Important Questions
 

1. Write the main points of Electronic Theory of Valence by Lewis and Kossel.    (AS1) 4 Marks

A: ‣ The basis for their theory was valence in terms of electrons.

‣ They provided logical explanation of valence on the basis of the lack of chemical activity of noble gases which led to the proposal of octet rule.

‣ When the atoms of main group elements, such as IA, IIA, IIIA, IVA, VA, VIA, VIIA and zero or VIIIA group elements, are allowed to undergo chemical changes. We notice that they try to get octet electronic configuration in the outer shells.

‣ Group IA elements (Li to Cs) try to lose one valence shell electron from their atoms to form corresponding uni-positive ions which get octet in their outer shells.

e.g.: ₁₁Na  2, 8, 1; ₁₁Na⁺   2, 8

‣ Group IIA elements (Mg to Ba) try to lose two valence electrons from their atoms during chemical changes and form di-positive ions with the octet in the outer shells.

e.g.: ₁₂Mg  2, 8, 2; ₁₂Mg⁺   2, 8

‣ Group IIIA elements try to lose three valence electrons from their atoms and form corresponding tri positive ions with octaves in the outer shells.

e.g.: ₁₃Al    2, 8, 3; ₁₃Al 3⁺   2, 8

‣ Group VIA elements try to gain two electrons into the valence shells of their atoms during the chemical changes and form corresponding di-negative anions which get octet in their outer shells.

e.g.: ₈O 

2, 6; ₈O^2-   2, 8

‣ Group VIIA elements try to gain one electron into the valence shells of their atoms during the chemical changes and form corresponding uni-anions which get octet in their outer shells.

e.g.: ₉F   2, 7; ₉F^-   2, 8

‣ Group VIIIA elements, the noble gases do not try to lose or gain electrons. Generally, Helium and Neon do not participate in chemical changes. Even other elements of VIII A do not gain or lose electrons from their atoms when they participate in a very few chemical changes.

e.g.: ₁₀Ne   2, 8;

No electron gain or loss from the Neon atom.

2. Explain the formation of any two compounds according to Kossel's theory (Electron transfer theory).    (AS1) 4 Marks
                                                                           (OR)
Explain the formation of Ionic compounds NaCl, MgCl₂ through Lewis electron dot symbol.  (AS1) 4 Marks
A: i) Formation of Sodium Chloride (NaCl):
Sodium Chloride is formed from the elements Sodium and Chlorine.
      Na + Cl   Na⁺ + Cl ^-   NaCl
Cation Formation: When Sodium (Na) atom loses one electron to get Octet Electron configuration it forms a cation (Na⁺) and gets electron configuration that of Neon (Ne) atom.
                    Na   Na⁺ + e^-
Electronic Configuration (2, 8, 1) (2, 8)
                        or [Ne] 3s¹ [Ne]
Anion formation: Chlorine has shortage of one electron to get octet in its valence shell. So, it gains the electron from Na atom to form anion and gets electron configuration as that of Argon (Ar)
                  Cl + e^-   Cl ^-
Electronic Configuration (2, 8, 7) (2, 8, 8)
                    or [Ne] 3s² 3p⁵ [Ne] 3s² 3p⁶ or [Ar]
Formation of the compound NaCl from its ions:
    Transfer of electrons between Na and Cl atoms results in the formation of Na⁺ and Cl^- ions. These oppositely charged ions get attracted towards each other due to electrostatic forces and form the compound Sodium Chloride (NaCl)
            Na⁺ + Cl ^-   Na⁺ Cl^- or NaCl
Electron transfer

II) Formation of Magnesium Chloride (MgCl₂):
Magnesium Chloride is formed from the elements Magnesium and Chlorine.
            Mg + Cl₂  MgCl₂
Cation formation:
           Mg

 Mg²⁺ + 2e^-
Electronic configuration (2, 8, 2) (2, 8)
         or [Ne] 3s² [Ne]
Anion formation:
                  2 Cl + 2e ^-   2 Cl ^-
Electronic Configuration (2, 8, 7)                (2, 8, 8)
                                  or [Ne] 3s² 3p⁵     [Ne] 3s² 3p⁶    or    [Ar]

The Compound MgCl₂ formation from its ions:
   Mg²⁺ gets 'Ne' configuration and each Cl^- gets 'Ar ' configuration.
    One Mg atom transfers two electrons one each to two Cl atoms and so formed Mg²⁺ and 2 Cl^- attract to form MgCl₂.
     Mg²⁺ + 2 Cl ^-   MgCl₂
Electronic transfer

3. Mention the factors affecting the formation of cations and anions?   (AS1) 4 Marks
A: » The tendency of elements by losing electrons is called the metallic character or electro positively.

» Elements with more electropositive character form cations.
» The tendency of elements by gaining electrons is called the non-metallic character or electro negativity.
» Elements with more electronegative character form anion.
» Ionic bond is formed between atoms of elements with electronegativity difference equal to or greater than 1.9.
» Ionic bond is formed by transfer of electrons from low ionisation potential atom to high electronegativity atom.
» The tendency of losing electrons to form cations or gaining electrons to form anions depends on
  1) Atomic Size
  2) Ionisation Potential
  3) Electron affinity
  4) Electronegativity
» The atoms of elements with low ionisation potential, low electron affinity, high atomic size and low electronegativity from cations.
» The atom of elements with high ionisation potential, high electron affinity, small atomic size and high electronegativity form anions.

4. What do you understand about Ionic bond? Explain. (AS1) 4 Marks
A: » Ionic bond was proposed by Kossel.
» The electrostatic force of attraction between oppositely charged ions is called ionic bond or electrostatic bond or electro valent bond.
» Ionic bond is formed by the transfer of electrons from one atom to another atom.
» The number of electrons lost or gained by an atom in the formation of ionic bond is called electrovalency.
» Ionic bond is a non directional bond.
» The number of electrons lost from a metal atom is the valence of its element, which is equal to its group number.
» The number of electrons gained by a non metal element for its atom is its valency, which is equal to (8-its group number).
» The electrostatic attractive force that keeps cation and anion together to form a new electrically neutral compound.
» NaCl, MgCl₂, Na₂O etc are ionic compounds.

» Sodium atom loses one electron to form Na⁺ ion and Chlorine atom gains one electron to form Cl ^- ion. Both ions get stable inert gas configuration. These oppositely charged ions get attracted towards each other due to electrostatic forces to form Sodium-Chloride (NaCl).
    Na → Na⁺ + e^-
    Cl + e ^- → Cl ^-
    Na⁺ + Cl ^- → Na⁺ Cl ^- or NaCl
 

5. What do you understand about covalent bond? Explain.  (AS1) 4 Marks
A: » The formation of covalent bond was proposed by G.N. Lewis.

» The chemical bond formed by sharing of electrons between two atoms is called covalent bond.
» In the covalent bond formation both the atoms contribute equal number of electrons.
» Bonded atoms are short of one or more electrons than the inert gas configuration.
» The total number of covalent bonds that an atom of an element forms is called its covalency.
» The number of electrons contributed by an atom in the covalent bond formation is equal to its covalency.

» The number of covalent bonds formed by an atom depends on the number of unpaired electrons in their valency shell and the number of electrons required to attain inertgas configuration.
» The bond formed by sharing of one, two or three electron pairs between two atoms is called single bond (-), double bond (=) or triple bond (≡) respectively.
» F₂, O₂, N₂, CH₄, NH₃, H₂O etc are covalent compounds.
» In the formation of F₂ molecule, each Fluorine atom contributes one electron for bond formation and one electron pair is shared between two atoms.
                 
» In the formation of O₂ molecule, each Oxygen atom contributes two electrons for bond formation and two electron pairs are shared between two atoms to form a double bond between them.
            

» In the formation of N₂ molecule each Nitrogen atom contributes three electrons for bond formation and three electron pairs are shared between two atoms to form a triple bond between them.
            
 

6. Write the main points of Valence Shell Electron Pair Repulsion Theory (VSEPRT). (AS1) 4 Marks
A: » VSEPRT considers electrons in the valence shell which are in covalent bonds and in lone pairs as charge clouds that repel one another and stay as far apart as possible. This is the reason why molecules get specific shapes.
» If we know the total number of electron pairs in the valence shell as covalent bonds and lone pairs in the central atom, it will help us to predict the arrangement of those pairs around nucleus of the central atom and from that the shape of the molecule.
» Lone pairs occupy more space around the central atom than bond pairs. Lone pair means unshared electron pair or non-bonding electron pair.

These are attracted to only one nucleus where as the bond pair is shared by two nuclei. Thus, the presence of lone pairs on the central atom causes slight distortion of the bond angles from the expected regular shape. If the angle between lone pair and bond pair increases at the central atom due to more repulsion, the actual bond angles between atoms must be decreased.
» If two bond pairs are present in two covalent bonds around the nucleus of the central atom without any lone pairs in the valence shell, they must be separated by 180° to have minimum repulsion between them. Thus, the molecule would be linear.
             e.g.: Beryllium Chloride
                   
» If three bond pairs are there in three covalent bonds around the nucleus of the central atom, without any Lone pairs they get separated by 120° along three corners of a triangle. Therefore, the shape of the molecule is trigonal-planar.
e.g.: Boran trifluaride
         

» If there are four bond pairs in the valence shell of the central atom, the four bond pairs will orient along the four corners of a tetrahedron and the bond angle expected is 109° 28'
      e.g.: Methane
                  
» If there are three bond pairs and one lone pair i.e. unshared electron pair, then the lone pair occupies more space around the nucleus of the central atom. The remaining three bond pairs come relatively closer as in NH₃ molecule.
     e.g.: Ammonia
                   

» If there are two bond pairs and two lone pairs of electrons around the nucleus of the central atom in its valence shell, lone pair - lone pair repulsion is greater than lone pair - bond pair repulsion. Therefore, the angle between bond pairs further decreases.
    e.g.: Water
                  
» Valence Shell Electron Pair Repulsion Theory (VSEPRT) mainly fails in explaining the strengths of the bonds.
 

7. Write the main points of Valence Bond Theory    (AS1) 4 Marks
A: Main points of Valence Bond Theory:
» A covalent bond between two atoms is formed when the two atoms approach each other closely and one atom overlaps its valence orbital containing unpaired electron, the valence orbital of the other atom that contains the unpaired electron of opposite spin. The so formed paired electrons in the overlapping orbitals are attracted to the nuclei of both the atoms. This bonds the two atoms together.

e.g.: 
                   
» The greater the overlapping of the orbitals that form the bond, the stronger will be the bond. This gives a directional character to the bond when other than 'S' orbitals are involved.
» Each bonded atom maintains its own atomic orbitals but the electron pair in the overlapping orbitals is shared by both the atoms involved in the overlapping.
» If two atoms form multiple bonds between them the first bond is due to the overlap of orbitals along the inter-nuclear axis giving a stronger sigma (σ) bond. After formation of (σ) bond the other bonds are formed due to the overlap of orbitals side wise or laterally giving weaker Π bonds. The 'σ' bond is stronger because the electron pair shared is concentrated more between the two nuclei due to end-end or head on overlap and attracted to both the nuclei. The Π bond overlap gives a weaker bond due to the lateral overlap of 'p' orbitals which is not to greater extent.

8. Write the formation of double bond and triple bond according to Valence Bond Theory.  (AS1) 4 Marks
                                                                   (OR)
Explain the formation of Oxygen and Nitrogen molecules according to Valency Bond Theory.
A: i) Formation of double bond (Formation of O₂ molecule):
           ₈O has electronic configuration 1s² 2s² 2p_x² 2p_y¹ 2p_z¹.
» If the py orbital of one 'O' atom overlaps the 'p_y' orbital of other 'O' atom along the in inter nuclear axis, a sigma p_y - p_y bond (σp_y - p_y) is formed. p_z orbital of one 'O' atom overlaps the p_z orbital of other 'O' laterally, perpendicularly to the internuclear axis giving a π p_z - p_z bond. O₂ molecule has a double bond between two Oxygen atoms.
    

ii) Formation of triple bond (Formation of N₂ molecule):
» ₇N has electronic configuration 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹. Suppose that p_x orbital of one 'N' atom overlaps the 'p_x 'orbital of the other 'N' atom giving σp_x - p_x bond along the inter-nuclear axis. The p_y and p_z orbitals of one 'N' atom overlap the p_y and p_z orbital of other 'N' atom laterally, respectively perpendicular to inter-nuclear axis giving πp_y - p_y and πp_z - p_z bonds. Therefore, N₂ molecule has a triple bond between two Nitrogen atoms.
  

9. Explain the formation of any two molecules using hybridisation.   (AS1) 4 Marks
A: i) Formation of BeCl₂ molecule:
» ₄Be has electronic configuration 1s² 2s². An excited state is suggested for Beryllium in which an electron from '2s' shifts to 2p_x level. Electronic configuration of ₄Be is 1s² 2s¹ 2p_x¹ and 17^Cl is 1s² 2s² 2p⁶ 3s² 3p_x² 3p_y² 3p_z¹
» Beryllium atom in its excited state allows its 2s orbital and 2p_x orbital which contain unpaired electrons to intermix and redistribute to two identical orbitals. As per Hund's rule each orbital gets one electron. The new orbitals based on the types of orbitals that have undergone hybridisation are called sp orbitals. The two sp orbitals of Be get separated by 180°.
» Now each Chlorine atom comes with its 3p_z¹ orbital and overlaps it the sp orbitals of Beryllium forming two identical Be - Cl bonds (σsp - p bond s).
    

ii) Formation of BF₃ molecule:
» ₅B has electronic configuration 1s² 2s²  2p_x¹
» Boron (B) first undergoes excitation to get electronic configuration 1s² 2s¹ 2p_x¹ 2p_y¹.
» As it forms three identical B - F bonds in BF₃. It is suggested that excited 'B' atom undergoes hybridisation. There is an inter mixing of 2s, 2p_x, 2p_y orbitals and their redistribution into three identical orbitals called sp² hybrid orbitals. For three sp² orbitals to get separated to have minimum repulsion the angle between any orbitals is 120° at the central atom and each sp² orbital gets one electron. Now three Fluorine atoms overlap their 2p_z orbitals containing unpaired electrons (₉F 1s² 2s² 2p_x² 2p_y² 2p_z¹).
» The three sp² orbitals of 'B' that contain unpaired electrons to form three σ sp² - p bonds.
                         

10. Draw the figures of BeCl₂, BF₃, H₂O, HCl molecules based on Hybridisation.   (AS5) 4 Marks
A:

11. Draw the Lewis dot structures of B, C, N and O atoms   (AS 5) 2 Marks
A:
        
 

12. Draw the Lewis dot structures of Valence electrons in He, Ne, Ar, Kr atoms.  (AS5) 2 Marks
A:
        
 

13. Draw the dot diagrams of (i) Methane (ii) Water       (AS5) 2 Marks
A: (i) Methane (CH₄) molecule
                 
(ii) Water (H₂O) molecule
                      

14. Draw the Lewis dot structure of the following molecules. (AS5) 2 Marks
      i) F₂      ii) O₂
A: i) F₂ molecule.
       
    ii) O₂ molecule.
           
 

15. Draw the dot diagrams of formation of the molecules    (AS5) 2 Marks
     i) Ammonia      ii) Water
A: i) Ammonia (NH₃) molecule:
     

ii) Water (H₂O) molecule:
                  
 

16. Draw the electron dot structures of the molecules.     (AS5) 2 Marks
                 i) BeCl₂             ii) BF₃
A: i) BeCl₂
              
     ii) BF₃:
             

17. Draw the figure showing the formation of N₂ molecules based on valence bond theory.   (AS5) 2 Marks
A:
     
 

18. Observe the figure given below and answer the following questions.       (AS4) 2 Marks
                        
     i) How many valence electrons are there in Nitrogen atom?
     ii) Which bond is formed between Nitrogen atoms?
A: i) There are 5 valency electrons in Nitrogen atom.
     ii) There is covalent and triple bond in between of Nitrogen atoms.

19. Answer the questions given with the help of the figure.                              (AS4) 2 Marks
                             

   i) What is the shape of NH₃ molecule?
   ii) How many σs - sp³ bonds are there in NH₃ molecule?
A: i) The shape of the NH₃ molecule is triagonal pyramidal
     ii) Three σs - sp³ bonds
20. Fill the table with suitable answer.                                                                  (AS4) 2 Marks

A:

21. When the teacher asked "How many bonds are there in Ammonia?" The student by looking at the diagram answered ''There are 3 bonds". Explain the bonds in this molecule as per hybridisation    (AS1) 2 Marks
A: In Ammonia molecule, N participate in sp³ hybridisation. In this process one '2s' orbital and three '2p' orbitals of Nitrogen intermix and form four sp³ hybrid orbitals. One of the four sp³ orbitals get a pair of electrons and the other three sp³ orbitals get one electron each. These three sp³ orbitals overlap with 1S orbital of Hydrogen atoms and form three σs - sp³ bonds.
 

22. In Ammonia molecule, the bond angle is 107° 48' instead of 109° 28' and in water molecule the bond angle is 104° 31' instead of 109° 28'. Give reasons for this. (AS1) 2 Marks
A: » sp³ hybridisation is present in NH₃ molecule.  should be 109° 28' for sp3 hybridisation. As there is a lone pair in one of the sp3 orbitals. There is a greater lone pair repulsion which decreases the bond angle  from 109° 28' to 107° 48'
» sp³ hybridisation is present in water (H₂O) molecule.  should be 109° 28' for sp3 hybridisation. Due to the lone pair- lone pair repulsions and lone pair- bond pair repulsions,  decreases from 109° 28' to 104° 31'.

23. Draw the Lewis electron dot structure which shows the formation of Aluminium chloride, Di Sodium monoxide molecules.    (AS5) 2 Marks
A: i) Formation of Aluminium Chloride:

ii) Formation of Di Sodium Monoxide:
  
 

24. Draw the Lewis electron dot structure that shows the formation of Sodium Chloride, Magnesium Chloride molecules.   (AS5) 2 Marks
A: i) Formation of Sodium Chloride: