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Chapter 4 Chemical Bonding and Molecular Structure

Through this chapter, you can understand Kossel-Lewis approach to chemical Bonding. You can also explain the octet rule and its limitations, draw Lewis structures of simple molecules and the formation of different types of bonds. You will also be able to describe the VSEPR theory and predict the geometry of simple molecules. You can also explain the valence bond approach for the formation of covalent bonds. You can also predict the directional properties of covalent bonds and explain the concept of hydrogen bond. Description of molecular orbital theory of homo-nuclear diatomic molecules is also given. Explanation of different types of hybridisation involving s, p and d orbitals is also given.

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Exercise 1 ( Page No. : 135 )

  • Q1 Explain the formation of a chemical bond.
    Ans:

    A chemical bond is defined as an attractive force that holds the constituents (atoms, ions etc.) together in a chemical species.

    Various theories have been suggested for the formation of chemical bonds such as the electronic theory, valence shell electron pair repulsion theory, valence bond theory, and molecular orbital theory.

    A chemical bond formation is attributed to the tendency of a system to attain stability. It was observed that the inertness of noble-gases was because of their fully filled outermost orbitals. Hence, it was postulated that the elements having incomplete outermost shells are unstable (reactive). Atoms, therefore, combine with each other and complete their respective octets or duplets to attain the stable configuration of the nearest noble gases. This combination can occur either by sharing of electrons or by transferring one or more electrons from one atom to another. The chemical bond formed as a result of sharing of electrons between atoms is called a covalent bond. An ionic bond is formed as a result of the transference of electrons from one atom to another.


    Q2 Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.
    Ans:

    G.N Lewis introduced symbols to represent the valence electrons in the atoms.They are called Lewis symbols, where valence electrons are shown as dots surrounding the symbol of atom or ion.

    Mg: Mg belongs to 3rd period & 2nd group of periodic table & has 2 electrons in the valence shell, therefore lewis structure is  .

     

    Na: Na has electronic configuration of [Ne]10 3s1, i.e it has 1 electron in outer most shell, Hence, the Lewis dot structure is:

     

    B: Boron belongs to 2nd period & 13th group with 3 electron in valence shell. Hence, the Lewis dot structure is:

     

    O: Oxygen has electronic configuration of [He]2 2s2 2p4,i.e having 6 electrons in valence shell. Hence, the Lewis dot structure is: 

     

    N: Nitrogen has electronic configuration of [He]2 2s2 2p3. Hence, the Lewis dot structure is:

     

    Br: There are seven valence electrons in bromine. Hence, the Lewis dot structure is:


    Q3 Write Lewis symbols for the following atoms and ions: S and S2–; Al and Al3+; H and H–
    Ans:

    (i) S and S2– Electronic configuration of S is [Ne]10 3s23p4, i.e it has 6 electrons in its valence shell,therefore The Lewis dot symbol of sulphur (S) is . The dinegative charge shows that there will be two electrons more in addition to the six valence electrons. The electronic configuration can be [Ne]10 3s23p6, Hence, the Lewis dot symbol of S2– is .

     

    (ii) Al and Al3+ The electronic configuration of Aluminium is [Ne]10 3s2 3p1, i.e the valence electron is 3,therefore The Lewis dot symbol of aluminium (Al) is . The tripositive charge on a species infers that it has donated its three electrons to other atom & its possible electronic configuration is [Ne]10 3s1. Hence, the Lewis dot symbol is .

     

    (iii) H and H– Hydrogen has electronic configuration of 1s1,it has 1 electron in its valence shell. The Lewis dot symbol of hydrogen (H) is . The uninegative charge infers that there will be one electron more in addition to the one valence electron. The possible configuration may be 1s2. Hence, the Lewis dot symbol is .


    Q4 Draw the Lewis structures for the following molecules and ions: H2S, SiCl4, BeF2, , HCOOH
    Ans:

    1) H2S H has 1 electron in its valence shell & sulphur has 6 electron its valence shell. A planar structure will be formed

     

    2) Si has 4 electrons its valence shell & Cl need only 1 electron to complete its octet, therefore 4 Cl will make 4 bonds with Si.A tetrahedral shape will be formed    

                 

     

    3) Be has 2 electrons in its valence shell & Fe has 2 electrons in its valence shell, so they make a planar structure

     

    4) here C has 4 electron its valence shell, therefore it can make 4 bond with other atoms, whereas O has 6 electrons in its valence shell, it can accept 2 electron easily to complete its octet, therefore one O will make a double bond with C , while other two O will make 2 single bond with C.

     

    5) HCOOH- C has 4 electron its valence shell, therefore it can make 4 bond with other atoms, whereas O has 6 electrons in its valence shell, it can accept 2 electron easily to complete its octet,


    Q5 Define octet rule. Write its significance and limitations.
    Ans:

    The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell. When atoms have fewer than eight electrons, they tend to react and form more stable compounds. It was postulated by Kossel & Lewis. The atoms complete their octet by either loosing or gaining of electrons. They do so to attain the nearest noble gas stable electronic configuration for example, the octet rule successfully explained the formation of chemical bonds depending upon the nature of the element.

    The octet rule successfully explained the formation of chemical bonds depending upon the nature of the element.

     

    Limitations of the octet theory:

     

    1) The main limitation to the rule is hydrogen, which is at its lowest energy when it has two electrons in its valence shell. Helium (He) is similar in that, it too only has room for two electrons in its valence shell.

    Hydrogen and helium have only one electron shell. The first shell has only one- s orbital and no p orbital, so it holds only two electrons. Therefore, these elements are most stable when they have two electrons.

     

    2) Second limitation are aluminium and boron, which can function well with six valence electrons. Consider BF3, The boron shares its three electrons with three fluorine atoms. The fluorine atoms follow the octet rule, but boron has only six electrons. Most elements to the left of the carbon group have so few valence electrons that they are in the same situation as boron, they are electron deficient.

     

    3) In Period 3, the elements on the right side of the periodic table have empty d orbitals. The d orbitals may accept electrons, allowing elements like sulphur and phosphorus to have more than an octet. Compounds such as PCl5 and SF6 can form. These compounds have 10 and 12 electrons around their central atoms, respectively.

     

    Xenon hexafluoride uses d-electrons to form more than an octet. This compound shows another exception: a noble gas compound.

    4) The octet rule is not satisfied for all atoms in a molecule having an odd number of electrons. For example, NO and NO2 do not satisfy the octet rule.

    5) The rule failed to predict the shape and relative stability of molecules.

    6) It is based upon the inert nature of noble gases. However, some noble gases like xenon and krypton form compounds such as XeF2, KrF2 etc.

    7) This rule cannot be applied to those compounds in which the number of electrons surrounding the central atom is less than eight. For example, LiCl, BeH2, AlCl3 etc. do not obey the octet rule.


    Q6 Write the favourable factors for the formation of ionic bond.
    Ans:

    Ionic bond is a type of chemical bond which involves the transfer of one or more electrons from one ion to another ions of opposite charges. The ion which gains an electron is called as anions, whereas the ion which has loose an electron is called as cation. Mostly the cations are metals and anions are non-metal. For example:

    Na + Cl → Na+ + Cl- → NaCl

    Hence, favourable factors for ionic bond formation are as follows:

    (i) Low ionization enthalpy of metal atom.

    (ii) The two ion must be different and should possess opposite charge.

    (iii) High electron gain enthalpy (Δeg H) of a non-metal atom.

    (iv) High lattice energy of the compound formed.

    (v) The electron negativity between two ions should be greater than 1.7.


    Q7 Discuss the shape of the following molecules using the VSEPR model: BeCl2, BCl3, SiCl4, AsF5, H2S, PH3
    Ans:

    Valence shell electron pair repulsion (VSEPR) theory:

    It is a model used to predict the 3D geometry of individual molecules from the number of electron pairs surrounding their central atoms. It was developed by Gillespie & Nyholm. The various geometries are as follows according to number of lone pair present:

     

    BeCl2:

    The central atom has no lone pair and there are two bond pairs. i.e., BeCl2 is of the type EX2. Hence, it has a linear shape.

     

    BCl3:

    The central atom has no lone pair and there are three bond pairs. Hence, it is of the type EX3. Hence, it is trigonal planar.

     

    SiCl4:

    The central atom has no lone pair and there are four bond pairs. Hence, the shape of SiCl4 is tetrahedral being the EX4 type molecule.

     

    AsF5:

    The central atom has no lone pair and there are five bond pairs. Hence, AsF5 is of the type EX5. Therefore, the shape is trigonal bipyramidal.

     

    H2S:

    The central atom has one lone pair and there are two bond pairs. Hence, H2S is of the type AB2E. The shape is Bent.

     

    PH3:

    The central atom has one lone pair and there are three bond pairs. Hence, PH3 is of the EX5 type. Therefore, the shape is trigonal bipyramidal.


    Q8 Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
    Ans:

    The tetrahedral geometry shown by NH3 & H2O is

     

    In NH3, the N atom is surrounded by three bond pairs (N—H) and one lone pair of electrons, which is present on the N atom. The repulsion shown by one lone pair in NH3 towards the three bond pair is very less because the number of bonding electrons is more as compared to lone pair of electron, as a result there will be less repulsion between the lone pair and bonding electrons. So, the bond angle is 107, nearer to normal tetrahedral bond angle of 109°.5’.

    But in case of H2O, the O atom is surrounded by two bond pairs and two lone pair of electrons. Consequently, the repulsion will be stronger and greater between the lone pairs and bond pairs. The bond pairs will be repelled more by the lone pairs, resulting in decrease of bond angle to 104°, as compared to normal tetrahedral bond angle 109°.5’.

    Therefore instead NH3 and H2O have tetrahedral geometry, still they have different bond angles.


    Q9 How do you express the bond strength in terms of bond order?
    Ans:

    In a molecule, as we increase the number of electrons shared between two atoms, there will be an increase in bond order, also there will be an increase in the strength of the bond and decrease the distance between nuclei.

    Bond Number of electrons Bond Order Bond Strength Bond Length
    Single                   2        1    Weakest    Longest
    Double                   4        2          -        -
    Triple                   6        3    Strongest    Shortest

     

     


    Q10 Define the bond length.
    Ans:

    Bond length is defined as the distance between the centres of the nuclei of two bonded atoms in a molecule. It represents equilibrium inter-nuclear separation distance of the bonded atoms in a molecule. The length of the bond is determined by the number of bonded electrons (the bond order). The higher the bond order, the stronger the pull between the two atoms and the shorter the bond length. Bond lengths are expressed in terms of Angstrom (10-10 m) or picometer (10-12 m) and are measured by spectroscopic X-ray diffractions and electron-diffraction techniques.

    In an ionic compound, the bond length is the sum of the ionic radii of the constituting atoms (d = r+ + r-). In a covalent compound, it is the sum of their covalent radii (d = rA + rB).


    Q11 Explain the important aspects of resonance with reference to the  ion.
    Ans:

    In some molecules single Lewis structure fails to explain all the characteristics of the molecule. As a result a number of structures can be drawn to explain all the characteristic .Such structures are called resonating structures & resonance is defined as the phenomenon as a result of which a molecule can be expressed in different forms none of which can explain all the properties of the molecule. As in case of  ion all the C---- O bonds are equivalent. Therefore it is not correct to represent  ion through one Lewis strucuture. Instead three resonating structures can be drawn ,which are as follows:

     


    Q12 H3PO3 can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H3PO3? If not, give reasons for the same.  
    Ans:

    The most favourable condition for writing resonating structures is that the different resonating structures must have same position of atoms.

    Here in the given example the H atom has changed its position in the second structure as compared to first, therefore these two structures cannot be considered as the resonating structures for the given compound.


    Q13 Write the resonance structures for SO3, NO2 and .
    Ans:

    Resonance is defined as the phenomenon as a result of which a molecule can be expressed in different forms, none of which can explain all the properties of the molecules. The actual structure of the molecule is called resonance hybrid.

    The resonating structures must have same position of the atoms, they must have same number of paired & unpaired electrons, they should have nearlt same energy. The resonance structures are:

    (a) SO3:

     

    (b) NO2

     

    (c)

     


    Q14 Use Lewis symbols to show electron transfer between the following atoms to form cations and anions: (a) K and S (b) Ca and O (c) Al and N.
    Ans:

    (a) K and S:

    The electronic configurations of K and S are as follows:

    K: 2, 8, 8, 1

    S: 2, 8, 6

    Sulphur (S) requires 2 more electrons to complete its octet. Potassium (K) requires one electron more than the nearest noble gas i.e., Argon. Hence, the electron transfer can be shown as:

    (b) Ca and O: The electronic configurations of Ca and O are as follows:

    Ca: 2, 8, 8, 2

    O: 2, 6

    Oxygen requires two electrons more to complete its octet, whereas calcium has two electrons more than the nearest noble gas i.e., Argon. Hence, the electron transfer takes place as:

    (c) Al and N: The electronic configurations of Al and N are as follows:

    Al: 2, 8, 3

    N: 2, 5

    Nitrogen is three electrons short of the nearest noble gas (Neon), whereas aluminium has three electrons more than Neon. Hence, the electron transference can be shown as:


    Q15 Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is bent while that of CO2 is linear. Explain this on the basis of dipole moment.
    Ans:

    Dipole moment is the product of the magnitude of the positive or negative charge (q) and the distance (d) between the charges, i.e.

    μ = q x d.

    The SI unit is coulomb metre. The molecule with symmetrical and linear geometries have zero dipole moment because they are vector in nature and the dipole of different bonds cancel with one another. CO2 is symmetrical and has dipole moment 0. Therefore it will have a linear shape with bond angle 180°.

    Resultant μ = 0 D

     

    H2O is a unsymmetrical molecule and bent geometries, the geometries have specific dipole moment (1.84D) because the bond polarities do not cancel each other. Therefore H2O will have a bent structure with bond angle 104°.

     


    Q16 Write the significance/applications of dipole moment.
    Ans:

    The important significance of dipole moment are as follows:

    1. For predicting the nature of the molecules- molecules with zero dipole moment are non- polar while the molecule with specific dipole moment are polar in nature.

    2. For determining the shape of the molecule-If a molecule has a specific dipole moment then its shape will not be symmetrical, they may be bent or angular and the molecule with zero dipole moment will be symmetrical and has linear shape.

    3. For comparing the polarities of the molecules- greater the dipole moment value, more will be the polarity and vice versa.


    Q17 Define electronegativity. How does it differ from electron gain enthalpy?
    Ans:

    Electronegativity is defined as the tendency of an element to attract the shared pair of electrons towards itself in a covalent bond. There is no specific units for electronegativity. It is only a relative tendency.

    Electron gain enthalpy is the energy released when one mole of electron are added to gaseous atoms of an element. It can be negative or positive depending upon whether the electron is added or removed. An element has a constant value of the electron gain enthalpy that can be measured experimentally.


    Q18 Explain with the help of suitable example polar covalent bond.
    Ans:

    If two different atoms are linked to each other by covalent bond, then the shared electron pair will not lie in the centre because the bonding atoms differ in electro-negativities. Such a bond is called as polar covalent bond. In NaCl, chlorine is more electronegative than sodium. Hence it will have more control over the shared pair of electrons and will develops a partial negative charge and sodium will acquire partial positive charge and a polar covalent bond is formed. Greater the difference in electronegativity of bonding atoms, more will be the polarity of the bond.


    Q19 Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3.
    Ans:

    Dipole moment helps in calculating the percentage ionic character of polar bonds. It is the ratio of observed dipole moment to the dipole moment for the complete transfer of electrons. Greater the difference in electronegativity of bonding atoms, the greater will be the ionic character.

    On this basis, the order of increasing ionic character in the given molecules is

    N2 < SO2 < ClF3 < K2O < LiF.


    Q20 The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.
    Ans:

    Since all the atoms are in either period 1 or 2, this molecule will adhere to the octet rule. The exception of course being the hydrogen’s, they follow the duet rule (2 electrons). In the given structure, the H atom is bonded to carbon with a double bond, which is not possible because H has only one electron to share with carbon. Also the second carbon atom is not having complete valency, which is four, so it has to make one more bond with Oxygen. Therefore the correct Lewis structure for acetic acid is as follows:


    Q21 Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH4 is not square planar?
    Ans:

    The phenomenon of mixing of orbitals of the same atom with slight difference in energies so as to redistribute their energies and give new orbitals of equivalent energy and shape. The new orbitals formed are known as hybrid orbitals, it is called as hybridization.

    Now, Electronic configuration of carbon atom:

    C: 1s2 2s2 2p2

    In the excited state, the orbital picture of carbon can be represented as:

    Here 1 s orbital & 3 p orbitals under goes hybridization to form sp3 hybrid orbitals .Hence, carbon atom undergoes sp3 hybridization in CH4 molecule and takes a tetrahedral shape.

                                       

    For a square planar shape, the hybridization of the central atom has to be dsp2, ie, 1 s orbital, 3 p orbitals & 1 d orbitals has to undergo hybridization .However, an atom of carbon does not have d-orbitals to undergo dsp2 hybridization. Hence, the structure of CH4 cannot be square planar.

    Moreover, with a bond angle of 90° in square planar, the stability of CH4 will be very less because of the repulsion existing between the bond pairs. Hence, VSEPR theory also supports a tetrahedral structure for CH4.


    Q22 Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are polar.
    Ans:

    The molecule with symmetrical and linear geometries have zero dipole moment because they  are vector in nature and the dipole of different bonds cancel with one another and in a unsymmetrical molecule and bent geometries, the geometries have specific dipole moment because the bond polarities do not cancel each other.

    The Lewis structure for BeH2 is as follows:

    Since there is no lone pair at the central atom (Be) and there are two bond pairs, BeH2 is of the type AB2. It has a linear structure.

    Dipole moments of each H–Be bond are equal and are in opposite directions. Therefore, they nullify each other. Hence, BeH2 molecule has zero dipole moment.

    While in Be---H , dipole moment of Be  & H are not equal & they donot nullify each other.hence has a specific dipole moment.


    Q23 Which out of NH3 and NF3 has higher dipole moment and why?
    Ans:

    In both molecules i.e., NH3 and NF3, the central atom (N) has a lone pair electron and there are three bond pairs. Hence, both molecules have a pyramidal shape. Since fluorine is more electronegative than hydrogen, it is expected that the net dipole moment of NF3 is greater than NH3. However, the net dipole moment of NH3 (1.46 D) is greater than that of NF3 (0.24 D).

    This can be explained on the basis of the directions of the dipole moments of each individual bond in NF3 and NH3.

    Since in case of NF3, fluorine is more electronegative than nitrogen, so it will attract the shared pair of electron towards itself from nitrogen. As a result dipole due to all F is in opposite direction to the resultant dipole due to N-F bonds, therefore the net dipole decreases.

    In NH3, Nitrogen is more electronegative than H, so it will attract the shared pair of electrons towards itself. As a result dipole due to Nitrogen is in same direction as the resultant dipole moment due to N-H bond and hence the net dipole increases.

    Hence, the net dipole moment of NF3 is less than that of NH3.


    Q24 What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals.
    Ans:

    The phenomenon of mixing of orbitals of the same atom with slight difference in energies so as to redisitribute their energies & give new orbitals of equivalent energy & shape.The new orbitals formed are known as hybrid orbitals & is called as hybridization.These hybrid orbitals have minimum repulsion between their electron pairs and thus, are more stable. Hybridization helps indicate the geometry of the molecule.

    Shape of different hybridization are :

     


    Q25 Describe the change in hybridisation (if any) of the Al atom in the following reaction.
    Ans:

    Here in the given equation AlCl3 , Al is shifting its hybridization state from sp2 to sp3 because if we see the electronic configuration of Al it is [Ne] 3s2 3p1, i.e it has 3 electrons in its valence shell during ground state.In its excited state its one s orbital & two p orbitals will take part in hybridization to form sp2 hybrid orbital.

    The orbital picture of aluminium in the excited state can be represented as:

    Now according to the given reaction the product formed AlCl4- , will only be formed if its 3pz orbital is also filled by one electron in its excited state to form a sp3 hybrid orbital .therefore Al will shift its hybridization form sp2 to sp3 to form a tetrahedral shape


    Q26 Is there any change in the hybridisation of B and N atoms as a result of the following reaction? BF3 + NH3 → F3B.NH3  
    Ans:

    The atomic number of boron is 5 & its electronic configuration in ground state  is 1s2 2s2 2p1, & in excited state is 1s2 2s1 2p2 , this means its one s & 2 p orbital will take part in hybridization to form sp2  hybrid orbital.

    Similarly N atom has atomic number of 7, with electronic configuration of 1s2 2s2 2p3 in its ground state & in it excited state it has sp3 hybridization.

    Now according to the question Boron & Nitrogen in reactant stage has sp2 & sp3  hybridization respectively, but on the product side an adduct is formed, wherein Boron has changed it s hybridization to sp3 while Nitrogen remains in same sp3 hybridization


    Q27 Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules.
    Ans:

    C2H4 :

    The electronic configuration of C-atom in the excited state is:

    C has 4 unpaired electrons in the valence shell,therefore in the formation of C2H4 , each Carbon atom undergoes sp2 hybridization leaving 2pz orbital unhybridized.

    One sp2 hybrid orbital of one carbon overlaps with sp2 hybrid orbital of other carbon atom to form sigma c---c bond.Now the unhybridized 2pz orbital of one carbon atom overlaps sidewise with similar orbital of the other carbon atom to form pie bond.The remaining 2 sp2 hybrid orbitals of both carbon atoms overlap axially with the half filled 1s orbitals of hydrogen atom forming C---H sigma bond

     

               ORBITAL STRUCTURE OF C2H4

     

    C2H2 :

    Here both the carbon atoms undergoes sp hybridization leaving 2 unhybridized (2py & 2pz) orbitals.One sp hybrid orbital of one carbon atom overlaps axially with sp hybrid orbital of other carbon forming C----C sigma bond. The two unhybridised orbital of one carbon overlaps sidewise with similar orbital of other carbon atom to form two pie bonds. The remaining hybrid orbitals of each carbon overlaps axially with half filled orbitals of Hydrogen forming sigma bond.

          ORBITAL PICTURE OF C2H2


    Q28 What is the total number of sigma and pi bonds in the following molecules? (a) C2H2 (b) C2H4
    Ans:

     

    Sigma (σ) bond is the bond formed when the two half -filled atomic orbitals overlap along their inter-nuclear axes. Sigma bond will always represent a single bond.

    Pie (π) bond is formed when the two half- filled atomic orbitals belonging to the bonding atoms overlap along sides (sidewise or lateral overlap).

    A double bond will always have one sigma bond and one pie bond.

    A triple bond will always have one sigma bond and two pie bond. So from the structure of C2H2, it is clear that it has three sigma bonds and two pie bonds.

    From the structure of C2H4,it is clear that it has five sigma bonds & one pie bond.

     


    Q29 Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why? (a) 1s and 1s (b) 1s and 2px (c) 2py and 2py (d) 1s and 2s.
    Ans:

    Sigma bond is always formed between two half filled atomic orbitals along their inter- nuclear axis ,i.e the line joining the centers of the nuclei of two atoms (axial overlapping).

    From the given data if we consider x axis as the internuclear axis 2py and 2py overlapping is not possible as both the 2py orbital will lie on y axis.While the other 1s, 2s, 2px all lie on X axis internuclear axis.


    Q30 Which hybrid orbitals are used by carbon atoms in the following molecules? (a)CH3–CH3; (b) CH3–CH=CH2; (c) CH3-CH2-OH; (d) CH3-CHO (e) CH3COOH
    Ans:

    Hybrid orbital are formed by the overlapping of orbitals having almost same energy.

    Single bond= one sigma bond = sp3 hybridization

    Double bond = one sigma bond + one pie bond = sp2 hybridization

    Triple bond = one sigma bond + two pie bond = sp hybridization

    (a)

     According to structure, C1 & C2 are making 4 sigma bonds(single bond) each with the help of one s hybrid orbital & 3 p hybrid orbital, hence C1& C2 are sp3 hybridised

    (b)

     

    Here C1 is making 4 sigma bond therefore it s sp3 hybridised  , while C2 and C3  are making a double bond( 1 sigma + 1 pie bond ) therefore they both are sp2 hybridized.

    (c)

    Both the carbons C1 & C2 are making single bond(sigma bond),therefore they are sp3hybridized.

    (d)

    From the structure it is clear that C1 is making sigma bonds only, therefore it is sp3 hybridised.C2 is making a double bond therefore it is sp2 hybridised.

    C1 is sp3 hybridized and C2 is sp2 hybridized.

    (e)

    Here C1 is making a sigma bond therefore it is in sp3 hybridization state, while C2 is making a double bond ,it  is in sp2 hybridised state.


    Q31 What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
    Ans:

    Bond pairs – the pair of electrons which are involved in the formation of bond between bonded atoms are known as bond pairs.

    Lone pairs – the pair of electrons present on bonded atoms which do not take part in bond formation are known as lone pairs.

    The bond pair of electron can either be shared or transferred between two bonded atoms.

    For example:


    Q32 Distinguish between a sigma and a pi bond.
    Ans:

    The following are the differences between sigma and pi-bonds:

    Sigma (σ) Bond Pi (π) Bond
    (a)  the bond is formed by the axial overlap of the atomic orbitals The bond is formed by the sidewise overlap of  atomic orbitals.
    (b)  the overlap is large & a strong bond is formed The overlap is small & weak bond is formed
    (c ) all the s orbitals & only one lobe of p orbital is involved in overlapping S orbital donot take part in bond formation,while both lobes of p orbital take part in overlapping
    (d) The electron cloud is symmetrical about the internuclear axis The electron cloud is unsymmetrical.
    (e)  the bond can also be formed in the absence of pie bond. The bond cannot be formed in the absence of sigma bond..
    (f) Free rotation about σ bonds is possible. Free Rotation of the atoms around pie bond is not possible

     


    Q33 Explain the formation of H2 molecule on the basis of valence bond theory.
    Ans:

    When hydrogen atoms combine to form molecules of hydrogen, 433 KJ/mol. energy is released.

    Whenever atoms combine to form a molecule, there is always decrease in the energy. Actually, when two H atoms are far separated, they do not have any force of interaction (attractive or repulsive). As they tend to come closer, two different forces operates:

    (a) The nucleus of one H is attracted towards the electrons of the other H atoms and Vice-versa. The energy is released in attraction.

    (b) The nuclei of the atoms as well as their electrons repel each other. Energy is needed to overcome the force of repulsion.

    Now in case, if the magnitude of the attractive forces is more than that of the repulsive forces, a stable molecule will be formed.

    However, if the repulsive forces are more than the attractive forces, then the atoms will not combine.

    If for example HA & HB are two hydrogen atoms & eA & eB are their respective electrons, then attractive & repulsive forces may be shown as follows:

     

                                                             ------------ old attractive forces

                                               _________ new attractive forces

    The number of new attractive and repulsive forces is same, but the magnitude of attractive forces is more. Thus, when two hydrogen atoms approach each other, the overall potential energy decreases, hence a stable hydrogen molecule is formed.


    Q34 Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
    Ans:

    The overlap of atomic orbitals means their partial merger, because of which their certain portion becomes common to both of them. The important conditions required for the linear combination of atomic orbitals to form molecular orbital are as follows:

    a) The atomic orbital with one electron (completely half filled) can take part in combination or overlapping.

    b) During the time of overlap, the combining orbitals must have opposite spins of electrons.

    c) The combining atomic orbitals must have the same or nearly the same energy. This means that in a homo-nuclear molecule, the 1s-atomic orbital of an atom can combine with the 1s-atomic orbital of another atom and not with the 2s-orbital.

    d) The overlapping must be maximum as far as possible.


    Q35 Use molecular orbital theory to explain why the Be2 molecule does not exist.
    Ans:

    The electronic configuration of Beryllium is 1s2 2s2.

    From the electronic configuration it is clear that there is no singly filled atomic orbital present in beryllium.

    Without the half- filled orbital, the overlapping is not possible, therefore Be2 molecule does not exist.


    Q36 Compare the relative stability of the following species and indicate their magnetic properties: O2,O2+,O2- (superoxide), O22-(peroxide)
    Ans:

    The stability of following species can be decided on the basis of bond order as follows:

    O2 : KK

    Bond order =1/2(Nb-Na)

    = 1/2(8-4)

    = 2 = Paramagnetic

     

    Similarly, the electronic configuration of O2+ can be written as:

    Bond order of  O2= 1/2(8-3)

    = 2.5 = paramagnetic

     

    Electronic configuration of O2- ion will be:

    Bond order of = 1/2(8-5)

    = 1.5 = paramagnetic

     

    Electronic configuration of O22- ion will be:

    Bond order of O22- =1/2(8-6) = 1  diamagnetic

     

    The larger the bond order,greater is the stability

    Therefore the stability decreases as O2> O2 > O2- > O22-.


    Q37 Write the significance of a plus and a minus sign shown in representing the orbitals.
    Ans:

    The orbital is the maximum probability of finding an electron around the nucleus. This probability is measured in terms of wave function. The wave function can have a positive or negative values. Therefore for defining the wave function, plus sign is used for positive wave function while a minus sign is used for negative wave function of an orbital.


    Q38 Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds?
    Ans:

    Atomic number of Phosphorous is 15.

    The Ground state configuration is

    & the excited state configuration is

    The 5 electrons present in valence shell form bond pairs with the electrons of five chlorine atoms. Phosphorus atom is sp3d hybridized in the excited state.

    PCl5

    The geometry of PCl5 is trigonal bipyramidal .The P atom lies in the centre of an equatorial triangle & three P-Cl bonds (equatorial bonds) are directed towards its three corners with 120° bond angle. The remaining two P-Cl bonds (axial bonds) lie above & below the plane of the triangle at bond angle 90° .

     

    The axial bonds are longer than equatorial bonds because the axial Cl atoms suffer from more repulsion then the equatorial Cl atoms,as a result the axial Cl atoms tries to reside far away from the equatorial Cl atoms, & hence axial bond are longer than equatorial bonds.


    Q39 Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
    Ans:

    A hydrogen bond represents a dipolar attraction between hydrogen atom and highly electronegative atom. The hydrogen bond is represented by ----------(dotted lines).

    For example: H -----------X -----------X -------- X------

     

    X = O, N, S, F, Cl….

    The bond pair involved in hydrogen bonding was attracted more towards highly electronegative atom, as a result H atom will start acquiring a partial positive charge (δ+) and the electronegative atom will acquire a partial negative charge (δ-).

    The magnitude of H-bonding is maximum in the solid state and minimum in the gaseous state.

    Hydrogen bonds are of two types

    (i) Intermolecular H-bond e.g., HF, H2O etc.

    (ii) Intramolecular H-bond e.g., o-nitrophenol

    Vander waal’s forces are weak forces and exist in non -polar compounds and noble gases .As compared to hydrogen bonds they are weak in nature because of strong dipole -dipole interaction in hydrogen bonding.


    Q40 What is meant by the term bond order? Calculate the bond order of: N2, O2,O2+,and O2-.
    Ans:

    Bond order is defined as the number of covalent bonds in a covalent molecule. It is equal to one half of the difference between the number of electrons in the bonding and antibonding molecular orbitals.

    Bond order = [no of electrons in bonding MO – no of electrons in antibonding MO] / 2

    BOND ORDER OF N2

    Electronic configuration of N

    Number of bonding electrons = 10

    Number of anti-bonding electrons = 4

    Bond order of nitrogen molecule = 1/2 (10-4) = 3

     

    Bond order of O2

    Electronic configuration of O

    Number of bonding electrons = 8

    Number of anti-bonding electrons = 4 

    Bond order = 1/2 (8-4) =2

    Hence, the bond order of oxygen molecule is 2.

     

    Bond order of O2+

    Electronic configuration of can be written as:

    Bond order of O2+=1/2 (8-3)= 2.5

    Thus, the bond order of  O2+ is 2.5.

     

    Bond order of O2-

    The electronic configuration of O2ion will be:

    Bond order of = O2- =1/2 (8-5) = 1.5

    Thus, the bond order of ion is  O2- = 1.5.


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