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Chemical Bonding Past Papers
Solved past paper MCQs for Chemical Bonding from official UHS, NUMS, SZABMU, DUHS, and KMU examinations. Includes verified distractor autopsies and step-by-step cognitive explanations.
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Official printable A4 booklet containing all 75 verified past paper MCQs with unbroken options, 5-column rapid answer key, and Propolis cognitive error autopsies.
Which of the following compound shows shortest C-H bond length?
A
Methane
B
Acetylene
C
Ethylene
D
Ethane
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The length of a C-H bond decreases as the s-character of the carbon's hybrid orbital increases, because s-orbitals are spherical and held closer to the nucleus.
In ethane and methane (\( \text{sp}^3 \)), the s-character is 25%.
In ethylene (\( \text{sp}^2 \)), the s-character is 33.3%.
In acetylene (\( \text{HC}\equiv\text{CH} \), which is \( \text{sp} \) hybridized), the s-character is 50%.
Because an \( \text{sp} \) orbital holds electrons closest to the nucleus, the resulting C-H bond in acetylene is the shortest and tightest among the options.
Why other options are incorrect:
Option A, Option C, Option D: These have lower s-character (\( \text{sp}^2 \) and \( \text{sp}^3 \)), resulting in more elongated, p-character dominant hybrid orbitals, leading to longer C-H bonds.
Bond energy decreases as the size of the bonding atoms increases, because a longer bond distance results in weaker electrostatic attraction between the nuclei and the shared electrons.
The paramagnetic behavior of molecules such as oxygen molecules cannot be explained by:
A
Molecular orbital theory
B
Hybridization
C
Valence bond theory
D
Valence shell electron pair repulsion theory
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Liquid oxygen is attracted to a magnet (paramagnetism) because it possesses unpaired electrons. Classical Valence Bond Theory (VBT) completely fails to predict this.
According to Valence Bond Theory and simple Lewis structures, the two Oxygen atoms share two pairs of electrons to form a double bond, meaning all electrons are paired. This falsely predicts oxygen should be diamagnetic.
Molecular Orbital Theory (MOT) correctly maps the electrons into bonding and antibonding orbitals, revealing two unpaired electrons in the degenerate \( \pi^* \) antibonding orbitals, perfectly explaining the paramagnetism.
Therefore, VBT is the classic textbook theory that fails this test. (Note: While the provided answer key listed B, VBT is the universally accepted and tested scientific limitation).
Why other options are incorrect:
Option A: MOT is specifically the theory that successfully explains it.
Option B & Option D: While these also don't explain magnetism, VBT is the direct theoretical framework regarding bond pairing that famously fails the Oxygen test.
Primary amines (\( \text{R-NH}_2 \)) share the same core structural geometry as ammonia (\( \text{NH}_3 \)), driven by \( \text{sp}^3 \) hybridization with one lone pair.
Formula:
$$ \text{Type} = \text{AB}_3\text{E} $$
Solution:
In a primary amine, the central Nitrogen atom bonds to two Hydrogens and one Carbon group.
It retains 1 lone pair of electrons.
The four total domains adopt a tetrahedral electron geometry, but because one is a lone pair, the atoms form a pyramid shape.
This is often termed trigonal pyramidal (or loosely "tetrahedral pyramidal" to denote its tetrahedral domain origin).
Why other options are incorrect:
Option A, Option C, Option D: These geometries ignore the physical space required by the nitrogen's lone pair, which forces the bonds out of a flat plane.
This shape is found in \( \text{AX}_2 \) when the bond angle is 180°:
A
Linear
B
Bent
C
Triangle
D
Tetrahedral
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
According to VSEPR theory, a central atom bonded to exactly two domains with zero lone pairs will repel those domains to the maximum possible geometric distance.
Predict the highest bond energy of the following single bond:
A
C - H
B
C - N
C
C - O
D
C - C
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Bond energy is strongly dependent on bond length. Shorter bonds are generally stronger. A bond involving the tiny Hydrogen atom will be significantly shorter than bonds between two Period 2 elements.
Hydrogen is the smallest atom (only a 1s orbital). Because its radius is so small, the Carbon and Hydrogen nuclei sit very close together.
This short distance (approx. 109 pm) results in a highly concentrated electron density between the nuclei, creating a very strong, stable bond (approx. 413 kJ/mol).
Bonds like C-C, C-N, and C-O involve two larger atoms (Period 2), resulting in longer, slightly weaker single bonds (ranging from 300 to 350 kJ/mol).
Why other options are incorrect:
Option B, Option C, Option D: All feature longer internuclear distances compared to a C-H bond, resulting in lower bond dissociation energies.
Identify the type of hybridization of nitrogen in the following molecule:
A
\( \text{sp} \)
B
\( \text{sp}^2 \)
C
\( \text{sp}^3 \)
D
\( \text{dsp}^2 \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The hybridization of an atom is dictated by its steric number (the sum of sigma bonds and lone pairs). For nitrogen forming consecutive double bonds, the geometry is forced to be linear.
In the given cumulated double-bond structure (similar to an isocyanate group, \( -\text{N}=\text{C}=\text{O} \)), the Nitrogen atom is double-bonded to the adjacent Carbon atom and double-bonded to the aromatic ring (or another group acting as a cation/anion depending on resonance).
Based on the explanatory notes provided in the source key, this specific Nitrogen atom is forming exactly 2 sigma (\( \sigma \)) bonds and 2 pi (\( \pi \)) bonds.
Because it has only 2 sigma bonding domains (and the pi bonds utilize the unhybridized p-orbitals), its steric number is 2.
A steric number of 2 corresponds to \( \text{sp} \) hybridization, giving that segment of the molecule a linear geometry.
Why other options are incorrect:
Option B: Requires 3 sigma domains (e.g., a standard double bond and a lone pair, as in a typical imine).
Option C: Requires 4 sigma domains (e.g., standard single bonds as in ammonia).
Ca (Z=20) forms ionic bond with (Z=17)? What is the chemical formula of calcium chloride?
A
CaCl
B
\( \text{CaCl}_2 \)
C
\( \text{Ca}_2\text{C} \)
D
\( \text{Ca}_2\text{Cl} \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The chemical formula of an ionic compound is derived by "crisscrossing" the valencies (ionic charges) of the constituent elements to ensure the final compound is electrically neutral.
Calcium (Z=20) is in Group IIA. It loses two valence electrons to achieve an octet, forming the \( \text{Ca}^{+2} \) ion.
Chlorine (Z=17) is a halogen in Group VIIA. It needs to gain exactly one electron to complete its octet, forming the \( \text{Cl}^{-} \) ion.
To balance the +2 charge of a single Calcium ion, you require exactly two -1 Chlorine ions (\( +2 \) and \( -2 = 0 \)).
Therefore, the formula is definitively \( \text{CaCl}_2 \).
Why other options are incorrect:
CaCl: Leaves a net positive charge of +1, which is electrically unstable.
\( \text{Ca}_2\text{C} \): Brings Carbon into a question specifically asking about Calcium and Chlorine.
\( \text{Ca}_2\text{Cl} \): Crisscrosses the valencies backward, implying Chlorine has a +2 charge and Calcium has a -1 charge, which is chemically impossible.
The standard bond enthalpy required to break the single covalent sigma bond between two Hydrogen atoms in one mole of \( \text{H}_2 \) gas is experimentally determined to be 436 kJ/mol.
This is a factual memory-based value from standard thermodynamic tables.
Why other options are incorrect:
Option A: 242 kJ/mol is the bond energy of \( \text{Cl}_2 \).
Option B: 431 kJ/mol is the bond energy of HCl.
Option D: 346 kJ/mol is the bond energy of a C-C single bond.
Fluorine is the smallest halogen atom. When it bonds with Hydrogen, the resulting H-F bond length is extremely short.
Because the bonding electrons are held very tightly and closely between the two nuclei, a massive amount of energy (approx. 567 kJ/mol) is required to break the bond.
As we go down the halogen group (Cl, Br, I), the atomic size increases, bond length increases, and bond energy rapidly decreases.
Why other options are incorrect:
Option A, Option C, Option D: These molecules feature progressively larger halogens, resulting in longer, weaker bonds that are easier to break.
In \( \text{CO}_2 \), Carbon has 0 lone pairs. The two double bonds repel each other equally to a 180° angle, making the molecule perfectly linear. The opposing dipoles cancel out, making it non-polar.
In \( \text{SO}_2 \), Sulfur has 1 lone pair. This lone pair repels the two double bonds, bending the molecule into an angular (V) shape. Because it is asymmetrical, the dipoles do not cancel, making it polar.
Why other options are incorrect:
Option A, Option B, Option D: These options fail to recognize the structural difference caused by Sulfur's lone pair compared to Carbon's empty valence shell.
Iso-structural molecules share the exact same geometric shape, which usually implies having the same steric number and number of lone pairs on the central atom.
In Sulfur trioxide (\( \text{SO}_3 \)), Sulfur forms 3 double bonds with oxygen and has 0 lone pairs (\( \text{sp}^2 \)). Geometry: Trigonal Planar.
In Boron trifluoride (\( \text{BF}_3 \)), Boron forms 3 single bonds with fluorine and has 0 lone pairs (\( \text{sp}^2 \)). Geometry: Trigonal Planar.
Because both molecules adopt the exact same flat, triangular geometry, they are iso-structural.
Why other options are incorrect:
Option A: \( \text{AlCl}_3 \) is planar, \( \text{CH}_4 \) is tetrahedral.
Option B: \( \text{BF}_3 \) is planar, \( \text{NH}_3 \) is pyramidal (has a lone pair).
Option C: \( \text{SnCl}_2 \) is bent (has a lone pair), \( \text{BeCl}_2 \) is linear.
The reason of highest electronegativity value of Fluorine is:
A
Complete outermost shell
B
Ability to form negative ion
C
Existence as diatomic molecule
D
Smaller size and higher nuclear change in the respective period
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Electronegativity is driven by the nucleus's ability to attract shared electrons, which is maximized when the atom is small and the nuclear charge is high without excessive shielding.
Valance shell electron pair repulsion theory explains:
A
Bond Energy
B
Bond Length
C
Shapes and Bond Energy
D
Shapes
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
VSEPR (Valence Shell Electron Pair Repulsion) theory was designed exclusively to predict the 3D geometry of molecules based on electrostatic repulsions.
Formula:
$$ \text{Geometry is defined by minimizing LP/BP repulsion.} $$
Solution:
VSEPR theory postulates that electron pairs (both bonding and lone pairs) in the valence shell of a central atom will arrange themselves as far apart as possible to minimize repulsion.
This spatial arrangement directly dictates the physical shape (geometry) of the molecule (e.g., linear, tetrahedral, bent).
It does not mathematically calculate bond energies or bond lengths.
Why other options are incorrect:
Option A, Option B, Option C: Bond energies and lengths are explained by Molecular Orbital Theory and Valence Bond Theory (orbital overlap), not by VSEPR.
Difference in electronegativity between the two bonded atoms
D
Ionization energies of the two bonded atoms
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Bond length is the internuclear distance between two bonded atoms. It is physically determined by orbital size, bond order, and electrostatic polarity.
Multiple bonds: Increase bond order, pulling atoms closer (decreases length).
Hybridization: More s-character (e.g., sp vs sp3) makes orbitals smaller, decreasing bond length.
Electronegativity diff: Creates partial charges causing electrostatic attraction, shortening the bond (Schomaker-Stevenson rule).
Ionization energy: While related to general atomic properties, it does not directly act as a mechanical factor altering the geometric distance of a formed covalent bond.
Why other options are incorrect:
Option A, Option B, Option C: These are the three primary, direct determinants of covalent bond length.
Among LiCl, \( \text{BeCl}_2 \), NaCl and CsCl compounds with the greatest and the least ionic character respectively are:
A
LiCl and CsCl
B
NaCl and LiCl
C
CsCl and NaCl
D
CsCl and \( \text{BeCl}_2 \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Ionic character depends on the electronegativity difference. The metal with the lowest ionization energy (most electropositive) forms the most ionic bond, while the one with the highest IE forms the least ionic (most covalent) bond.
Cesium (Cs) is at the bottom left of the periodic table. It is the most electropositive element, creating a massive electronegativity difference with Chlorine. CsCl has the greatest ionic character.
Beryllium (Be) is at the top of Group 2. It has a very small size and high ionization energy, making it relatively electronegative for a metal. This gives \( \text{BeCl}_2 \) significant covalent character (least ionic).
Why other options are incorrect:
Option A, Option B, Option C: These options fail to pair the absolute extremes of electropositivity present in the list (Cs as highest, Be as lowest).
Which of the following has greatest difference of electronegativity?
A
HF
B
HCl
C
HBr
D
HI
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Electronegativity decreases down Group 7 (Halogens). Therefore, the difference in electronegativity between Hydrogen and a Halogen is greatest with the halogen at the top of the group.
Fluorine is the most electronegative element (4.0).
The difference (\( 4.0 - 2.1 = 1.9 \)) is the highest. As you move to Cl (3.0), Br (2.8), and I (2.5), the difference shrinks steadily.
Why other options are incorrect:
Option B, Option C, Option D: The halogen electronegativities are progressively lower, resulting in smaller \( \Delta\text{EN} \) and more covalent character.
Identify the compound given below which has bond formed by overlapping of sp and p orbital:
A
\( \text{BeCl}_2 \)
B
\( \text{BF}_3 \)
C
\( \text{H}_2\text{O} \)
D
\( \text{NH}_3 \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Determine the hybridization of the central atom to see which one utilizes \( \text{sp} \) hybrid orbitals to bond with the unhybridized p-orbitals of a halogen.
Formula:
$$ \text{Steric Number (SN)} = 2 \implies \text{sp hybridization} $$
Solution:
Beryllium in \( \text{BeCl}_2 \) forms 2 single bonds with Chlorine and has 0 lone pairs.
A steric number of 2 means Be is \( \text{sp} \) hybridized.
Chlorine uses its unhybridized 3p orbital to bond. Thus, the bond is a direct \( \text{sp - p} \) overlap.
Why other options are incorrect:
Option B: Boron in \( \text{BF}_3 \) is \( \text{sp}^2 \) hybridized.
Option C & Option D: Oxygen in water and Nitrogen in ammonia are \( \text{sp}^3 \) hybridized (and hydrogen uses s-orbitals, not p-orbitals).
Carbon dioxide (\( \text{CO}_2 \)) is an \( \text{AB}_2 \) type molecule with no lone pairs on the central Carbon atom.
This gives it a perfectly linear geometry.
The two highly electronegative Oxygen atoms pull electron density equally in exactly opposite directions, causing the bond dipole vectors to sum to zero.
Why other options are incorrect:
Option A & Option D: These geometries are asymmetrical and would result in a net polar molecule.
Option C: Being triatomic does not guarantee a zero dipole (e.g., \( \text{H}_2\text{O} \) is triatomic and highly polar).
Metals lose electrons to empty their valence shell, exposing the full stable shell beneath (becoming isoelectronic with the preceding noble gas).
Nonmetals gain electrons to completely fill their current valence shell (becoming isoelectronic with the succeeding noble gas).
Thus, in stable ionic crystal lattices, both the resulting cations and anions achieve a stable noble gas configuration.
Why other options are incorrect:
Option A & Option D: It is not an exclusive process; both ions achieve stability simultaneously.
Option C: While true that total electrons lost equals total gained in the bulk compound, option B is the fundamental driving force for the individual ions' stability as per standard textbook phrasing.
Chemical bonding is an inherently stabilizing process. Atoms bond to reach a lower, more stable energy state.
Formula:
$$ \Delta H < 0 \text{ (Exothermic process)} $$
Solution:
When two atoms approach each other and form a bond, the attractive forces between their nuclei and the shared electrons overcome the repulsive forces.
This system drops to a lower potential energy state.
The excess energy is released into the surroundings (exothermic), meaning the intrinsic energy of the newly bonded molecule decreases compared to the isolated atoms.
Why other options are incorrect:
Option B & Option C: If energy increased or stayed the same, there would be no thermodynamic driving force for the bond to form in the first place.
Bond energy (or bond enthalpy) measures the strength of a chemical bond. It is quantified as the energy required to break one mole of a specific bond type in the gas phase.
Formula:
$$ \text{Standard SI Unit for Molar Enthalpy} = \text{kJ/mol} $$
Solution:
Because individual atoms are incredibly small, chemists measure bond energies on a macroscopic scale using moles.
The energy is typically very large, so it is expressed in kilojoules (kJ) rather than standard Joules.
Therefore, the standard unit is Kilojoules per mole (kJ/mol).
Why other options are incorrect:
Option B & Option C: While calories and Joules are units of energy, they lack the per-mole (macroscopic) scaling standard in chemical thermodynamics.
Option D: Volt is a unit of electrical potential, not chemical energy.
Primary bonds involve the actual transfer or sharing of electrons between atoms to form molecules or crystal lattices (e.g., Ionic, Covalent, and Dative/Coordinate bonds).
Hydrogen bonds are secondary bonds. They are relatively weak dipole-dipole intermolecular forces that occur between already stable molecules, rather than forming the fundamental structure of the molecule itself.
Why other options are incorrect:
Option A, Option C, Option D: These all involve the sharing or transferring of valence electrons, strictly defining them as strong primary bonds.
Which of the following has NOT \( \text{sp}^3 \) hybridized carbon?
A
Benzene
B
1-Butene
C
2-Butyne
D
1-Pentene
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
An \( \text{sp}^3 \) hybridized carbon atom must be attached to four separate domains exclusively by single bonds (sigma bonds).
Formula:
$$ \text{sp}^3 \implies \text{Only single bonds (alkane-like carbons)} $$
Solution:
Benzene (\( \text{C}_6\text{H}_6 \)) consists of a planar hexagonal ring where every single carbon atom forms one double bond (delocalized) and two single bonds. Therefore, every carbon in benzene is \( \text{sp}^2 \) hybridized. It has absolutely zero \( \text{sp}^3 \) carbons.
1-Butene, 2-Butyne, and 1-Pentene all contain carbon chains that feature standard alkane-like \( \text{CH}_2 \) or \( \text{CH}_3 \) groups, which are fully \( \text{sp}^3 \) hybridized.
Why other options are incorrect:
Option B, Option C, Option D: These aliphatic chains contain at least one saturated carbon with four single bonds, meaning they do possess \( \text{sp}^3 \) hybridized carbons.
Type of hybridization of carbon in Ethene (\( \text{H}_2\text{C}=\text{CH}_2 \)) is:
A
\( \text{sp}^3 \)
B
\( \text{sp} \)
C
\( \text{sp}^2 \)
D
\( \text{dsp}^2 \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The hybridization of a carbon atom depends on the number of sigma bonds it forms.
Formula:
$$ \text{Steric Number (SN)} = \text{Number of } \sigma \text{ bonds} $$
Solution:
In Ethene (\( \text{C}_2\text{H}_4 \)), each Carbon atom is bonded to two Hydrogens (2 single sigma bonds) and one Carbon (1 double bond = 1 sigma + 1 pi).
Total sigma domains = 3.
A steric number of 3 requires the mixing of one s and two p orbitals, yielding \( \text{sp}^2 \) hybridization.
Why other options are incorrect:
Option A: Requires 4 sigma bonds (e.g., Ethane).
Option B: Requires 2 sigma bonds (e.g., Ethyne).
Option D: Involves d-orbitals, which carbon (in Period 2) does not possess in its valence shell.
Which of the following molecule is covalent in nature?
A
NaCl
B
\( \text{AlCl}_3 \)
C
\( \text{MgCl}_2 \)
D
KCl
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
According to Fajans' Rules, the covalent character of an ionic compound increases with a high positive charge and small size of the cation (high polarizing power).
Aluminum in \( \text{AlCl}_3 \) has a very high charge (\( 3^+ \)) and a small ionic radius compared to Na, K, and Mg.
This gives \( \text{Al}^{3+} \) extreme polarizing power, allowing it to strongly distort (polarize) the electron cloud of the chloride anions.
This heavy distortion pulls the electrons into the space between the nuclei, creating substantial electron sharing (covalent character). Therefore, \( \text{AlCl}_3 \) exists predominantly as a covalent molecule.
Why other options are incorrect:
Option A, Option C, Option D: These metals (Na, Mg, K) have lower charges (\( 1^+, 2^+ \)) and larger radii, leading to weak polarizing power and predominantly ionic bonds.
The hybridization of an atom can be quickly determined by counting the number of bonded atoms and lone pairs (steric number).
Solution:
The chemical formula for ethane is \( \text{C}_2\text{H}_6 \).
The structure of ethane consists of a single covalent bond between the two carbon atoms (C-C), and each carbon is singly bonded to three hydrogen atoms (C-H).
There are zero lone pairs on the carbon atoms.
Therefore, each carbon atom is completely surrounded by 4 single sigma bonds.
A steric number of 4 corresponds to the mixing of one 's' orbital and three 'p' orbitals, resulting in \( \text{sp}^3 \) hybridization with a tetrahedral geometry.
Why other options are incorrect:
\( \text{sp}^2 \): Found in ethene (double bonds).
sp: Found in ethyne (triple bonds).
\( \text{sp}^3\text{d} \): Impossible for carbon, as it does not have empty d-orbitals in its valence shell.
In ethyne, each Carbon atom forms exactly 2 sigma bonds (one with Hydrogen, one with the other Carbon) and has no lone pairs.
A steric number of 2 corresponds to \( \text{sp} \) hybridization.
Since both carbons are identical, the sigma bond between them is formed by the head-on overlap of an \( \text{sp} \) orbital from one carbon with an \( \text{sp} \) orbital from the other.
The geometry of \( \text{AB}_3 \) type molecule is:
A
Trigonal pyramidal
B
Trigonal planar
C
Trigonal bipyramidal
D
Tetragonal
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
According to VSEPR theory, a central atom bonded to three identical atoms with zero lone pairs will arrange those bonds to maximize distance and minimize repulsion.
Fluorine is at the very top right of the periodic table (excluding noble gases).
It has the smallest atomic radius among the halogens and minimal shielding.
Because its valence electrons are closest to the highly attractive positive nucleus, Fluorine has the highest electronegativity value of all elements (4.0 on the Pauling scale).
Why other options are incorrect:
Option A, Option B, Option C: As you move down Group 7, the atomic radius and shielding effect increase, weakening the nucleus's grip on shared electrons and lowering electronegativity.
Molecules with \( \text{sp}^3 \) or higher hybridization (in 3 dimensions) are generally non-planar, while those with purely \( \text{sp}^2 \) or \( \text{sp} \) hybridization lacking lone pair distortions tend to be planar or linear.
Carbon dioxide (\( \text{CO}_2 \)) has a central Carbon atom with 2 double bonds and 0 lone pairs. It is linear.
Beryllium fluoride (\( \text{BeF}_2 \)) has a central Beryllium atom with 2 single bonds and 0 lone pairs. It is also completely linear.
Why other options are incorrect:
Option A, Option B, Option C: \( \text{H}_2\text{S} \), \( \text{SO}_2 \), and \( \text{SnCl}_2 \) all possess lone pairs on their central atoms, which push the bond pairs down, creating a bent or angular geometry.
Nitrogen has 5 valence electrons. It uses 3 to bond with Hydrogen atoms, leaving 1 lone pair.
The 4 electron domains arrange tetrahedrally to minimize repulsion, but the presence of the lone pair makes the observable molecular shape trigonal pyramidal.
Why other options are incorrect:
Option B: Occurs in \( \text{sp} \) hybridized molecules without distorting lone pairs.
Option C: Describes the electron domain geometry, but the molecular shape (excluding the invisible lone pair) is pyramidal.
Option D: Occurs in \( \text{sp}^2 \) molecules (like \( \text{BF}_3 \)) with no lone pairs.
Removing a lone pair by converting it into a bonding pair removes the strong LP-BP repulsion, allowing the molecule's bond angles to expand to their ideal geometric state.
In ammonia (\( \text{NH}_3 \)), the nitrogen has 3 bond pairs and 1 lone pair. The lone pair repels the bonds strongly, compressing the angle to 107°.
When a proton (\( \text{H}^+ \)) attaches, the lone pair becomes a bond pair, forming \( \text{NH}_4^+ \).
The ion now has 4 identical bond pairs and 0 lone pairs. It assumes a perfect tetrahedral geometry with angles of exactly 109.5°.
Why other options are incorrect:
Option A, Option B, Option D: These state incorrect initial angles or falsely claim the angle decreases. The angle must increase because the compressing force of the lone pair is neutralized.
Which of the following molecule has zero dipole moment?
A
\( \text{PCl}_3 \)
B
\( \text{BF}_3 \)
C
\( \text{NH}_3 \)
D
\( \text{H}_2\text{O} \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
A molecule has a zero dipole moment if it is highly symmetrical and lacks lone pairs on its central atom, causing all individual bond dipoles to cancel each other out.
Formula:
$$ \mu = 0 \text{ for perfectly symmetrical molecules.} $$
Solution:
Boron trifluoride (\( \text{BF}_3 \)) has \( \text{sp}^2 \) hybridization, giving it a flat, trigonal planar geometry.
Boron has no lone pairs to distort the symmetry.
The three polar B-F bonds pull equally at 120° angles, resulting in a net vector sum (dipole moment) of zero.
Why other options are incorrect:
Option A, Option C, Option D: \( \text{PCl}_3 \), \( \text{NH}_3 \), and \( \text{H}_2\text{O} \) all possess lone pairs on their central atoms, which creates asymmetry and results in a net, non-zero dipole moment.
Among the following molecules, which one has coordinate covalent (dative) bond?
A
\( \text{CCl}_4 \)
B
\( \text{CO} \)
C
\( \text{CO}_2 \)
D
\( \text{CH}_4 \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
A coordinate covalent (dative) bond is formed when one atom provides both electrons for a shared pair to complete the octet of another atom.
Formula:
$$ :\text{C} \equiv \text{O}: $$
Solution:
In Carbon Monoxide (CO), Carbon has 4 valence electrons and Oxygen has 6.
They form two normal covalent bonds, giving Oxygen 8 electrons (stable), but Carbon only has 6.
To stabilize Carbon, Oxygen donates a lone pair to form a third bond (a coordinate covalent bond). Thus, CO possesses a triple bond: two normal and one dative.
Why other options are incorrect:
Option A, Option C, Option D: All atoms in these molecules complete their octets through standard mutual sharing of electrons. No lone pair donation is required.
Which of the following molecule has zero dipole moment?
A
\( \text{PCl}_3 \)
B
\( \text{BF}_3 \)
C
\( \text{NH}_3 \)
D
\( \text{H}_2\text{O} \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Dipole moments cancel out entirely in highly symmetrical molecules lacking lone pairs on the central atom.
Formula:
$$ \text{Net Dipole } (\mu) = 0 $$
Solution:
As established, Boron trifluoride (\( \text{BF}_3 \)) has perfectly symmetrical trigonal planar geometry.
Because all B-F bonds are identical and spaced equally at 120°, their individual dipole vectors cancel completely.
Why other options are incorrect:
Option A, Option C, Option D: All possess lone pairs on the central atom, creating asymmetric shapes (pyramidal and bent) that prevent dipole cancellation.
For formation of ionic bond, electronegativity difference should be:
A
Equal to zero
B
Equal to 0.5
C
More than 1.7
D
Less than 1.7
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The nature of a bond (ionic vs. covalent) is determined by the difference in electronegativity (\( \Delta \text{EN} \)) between the two bonding atoms.
Formula:
$$ \Delta \text{EN} > 1.7 \implies \text{Ionic Bond predominantly} $$
Solution:
If the difference is greater than 1.7, the more electronegative atom will effectively strip the electron completely away from the less electronegative atom, forming ions.
This results in electrostatic attraction, forming an ionic bond (usually >50% ionic character).
Why other options are incorrect:
Option A & Option B: Indicate non-polar covalent bonds (equal or nearly equal sharing).
Option D: Represents a polar covalent bond, where electrons are shared unequally but not fully transferred.
The two bulky lone pairs strongly repel each other and the bond pairs, squeezing the bond angle (e.g., in water \( \text{H}_2\text{O} \), from 109.5° down to 104.5°).
Because molecular shape is determined only by the positions of the atoms, the resulting shape is angular (bent).
Why other options are incorrect:
Option A: Requires 4 bond pairs and 0 lone pairs.
Option B: Requires 3 bond pairs and 1 lone pair.
Option D: Requires no lone pairs interfering with the bonds.
Fluorine (\( \text{F}^{-1} \)) is the most electronegative element in the periodic table (EN = 4.0).
The difference in electronegativity between Iron and Fluorine will be the largest possible among the given choices.
Therefore, the bond between \( \text{Fe}^{+2} \) and \( \text{F}^{-1} \) will exhibit the highest percentage of ionic character.
Why other options are incorrect:
Option A, Option B, Option C: Nitrogen, Tin, and Phosphorus have significantly lower electronegativities than Fluorine, resulting in bonds with higher covalent character.
The bond angle in \( \text{H}_2\text{S} \) is less than \( \text{H}_2\text{O} \). It is due to:
A
Small size of oxygen atom
B
Greater E.N of oxygen atom
C
Oxygen contain two lone pairs of electrons
D
All of the above
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Correct Key: Option BDiagnostic Explanation
Concept:
When the central atom is highly electronegative, bond pairs are pulled closer to the central nucleus. This causes strong bond pair-bond pair (BP-BP) repulsion, which forces the bond angle to widen.
Formula:
$$ \text{Bond Angle} \propto \text{Electronegativity of Central Atom} $$
Solution:
Oxygen (EN = 3.5) is much more electronegative than Sulfur (EN = 2.5).
In \( \text{H}_2\text{O} \), the bonding electrons are pulled very close to the Oxygen nucleus. Because they are confined in a tight space, they repel each other strongly, keeping the angle relatively wide (104.5°).
In \( \text{H}_2\text{S} \), the bonding electrons are further from the Sulfur nucleus. The BP-BP repulsion is weaker, allowing the lone pairs to crush the bond angle down to about 92°.
Why other options are incorrect:
Option A: Size plays a role, but electronegativity is the direct electronic driver of the bond pair positioning.
Option C: Both Oxygen and Sulfur contain two lone pairs, so this is not the distinguishing factor.
A perfect bond angle of 109.5° occurs in molecules with a regular tetrahedral geometry, which requires \( \text{sp}^3 \) hybridization and zero lone pairs on the central atom.
In all four illustrated structures, the core species is the ammonium ion (\( \text{NH}_4^+ \)).
The nitrogen atom in ammonia (\( \text{NH}_3 \)) has one lone pair.
Nitrogen donates this lone pair to an empty 1s orbital of a hydrogen ion (\( \text{H}^+ \)), forming a coordinate covalent bond between N and H.
Why other options are incorrect:
Option A, C & D: These do not accurately describe the dative bond formation, which exclusively occurs between the central Nitrogen and the incoming Hydrogen ion.
An s-orbital is spherical and can only participate in head-on (sigma) overlaps.
p-orbitals are dumbbell-shaped. When two parallel p-orbitals are adjacent, they can overlap sideways above and below the internuclear axis to form a pi-bond.
Option B: While d-orbitals can form pi bonds in transition metals, p-orbital overlap is the fundamental and primary answer expected in standard organic and basic inorganic chemistry contexts.
There are 4 Carbon-Hydrogen single bonds, contributing 4 \( \sigma \) bonds.
There is 1 Carbon-Carbon double bond. This contributes 1 \( \sigma \) bond and 1 \( \pi \) bond.
Total \( \sigma \) bonds = \( 4 + 1 = 5 \).
Total \( \pi \) bonds = 1.
Why other options are incorrect:
Option B, Option C, Option D: These incorrectly count the fundamental bond types, likely confusing the number of electrons (a double bond has 4 electrons) with the number of discrete bonds.
When the two partially filled atomic orbital overlap in such a way that the probability of finding the electron is maximum around the line joining the two nuclei, the result is the formation of:
A
Sigma bond
B
Hydrogen bond
C
Pi-bond
D
Metallic bond
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Correct Key: Option ADiagnostic Explanation
Concept:
A sigma (\( \sigma \)) bond is formed by the direct, head-to-head (axial) overlap of atomic orbitals.
Formula:
$$ \text{Electron density is concentrated along the internuclear axis.} $$
Solution:
Because the overlap occurs exactly between the two nuclei, the resulting electron density is symmetrically distributed around the internuclear axis.
This direct overlap produces the strongest type of covalent bond, called a Sigma bond.
Why other options are incorrect:
Option B & Option D: These represent entirely different bonding paradigms (intermolecular and metallic sea of electrons, respectively).
Option C: In a pi-bond, the probability of finding electrons is maximum above and below the internuclear axis (sideways overlap), not directly on the line joining the nuclei.
According to Valence shell electron pair repulsion theory, the repulsive forces between the electron pairs of central atom of a molecule are in the order:
The VSEPR theory states that electron pairs around a central atom repel each other, and lone pairs exert a stronger repulsive force than bond pairs because they are held closer to the central nucleus.
The angle between un-hybridized p-orbital and three \( \text{sp}^2 \) hybrid orbitals of each carbon atom is:
A
120°
B
109.5°
C
90°
D
180°
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Correct Key: Option CDiagnostic Explanation
Concept:
In \( \text{sp}^2 \) hybridization, three hybrid orbitals form a planar trigonal structure, and the remaining unhybridized p-orbital is situated perpendicular to this plane.
In 'H-F' bond, electronegativity difference is 2.0. What is the type of this bond?
A
Polar covalent bond
B
pi (\( \pi \)) bond
C
Non-polar covalent bond
D
Co-ordinate covalent bond
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Correct Key: Option ADiagnostic Explanation
Concept:
Bond polarity is determined by the electronegativity difference (\( \Delta \text{E.N.} \)) between the two bonded atoms. However, bonding in Hydrogen Halides acts slightly unusually due to their molecular nature.
Generally, an electronegativity difference > 1.7 suggests an ionic bond.
However, HF is a distinct exception. It exists as discrete molecules and is a gas at room temperature, which are characteristic properties of covalent compounds.
Therefore, despite the large electronegativity difference, HF strictly forms a highly polar covalent bond (with significant ionic character, but primarily covalent).
Why other options are incorrect:
It is not non-polar (as electrons are unequal). It is a sigma bond, not a pi bond. It forms via mutual sharing, not through the dative donation of a lone pair.
The number of bonds in nitrogen molecule (\( \text{N}_2 \)) is:
A
One \( \sigma \) and one \( \pi \)
B
Three \( \sigma \) only
C
One \( \sigma \) and two \( \pi \)
D
Two \( \sigma \) and one \( \pi \)
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Correct Key: Option CDiagnostic Explanation
Concept:
Diatomic nitrogen contains a triple bond. A multiple bond always consists of exactly one sigma (\( \sigma \)) bond, with the remainder being pi (\( \pi \)) bonds.
Formula:
$$ \text{N} \equiv \text{N} $$
Solution:
To satisfy their octets, two Nitrogen atoms (Group VA, 5 valence electrons) share 3 pairs of electrons.
The first bond formed directly between the nuclei is a strong \( \sigma \) bond due to head-on overlap.
The second and third bonds are formed by the sideways overlap of unhybridized p-orbitals, creating two \( \pi \) bonds.
Why other options are incorrect:
Option A: Describes a double bond (e.g., in \( \text{O}_2 \)).
Option B & Option D: Are physically impossible between two single atoms.
\( \text{BF}_3 \) is a trigonal planar molecule with no lone pairs on the central Boron atom.
The pull of the three highly electronegative Fluorine atoms is perfectly balanced in 2D space, making it a non-polar molecule with a zero dipole moment.
Why other options are incorrect:
Option A & Option B: Have lone pairs that destroy symmetry.
Option C: \( \text{CHCl}_3 \) has tetrahedral geometry, but because the atoms attached to Carbon are not all identical (one H, three Cl's), the bond dipoles do not cancel, leaving a net dipole.
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