Model Test Questions XI Chemistry Chemical Bond Test # 2 – (2025 Updated Questions & Answers)

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In this post, we’re sharing XI Chemistry Model Test #2 for Chapter 3: Chemical Bond. These questions are carefully designed to help you understand key bonding concepts, prepare for exams, and practice like a topper. Perfect for revision before your class tests or board exams 2025.

Prepare for your Class 11 Chemistry exam with our Model Test #2 for Chapter 3: Chemical Bond. Includes important conceptual and past paper questions according to the latest 2025 syllabus. Perfect for Sindh, Punjab, and Federal Board students.

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🧪✨ Model Test Questions XI Chemistry: Chemical Bond Test #2 – 2025 Updated Questions & Answers 🔥📘

Q1.  Differentiate between the following.

(i) Sigma and pi bond                         

(ii)Bonding & anti-bonding molecular orbitals (BMO and AMO) 

(iii) VBT and MOT

Q2.  Identify the type of hybridization in the central atom of following molecules:

Q3. Describe strength of covalent bond in term of VBT

Q4. Write down limitations of VBT

Q5. Predict the shape of the following molecules on the basis of AOH and VSEPR theory.

(i)    BeCl₂     (linear, two AEP, No LP) 

(ii)  NH₃        (pyramidal, 4 AEP with 1 LP, LP shrinks BA)

(iii)  H₂O        (angular, 4 AEP with 2 LP, LP shrinks BA)        

Q6.  Write down the main postulates of VSEPR theory and Predict and explain the shape of following molecules on the basis of VSEPR theory shape of AlCl₃, CCl₄, PH₃, H₂S

Q7. What is Hybridization? Explain this process with general example of Be, B and carbon. Explain sp³hybridization in CH₄ molecule and sp² hybridization in C₂H₄ molecule

Q8.  Write down basic postulates of valence bond theory. Explain various types of overlapping lead to sigma and pi bond.

Q9. Draw a molecular orbital diagram of O₂ molecules and explain following

(i)           Write down MO electronic configuration

 (ii)         Determine the bond order

(iii)        Explain the paramagnetic behaviour.

Q10. Differentiate between sp³ and sp² hybridization and explain sp hybridization. Discus sp hybridization and the shape of ethyne molecule.

Q1.Differentiate between the following.

(i)    Sigma and pi bond                        

(ii) Bonding and anti-bonding molecular orbitals (BMO & AMO)   

(iii)  VBT and MOT

Answer (Learn any 5 points)

Difference between Bonding & Anti-bonding Molecular Orbitals

Difference between Sigma and Pi Bond

Difference between VBT and MOT 

Q2. Identify the type of hybridization in the central atom of following molecules:       

Answer

Q3. Describe strength of covalent bond in term of VBT

Answer

Strengths of Bond in terms of Valence Bond Theory

According to VBT, the strength of covalent bond is proportional to the extent of overlapping o atomic orbitals. The grater the overlapping, the stronger the bond between the atoms and the higher the bond energy required to break the bond.

The extent of overlapping in sigma bond is sufficient. This results in a strong bond between two atoms. In pi bond the orbital has two regions of electronic density, one above the nuclear axis and the other is below the nuclear axis. The lateral overlapping of atomic orbitals is not maximum that is why pi bond is weaker than sigma bond.

The relative strength of s sigma is related to the overlap of atomic orbitals. This is known as the Principle of Maximum Overlap. The relative bond strength is   p–p > s–p > s–s.

Due to spherical charge distribution in s-orbital, generally s-s overlap is not so effective as s-p and p-p overlap. Since p-orbitals have directional charge distribution and longer lobes which cause more effective overlap. Thus s-s sigma bond is relatively weak than s-p and p-p sigma bond. 

Q4.  Write down limitations of VBT

Answer

1.  It fails to describe the valencies in carbon, boron and beryllium.

2. It does not tell about the para magnetic behaviour of oxygen.

3. It fails to describe the delocalization of electrons in the molecules.

Q5. Predict the shape of the following molecules on the basis of AOH and VSEPR theory.

(i)  BeCl₂ (linear, two AEP, No LP) 

(ii) NH₃ (pyramidal, 4 AEP with 1 LP, LP shrinks BA)

(iii) H₂O (angular, 4 AEP with 2 LP, LP shrinks BA)  

Answer

Shape of Beryllium Chloride (BeCl₂) predicted by VSEPR theory

From Lewis structure of BeCl₂, there are two active set of electron (bond pairs) surrounding the central atom beryllium and the central atom has no lone pair i.e. BeCl₂ is of the AB₂ type molecule. According VESPER model, two active electron pairs must be arranged as far apart as possible to minimize repulsion which is achieved by placing two chlorine atoms on the opposite site of beryllium atom giving the linear geometry to BeCl₂. 

Shape of Beryllium Chloride (BeCl₂) predicted by Hybridization

Details

Since in BeCl₂, beryllium (central atom) is attached with two Cl atoms, thus according to hybridization, Be uses its sp hybrid orbitals to make its bonds i.e. in BeCl₂, Beryllium is sp hybridized. This can be proved by considering its ground state, excited state and hybridized electronic configurations:

Ground State Configuration of ₄Be = 1s^↿⇂ , 2s^↿⇂, 2p_x⁰ 2p_y⁰ 2p_z⁰ 

(1 unpaired electrons; Zero-valent)

Excited State Configuration of ₄Be = 1s^↿⇂ , 2s^↿, 2p_x^↿ 2p_y⁰ 2p_z⁰ 

(2 unpaired electrons; divalent)

Hybridized State Configuration ₄Be = 2s^↿ + 2p_x^↿ Þ sp^↿ + sp^↿           

(Equivalent Valencies)

(These two half-filled orbitals of beryllium (one s and one p orbital) undergoes mixing to produce two new equivalent sp hybrid orbitals of equal energy and are co-linear at an angle of 180° which provide maximum separation and overlap assuming a linear geometry).

The two sp hybrid orbitals of beryllium overlap with 3p_z orbital of two chlorine atoms in straight line to give two Be–Cl  sigma bonds. Thus BeCl₂ molecule acquires linear structure. The linear geometry of molecule is due to the maximum repulsion of two electron pairs.

Shape of NH₃ predicted by VESPER Model 

Shape of Ammonia (NH₃) predicted by Hybrid Orbital Model

Shape of H₂O predicted by VESPER Model 

Shape of Water (H₂O) predicted by Hybrid Orbital Model

Details 

In water i.e. H₂O, oxygen is bonded to two hydrogen atoms with two lone pair on O atom, thus oxygen gets sp³-hybridized and uses sp³ hybrid orbitals to make its bonds i.e. in H₂O, oxygen is sp³-hybridized. This can be explained by considering its electronic configuration.

The ground state electronic configuration of oxygen is 1s^↿⇂ 2s^↿⇂ 2p_x^↿ 2p_y^↿ 2p_z^↿. But it is assumed that one electron from 2s orbital get promoted to 2pz orbital resulting in its excited state electronic configuration of oxygen is 1s^↿⇂ 2s^↿ 2p_x^↿ 2p_y^↿ 2p_z^↿⇂. These two half-filled and two fully filled orbitals of valence shell of oxygen  undergoes mixing to produce four new equivalent sp³ hybrid orbitals of equal energy and shape directed towards the corner of a regular tetrahedron at angle of 109.5°.

Out of four sp³ hybrid orbitals of oxygen, two have single electrons which overlap with 1s orbital of hydrogen atoms to form two O–H sigma bonds. The remaining two non-bonding sp³ orbitals with two lone pairs on oxygen remains unbounded. Because the repulsion of lone pair is grater than bond pairs, the shape of water molecule is not regular tetrahedron. The stronger repulsion of non-bonding orbital on O deviates the bond angle from 109.5° to 104.5°. The distortion of bond gives rise to angular geometry in water molecule. 

Q6.  Write down the main postulates of VSEPR theory and Predict and explain the shape of following molecules on the basis of VSEPR theory shape of

                AlCl₃, CCl₄, PH₃, H₂S

Answer

Theories of Shape of the Molecules

Simple polyatomic molecules and ions generally acquire linear, trigonal, tetrahedral, pyramidal and angular shapes etc. These shapes can be determined experimentally, however also predicted on theoretical basis. Beside VBT, there are two more significant theories which describe the shape of molecules

1.  Valence shell electron pair repulsion theory (VSEPR/VESPER) 

2. Hybridization or Hybrid orbital model

Definition

This theory was put forward by Sigwick and Powell in 1940. This theory was based on electron pairs repulsion of the valence shell of central atom which is responsible to give characteristic shape of molecules.

Assumptions

1. There may be two types of electron pairs surrounding the central atom.

(a) Bond Pairs; 

These are the result of the sharing of unpaired electrons of central atom with unpaired electrons of surrounding atoms. They are also active set of electrons. (The electrons which take part either in single, double or triple bond formation (bond pair) between the central atom and surrounding atom are considered to be one pair of Active Electrons).

(b)Lone Pairs; 

These are the paired electrons of central atom which do not take part in sharing. They are also called non-bonding pairs. It is also considered as active set of electrons.
The sum of bond pair and lone pair are collectively called active pair or steric number.

In case of molecules with multiple bonds in the form of double and triple bonds, the pi electrons are considered to be inactive set of electrons in VESPER because pi bonds do not alter the basic idealized geometry of a molecules. Hence pi electrons are not included in the count of total active electron pairs.

2.    being similarly charged i.e. negative, the bond pairs as well as the lone pairs repel each other.

3.    Due to repulsion, the electrons pairs of central atom try to be as far as apart as possible; hence the orient themselves in space in such a manner that force of repulsion between them is minimized.

4.      The force of repulsion between lone pairs and bond pairs is not the same. (Since lone pairs are spread out more broadly than the bonding pairs, repulsion is greatest between two lone pairs, intermediate between the lone pair and bonding pair and weakest between two bonding pairs).  The order of repulsion is:

Lone Pair-Lone Pair repulsion > Lone Pair-Bond Pair repulsion> Bond Pair-Bond Pair repulsion

5.   The shape of molecule depends upon total no. of active electron pairs (bonding and lone pairs) or (steric number). It is summarized as follows

Q7. What is Hybridization? Explain this process with general example of Be, B and C. Explain sp³ hybridization in CH₄ molecule and sp² hybridization in C₂H₄ molecule

Answer

Definition

The word HYBRIDIZATION means mixing or blending. The blending or mixing of different atomic orbitals (belonging to an atom) having small energy differences to give the same number of new equivalent orbitals of same shape, size and energy is called Hybridization and the new equivalent orbitals are called HYBRID ORBITALS.

Explanation with special reference to orbital hybridization in Carbon and Boron (Brief)

The number of half-filled valence orbitals or unpaired electrons in the valence shell of an atom constitutes its valency. However, this rule is violated in compounds of group IIA (Be), IIIA (B) and IVA (C) etc.

Hybridized State (Equivalent Valencies)

On the basis of excited state electronic configuration, it might be expected that these elements would form different type of single covalent bond with different bond lengths, bond energy and directional nature. e.g. One expect that in CH₄, three C–H bonds formed by the overlap of three 2p–orbitals would be identical (having directional nature and higher energy) while the fourth C–H bond formed due to 2s-orbtial would be different (having non-directional nature and lower energy). In actual practice, all four C–H bonds in CH₄ are identical in all respects (i.e. in bond energy and bond length).

Linus Pauling solved this discrepancy by suggesting the idea of hybridization involving the mixing of atomic orbitals having nearly equal energies in various ways within an atom to form equivalent hybrid orbitals. It means that atomic orbitals of Be, B and C have equalized their energies, size and shape. This process is called Hybridization and these atoms are said to be HYBRIDIZED.

Detailed Description of Explanation with special reference to orbital hybridization in Carbon and Boron

Ground State Configuration

The number of half-filled valence orbitals or unpaired electrons in the valence shell of an atom constitutes its valency. However, this rule is violated in compounds of group IIA (Be), IIIA (B) and IVA (C) etc.

from their ground or atomic states electronic configuration in terms of unpaired electrons, Be  appear to behave as an inert gas (valency = 0), boron might be expected to be monovalent (valency =1) and carbon would be divalent (valency = 2) [due to the presence of one and two half-filled orbitals respectively]. In actual practice, however, Be, B and C are divalent, trivalent and tetravalent in most of its compounds respectively.

Electronic Configuration of ₄Be = 1s^↿⇂ , 2s^↿⇂, 2p_x⁰ 2p_y⁰ 2p_z⁰  (1 unpaired electrons; monovalent)

Electronic Configuration of ₅B = 1s^↿⇂ , 2s^↿⇂, 2p_x^↿⇂ 2p_y⁰ 2p_z⁰ (2 unpaired electrons; divalent)

Electronic Configuration of ₆C = 1s^↿⇂ , 2s^↿⇂, 2p_x^↿ 2p_y^↿ 2p_z⁰   (3 unpaired electrons; trivalent)

Excited State Configuration

To account for these discrepancies (anomalies) in the valency of such elements, it is assumed that some of the paired electrons are uncoupled and one of the electron from the lower energy orbital belonging to the ground state (2s) is promoted to empty orbital of slightly higher energy (p_y and p_z) before bond formation achieving the excited state. This arrangement of electrons after promotion is referred to as an excited state. The excited state electronic configuration results in an increase in the number of unpaired electrons.

The promotion will require an input of energy. The energy required for the excitation (promotion) and unpairing the electron (of 2s) is compensated by the heat of reaction (energy) released during hybridization and the process of additional covalent bond formation. According to these excited state electronic configurations, Be becomes divalent, boron becomes trivalent while carbon becomes tetravalent.

Electronic Configuration of ₄Be = 1s^↿⇂ , 2s^↿, 2p_x^↿ 2p_y⁰ 2p_z⁰      (2 unpaired electrons; divalent)

Electronic Configuration of ₅B   = 1s^↿⇂ , 2s^↿⇂, 2p_x^↿ 2p_y^↿ 2p_z⁰       (3 unpaired electrons; trivalent)

Electronic Configuration of ₆C   = 1s^↿⇂ , 2s^↿, 2p_x^↿ 2p_y^↿ 2p_z^↿             (4 unpaired electrons; tetravalent)

Hybridized State

On the basis of excited state electronic configuration, it might be expected that Be would form 2 covalent bonds, boron would form 3 and carbon would form 4 covalent bonds. One expect that in CH₄, three C–H bonds formed by the overlap of three 2p–orbitals would be identical (having directional nature and higher energy) while the fourth C–H bond formed due to 2s-orbtial would be different (having non-directional nature and lower energy). In actual practice, all four C–H bonds in CH₄ are identical in all respects (i.e. in bond energy and bond length).

Linus Pauling settled this disparity by suggesting the idea of hybridization involving the mixing of atomic orbitals having nearly equal energies in various ways within an atom to form equivalent hybrid orbitals. It means that atomic orbitals of Be, B and C have equalized their energies, size and shape. This process is called Hybridization and these atoms are said to be HYBRIDIZED.

Explanation of sp³ hybridization in CH₄ molecule

In methane i.e. CH₄, carbon is bonded to four hydrogen atoms, thus carbon gets sp³-hybridized and uses sp³ hybrid orbitals to make its bonds. The ground state electronic configuration of carbon is 1s^↿⇂ 2s^↿⇂ 2p_x^↿ 2p_y^↿ 2p_z⁰ showing that it is divalent. But it is assumed that one electron from 2s orbital get promoted to 2pz orbital to make it tetravalent resulting in its excited state electronic configuration of carbon is 1s^↿⇂ 2s^↿ 2p_x^↿ 2p_y^↿ 2p_z^↿. These four half-filled orbitals of carbon (one s and three p orbitals) undergoes mixing to produce four new equivalent sp³ hybrid orbitals of equal energy and shape directed towards the corner of a regular tetrahedron at angle of 109.5°.

Each sp³ hybrid orbital with one electrons overlaps with 1s orbital of H atom on linear axis to form four C–H sigma bonds. (Each H–C bond is sigma bond which is formed due to s-sp³ overlapping. Thus CH₄ has tetrahedral geometry. Each bond angle in CH₄ is 109.5°. The bond length between C–H is 1.09Å).

Explanation of sp² hybridization in C₂H₄ molecule

Ethene molecules consists of two central carbon atoms. In ethene i.e. C₂H₄, each carbon is bonded to three other atoms (i.e. 2 H and 1 C atom) thus each carbon uses sp² hybrid orbitals plus unhybrid p_z orbital to make its bonds.

The three sp² hybrid orbitals arrange themselves in trigonal planar geometry at 120°. One sp²-hybrid orbital of each carbon atom overlaps linearly with other carbon to form a C–C sigma bond. The remaining two sp²-hybrid orbitals of each carbon overlaps linearly with 1s orbitals of two hydrogen atoms to form four C–H sigma bonds. Now unhybrid p_z orbitals of both carbon atoms which lie at right angle overlap side wise to form a pi bond between carbon to carbon.

Q8.  Write down basic postulates of valence bond theory. Explain various types of overlapping lead to sigma and pi bond.

Answer

In 1927, Heitler and London put forward Valence Bond Theory (VBT). According to VBT, the bonding electrons occupy the atomic orbitals of the bonded atoms, and the shared electron is influenced by one nucleus (monocentric).

Main Postulates

1.   A covalent bond is formed due to the overlapping of half-filled atomic orbitals of combing atoms (Sharing of some common regions in space by atomic orbitals is called overlapping)

2.    As a result of overlapping there is maximum electron cloud between the two overlapped orbitals.

3.  The electron present in both overlapped orbital should be in opposite spin.

4. The strength of covalent is determined by the extent of overlap. The greater the overlap, the stronger is the bond.

5.    Each atom involves in the overlapping keep its own atomic orbital but the electron pair is shared up by both atoms which take part in the overlapping. (atomic orbitals which are involved in the bond formation maintain their individual nature and identities).

6.   (In VBT, some of the valence electrons are indicated as unshared and uninvolved in the formation of molecule).

Types of Overlapping

There are two types of overlapping:

(a) Head-on (i.e. End to end) overlapping (results in s-bond).

(b) Side-ways (i.e. lateral) overlapping (results in p-bond).

Sigma bond formation due to s-s linear overlapping in H₂ Molecule

s-s overlapping involves the overlap of two half-filled s orbitals of two combining atoms. This type of overlap exist in H₂ molecule.

Each hydrogen atom has one half-filled 1s-orbital containing one electron. In the formation of H₂ molecule, the two H atoms come closer to each other, 1s-orbitals of both hydrogen atoms containing an unpaired electron overlap with each other along inter-nuclear axis (s-s- overlap) to form a H–H sigma bond (which constitutes a single covalent bond in H₂ molecule) in which the electron cloud is rich along the axis between two nuclei.

Sigma bond formation due to s-p Linear overlapping in HX (HF, HCl, HBr, HI) Molecule

s-p overlapping involves the overlap of half-filled s orbital and half-filled p orbital of two combining atoms. This s-p type of overlap exist in HX molecules like HF, HCl, HBr, HI.

HF molecule is formed by the linear overlapping of half-filled 1s orbital of hydrogen atom and half-filled 2p_z orbital of fluorine atom.

Sigma bond formation due to p-p Linear overlapping in F₂ Molecule

p-p overlapping involves the overlap of two half-filled p orbitals of two combining atoms. This p-p type of overlap exist in X₂ molecules like F₂, Cl₂, Br₂, I₂.

The electronic configuration of F is 1s², 2s² 2p_x² 2p_y² 2p_z¹ showing that each fluorine atom has one half-filled 2p_z-orbital available for overlap. F₂ molecule is formed by the linear overlapping of half-filled 2p_z orbitals of F atoms.

According to VBT, in the formation of F₂ molecule, half-filled 2p_z-orbitals of both fluorine atoms containing an unpaired electron overlap linearly with each other along inter-nuclear axis (p_z-p_z-overlap) sharing the electron pair to form a F–F sigma bond (which constitutes a single covalent bond in F₂ molecule).

Pi bond formation due to p-p Lateral overlapping in O₂ Molecule

The electronic configuration of oxygen (Z = 8) is 1s², 2s² 2p_x² 2p_y¹ 2p_z¹. There are two partially filled p-orbitals on each oxygen atom i.e. 2p_y and 2p_z. The head to head overlapping of 2p_y orbitals of two oxygen atoms lead to form a sigma bond while lateral overlapping of 2p_z orbitals of two oxygen atoms results in the formation of a p-bond.

Q9.  Draw a molecular orbital diagram of O₂ molecules and explain following

(i)    Write down MO electronic configuration

(ii)   Determine the bond order

(iii)  Explain the paramagnetic behaviour.

Short Answer

Summary of MO Diagram of O₂

Given element  – Oxygen

Atomic number of Oxygen = 8

Electronic configuration of O atom = 1s^↿⇂, 2s^↿⇂, 2p_x^↿⇂ 2p_y^↿ 2p_z^↿

Total number of electrons in Oxygen molecule = 16 (8 by each O)

Electronic configuration of O₂ =

𝜎1s^↿⇂ < 𝜎*1s^↿⇂ < 𝜎2s^↿⇂ < 𝜎*2s^↿⇂ < < 𝜎 2px^↿⇂ < 𝜋2py^↿⇂ = 𝜋2pz^↿⇂ < 𝜋*2py^↿ = 𝜋*2pz^↿ < 𝜋*2px

Bond order of O₂ molecules = N_b – N_a/2 = 10–6 /2 = 2 (Double bond, O=O)

Paramagnetic nature = Two unpaired electrons in anti-bonding orbitals, hence O₂ is paramagnetic.

Detailed Answer

Molecular orbital Diagram for O₂ molecule

The electronic configuration of oxygen (Z=8) is 1s², 2s², 2p⁴ (1s^↿⇂, 2s^↿⇂, 2p_x^↿⇂ 2p_y^↿ 2p_z^↿) Thus there are five atomic orbitals with 8 electrons in each oxygen atom. The two participating oxygen atoms contribute a total 16 valence electrons. These five atomic orbitals of both oxygen atoms combine to form ten molecular orbitals as shown in molecular orbital energy diagram.

Two p-atomic orbitals (one from each oxygen) atom combine to form two molecular orbitals, the bonding molecular orbital σ2p_x and antibonding molecular orbital σ*2p_x. The other four p-atomic orbitals (two from each oxygen) atom combines to give four molecular orbitals, two bonding molecular orbitals i.e. π2p_y and π2p_z, while two antibonding molecular orbitals i.e. π*2p_y and π*2p_z. The electron filling in these molecular orbitals follows Aufbau, Pauli exclusion principle and Hund’s rule.

Out of eight electrons, six go to bonding molecular orbitals and two to the antibonding molecular orbitals. As electrons are also present in antibonding molecular orbitals, so weak bonds will be formed.

Bond order

Bond order of O₂ molecules is determined as

Bond order of O₂ molecules = N_b – N_a/2 = 10–6 /2 = 2 (Double bond, O=O)

As the bond order in Oxygen is 2 so two bonds i.e. double bonds formed between two oxygen atoms (O=O). 

Reason of Paramagnetic Nature of O₂ molecule

Since there are two unpaired electrons in degenerate anti-bonding molecular orbital (π*2p_y and π*2p_z), O₂ molecule is paramagnetic in nature.

molecular orbital configuration with increasing energy order

𝜎1s^↿⇂ < 𝜎*1s^↿⇂ < 𝜎2s^↿⇂ < 𝜎*2s^↿⇂ < < 𝜎 2px^↿⇂ < 𝜋2py^↿⇂ = 𝜋2pz^↿⇂ < 𝜋*2py^↿ = 𝜋*2pz^↿ < 𝜋*2px

Q10. Differentiate between sp³ and sp² hybridization and explain sp hybridization. Discus sp hybridization and the shape of ethyne molecule.

Answer

Definition of “sp”-Hybridization

The type of hybridization involving combination of one s (2s) and one “p” (2p) atomic orbitals to produce two new equivalent sp hybrid orbitals of identical shape and energy which are arranged co-linearly with a bond angle of 180° is called sp-hybridization. The two 2p (2p_y and 2p_z) orbitals remains unhybridized which lie perpendicularly to the plane of sp hybrid orbitals.

The sp hybrid orbitals are co-linear at an angle of 180° which provide maximum separation and overlap. Each sp hybrid orbital has 50% s-character and 50% p character (1:1 ratio).

Occurrence of “sp”-Hybridization

“sp” hybridization is found in those compounds where central atom (carbon) is bonded by two other atoms or groups e.g. alkyne i.e. ethyne (C₂H₂), BeCl₂, CO₂, CS₂.

Hybrid Molecular Structure of Ethyne (Acetylene)

In ethyne (acetylene), each carbon is bonded to two atoms (i.e. one H atom and one C atom) thus each carbon uses sp hybrid orbitals plus two unhybrid p_y and p_z orbitals to make its bonds. In sp hybridization of each carbon in ethyne there is a mixing of only s and p_x orbitals to give two sp hybrid orbitals that arrange linearly  at 180° angles while the two unhybrid orbitals p_y and p_z are inclined at right angle to the sp hybrid orbitals as well as to each other. The two hybrid sp orbitals and the two unhybrid atomic orbitals p_y and p_z contain single electron.

One sp-hybrid orbital of each carbon atom overlaps linearly with sp orbital of other carbon to form a C–C sigma bond. The remaining sp-hybrid orbital of each carbon overlap with 1s orbital of Hydrogen atoms to form two C–H sigma bond. Thus overall three sigma bonds are formed at 180^○ angles which makes linear geometry. Two unhybrid atomic orbitals (2p_y and 2p_z) of each carbon lying parallel to each other undergoes side wise overlapping to form two pi bonds between C to C. One sigma and two pi bonds between the carbon atoms of ethyne molecules is termed as triple bond. 

🔷🔥 MDCAT/ECAT VSEPR Theory MCQs

1️⃣ Electron-pair geometry of CO₂ is:
🟥 A. Bent
🟦 B. Linear
🟩 C. Trigonal planar
🟨 D. Tetrahedral

2️⃣ Shape of SO₂ according to VSEPR:
🟥 A. Linear
🟦 B. Bent
🟩 C. Trigonal planar
🟨 D. Pyramidal

3️⃣ H₂O has bond angle less than 109.5° because:
🟥 A. Double bonds
🟦 B. Two lone pairs
🟩 C. More electronegativity
🟨 D. Four bonding pairs

4️⃣ NH₃ has which shape?
🟥 A. Bent
🟦 B. Trigonal planar
🟩 C. Trigonal pyramidal
🟨 D. Linear

5️⃣ Electron domains in BF₃:
🟥 A. Two
🟦 B. Three
🟩 C. Four
🟨 D. Five

6️⃣ Shape of PCl₅ according to VSEPR:
🟥 A. Tetrahedral
🟦 B. Trigonal bipyramidal
🟩 C. Octahedral
🟨 D. Square pyramidal

7️⃣ SF₆ has geometry:
🟥 A. Square planar
🟦 B. Trigonal bipyramidal
🟩 C. Octahedral
🟨 D. T-shaped

8️⃣ Lone pairs always prefer which position in 5-electron pair geometry?
🟥 A. Axial
🟦 B. Equatorial
🟩 C. Random
🟨 D. Peripheral

9️⃣ Shape of XeF₂ is:
🟥 A. Linear
🟦 B. Bent
🟩 C. See-saw
🟨 D. T-shaped

🔟 Electron-pair geometry of NH₄⁺:
🟥 A. Tetrahedral
🟦 B. Trigonal planar
🟩 C. Linear
🟨 D. Octahedral

1️⃣1️⃣ Shape of CH₄ is:
🟥 A. Trigonal pyramidal
🟦 B. Tetrahedral
🟩 C. Trigonal planar
🟨 D. Bent

1️⃣2️⃣ Which has maximum bond angle?
🟥 A. H₂O
🟦 B. NH₃
🟩 C. BF₃
🟨 D. CH₄

1️⃣3️⃣ NO₂⁻ has shape:
🟥 A. Bent
🟦 B. Linear
🟩 C. T-shaped
🟨 D. Pyramidal

1️⃣4️⃣ A molecule with 4 bonding pairs and 0 lone pairs is:
🟥 A. Linear
🟦 B. Tetrahedral
🟩 C. Bent
🟨 D. Square planar

1️⃣5️⃣ Regions of electron density in CO₃²⁻ are:
🟥 A. Four
🟦 B. Two
🟩 C. Three
🟨 D. Five

1️⃣6️⃣ Which molecule is see-saw shaped?
🟥 A. SF₄
🟦 B. SF₆
🟩 C. CO₂
🟨 D. NH₃

1️⃣7️⃣ Which has T-shaped geometry?
🟥 A. BrF₃
🟦 B. BF₃
🟩 C. NH₄⁺
🟨 D. CO₂

1️⃣8️⃣ A molecule with one lone pair and three bonding pairs is:
🟥 A. Linear
🟦 B. Bent
🟩 C. Trigonal pyramidal
🟨 D. Tetrahedral

1️⃣9️⃣ Central atom of XeF₄ is:
🟥 A. Trigonal planar
🟦 B. Octahedral
🟩 C. Square planar
🟨 D. Bent

2️⃣0️⃣ In VSEPR, double and triple bonds are counted as:
🟥 A. Two regions
🟦 B. Three regions
🟩 C. One region
🟨 D. Zero region

🎉 Answers
1️⃣ 🟦 B
2️⃣ 🟦 B
3️⃣ 🟦 B
4️⃣ 🟩 C
5️⃣ 🟦 B
6️⃣ 🟦 B
7️⃣ 🟩 C
8️⃣ 🟦 B
9️⃣ 🟥 A
🔟 🟥 A
1️⃣1️⃣ 🟦 B
1️⃣2️⃣ 🟩 C
1️⃣3️⃣ 🟥 A
1️⃣4️⃣ 🟦 B
1️⃣5️⃣ 🟩 C
1️⃣6️⃣ 🟥 A
1️⃣7️⃣ 🟥 A
1️⃣8️⃣ 🟩 C
1️⃣9️⃣ 🟩 C
2️⃣0️⃣ 🟩 C

🔹 Chemical Bonding Test #1 – XI Chemistry 🧪✨

Q1. The highest bond energy in the following is:
🟨 A Cl–Cl
🟩 B H–O
🟦 C H–F
🟪 D H–N

Q2. The molecule which has zero dipole moment is:
🟨 A NH₃
🟩 B HCl
🟦 C CCl₄
🟪 D H₂O

Q3. The correct relation between Debye and coulomb meter is:
🟨 A 1 D = 3.33 × 10⁻³⁰ C·m
🟩 B 1 D = 1.6 × 10⁻¹⁹ C·m
🟦 C 1 D = 1.88 × 10⁻¹² C·m
🟪 D 1 D = 1.23 × 10⁻⁸ C·m

Q4. Octet rule is not followed in the formation of:
🟨 A CH₄
🟩 B BCl₃
🟦 C NF₃
🟪 D CO₂

Q5. Which factor is not responsible for the formation of ionic bond?
🟨 A Crystal lattice energy
🟩 B Ionization enthalpy
🟦 C Electron affinity
🟪 D Density

Q6. Which of the following compound possesses covalent bond?
🟨 A MgCl₂
🟩 B BF₃
🟦 C NaH
🟪 D CsCl

Q7. Which of the following molecule has not zero dipole moment?
🟨 A SO₂
🟩 B CO₂
🟦 C BF₃
🟪 D BeF₂

Q8. Maximum how many numbers of hydrogen bond can be formed by H₂O molecule?
🟨 A 2
🟩 B 3
🟦 C 4
🟪 D 1

Q9. Which of the following has co-ordinate covalent bond?
🟨 A NH₄⁺
🟩 B H₂O
🟦 C H₃O⁺
🟪 D Both A and C

Q10. Which of the following has the highest bond energy?
🟨 A C₂H₂
🟩 B C₂H₄
🟦 C C₂H₆
🟪 D C₂H₅OH

Q11. Which one of the following bond has the most polar character?
🟨 A C–O
🟩 B C–Br
🟦 C C–F
🟪 D C–S

Q12. Generally an ionic bond is formed between elements of group:
🟨 A IA and VIIA
🟩 B IIIA and VIIA
🟦 C VA and VIIA
🟪 D None of them

Q13. Chemical bond formation takes place due to interaction of ……… electrons.
🟨 A Valence
🟩 B Inner
🟦 C Free
🟪 D Paired

Q14. Which molecule does not have single covalent bond?
🟨 A O₂
🟩 B N₂
🟦 C CCl₄
🟪 D HF

Q15. Which molecule has single covalent bond?
🟨 A O₂
🟩 B N₂
🟦 C C₂H₂
🟪 D None of them

Q16. Which one is incorrect regarding the examples of Non-Polar bond?
🟨 A S=C=S
🟩 B PH₃
🟦 C CH₄
🟪 D CH₃Cl

Q17. Which molecule contains three shared pairs of electrons between two of its atoms?
🟨 A N₂
🟩 B C₂H₄
🟦 C CO₂
🟪 D H₂O

Q18. Which of the following is ionic?
🟨 A HCl
🟩 B CHCl₃
🟦 C KI
🟪 D IF₅

Q19. In which molecule intra molecular hydrogen bond can be formed?
🟨 A o-nitro phenol
🟩 B Aniline
🟦 C Ethylene glycol
🟪 D All of these

Q20. How many core electrons does a fluorine (F) atom contain?
🟨 A 2
🟩 B 5
🟦 C 7
🟪 D 10

Q21. The degree or extent of polarity of a molecule is called:
🟨 A Bond energy
🟩 B Moment arm
🟦 C Dipole moment
🟪 D Ionic character

Q22. The one sided sharing of electrons b/w two atoms produces which type of bond?
🟨 A Covalent bond
🟩 B Hydrogen bond
🟦 C Dative bond
🟪 D Ionic bond

Q23. Which of the following has the shortest C–C bond length?
🟨 A C₂H₂
🟩 B C₂H₄
🟦 C C₂H₆
🟪 D C₂H₅OH

Q24. Which molecule has single covalent bond?
🟨 A O₂
🟩 B CH₄
🟦 C CO₂
🟪 D N₂

Q25. Which molecule has double covalent bond?
🟨 A HF
🟩 B N₂
🟦 C CCl₄
🟪 D O₂

Q26. Bond making is always:
🟨 A Endothermic
🟩 B Exothermic
🟦 C Both A and B
🟪 D None of them

Q27. A covalent bond in which the shared electron pair is attracted equally by the two atoms is called:
🟨 A Polar bond
🟩 B Non-Polar bond
🟦 C Dative bond
🟪 D Hydrogen bond

Q28. Which molecule has greater bond energy?
🟨 A Polar molecule
🟩 B Non-Polar molecule
🟦 C None of them
🟪 D All of them

Q29. A bond is said to be 100% covalent, if it exists between atoms of:
🟨 A Same elements
🟩 B Different elements
🟦 C Same group
🟪 D Different group

Q30. Which one of the following bond has the least polar character?
🟨 A C–H
🟩 B C–Cl
🟦 C C–O
🟪 D C–N

Q31. Which one of the following has zero dipole moments?
🟨 A Diatomic polar molecules
🟩 B Non-linear (bent) polyatomic molecules
🟦 C Polyatomic polar symmetrical molecules
🟪 D Polyatomic polar unsymmetrical molecules

Q32. The b.p of water is higher than H₂S because:
🟨 A It has ionic bonding
🟩 B Water can form 3D hydrogen bonding of varying length
🟦 C Water is non-polar
🟪 D None of the above

Q33. In which one of the following does ionic bonding occur between the named atoms?
🟨 A Silicon and chlorine in silicon tetrachloride
🟩 B Hydrogen and aluminium in aluminum hydride
🟦 C Cesium and fluorine in cesium fluoride
🟪 D Iron and chlorine in tetrachloroferrate ion

Q34. Which equation defines the lattice energy of the ionic compound XY?
🟨 A X(s) + Y(s) → XY(s)
🟩 B X⁺(s) + Y⁻(s) → XY(s)
🟦 C X(g) + Y(g) → XY(s)
🟪 D X⁺(g) + Y⁻(g) → XY(s)

Q35. In order to form a compound with oxygen, an atom of a Group II element must:
🟨 A Transfer two electrons to an atom of oxygen
🟩 B Receive two electrons from an atom of oxygen
🟦 C Share two electrons with an atom of oxygen
🟪 D Bond with two atoms of oxygen

Q36. Which of the following relation is correct?
🟨 A Bond order ∝ bond energy ∝ bond length ∝ stability
🟩 B Bond order ∝ bond energy ∝ 1/bond length ∝ stability
🟦 C Bond order ∝ 1/bond length ∝ 1/bond energy ∝ stability
🟪 D Bond order ∝ bond energy ∝ bond length ∝ 1/stability

Q37. What is formed when an element X of atomic number 19 reacts with an element Y of atomic number 17?
🟨 A A covalent compound of formula XY₂
🟩 B A covalent compound of formula XY
🟦 C An ionic compound of formula XY
🟪 D An ionic compound of formula X₂Y

Q38. Which one has zero dipole moment?
🟨 A Linear polyatomic polar molecule
🟩 B Diatomic polar molecule
🟦 C Non-linear (angular) polyatomic polar molecule
🟪 D None of the above

Q39. Element X and Y combine to form the gas XY₂. What are X and Y?
🟨 A Calcium & Chlorine
🟩 B Carbon & Hydrogen
🟦 C Carbon & Oxygen
🟪 D Hydrogen & Oxygen

---

✅ Chemical Bonding Test #1 – Answers 🎨

1. C 🟦 H–F

2. C 🟦 CCl₄

3. A 🟨 1 D = 3.33 × 10⁻³⁰ C·m

4. B 🟩 BCl₃

5. D 🟪 Density

6. B 🟩 BF₃

7. A 🟨 SO₂

8. C 🟦 4

9. D 🟪 Both A and C

10. A 🟨 C₂H₂

11. C 🟦 C–F

12. A 🟨 IA and VIIA

13. A 🟨 Valence

14. B 🟩 N₂

15. D 🟪 None of them

16. D 🟪 CH₃Cl

17. A 🟨 N₂

18. C 🟦 KI

19. D 🟪 All of these

20. A 🟨 2

21. C 🟦 Dipole moment

22. C 🟦 Dative bond

23. A 🟨 C₂H₂

24. B 🟩 CH₄

25. B 🟩 N₂

26. B 🟩 Exothermic

27. B 🟩 Non-Polar bond

28. B 🟩 Non-Polar molecule

29. A 🟨 Same elements

30. A 🟨 C–H

31. C 🟦 Polyatomic polar symmetrical molecules

32. B 🟩 Water can form 3D hydrogen bonding of varying length

33. C 🟦 Cesium and fluorine

34. D 🟪 X⁺(g) + Y⁻(g) → XY(s)

35. A 🟨 Transfer two electrons to an atom of oxygen

36. A 🟨 Bond order ∝ bond energy ∝ bond length ∝ stability

37. C 🟦 An ionic compound of formula XY

38. D 🟪 None of the above

39. D 🟨 Hydrogen & Oxygen

🧪 Chemical Bonding Test #2 – MCQs 🎨

1️⃣ If the bond angle in AB₂ type molecule is 104.5°, its geometry should be:
🟨 A Linear
🟩 B Pyramidal
🟦 C Bent
🟪 D Planar Trigonal

2️⃣ The molecule which has maximum bond angle:
🟨 A CS₂
🟩 B H₂O
🟦 C NH₃
🟪 D BF₃

3️⃣ The shape and hybridization of BCl₃ molecule is:
🟨 A Tetrahedral & sp³
🟩 B Linear & sp
🟦 C Planar trigonal & sp²
🟪 D Angular & sp³

4️⃣ Among the following molecules which has trigonal pyramidal shape?
🟨 A SO₂
🟩 B CO₂
🟦 C NH₃
🟪 D C₂H₄

5️⃣ A simple covalent molecule with 2 bond pairs and 2 lone pairs around central atom, its shape is:
🟨 A Linear
🟩 B Planar trigonal
🟦 C Angular
🟪 D Tetrahedral

6️⃣ The bond order of N₂ molecule is:
🟨 A 3
🟩 B 1
🟦 C 2
🟪 D 0

7️⃣ The number of sigma and pi bonds in C₂H₄ molecules are respectively:
🟨 A 5 & 1
🟩 B 2 & 2
🟦 C 3 & 1
🟪 D 4 & 2

8️⃣ The molecular orbitals which have higher energy than parent atomic orbitals are called:
🟨 A BMO
🟩 B Hybrid orbital
🟦 C AMO
🟪 D All of them

9️⃣ Sigma bond has ……….. electron density:
🟨 A Highest
🟩 B Lowest
🟦 C Intermediate
🟪 D All of them

🔟 A molecule of acetylene (C₂H₂) contains total:
🟨 A 2 σ & 2 π bonds
🟩 B 4 σ & 1 π bond
🟦 C 3 σ & 3 π bonds
🟪 D 5 σ & no π bond

1️⃣1️⃣ Inter-electronic repulsion forces decrease sharply with increasing:
🟨 A Bond angle
🟩 B Bond length
🟦 C Bond energy
🟪 D All of them

1️⃣2️⃣ Benzene has bond angles same as:
🟨 A Tetrahedral geometry
🟩 B Linear geometry
🟦 C Triangular geometry
🟪 D None of these

1️⃣3️⃣ Which one pair has same type of hybridization of central atom?
🟨 A BF₃, SO₃
🟩 B BF₃, NH₃
🟦 C CH₃⁺, BF₃
🟪 D H₂S, SO₂

1️⃣4️⃣ How many sigma bonds can be predicted between two atoms?
🟨 A 3
🟩 B 1
🟦 C 2
🟪 D 4

1️⃣5️⃣ The presence of lone pair on the central atom ………… bond angle:
🟨 A Contract
🟩 B Expand
🟦 C Has no effect
🟪 D All of them

1️⃣6️⃣ Which of the following has pyramidal shape?
🟨 A NH₃
🟩 B PH₃
🟦 C PCl₃
🟪 D All of them

1️⃣7️⃣ The shape of NHCl₂ molecule is:
🟨 A Tetrahedral
🟩 B Angular
🟦 C Planar Trigonal
🟪 D Linear

1️⃣8️⃣ Bond length is largest in case of ………. overlapping:
🟨 A sp²-s
🟩 B sp²-sp²
🟦 C sp³-sp³
🟪 D sp-sp

1️⃣9️⃣ In CH₄ molecule, a sigma bond is formed by axial overlapping of half-filled:
🟨 A s-s orbitals
🟩 B s-sp³ orbitals
🟦 C d-d orbitals
🟪 D p-p orbitals

2️⃣0️⃣ Strongest bond is formed by the overlap of:
🟨 A s-s orbitals
🟩 B p-p orbitals
🟦 C s and p orbitals
🟪 D None of them

2️⃣1️⃣ In which molecule carbon atom is sp³ hybridized?
🟨 A C₂H₂
🟩 B C₂H₆
🟦 C C₃H₆
🟪 D C₂H₄

2️⃣2️⃣ A molecule of ethylene (C₂H₄) contains total:
🟨 A 2 σ & 1 π bond
🟩 B 5 σ & 1 π bond
🟦 C 4 σ & 2 π bond
🟪 D 3 σ & 1 π bond

2️⃣3️⃣ In which of the following strong H-bond is present?
🟨 A F-H…F
🟩 B O-H…O
🟦 C O-H…N
🟪 D O-H…F

2️⃣4️⃣ What is bond energy of H-bond?
🟨 A 40 J mol⁻¹
🟩 B 40 kJ mol⁻¹
🟦 C 40 kg J mol⁻¹
🟪 D 40 cal mol⁻¹

2️⃣5️⃣ In OF₂, number of bond pairs and lone pairs of electrons are respectively:
🟨 A 2, 8
🟩 B 2, 9
🟦 C 2, 8
🟪 D 2, 10

2️⃣6️⃣ The shape of CO₂ molecule is similar to:
🟨 A H₂S
🟩 B SnCl₂
🟦 C SO₂
🟪 D BeF₂

2️⃣7️⃣ How many core electrons does a Sulfur (S) atom contain?
🟨 A 4
🟩 B 6
🟦 C 8
🟪 D 10

2️⃣8️⃣ Which one of the following ions has more electrons than protons and more protons than neutrons?
🟨 A D⁻
🟩 B D₃O⁺
🟦 C OD⁻
🟪 D OH⁻

2️⃣9️⃣ Which one has the same number of electrons as an alpha particle?
🟨 A H
🟩 B H⁺
🟦 C He
🟪 D Li⁺

3️⃣0️⃣ Which one does not contain delocalized electrons?
🟨 A Copper
🟩 B Benzene
🟦 C Diamond
🟪 D Graphite

3️⃣1️⃣ Which one conducts electricity by movement of ions?
🟨 A Copper metal
🟩 B Mercury
🟦 C Molten sodium chloride
🟪 D Graphite

3️⃣2️⃣ The molecular orbitals which have higher energy than parent atomic orbital are called:
🟨 A BMO
🟩 B Hybrid orbital
🟦 C AMO
🟪 D All of them

3️⃣3️⃣ Sigma bond has ……….. electron density:
🟨 A Highest
🟩 B Lowest
🟦 C Intermediate
🟪 D All of them

3️⃣4️⃣ A molecule of acetylene (C₂H₂) contains total:
🟨 A 2 σ & 2 π bonds
🟩 B 4 σ & 1 π bond
🟦 C 3 σ & 3 π bonds
🟪 D 5 σ & no π bond

3️⃣5️⃣ Inter-electronic repulsion forces decrease sharply with increasing:
🟨 A Bond angle
🟩 B Bond length
🟦 C Bond energy
🟪 D All of them

3️⃣6️⃣ Benzene has bond angles same as:
🟨 A Tetrahedral geometry
🟩 B Linear geometry
🟦 C Triangular geometry
🟪 D None of these

3️⃣7️⃣ Which one pair has same type of hybridization of central atom?
🟨 A BF₃, SO₃
🟩 B BF₃, NH₃
🟦 C CH₃⁺, BF₃
🟪 D H₂S, SO₂

3️⃣8️⃣ In O₂, each oxygen atom is …………. Hybridized.
🟨 A sp²
🟩 B sp³
🟦 C sp
🟪 D None of them

3️⃣9️⃣ How many sigma bonds can be predicted between two atoms?
🟨 A 3
🟩 B 1
🟦 C 2
🟪 D 4

4️⃣0️⃣ The presence of lone pair on the central atom ………… bond angle:
🟨 A Contract
🟩 B Expand
🟦 C Has no effect
🟪 D All of them

4️⃣1️⃣ Which of the following has pyramidal shape?
🟨 A NH₃
🟩 B PH₃
🟦 C PCl₃
🟪 D All of them

4️⃣2️⃣ The shape of NHCl₂ molecule is:
🟨 A Tetrahedral
🟩 B Angular
🟦 C Planar Trigonal
🟪 D Linear

4️⃣3️⃣ Bond length is largest in case of ………. overlapping:
🟨 A sp²-s
🟩 B sp²-sp²
🟦 C sp³-sp³
🟪 D sp-sp

4️⃣4️⃣ In CH₄ molecule a sigma bond is formed by axial overlapping of half filled:
🟨 A s-s orbitals
🟩 B s-sp³ orbitals
🟦 C d-d orbitals
🟪 D p-p orbitals

4️⃣5️⃣ Strongest bond is formed by the overlap of:
🟨 A s-s orbitals
🟩 B p-p orbitals
🟦 C s and p orbitals
🟪 D None of them

4️⃣6️⃣ In which molecule carbon atom is sp³ hybridized?
🟨 A C₂H₂
🟩 B C₂H₆
🟦 C C₃H₆
🟪 D C₂H₄

4️⃣7️⃣ A molecule of ethylene (C₂H₄) contains total:
🟨 A 2 σ & 1 π bond
🟩 B 5 σ & 1 π bond
🟦 C 4 σ & 2 π bond
🟪 D 3 σ & 1 π bond

4️⃣8️⃣ In OF₂, number of bond pairs and lone pairs of electrons are respectively:
🟨 A 2, 8
🟩 B 2, 9
🟦 C 2, 8
🟪 D 2, 10

4️⃣9️⃣ The shape of CO₂ molecule is similar to:
🟨 A H₂S
🟩 B SnCl₂
🟦 C SO₂
🟪 D BeF₂

📝 Chemical Bonding Test #2 – Answers 🎨

1️⃣ 🟦 C Bent
2️⃣ 🟪 D BF₃
3️⃣ 🟦 C Planar trigonal & sp²
4️⃣ 🟦 C NH₃
5️⃣ 🟦 C Angular
6️⃣ 🟨 A 3
7️⃣ 🟨 A 5 & 1
8️⃣ 🟩 B Hybrid orbital
9️⃣ 🟨 A Highest
🔟 🟨 A 2 σ & 2 π bonds
1️⃣1️⃣ 🟩 B Bond length
1️⃣2️⃣ 🟨 A Tetrahedral geometry
1️⃣3️⃣ 🟨 A BF₃, SO₃
1️⃣4️⃣ 🟩 B 1
1️⃣5️⃣ 🟨 A Contract
1️⃣6️⃣ 🟪 D All of them
1️⃣7️⃣ 🟩 B Angular
1️⃣8️⃣ 🟦 C sp³-sp³
1️⃣9️⃣ 🟩 B s-sp³ orbitals
2️⃣0️⃣ 🟩 B p-p orbitals
2️⃣1️⃣ 🟩 B C₂H₆
2️⃣2️⃣ 🟦 C 4 σ & 2 π bond
2️⃣3️⃣ 🟩 B O-H…O
2️⃣4️⃣ 🟩 B 40 kJ mol⁻¹
2️⃣5️⃣ 🟪 D 2, 10
2️⃣6️⃣ 🟪 D BeF₂
2️⃣7️⃣ 🟦 C 8
2️⃣8️⃣ 🟨 A D⁻
2️⃣9️⃣ 🟦 C He
3️⃣0️⃣ 🟨 A Copper
3️⃣1️⃣ 🟦 C Molten sodium chloride
3️⃣2️⃣ 🟩 B Hybrid orbital
3️⃣3️⃣ 🟨 A Highest
3️⃣4️⃣ 🟨 A 2 σ & 2 π bonds
3️⃣5️⃣ 🟩 B Bond length
3️⃣6️⃣ 🟨 A Tetrahedral geometry
3️⃣7️⃣ 🟨 A BF₃, SO₃
3️⃣8️⃣ 🟦 C sp
3️⃣9️⃣ 🟩 B 1
4️⃣0️⃣ 🟨 A Contract
4️⃣1️⃣ 🟪 D All of them
4️⃣2️⃣ 🟩 B Angular
4️⃣3️⃣ 🟦 C sp³-sp³
4️⃣4️⃣ 🟩 B s-sp³ orbitals
4️⃣5️⃣ 🟩 B p-p orbitals
4️⃣6️⃣ 🟩 B C₂H₆
4️⃣7️⃣ 🟦 C 4 σ & 2 π bond
4️⃣8️⃣ 🟪 D 2, 10
4️⃣9️⃣ 🟪 D BeF₂

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