XI Chemistry Model Test Questions Test # 2 for Chapter # 3 (Chemical bond)

 

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 and AMO)           

(iii)  VBT and MOT

Answer (Learn any 5 points)

Difference between Bonding and 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 (Breif)

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.