absolute term of the element?

A. Ionization energy

B. Electro negativity

C. Electron affinity

D. Atomic size

Solution: B
Electronegativity is the relative tendency of an atom of an element to draw the electron density towards itself when bonded to another atom of a different element or same element with a different hybridization/oxidation state. It is not a measurable quantity. So, it is not an absolute.

10 / 50

10. The process requiring absorption of energy is
A. F→ F⁻
B. O → O₂⁻
C. Cl → Cl⁻
D. H → H⁻

Solution: B
The second Ionization energy is greater than the first ionization energy similarly second electron affinity is greater than the first Electron affinity the energy is to be supplied to force the second e⁻ into the anion.

11 / 50

11. An atom with electronegativity has

A. Large Size

B. High ionization potential

C. Low electron affinity

D. Low ionization potential

Solution: B
High electronegativity gives high ionization potential. Electrons with low ionization energies have low electronegativities because their nuclei do not exert a strong attractive force on electrons. Elements with high ionization energies have high electronegativities due to the strong pull exerted on electrons by the nucleus.

12 / 50

12. Which of the following properties can be attributed to water?

A. It is a good conductor of electricity

B. It is attracted by a magnet

C. It has a polar covalent bond with an angular structure

D. Both A and B

Solution: C
Water has polar covalent bonds.
Magnet cannot attract water because a magnet makes a magnetic field, and water molecules are electric dipoles. Electric and magnetic fields, though aspects of the same fundamental force, are not the same thing.
Pure water is not a good conductor of electricity. Ordinary distilled water in equilibrium with carbon dioxide of the air has a conductivity of about 10 x 10^-6 W¹m^-1.

13 / 50

13. The electron affinities of N, O, S and Cl are

A. O ≈ Cl < N ≈ S

B. O < S < Cl < N

C. N < O < S < Cl

D. O < N < C < S

Solution: C
Chlorine is a group of VIIA that has max electron affinity. Nitrogen has zero electron affinity as its extra stability is associated with its half-filled orbitals. Sulphur has more electron affinity than oxygen because the effect of the small size of an oxygen atom is more dominated by the repulsion of electrons already present in 2p orbitals of O atoms.

14 / 50

14. Which one has the lowest electron affinity

A. S

B. Se

C. Mg

D. O

Solution: C
Electron affinity refers to the energy change that occurs when an atom gains an electron to form a negative ion. It is a measure of the tendency of an atom to attract an additional electron.

Among the given options, magnesium (Mg) has the lowest electron affinity. This is because magnesium is a metal, and metals generally have low electron affinities. Metals tend to lose electrons to form positive ions (cations) rather than gain electrons to form negative ions (anions).

15 / 50

15. The percentage of coordinate covalent bond in NH₄⁺

A. 33%

B. 66%

C. 25%

D. 50%

Solution: C
In NH₄⁺ one coordinate covalent bond and three covalent bonds present.
Percentage coordinate covalent bond = 1/4 × 100 = 25

16 / 50

16. The electronegativity of cesium is 0.7 and that of fluorine is 4.0. The bond formed between cesium and fluorine is

A. Covalent

B. Electrovalent

C. Coordinate

D. Metallic

Solution: B
The bond between two atoms is electrovalent if the electronegativity difference between the atoms is 1.7 or greater. Since the electronegativity difference between cesium and fluorine (4-0.7) is greater than 1.7 the bond formed will be electrovalent.

17 / 50

17. Ionization energies of three elements A = 200Kjmol^-1 B= 150 kjmol^-1 C= 400 KJmol^-1 Correct one order for metallic character is

A. B > A > C

B. A > B > C

C. C > A > B

D. C > B > A

Solution: A
The greater the ionization energy, the lesser will be the metallic character and vice versa. Metallic character refers to the level of reactivity of a metal. The metallic character increases as you go down a group. Since the ionization energy decreases going down a group.

18 / 50

18. A chemical bond between the most electropositive and most electronegative elements shell be _____ in nature

A. Covalent

B. Coordinate covalent

C. Hydrogen

D. Ionic

Solution: D
A bond formed b/w electropositive & electronegative atom is ionic in nature. Electropositive atoms have a tendency to donate electron while electronegative atoms have a tendency to gain electrons. A bond formed by complete transfer of electron is called the ionic bond.

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19. Ionization energies of the elements of group VA greater than those of group VI A are the reason is that

A. Group VI A elements have gratin shielding effect

B. Group V A elements half-filled atomic orbitals

C. Group V A a element have less electromagnetic

D. Group V A elements have a greater shielding effect

Solution: B
Atoms with half-filled and completely filled orbitals in the outer shell are more stable thus electrons are held more tightly and require more energy to be ionized, the ionization energy of an outer electron is reduced by the effects of shielding.

20 / 50

20. Which among the following have triple bonds
(I) NO^3- (II) HCN (III) CO (IV) CH₃CN

A. Only (I), (II) and (III)

B. Only (II), (III) and (IV)

C. Only (I), (II) and (IV)

D. Only (I), (III) and (IV)

Solution: B

21 / 50

21. In which of the following central atom can form a co-ordinate covalent bond?
A. BF₃
B. NH₃
C. H₃O⁺
D. Both A and B

Solution: D

In ammonia, nitrogen has 5 valence electrons. Out of these, 3 electrons are shared by 3 hydrogen atoms to form ammonia molecule while the remaining two electrons form a lone pair, which can be donated to any electron deficient atom or molecule.

In BF₃ molecule, boron is short of 2 electrons. So, to complete its octet, it shares the lone pair of nitrogen, forming a dative bond.

22 / 50

22. Fluorine has an atomic number 9 and chlorine have 17. Which has the highest electron affinity?

A. Both are equal

B. Chlorine

C. Fluorine

D. Cannot be determined

Solution: B
Due to the small size of fluorine, the incoming electron is repelled by the electronic cloud of Fluorine thus, chlorine has greater electron affinity due to its larger size as compared to fluorine.

23 / 50

23. Each of the following pairs of chemical species will combine. Which of the following combinations is best explained by the coordinate covalent bond?
A. H2, +l₂
B. Mg, +O₂
C. Cl, +Cl
D. H₂O, H⁺

Solution: D
The H⁺ ion attacks on lone pair of oxygen atom in H₂O and forms coordinate covalent with oxygen, where oxygen is the donor.

As a result, H₃O⁺ ion is formed which is known as hydronium ion.

It has a pyramidal shape like ammonia molecule.

24 / 50

24. Bonds in silica are

A. Electrovalent

B. Metallic

C. Dative

D. Covalent

Solution: D
Silica has a covalent bond in it. SiO₂ is called silica. In silica, every silicon atom is attached tetrahedrally to four oxygen atoms by a covalent bond.

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25. Arrange 'S', 'P' and 'As' in order of increasing ionization energy

A. S < P < As

B. As < S < P

C. P < S < As

D. As < P < S

Solution: B
As (IE: 947 kJ/mol) < S (IE: 999.6 kJ/mol) < P (IE:1011.81 kJ/mol)

The increasing order of ionization enthalpy is As < S < P.

On moving from left to right in a period, the ionization energy increases with few irregularities where half-filled and completely filled orbitals are involved.
On moving down the group, the ionization energy decreases.
Hence, the ionization enthalpy of As is lowest.
The ionization energy of P is higher than the ionization energy of S as in P, the electron is to be removed from half-filled 3p orbital whereas in case of S, the electron is to be removed from 3p orbital containing 4 electrons

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26. The force that an atom exerts on a pair of electrons are being shared with another atom for forming covalent bond is referred to as its

A. Electron affinity

B. Electronegativity

C. Ionization energy

D. Valency

Solution: B
The attraction that an atom exerts on a pair of electrons that have been shared with another atom for forming a covalent bond.

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27. Higher value of electron affinity means

A. Atom will lose electron easily

B. Atom will gain electron easily

C. Atom may form di-positive ion

D. The reason is unknown

Solution: B
Electron affinity is directly proportional to the ease of gain of electron.

28 / 50

28. A compound which has all three types of bonds
A. NH₄Cl
B. [Co (NH₃)₆] Cl₃
C. K₄[Fe (CN)₆]
D. All of these

Solution: D
=> Ammonia only has three covalent bonds between N and H atoms.

=> Sodium (+) ion and Hydroxyl (-) ions in NaOH have ionic bond only.

=> Calcium ion (+2) and Chloride ion (-) in Calcium chloride have ionic bond.

However:

In the formation of Ammonium (NH₄^⁺) ion, there are three Covalent bonds between N atom and three H atoms, and a Coordinate bond between the N atom and one H atom (as the lone pair of N atom is completely donated to H^⁺).

Further, NH₄^⁺ (cation) and Cl⁻ (anion) combines to give Ammonium chloride (NH₄Cl), due to which they have an ionic bond.

Thus, ammonium chloride has all three types of bonds, namely, Ionic bond, covalent bond, and coordinate bond.

Bonds are there in [Co (NH₃) 6] Cl₃

The six NH3 molecules are connected to Co by 6 coordinate-covalent bonds. Each NH3 molecule is linked to three H atoms by pure covalent bonds and the total no. of N-H bonds= 6 x 3 = 18.
Coordinate bond is a special type of covalent bond in which the shared electrons are contributed by only one atom.
Thus, there are 24 covalent bonds. The three Cl atoms exist as Cl^- ions.

The bonds present in K₄[Fe (CN)₆] are ionic, covalent and coordinate covalent bonds. Ionic bonds are present between K⁺ ion and [Fe (CN)₆]₄⁻ ion.

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29. A specie with a maximum percentage of coordinate covalent character in individual bonds
A. NH₄⁺
B. PH₄⁺
C. H₃O⁺
D. All have same

Solution: C
When water molecule combines with H⁺ion by the donation of lone pair electron from O-atom (in H₂O) to H⁺ ion results in the formation of Hydronium ion, H₃O⁺. A coordinate Covalent bond is a type of covalent bond that forms when one atom donates a pair of electrons (Nucleophile) to another atom (Electrophile). It has the most Coordinate Covalent bond characteristic in it.

30 / 50

30. Electron gas theory was proposed to explain the bonding in solids

A. Molecular

B. Ionic

C. Covalent

D. Metallic

Solution: D
Electron gas theory is a metallic bond theory that was proposed by drude and extended by Loren. According to this, each atom in a metal crystal loses all of its valence electrons and these electrons form gas or a pool.
So, option 'D' is the correct option.

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31. In which of the following molecule central atom is deficient in octane
A. NH₃
B. BCl₃
C. H₂O
D. PCl₅

Solution: B
From the Lewis dot structure of the given molecules, BCl₃ has 6 electrons at B while others have either 8 or more. BCl₃ is electron deficient compound because boron has only 6 valence electrons and is short of 2 electrons. Therefore, it can accept a pair of electrons from nucleophiles.

32 / 50

32. A compound which has all the three types of chemical bonds
A. NaOH
B. NH₄Cl
C. CaCl₂
D. All of these

Solution: B
Answer is Ammonium Chloride.

Reason:
=> Ammonia only has three covalent bonds between N and H atoms.

=> Sodium (+) ion and Hydroxyl (-) ions in NaOH have ionic bond only.

=> Calcium ion (+2) and Chloride ion (-) in Calcium chloride have ionic bond.
However:

In the formation of Ammonium (NH_₄^⁺) ion, there are three Covalent bonds between N atom and three H atoms, and a Coordinate bond between the N atom and one H atom (as the lone pair of N atom is completely donated to H^⁺).

Further, NH_₄^⁺ (cation) and Cl^⁻ (anion) combines to give Ammonium chloride (NH₄Cl), due to which they have an ionic bond.
Thus, ammonium chloride has all three types of bonds, namely, Ionic bond, covalent bond, and coordinate bond.

33 / 50

33. Which of the following configurations is associated with the biggest jump between 2nd and 3rd ionization energy?
A. 1s², 2s², 2p²
B. 1s², 2s², 2p⁶, 3s¹
C. 1s², 2s², 2p⁶, 3s²
D. 1s², 2s², 2p¹

Solution: C
Let us consider the configurations of the given elements after the loss of 2 electrons: -
A) 1s² 2s²
B) 1s²2s²2p⁵
C) 1s² 2s² 2p⁶
D) 1s² 2s¹
From the options above we notice that the greatest energy will be involved for option B as it has attained the noble gas configuration after the loss of 2 electrons and removal of a 3rd electron will require a lot of energy.

34 / 50

34. In a period, electronegativity from left to right

A. Increases

B. Decreases

C. Remain constant

D. Variable trend

Solution: A
Left to right E.N increase due to small size, large nuclear charge, high influence of nucleus.

35 / 50

35. The oxidizing power of an element depends upon its

A. Cell potential

B. Electron affinity

C. ionization energy

D. electrode potential

Solution: B
Oxidizing power of an element depends on the ability to accept electrons. If a atom has a low energy of dissociation, a high electron affinity and a higher hydration energy of its ions, it will have a high oxidizing power.

36 / 50

36. Which equation describes 1st ionization energy of chlorine?
A. Cl_(g) → Cl⁻_(g)− e⁻
B. Cl_(g) → Cl⁺_(g) + e⁻
C. ½ Cl_2(g) → Cl⁺_(g) + e⁻
D. Cl⁺_(g) + e⁻ → Cl⋅

Solution: B
The Chlorine Ionization Energy is the energy required to remove from atom one mole of electrons with subsequent production of positively charged ion of Chlorine.
Cl ------> Cl⁺ + e^-
This process can be repeated many times, but the energy cost is increased dramatically. The general equation for the Chlorine is:
Cl^N+ --------> Cl^(N+1)+ + e^-

37 / 50

37. The type of bond present in CuSO₄.5H₂O are

A. Electrovalent, covalent and coordinate

B. Electrovalent and covalent

C. Electrovalent and coordinate

D. Covalent and coordinate

Solution: A
Ionic bond (electrovalent bond) and the coordinate covalent bond between S and O, while the covalent bond is present between Cu and SO⁺⁺.

38 / 50

38. One of the following species would exhibit normal trend of ionization energy across the period

A. Al

B. P

C. O

D. S

Solution: B
Ionization energy generally increases moving from left to right across an element period (row). This is because the atomic radius generally decreases moving across a period, so
there is a greater effective attraction between the negatively charged electrons and positively-charged nucleus

39 / 50

39. Bond will be ionic when E.N difference of bonded atom is

A. Equal to 1.7

B. Greater than 1.7

C. Less than 1.7

D. No specificity exists

Solution: B
Electronegativity difference > 1.7 Ionic bond
Electronegativity difference < 1.7 Covalent bond

40 / 50

40. H-bonding is not present in

A. Glycerin

B. Water

C. Hydrogen Sulphate

D. Hydrogen fluorine

Solution: C
Hydrogen bonding occurs when hydrogen makes a Bond with an atom that has a relatively high electronegativity but sulfur does not have high electronegativity that's why H bonding is not present in it.

41 / 50

41. A pair which cannot form Co-ordinate covalent bond is
A. NH₃, AlCl₃
B. BF₃, F
C. NH₃, H⁺
D. BF₃, H⁺

Solution: D
Ammonium ions, NH₄⁺, are formed by the transfer of a hydrogen ion from the hydrogen chloride to the lone pair of electrons on the ammonia molecule but BF₃, H⁺ reaction is not possible because of the full octet of BF_3. So they have no reason to react and change their configuration.

42 / 50

42. Covalent compounds show the phenomena of isomerism because covalent bonds are

A. Directional and rigid

B. Non-directional and non-rigid

C. Non-directional and rigid

D. Directional and non-rigid

Solution: A
Ionic bond is non-directional because it is the electrostatic force between two opposite charges, hence bonding direction does not matter whereas covalent is directional as attraction is in a specific direction and at an angle relative to the bonding atoms.

43 / 50

43. Which one of the following has polar covalent bond:
A. HF
B. CH₄
C. H₂
D. N₂

Solution: A
HF = Electronegativity difference = 1.7
CH₄ = Electronegativity difference = 0.4
H₂ = Electronegativity difference= 0
N₂ = Electronegativity difference= 0

44 / 50

44. An ionic bond is produced after the complete transfer of

A. Nucleus

B. Neutrons

C. Electrons

D. Protons

Solution: C
An ionic bond is formed between opposite ions. Ions formed by transfer of electron.

45 / 50

45. Addition of 2^nd electron to a uni-negative ion is always:

A. Exothermic

B. Endothermic

C. Data is insufficient

D. Unpredictable

Solution: B
Due to the addition of a 2^nd electron to a uni-negative ion repulsion occurs. To overcome repulsion energy is needed so the process is endothermic.

46 / 50

46. In which of the following metallic bond is not present?

A. Ag

B. Zn

C. Se

D. Al

Solution: C
Although selenium has a metallic luster, it is not metallic in crystal structure or bonding characteristics. Its bonds are more covalent in nature.

47 / 50

47. Which one is not the absolute term of the element.

A. Ionization energy

B. Electron affinity

C. Electronegativity

D. Atomic size

Solution: C
The relative attraction of the nucleus for a bonded pair of electrons in a chemical bond is called electronegativity.

48 / 50

48. The substance with maximum ionic character is

A. NaCl

B. KCl

C. HF

D. CsF

Solution: D
CsF has maximum ionic character because of greater electronegativity difference between Cs and Fluorine.

49 / 50

49. A hydrogen bond is the strongest between the molecules of

A. HCl

B. HF

C. HBr

D. HI

Solution: B
HF has strongest hydrogen bonding. Fluorine is most electronegative, it attracts, the bond pair of electrons more strongly than other halogens. Resultantly, a strong dipole is created in HF molecule.

50 / 50

50. A large difference between the third and fourth ionization energies indicates the presence of:

A. 4 valence electron in atom

B. 5 valence electron in atom

C. 3 valence electron in atom

D. 8 valence electron in atom

Solution: C
For possible ns²np¹ configuration, the removal of fourth electron will be possibly from an inert gas electron configuration. So, there will be high jump in the fourth ionization enthalpy than the third ionization enthalpy which will take place from ns¹ electron configuration

Your score is

The time for this quiz has been finished.

Quiz-3 (Chemical Bonding)

1 / 50

1. Bond energy is increased by increasing all of the following except

A. % of s character

B. Bond polarity

C. Difference of E.N value

D. Atomic size

Solution: D
Atomic size does not affect the bond energy. It depends upon the bond strength.

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2. Bond order for O₂ molecule is:

A. 2

B. 1

C. 3

D. 4

Solution: A
Bond order of O₂ = N_b – N_a /2 = 8-4 /2 =2

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3. The tetrahedral geometry is represented by all except:
A. AlCl₃
B. NH₃
C. CH₄
D. none of these

Solution: A
Among the given choices, all have tetrahedral geometry except AlCl₃. The molecular geometry of AlCl₃ is trigonal planar with symmetric electron region distribution.

Choice D is the correct answer.

Choice A: H₃O⁺ has a molecular geometry of tetrahedral since there are 4 regions of electron density surrounding the central O atom (3 Hydrogens and 1 lone pair of electrons).

Choice B: Ammonia has 4 regions of electron density around the central nitrogen atom (3 bonds and one lone pair). These are arranged in a tetrahedral shape.

Choice C: In methane, the four hybrid orbitals are located in such a manner so as to decrease the force of repulsion between them. Nonetheless, the four orbitals do repel each other and get placed at the corners of a tetrahedron. CH₄ has a tetrahedral shape.

4 / 50

4. The dipole moment of CHCl₃ is 1.50D, while that of CCl₄ is zero Debye, because in CCl₄

A. Bond lengths are unequal

B. The structure is planar

C. Bonds cancel the effect of each other

D. Every bond is non-polar

Solution: C
There is equal E.N.D between all C-H so they cancel the effect of each other and geometry will be perfectly tetrahedral but in CHCl₃, C-H has different E.N.D than C-Cl so it's not a regular geometry and Cl = 1.05 D.

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5. Bond energy is increased by increasing all of the following except

A. % of s character

B. Small size of atoms

C. Difference of E.N value

D. Bond length

Solution: D
When the bond order is higher, bond length is shorter, and the shorter the bond length the greater the bond energy. ... A higher bond energy (or a higher bond order or shorter bond length) means that a bond is less likely to break apart. In other words, it is more stable than a molecule with a lower bond energy.

6 / 50

6. Which of the following compounds has no dipole moment?

A. Cis-2-Butene

B. Trans-2-Butene

C. 1-Butene

D. 2-Methyl-1-propene

Solution: B
Effect of dipoles cancel out due to structure. So, Trans-2-Butene has no dipole moment.

7 / 50

7. The bond order in N²⁺, N₂, and N^2- follows the order
A. N₂⁻ = N₂⁺ > N₂
B. N₂⁻ > N₂⁺ > N₂
C. N₂ > N₂⁻ = N₂⁺
D. N₂⁻ = N₂⁺= N₂

Solution: C
Bond order for N₂ is 3; bond order for N₂^- is 2.5 and bond order for N₂⁺ is 2.5.

8 / 50

8. Which one represents maximum deviation in bond length and bond energy?

A. HBr

B. HI

C. HF

D. HCl

Solution: C
Bond energy is a measure of the strength of a bond. The strength of a bond depends upon
the following factors.
(i) Electronegativity difference of bonded atoms
(ii) Sizes of the atoms
(iii) Bond length

Let us consider, first the part played by electronegativity difference. Look at the bond energies of H-X type of compounds, where X=F, Cl, Br, I. This data show that electrons are not equally shared between the bonded atoms i.e. HX. As halogen atom is more electronegative, the bonded pair is more attracted towards X atom and thereby polarity develops. This gives rise to additional attractive force for binding. From the difference between experimental bond energies and those calculated by assuming equal sharing, it is possible to estimate relative electronegativities. The comparison of these values shows that the discrepancy is the greatest for HF and the least for HI,

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9. By containing ''n'' atomic orbitals, the number of hybrids orbitals will be

A. 2n

B. n

C. 3n

D. Impossible to predict

Solution: B
The number of atomic orbitals combined always equals the number of hybrid orbitals formed.

10 / 50

10. A specie with dipole moment
A. Cis-1,2-dichloroethene
B. BeCl₂
C. CCl₄
D. Trans-1,2-dichloroethene

Solution: A
Both molecules contain polar chlorine-carbon bonds, but in the cis isomer, they are both on the same side of the molecule. That means that one side of the molecule will have a slight negative charge while the other is slightly positive. The molecule is therefore polar because of dipole moment.

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11. In the following reaction:

The bond energy H-H is 436 kJmol, H-F is 568kJmol^-1 and F-F is

A. 154 kJ/mol

B. 155 kJ/mol

C. 156 kJ/mol

D. 158 kJ/mol

Solution: B
2(B.E)_HF = (B.E)_F2 + (B.E)_H2 + ΔH
2(568) = (B.E)_F2 + 436 + 545
1136 = (B.E)_F2 + 981
(B.E)_F2 =1136-981
(B.E)_F2 =155kJmol^-1

12 / 50

12. Which of the followings have maximum bond order?
A. F₂
B. O₂¹⁺
C. O₂
D. N₂²⁻

Solution: B
Bond order of F₂ = 1
Bond order of O₂⁺¹ = 2.5
Bond order of O₂ = 2.0
Bond order of N₂ =2.0

Bond order = [(Number of electrons in bonding molecules) - (Number of electrons in antibonding molecules)]/2

13 / 50

13. Which orbitals have the same energy, symmetry, and shape?

A. 3p and 3d

B. Hybrid

C. 2s and 2p

D. 1s and 2s

Solution: B
All orbitals in a set of hybrid orbitals are equivalent in shape and energy. The type of hybrid orbitals formed in a bonded atom depends on its electron-pair geometry as predicted by the VSEPR theory. Hybrid orbitals overlap to form σ bonds.

14 / 50

14. The relationship between the bond energy of N₂ and N₂⁺ is
A. N₂⁺ > N₂
B. N₂ = N₂⁺
C. N₂ > N₂⁺
D. N₂ depends upon the condition

Solution: C
Bond energy is directly proportional to the stability of molecules.

15 / 50

15. Which one of the following bonds has the highest average bond energy (kcal/mole)?

A. S = O

B. C ≡ N

C. C = C

D. N ≡ N

Solution: D
N ≡ N has bond energy = 941KJmole^-1 = 225Kcal mole^-1
Choice A      S = O has bond energy = 523 KJmole^-1 = 125 Kcalmole^-1
Choice B      C = N has bond energy = 891 KJmole-1 = 213 Kcalmole^-1
Choice C     C = C has bond energy = 614 KJmole^-1 = 147 Kcalmole^-1

16 / 50

16. The formation of the coordinate covalent bond between BF₃ and NH₃ is due to:
A. Donation of lone pair by BF₃
B. Donation of lone pair by NH₃
C. The octet of BF₃ is complete
D. BF₃ has a planar structure

Solution: B
NH₃ has three covalent bonds and there is a lone pair of electrons on nitrogen atom. On the other hand, boron atom in BF₃ is deficient in electrons as its octet is incomplete. Therefore, nitrogen donates the pair of electrons to the acceptor BF₃ and that results in formation of co-ordinate covalent bond between BF₃ and NH₃.

17 / 50

17. The molecule which contains minimum bond angle between bond pairs

A. Water

B. Ammonia

C. Alkane with one carbon atom

D. Benzene

Solution: A
H₂O = 104.5^oNH₃ = 107.5^oCH₄ = 109.5^oC₆H₆ = 120^o

The difference in angles is due to presence of one and two lone pairs in ammonia and water. So, H₂O has the smallest bond angle. H₂O, 104.5 degrees, due to lone pair-lone pair repulsion.

The bond angles depend on the number of lone electron pairs. For example, boron trichloride has no lone pairs, a trigonal planar shape and bond angles of 120 degrees. The trioxygen molecule O₃ has one lone pair and forms a bent shape with bond angles of 118 degrees.

18 / 50

18. Identify a molecule with geometry similar to formaldehyde.
A. AlCl₃
B. NH₃
C. BeCl₂
D. SO₂

Solution: C
The molecule with geometry similar to formaldehyde is BeCl₂. Both formaldehyde and BeCl₂ have a trigonal planar molecular geometry. This is because the central atom (carbon in formaldehyde and beryllium in BeCl₂) has three electron domains (two single bonds and one lone pair in formaldehyde, and three single bonds in BeCl₂) and the electron domains are arranged in a plane with 120^° bond angles.

The other molecules listed do not have a trigonal planar molecular geometry. AlCl₃ has a tetrahedral molecular geometry, NH₃ has a trigonal pyramidal molecular geometry, and SO₂ has a bent molecular geometry.

19 / 50

19. ___________ has highest bond order
A. He₂
B. N₂
C. O₂
D. H₂

Solution: B
Bond order is the number of chemical bonds between a pair of atoms.
Bond order of N₂ is 3
Bond order of O₂ is 2
Bond order of H₂ is 1/2
Bond order of He₂ is 0.
Nitrogen has highest bond order.

20 / 50

20. Which of the following is the best explanation that CO₂ is a non-polar molecule?

A. Electronegativity different for C-O bond is zero

B. Dipole moment is zero

C. Absence of electrons

D. All of these

Solution: B
All the regular geometries (linear, tetrahedral, trigonal planar) with same atoms have zero dipole moment.

21 / 50

21. Electronic geometry of molecule of water is tetrahedral and bond angle is:
A. 180^∘
B. 104.5^∘
C. 109.5^∘
D. 120^∘

Solution: B
Due to presence of lone pairs, bond angle decreases from 109.5 to 104.5.

22 / 50

22. Which has the strongest bonding in the solid state?
A. Hydrogen Chloride (HCI)
B. Xenon (Xe)
C. Chlorine (Cl₂)
D. Sodium Chloride (NaCI)

Solution: D
As it is obvious, none of the other options than NaCl exist in a solid state So, NaCl has the strongest bond among all other options.
Hence, option 'D' is the correct option.

23 / 50

23. Among the following molecule, the highest dipole moment is for_____
A. CO₂
B. AlCl₃
C. SO₂
D. CS₂

Solution: C
Dipole moment (μ) is the measure of net molecular polarity, which is the magnitude of the charge Q at either end of the molecular dipole times the distance ''r'' between the charges.

μ = Q × r

Dipole moments tell us about the charge separation in a molecule. The larger the difference in electronegativities of bonded atoms, the larger the dipole moment.

Let's have a look at the individual options.

Choice A: CO₂ has zero dipole moment because it is a linear molecule.

Choice B: AlCl₃ is a nonpolar molecule because each Al-Cl bond is directed at the angle of 120° to each other in a plane, hence, canceling of dipole moment generated along these bonds is very easy. Therefore, no dipole moment is generated in the AlCl₃ molecule, hence, it is nonpolar in nature.

Choice D: CS₂ is a linear molecule in which two C = S bonds are oriented in the opposite directions at an angle of 180°. The bond dipoles of C = S have some dipole moment, but due to the linear structure of CS₂, the bond dipoles of two C = S bonds cancel each other. Therefore, the resultant dipole moment of CS₂ is zero.

Choice C: The sulfur dioxide molecule has a dipole moment value “of 1.61 Debye". It is a bent molecule with a sulfur atom placed in the center and two atoms of oxygen on the sides, one oxygen atom is connected by a double bond and the other oxygen ion by a single bond.

Therefore, Choice C is the best answer.

24 / 50

24. Given the bond energies N ≡ N, H−H and N−H bonds are 945, 436 and 391 KJ/mole respectively, the enthalpy of the following reaction

N_2(g) + 3H_2(g) → 2NH_3(g) is

A. −93Kj

B. 90kJ

C. 102kJ

D. 105kJ

Solution: A
In reactants, one N ≡ N and three H-H bonds are broken.
In products, six N-H bonds are formed.
Bond energy of N ≡ N bond is 945 kJ/mol.
The bond energy of the H-H bond is 436 kJ/mol.
The bond energy of the N-H bond is 391 kJ/mol.
The enthalpy change for the reaction is calculated by using the expression shown below.
ΔH_oreaction = Bond energy of reactants−Bond energy of products = [(N≡N) + 3(H−H)] − [6(N−H)]
Substitute the values in the above expression.
ΔH_oreaction = [945 + 3(436)] kJ/mol − [6(391)] kJ/mol = 2253kJ/mol − 2346kJ/mol = −93kJ/mol
Therefore, the enthalpy change for the reaction is −93kJ/mol

25 / 50

25. In CO₂, the central atom is hybridized.
A. sp³
B. sp²
C. sp
D. All of these

Solution: C
The type of hybridization in CO₂ is sp hybridization, and each carbon atom forms two sp hybrid orbitals. Out of two hybrid orbitals, one will be used to produce a bond with one oxygen atom, and the other will be used to produce a bond with another oxygen atom.

26 / 50

26. Orbitals of same energy produced after mixing of orbitals of different energy are called:

A. Degenerate orbitals

B. Molecular orbitals

C. Hybrid orbitals

D. All of these

Solution: A
For Example for d-shell for d-subshell L = 2, m = 0, ±1, ±2. It implies that it has five space orientations and are designated as dxy (m = -2), dyz (m = -1), dzx(m = +1), dx²-y²(m = +2) and dz²(m = 0). All these five d-orbitals are not identical in shape. In the absence of a magnetic field, all fived-orbitals have the same energy and they are said to be five fold degenerate orbitals.

Orbitals of same energy produced after mixing of orbitals of different energy are called degenerate orbitals.

27 / 50

27. The hybridization of central oxygen in ozone molecule is:
A. sp³
B. sp²
C. sp
D. All of these

Solution: B
A central atom with one lone pair and two atoms is sp² hybridized.

28 / 50

28. Unit of dipole moment is
A. Debye
B. Cm^-1
C. Poise
D. Both A and B

Solution: A
Unit of dipole moment is Debye, the S.I unit is mC.

29 / 50

29. The geometry of molecule in silica is

A. Trigonal

B. Cubic

C. Linear

D. Tetrahedral

Solution: D
In polymer form SiO₂ is tetrahedral and in the non-polymer form, it is linear. It is as simple as that.
It is tetrahedral in the solid state. Consider quartz SiO₂:

Silicon has 4 outer electrons so adopts a tetrahedral structure analogous to diamond.

30 / 50

30. In which of the followings central atom is not sp³ hybridized
A. NH₃
B. PH₃
C. BeF₃
D. H₃O⁺

Solution: C
In BeF₃^-, the central atom does not use sp³ hybrid orbitals in its bonding. Instead, the central atom Be is sp² hybridized.

31 / 50

31. Which C−X bond has the highest bond energy per mole?

A. C-F

B. C-Br

C. C-C

D. C-I

Solution: A
Bond energies depend on many factors: electron affinities, sizes of atoms involved in the bond, differences in their electronegativity, and the overall structure of the molecule. There is a general trend in that the shorter the bond length, the higher the bond energy.

32 / 50

32. Which of the following has a tetrahedral structure and the central atom is sp³ hybridized?
A. H₃O⁺
B. CH₃⁺
C. SiH₄
D. SO₂

Solution: C
H₃O⁺ = sp³, trigonal pyramidal CH⁺₃ = sp², trigonal planar SiH₄ = sp³, tetrahedral SO₂ = sp², bent

33 / 50

33. Which of the following theories explain ionic bond?

A. VSEPR theory

B. MOT theory

C. VBT theory

D. None of these

Solution: B
Valence bond (VB) theory assumes that all bonds are localized bonds formed between two atoms by the donation of an electron from each atom. Valence Bond theory describes covalent bond formation as well as the electronic structure of molecules not ionic bond.

VSEPR is a way of predicting the shapes of molecules. Ionic bonding does not form molecules (though of course there are molecular ions such as nitrate or hydroxide, and VSEPR applies within each molecular ion.

Molecular orbital theory can give us information about both ionic and covalent molecules and naturally predicts which molecules will be ionic and which will be covalent.

34 / 50

34. Which one of the following has the shortest carbon-carbon bond length?

A. Benzene

B. Ethyne

C. Ethene

D. Ethane

Solution: B
Triple bond has shortest C-C bond length.
C-C bond length in ethane = 1.54 A^∘
C-C bond length in ethene = 1.32 A^∘
C-C bond length in ethyne = 1.18 A^∘
C-C bond length in benzene = 1.37 A^∘

35 / 50

35. Which of the following is the best explanation that CO₂ is non-polar molecule?

A. Linear geometry

B. Dipole moment is zero

C. Sp hybridization

D. None of these

Solution: B
O^-δ = C^+δ = O^-δ linear geometry, μ = 0D

36 / 50

36. H₂S bond angle is

A. 109.5

B. 104.5

C. 102

D. 92.1

Solution: A
The bond angle of H₂S is 92.1°

37 / 50

37. Which of the following compounds contains an sp hybridized carbon atom?
A. CH₃-CO-CH₃
B. CH₂=CH-OH
C. CH₂=C=CH₂
D. CH₃-N=CH-CH₃

Solution: C
In this compound, carbon carries 2σ and 2π bond due to sp hybridization.

38 / 50

38. The pair likely to form the strongest hydrogen bonding
A. H₂O₂ and H₂O
B. CHI₃ and CH₃COCH₃
C. HCOH and CH₃COH
D. SiH₄ and SiCl₄

Solution: A
1-Oxygen in H₂O & H₂O₂ possesses two lone pairs of electrons so it can form strong intermolecular hydrogen bonding.

2- Iodoform test is given by those compounds which has COCH₃ group and also those alcohols which produces the CH₃CO and ketone do not form hydrogen bond
3- acetaldehyde and formaldehyde do not react
4- No bond hydrogen involves

39 / 50

39. Which of the following has lowest bond energy?
A. F₂
B. O₂
C. N₂
D. H₂

Solution: A
The bond energy of F-F is lowest = 155 KJ mole^-1The bond energy of O₂ is = 495 KJ mole^-1The bond energy of N₂ is = 941 KJ mole^-1The bond energy of H₂ is = 436 KJ mole^-1

40 / 50

40. The hybridization of C is in order of this in given compound
CH₃-CH=C=CH₂
A. sp³, sp², sp², sp³
B. sp³, sp², sp, sp²
C. sp³, sp², sp, sp³
D. sp³, sp, sp², sp³

Solution: B
1. sp³
2. sp²
3. Sp
4. sp²

Carbon - sp³ hybridization:
When carbon is bonded to four other atoms (with no lone electron pairs), the hybridization is sp³ and the arrangement is tetrahedral. Notice the tetrahedral arrangement of atoms around carbon in the two and three-dimensional representations of methane and ethane shown below.

Carbon - sp² hybridization:
A carbon atom bound to three atoms (two single bonds, one double bond) is sp² hybridized and forms a flat trigonal or triangular arrangement with 120° angles between bonds. Notice that acetic acid contains one sp² carbon atom and one sp³ carbon atom.
Carbon - sp hybridization:
The third possible arrangement for carbon is sp hybridization which occurs when carbon is bound to two other atoms (two double bonds or one single + one triple bond). This hybridization results in a linear arrangement with an angle of 180° between bonds.

41 / 50

41. Bond order of N₂⁻² molecule is

A. 2

B. 1

C. 3

D. 4

Solution: A
N₂ molecule contains 6 electrons in bonding molecular orbital only.

42 / 50

42. Choose the correct order of bond strength by overlapping of atomic orbitals:

A. Is-Is > Is-2s > 1s-2p

B. 2s-2s > 2s-2p > 2p-2p

C. 2s-2p >2s-2s> 2p-2p

D. Is-Is > Is-2p > 1s-2s

Solution: D
The correct order of bond strength by overlapping of atomic orbitals is
1s−1s >1s−2p >1s−2s.
When n is less, then overlapping is stronger.
Therefore, the bond formed by 1s−1s overlap is strongest.
When one set is totally non-directional and n is also different, then the bond with directional nature is stronger.
1s−2p>1s−2s

43 / 50

43. Which of the following has a bond formed by the overlap of sp-sp³hybrid orbitals?
A. CH₃C≡CH
B. CH₂=CHCH=CH₂
C. CH₃CH=CHCH₃
D. CH≡CH

Solution: A
In propyne, triply bonded carbon is sp hybridized which overlaps with sp³ hybridized Carbon of methyl group. Thus, sp-sp³ hybridization is present in CH=C≡CH₃.

44 / 50

44. The shape of the molecule will be trigonal planar if the outer most shell of the central atom has:

A. 3 bond pair, 0 lone pair

B. 3 bond pair, 1 lone pair

C. 3 bond pair, 2 lone pair

D. 4 bond pair, 0 lone pair

Solution: A
3 bond pair, zero lone pairs like in BF₃, SO₃ will be trigonal planar.

45 / 50

45. CO₂ and CH₄ have µ values

A. Zero and 1.85D

B. 1.7 D and 1.8 D

C. Both zero

D. None of these

Solution: C
CO₂ = Linear, CH₄= tetrahedral thus each bond pair are at equal distance that is they are symmetrically arranged hence each dipole moment of bond balance each other. So, its µ = 0D.

46 / 50

46. All of the following pairs have same shapes except:
A. SO₂ and NO₂⁺
B. NH₄⁺¹ and SiCl₄
C. NH₂⁻¹ and H₂O
D. NH₃ and PH₃

Solution: A
Bent= SO₂, Linear = NO₂⁺ Tetrahedral, NH₄⁺¹ and SiCl₄
Pyramidal, NH₃ and PH₃

47 / 50

47. Electron pool theory does not explain?

A. Hardness of metal

B. Metallic luster

C. Thermal and electrical conductance

D. None of these

Solution: A
Electron Pool Theory.
This theory states that each atom present in a metallic crystal loses all of its valence electrons. As a result, a "pool" of electrons is formed. It is believed that positively charged metal ions are held together by this "pool" of electrons.
It explains metallic luster, electricity conductivity, thermal conductivity, malleability and ductility.
It cannot explain hardness of the metal.

48 / 50

48. Higher the bond order, greater is

A. Bond dissociation energy

B. Bond Length

C. Covalent character

D. Para magnetism

Solution: A
The bond order shows the no. of bonds present in a molecule. Greater the no. of bonds b/w two bonded atoms Greater will be the bond energy.

Bond order = No. of electrons in Bonding orbital - No. of electrons in Antibonding orbital /2

49 / 50

49. Which one has a maximum bond dissociation energy:
A. F₂
B. Cl₂
C. Br₂
D. I₂

Solution: B
Bond dissociation energy order Cl₂> Br₂> F₂> I₂.

F₂ is expected to have the highest bond dissociation enthalpy but it is Cl₂. The correct decreasing order is Cl₂> Br₂> F₂> I₂is so because the fluorine atom has a very small size due to which there is a high interelectronic repulsion between two fluorine atoms, so the bond between two fluorine gets weaker and need less energy to dissociate.

The bond dissociation energy of F₂ is less than Cl₂ because F has high electronegativity due to which F−F repels each other having lesser bond dissociation enthalpy than Cl₂ where Cl is less electronegative than F.

50 / 50

50. Which statement regarding the ‘sp' hybridization in ethyne is incorrect?
A. It is shown with C≡C bond
B. It results in two hybrid orbitals
C. It results in a bond angle of 180^o
D. It shows triangular planar geometry

Solution: D
Ethyne contains sp hybridization and bond angle is 180^o, hence linear in shape.

Your score is

The time for this quiz has been finished.

Quiz-4 (Chemical Bonding)

1 / 50

1. The hybridization of atomic orbitals of nitrogen inNO₂⁺, NO₃⁻ and NH₄⁺ are
A. sp, sp³ and sp² respectively
B. sp, sp² and sp³ respectively
C. sp², sp and sp³ respectively
D. sp², sp³ and sp respectively

Solution: B
Here's a nice short trick.
1) Count and add all the electrons in valence shell
2) Divide it by 8
3) On dividing by 8 you will get a remainder and quotient, add quotient + (remainder/2)
4) Now get the hybridization corresponding to the number what you got

If 2 its sp, if 3 its sp², if 4 its sp³, if 5 its sp³d and so on.

EXP: NO₂⁺-> Total electrons in valence shell: 5+ 6 × 2-1 = 16
-> 16/8 = 2. Quotient = 2 and Remainder = 0
-> 2 + 0/2 = 2 i.e sp)

NH₄⁺
-> Total electrons in valence shell: 5+7 × 4-1 = 32
-> 32/8 = 4. Quotient =4 and Remainder =0
-> 4 + 0/2 = 4 i.e sp³)

NO₂^--> Total electrons in valence shell: 5+6× 3+1 = 24
-> 24/8 = 3. Quotient =3 and Remainder =0
-> 3 + 0/2 = 3 i.e sp²)

2 / 50

2. In which compound central atom contains two lone pairs and bond pairs.
A. SnCl₂
B. NH₃
C. H₂O
D. SO₂

Solution: C
To determine which compound has a central atom with two lone pairs and bond pairs, let's analyze the given options based on their molecular geometry and valence electron configurations:

Option A: SnCl₂

Central atom: Tin (Sn)
Tin has 4 valence electrons, and chlorine contributes 1 electron each.
The structure of SnCl₂ is bent (angular) due to two bond pairs and one lone pair.
Lone pairs on Sn: 1
Bond pairs on Sn: 2
Not correct (1 lone pair, not 2).

Option B: NH₃

Central atom: Nitrogen (N)
Nitrogen has 5 valence electrons, and each hydrogen contributes 1 electron.
The structure of NH₃ is trigonal pyramidal due to three bond pairs and one lone pair.
Lone pairs on N: 1
Bond pairs on N: 3
Not correct (1 lone pair, not 2).

Option C: H₂O

Central atom: Oxygen (O)
Oxygen has 6 valence electrons, and hydrogen contributes 1 electron each.
The structure of H₂O is bent (angular) due to two bond pairs and two lone pairs.
Lone pairs on O: 2
Bond pairs on O: 2
Correct (2 lone pairs and bond pairs).

Option D: SO₂

Central atom: Sulfur (S)
Sulfur has 6 valence electrons, and oxygen contributes 1 electron each.
The structure of SO₂ is bent (angular) with one lone pair and two bond pairs.
Lone pairs on S: 1
Bond pairs on S: 2
Not correct (1 lone pair, not 2).

3 / 50

3. The percentage of s-character in hybrid orbital which indicates the shortest bond length

A. 25% s-character

B. 50% s-character

C. 33.4% s-character

D. 75% s-character

Solution: B
The length and strength of a C−H bond depend on the hybridization of the carbon atom to which the hydrogen is attached. The more s character in the orbital used by carbon to form the bond, the shorter and stronger is the bond. So, a C−H bond formed by an sp hybridized carbon (50% s) is shorter and stronger than a C−H bond formed by an sp³ hybridized carbon (25% s).

4 / 50

4. Which of the following is not incorrect statement about hybridization?

A. It explains tetravalent nature of carbon

B. Pi-bonds and sigma-bonds involve in hybridization

C. Hybridization explains paramagnetic behavior of oxygen

D. All are incorrect

Solution: C
Hybridization failed to explain the paramagnetic behavior of the oxygen atom. This was explained by molecular orbital theory.

5 / 50

5. Type of hybridization of Copper in [Cu (NH³)₄] Cl₂ if it is square planner
A. sp³
B. sp²
C. dsp²
D. d2sp³

Solution: C
The complexes containing square planner geometry, dsp² hybridized.

6 / 50

6. In which of the following molecules central atom is sp³ hybridized?
A. 𝑁𝐻₄⁺
B. 𝑁𝐻₂⁻
C. 𝑁𝐻₃
D. All of these

Solution: D
Any molecule which contains total four sigma bonds possess sp³hybridized central atom. The lone pair of electrons also behaves as a bond pair.

7 / 50

7. Which substances consist of planar structure?

A. Benzene

B. Ethene

C. Cyclohexene

D. Both A and B

Solution: D
Both Benzene and Ethene are the planar structure. Each carbon is sp² hybridized.

8 / 50

8. Which compound contains an sp hybridized carbon atom?
A. CH₃CN
B. CH₃NH₂
C. CH₃CHO
D. None of these

Solution: A
The central C in CH₃CN is sp hybridized. In cyano group, carbon form only two sigma bonds, the triple bond between carbon and nitrogen behaves as a single bond. Pi bond does not participate in hybridization.

9 / 50

9. Out of CHCl₃, CH₄ and SF₄ the molecules having regular geometry are:
A. CH₄ and SF₄
B. CHCl₃ and SF₄
C. CH₄ only
D. CHCl₃ only

Solution: C
There is one lone pair in SF₄. There is a polar character in CHCl₃. There is no lone pair no polarity in CH₄. So, it's having regular geometry.

10 / 50

10. Sigma bond is formed by overlapping of

A. s-s

B. s-p

C. p-p linear

D. All of these

Solution: D
Sigma bond can be formed by overlapping of all of these.

A sigma bond is a covalent bond which is formed by the head on overlap of two atomic orbitals. The combination of overlapping orbitals can be s-s, s-p_z or p_z-p_z. Sigma bonding can be a bonding interaction or an antibonding interaction.

11 / 50

11. A set of orbitals that can undergo most efficient overlapping is
A. sp² sp²
B. sp³ sp³
C. sp sp
D. All of these

Solution: B
sp³ -sp³overlapping is more efficient because all bonds are sigma and strong. No pi bond (weak bond) is present.

12 / 50

12. A molecule of AX₄⁺type where A has electronic configuration 1s²,2s²,2p³ will have geometry.

A. Trigonal

B. Linear

C. Pyramidal

D. Tetrahedral

Solution: D
In NH₄⁺ ion four bond pairs are present. The repulsion between bond pairs is less as compared to the bond pair-lone pair repulsion. Therefore, the geometry is tetrahedral.

13 / 50

13. The angle between unhybridized p-orbital and three sp² hybrid orbital’s of each carbon atom in ethene is
A. 120⁰
B. 90⁰
C. 109.5⁰
D. 180⁰

Solution: B
Unhybridized p-orbital form an angle of 90^o with sp² hybridized orbital in ethane.

14 / 50

14. Which of the following has lone pair of electrons on central atoms?
A. 𝑁𝐻₄⁺
B. 𝐶𝐻𝐶𝑙₃
C. NF₃
D. 𝐶₆𝐻₆

Solution: C
NF₃ has three bond pairs and one lone pair. It has trigonal pyramidal geometry.

Choice A NH₄⁺ has a tetrahedral geometry
Choice B CHCl₃ is also tetrahedral geometry
Choice D C₆H₆has trigonal planar geometry

15 / 50

15. The geometry of the molecule will be regular if the central atom is surrounded by

A. Lone pairs only

B. Both lone pairs and bond pairs

C. Bond pairs only

D. All of these

Solution: C
When the central atom is surrounded by all Bond pairs its geometry will be regular e.g., in Methane carbon is surrounded by four bonded hydrogens. The main geometries without lone pair electrons are: linear, trigonal, tetrahedral, trigonal bipyramidal, and octahedral.

16 / 50

16. According to VSEPR theory, the shape of SO₃ is:

A. Planar triangular

B. Pyramidal

C. Tetrahedral

D. Square planar

Solution: A
A molecule with three bond pairs and no lone pair is trigonal planar.

17 / 50

17. Which one of the following is not planer?

A. Benzene

B. Boron trichloride

C. ammonia

D. Propane

E. Both C and D

Solution: E
A plane molecule is a molecule that has its geometry contained in a bidimensional plane.
All molecules with angular, trigonal or lineal geometries are considered planar. For example, H₂O; SO₃ and Cl₂.

Choice A: Benzene has a planar hexagonal shape. Each carbon atom is bonded to two other carbon atoms and one hydrogen atom using three of its valence electrons. Each of its C-C bonds is an intermediate, halfway between a single and double bond in length.

Choice B: BCl₃ has trigonal planar structure due to 3 bond pairs in the valence shell of boron.

Choice C: Ammonia has 4 regions of electron density around the central nitrogen atom (3 bonds and one lone pair). These are arranged in a tetrahedral shape. The resulting molecular shape is trigonal pyramidal with H-N-H angles of 106.7°.

Choice D: The molecular geometry of propane is tetrahedral not planar.

18 / 50

18. Which of the following mode of hybridization shows non-coplanar structure?
A. sp
B. sp³
C. sp²
D. All of these

Solution: B
One carbon atom is sp3 hybridized. It has tetrahedral geometry. Hence, in this molecule, all atoms are not coplanar.

19 / 50

19. Linear shape is associated with which set of hybrid orbitals?
A. sp
B. sp³
C. sp²
D. dsp²

Solution: A
In sp hybridization, one 's' and one 'p' orbitals intermix to form two sp-hybrid orbitals called sp hybrid orbitals. In sp hybridization, geometry is linear with the bond angle of 180^o.

20 / 50

20. Trigonal pyramid geometry is exhibited by

A. Phosphorous trichloride

B. Ammonium ion

C. Tin chloride

D. Phosphorous pentachloride

Solution: A
Phosphorus trichloride exhibit trigonal pyramid geometry. It has a bond angle of 109^o.

21 / 50

21. The geometry of CH₃^- is

A. Tetrahedral

B. Trigonal pyramidal

C. Planar

D. Linear

Solution: B
trigonal pyramidal geometry
The: CH₃ molecule retains this trigonal pyramidal geometry, having three bonding pairs and one lone pair around the C atom.

22 / 50

22. In which of the following hybrid orbitals, bond angle is maximum?
A. sp³
B. dsp²
C. sp²
D. sp

Solution: D
The bond angle is maximum in sp hybridization i.e., 180^oBond angle in sp³ hybridization = 109.5^oBond angle in sp²hybridization = 120^o

23 / 50

23. The shape of NH₂^-is

A. Linear

B. Pyramidal

C. Bent or Angular

D. Tetrahedral

Solution: C
NH₂^- has a bond angle of 104^o and Angular geometry.

If the amine group (NH₂) is part of a molecule which is pyramidal in structure. Ammonia, NH₃ is based on a tetrahedron with bond angles of 107 (reduced from 109.5 degrees to 107 degrees on account of the lone pair-bond pair repulsion), with the resulting shape being pyramidal. NH₂ negative is linear or angular in shape.

24 / 50

24. We observe different type of hybridization in one of the following molecules
A. BF₃
B. H₂O
C. CH₄
D. NH₃

Solution: A
BF₃ has a boron atom with three outer-shell electrons in its ground state and three fluorine atoms containing seven outer electrons. Further, if we observe closely, one boron electron is unpaired in the ground state. During the formation of this compound, the 2s orbital and two 2p orbitals hybridize. Only one of the empty p-orbital is left behind as the lone pair. In short, Boron needs 3 hybridized orbitals to make bonds with 3 atoms of F where the 2pz orbitals get overlapped with these hybridized sp₂ orbitals and bonds are formed.

The three hybridized sp² orbitals are usually arranged in a triangular shape.
BF₃molecule is formed by bonding between three sp² orbitals of B and p of 3 F atoms.
All the bonds in BF₃ are sigma bonds.

25 / 50

25. Pi bonds are produced by overlapping of

A. Un-hybrid orbitals

B. Hybrid orbitals

C. Hybrid and un-hybrid orbitals

D. Atomic orbital and hybrid orbital

Solution: A
σ bond is formed by overlapping of hybridized orbitals but π the bond is formed by parallel overlapping if unhybridized orbital e.g. py - py and p_z- p_z.

26 / 50

26. Which one of the following molecules has minimum repulsion between electron pairs in the valence shell of its central atom?
A. 𝐶𝐻₄
B. 𝐻₂𝑂
C. 𝑁𝐻₃
D. 𝐻₂𝑆

Solution: A
CH₄ has minimum repulsion between electron pairs in the valence shell of its central atom because it is surrounded by all bonded pairs.

27 / 50

27. The octet rule does not always hold for which of the following elements.

A. C

B. O

C. F

D. P

Solution: D
P in PCl₅ does not obey the octet rule.

28 / 50

28. Which of the following species has triangular pyramidal shape?
A. 𝐻₃𝑂⁺
B. 𝑁𝑂₃⁻
C. 𝐶𝑂₃⁻²
D. 𝐵𝐹₃

Solution: A
Hydronium Ion has three Bond pairs and lone pair. Due to lone pair, its structure is distorted and it becomes triangular pyramidal. H₃O⁺is isoelectronic with NH₃.

29 / 50

29. From the following given molecules which one will have different hybridization

A. Hydrogen sulphide

B. Phosphorus trichloride

C. Water

D. Stannous chloride

Solution: D
Tin (II) chloride, also known as stannous chloride, is a white crystalline solid with the formula Sn Cl₂. It forms a stable dihydrate, but aqueous solutions tend to undergo hydrolysis, particularly if hot. SnCl ₂ is widely used as a reducing agent (in acid solution), and in electrolytic baths for tin-plating. It has sp² hybridization.

Note:
PCl₃ is sp³ hybridized.
Water is sp³ hybridized.
Hydrogen sulphide is sp³ hybridized.

30 / 50

30. In which of the following molecules central atom is sp³ hybridized.
A. NH₄⁺
B. NH₃
C. NH₂⁻
D. All of these

Solution: D
NH₄⁺, NH₃, NH₂⁻¹ all show that tetrahedral geometry
In NH₄⁺, the nitrogen atom forms 4 bonds and has 0 lone pairs, so it needs 4 hybrid orbitals. Its hybridization state is therefore sp³.

During the formation of ammonia, one 2s orbital and three 2p orbitals of nitrogen combine to form four hybrid orbitals having equivalent energy which is then considered as an sp³ type of hybridization.

NH₂^- has an sp³ hybridization type. The central Nitrogen atom has four regions that are responsible for identifying hybridization. Here N has two unbounded electron pairs and two sigma bonds. And due to these four regions around the central nitrogen atom, NH₂^- has sp³ hybridization.

31 / 50

31. The shape of the molecule is decided by the

A. Number of hybrid and unhybrid orbitals

B. Number of hybrid orbitals only

C. Number of unhybrid orbitals

D. All of these

Solution: B
The number of hybrid orbitals in a set is equal to the number of atomic orbitals that were combined to produce the set. All orbitals in a set of hybrid orbitals are equivalent in shape and energy. The type of hybrid orbitals formed in a bonded atom depends on its electron-pair geometry as predicted by the VSEPR theory.

32 / 50

32. The energy of an atom in the compound is:

A. Higher than individual

B. Lesser than individual

C. No change

D. Impossible to predict

Solution: B
When atoms form chemical bond, they release energy

33 / 50

33. Which of the following has sp hybridized carbon atom?
A. CH₃ CN
B. CH₂=CH₂
C. CH₃CH₂CH₃
D. All of the above

34 / 50

34. How many electrons are shared between the atoms in the nitrogen molecule?

A. Three

B. Six

C. Four

D. Two

Solution: B
In N₂ molecule there is triple bond nitrogen atoms. So, six electrons are shared.

35 / 50

35. Which one of the following sub-shells also acts as an orbital

A. s

B. p

C. d

D. f

Solution: A
An atomic orbital contains a maximum of two electrons in it. So, s-subshell can also act as an atomic orbital because it has availability maximum for two electrons.

36 / 50

36. The structure of H₃O⁺ is:

A. Tetrahedral

B. Square planar

C. Trigonal planar

D. Trigonal pyramidal

Solution: D
Molecular geometry of H₃O⁺ is trigonal pyramidal because H₃O⁺ has three bond pair and one lone pair.

37 / 50

37. How many lone pairs are present in CO₂ molecule.

A. 2

B. 4

C. 6

D. 8

Solution: B
So each Oxygen has two pairs of unbinding electrons and there are 2 Oxygen atoms in the CO₂ molecule. This makes a total of four pairs of unbonded electrons.

38 / 50

38. Point out the hybridization of oxygen in ethers
A. sp²
B. dsp²
C. Sp
D. sp³

Solution: D
In ethers R – O – R, Oxygen is attached to two atoms directly. Hence it is sp³ hybridized.

39 / 50

39. In CuSO₄.5H₂O, _ _ _ _ _ water molecules are attached with Cu⁺² and _ _ _ _ _ _ _ _ _ water molecules is attached with SO₄^-2ion

A. 3, 2

B. 2, 3

C. 4, 1

D. 1, 4

Solution: C
Due to the large change to size ratio of cations (Cu⁺²) four water molecules are attached while one water molecules is attached to sulphate ion (SO₄^-2)

40 / 50

40. In graphite crystal, carbon is
A. sp hybridized
B. sp³ hybridized
C. sp² hybridized
D. None of these

Solution: C
In graphite, each carbon atom is covalently bonded to three carbon atoms to give trigonal geometry. Bond angle in graphite is 120^oC. Each carbon atom in graphite is sp² hybridized.

41 / 50

41. [Fe (CO)₅] shows one of the following geometries

A. Octahedral

B. Tetrahedral

C. Square planar

D. Trigonal bipyramidal

Solution: D
[Fe (CO)₅] contains five ligands. Which are arranged along five sides forming trigonal pyramidal shape.

42 / 50

42. Why is 1-butene more acidic than butane and butyne?

A. s character increase, acidic strength increase

B. s character increase, acidic strength decreases

C. s character increase, but no effect on acidity

D. None of the above

Solution: A
Acidic character increase with increase of S character. It means the compound that have sp hyberdization will be more acidic than the compound with other hyberdization.

s character increase, acidic strength increase
s character decrease then acidic strength decrease

C-H alkyne hydrogen is more acidic than alkene or alkane ones because the stability of the corresponding ion C-

The more acidic hydrogen of But-1-ene is one of the two allylic hydrogens in all position with the respect to the double bond C=C.

In fact exist two resonance structures that stabilize the intermediate negative ion.

Is is necessary to underline that no resonance forms are possible in simple alkane, alkene or alkyne ions.

The scale of acidity is then But-1-ene (allylic hydrogen) > alkyne >alkene > alkane

43 / 50

43. What is the hybridization of carbon atom in HCHO?
A. sp
B. sp³
C. sp²
D. dsp²

Solution: C
In formaldehyde, the carbon atom is directly attached with 3 atoms. Hence sp² hybridized.

44 / 50

44. The type of hybridization of boron in diborane is
A. sp-hybridization
B. sp³-hybridization
C. sp²-hybridization
D. sp³ d²-hybridization

Solution: B
The hybridization of B₂H₆ is sp³ because each molecule has a tetrahedral geometry.

₅B = 1s² 2s¹ 2p_x¹2p_y¹2p_z⁰(excited state configuration)

⁼1s²

(hybrid state configuration) four sp³ orbitals

45 / 50

45. The geometry of the molecule will be triangular pyramidal if the outer most shell of the central atom has:

A. 3 bond pair, one lone pair

B. 2 bond pair, 2 lone pair

C. 1 bond pair, 3 lone pair

D. 3 lone pair, 1 bond pair

Solution: A
H₃ have one lone pair and three bond pair and its geometry is triangular pyramidal.

The nitrogen in ammonia has 5 valence electrons and bonds with three hydrogen atoms to complete the octet. This would result in the geometry of a regular tetrahedron with each bond angle equal to 109.5°. However, the three hydrogen atoms are repelled by the electron lone pair in a way that the geometry is distorted to a trigonal pyramid (regular 3-sided pyramid) with bond angles of 107^o

# of bonding groups/domains on 'central' atom	# of lone pair electrons on 'central' atom	Molecular Geometry
3	0	trigonal planar
2	1	bent
4	0	tetrahedral
3	1	trigonal pyramidal

46 / 50

46. Which of the following does not show V-shaped structure?
A. SO₂
B. 𝑃𝑏𝐶𝑙₂
C. 𝑆𝑛𝐶𝑙₂
D. None of these

Solution: D
All of them belongs to AX₃type of molecule and has a V-shaped or bent structure.

47 / 50

47. The carbon atom of a carbonyl group is
A. Sp hybridized
B. Sp³ hybridized
C. Sp² hybridized
D. Not hybridized

Solution: C
Carbonyl carbon is directly attached to three atoms. Hence sp² hybridized.

48 / 50

48. Which type of bond is formed by the overlap of p orbitals?

A. Pi

B. Sigma

C. Both A and B

D. Neither A nor B

Solution: C
When p_x- p_x head-to-head overlapping occurs, σ-bond is formed while p_y- p_y or p_z - p_z parallel overlapping π-bond is formed.

49 / 50

49. According to VSEPR theory, the electron pairs around central atom arrange themselves in such a way that there is

A. Maximum repulsion and minimum distance

B. Maximum repulsion and maximum distance

C. Minimum repulsion and minimum distance

D. Minimum repulsion and maximum distance

Solution: D
According to VSEPR theory, the valence electron pairs are arranged around the central atom to remain at a maximum distance apart to keep repulsion minimum.

50 / 50

50. In octahedral complexes the central metal atom is ________hybridized
A. sp³
B. dsp²
C. dsp³
D. d²sp³

Solution: D
In octahedral complexes, 6 ligands are attached to the central metal atom and the hybridization is d²sp³or sp³d².

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Mock Tests of Chemical Bonding

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