Chemistry #expch001 Important Questions
Ch. 2. Chemical Bonding #expch_002
Note: Please use landscape mode on your cell phone or laptop for a better experience.

1. Find the pair with sp2 hybridization of the central molecule

(a)NH3 and NO2–

(b)BF3 and NH2–

(c)BF3 and NO2–

(d)none of these

Explanation

• To find the hybridization of central atoms:
• For a molecules,

•8 Totalno.ofvalenceelectronsofallatoms​

• i.e. for BF3​
• It is 24/8​=3
• This obtained quotient can be called as co ordination number.
• Co ordination number Hybridization type
• 2 sp
• 3 sp2
• 4 sp3
• 5 sp3d
• 6 sp3d2
• For Ions,
• Co ordination number =21​(v+x−c+a)
• where, v= no. of valence electrons in central atom
• x= no. of mono valent electrons around central atom
• c= +ve charge on Cation
• a= −ve charge on Anion
• For NO2−​
• It is 21​(5+0−0+1)=3
• Hence, According to the given table BF3​ and NO2−​ have sp2 hybridization

2. Which one has a pyramidal shape?

(a) SO3

(b) PCl3

(c) CO32-

(d) all of these

3. MX6 is a molecule with octahedral geometry. How many X – M – X bonds are at 180°?

(a) three

(b) two

(c) four

(d) six

Explanation

In octahedral molecule six hybrid orbitals directed towards the corners of a regular octahedron with a bond angle of 90°. According to this geometry, the number of X – M – X bonds at 180° must be three.

4. Molecules are held together in a crystal by

(a) hydrogen bond

(b) electrostatic attraction

(c) Van der Waal’s attraction

(d) dipole-dipole attraction

5. The structure of IF7 is (PPSC 2017)
   (a) Pentagonal bipyramid
   (b) Square pyramid
   (c) Trigonal bipyramid
   (d) Octahedral

Hi, If you feel any difficulty in understanding, Must watch the complete Lecture from the same Teacher, then start the assessment..

Full Lenth Lecture on CHEMICAL BONDING

6. Among the following mixtures, dipole-dipole as the major interaction, is present in
   (a) benzene and ethanol
   (b) acetonitrile and acetone
   (c) KCl and water
   (d)none of these

Explanation

Dipole-dipole force exist between polar molecules.

As all hydrocarbons are nonpolar so benzene (C6H6) is non-polar and carbon tetrachloride is nonpolar because of symmetrical tetrahedral geometry. So, here london force are the major interaction.

ethanol (C2H5OH) is polar. So here dipole-induce dipole force is the major interaction.

Acetonitrile (CH3CN) and acetone (CH3COCH3) both are polar. So here dipole-dipole force is the major interaction.

KCl is ionic compound while water is polar compound so here ion-dipole force is the major interaction.

Acetonitrile and acetone are polar molecules. Hence, dipole-dipole interaction exists between them. Between KCl and water, ion-dipole interaction is found and in benzene – ethanol and benzene – carbon tetrachloride, dispersion force is present.

7. Dipole-induced dipole interactions are present in which of the following pairs?
   (a) H2O and alcohol
   (b) Cl2 and CCl4
   (c) HCl and He atoms
   (d) SiF4 and He atoms

Explanation

The forces or interactions, which occur between a polar molecule and a nonpolar molecule or with an atom are termed as Dipole -induced dipole interactions.

Dipole -induced dipole forces/interactions are those which take place between a polar and a non polar molecule or with an atom.

H2​O and Alcohol both are polar, thus this is not an example of Dipole-Induce dipole.

While HCl is a polar while helium is non-polar, thus it is an example of Dipole induce dipole.

SiF4​ and He atom are both non polar.

Cl2​ and CCl4​ are both non-polar molecule.

( In CCl4​ chlorine atoms are bonded symmetrically with the carbon central atom thus it is non-polar similarly SiF4​ is also non-polar)

SB of Non Polar Molecules:

A molecule may be nonpolar either when there is an equal sharing of electrons between the two atoms of a diatomic molecule or because of the symmetrical arrangement of polar bonds in a more complex molecule.

Example molecules non polar

Toluene Gasoline

Helium (He) Neon (Ne)
Krypton (Kr) Xenon (Xe)

Hydrogen (H2) Nitrogen (N2)
Oxygen (O2) Carbon Dioxide (CO2)
Methane (CH4) Ethylene (C2H4)

SB of Polar molecules

Polar molecules interact through dipole–dipole intermolecular forces and hydrogen bonds. Polarity underlies a number of physical properties including surface tension, solubility, and melting and boiling points.

A polar molecule has a net dipole as a result of the opposing charges (i.e. having partial positive and partial negative charges) from polar bonds arranged asymmetrically.
Example polar molecules

Ammonia (NH3)

Sulfur Dioxide (SO2)

Hydrogen Sulfide (H2S

8. The number of types of bonds between two carbon atoms in calcium carbide is
   (a) Two sigma, two pi
   (b) One sigma, two pi
   (c) One sigma, one pi
   (d) Two sigma, one pi

Explanation

A single bond between two atoms is always considered as sigma bond.
A double bond between two atoms is always considered as one sigma and one pi bond
A triple bond between two atoms is always considered as one sigma bond and two pi bonds.
So according to the given structure CaC2 (Calcium carbide) has 1 sigma and 2 pi bonds

9. An atom of an element A has three electrons in its outermost orbit and that of B has six electrons in its outermost orbit. The formula of the compound between these two will be
   (a) A3B6
   (b) A2B3
   (c) A3B2
   (d) A2B

Explanation

A has 3 electrons in outermost orbit and B has 6 electrons in its outermost orbits. So A can give three electrons to complete its octet and B needs 2 electrons to complete its octet. So 2 atoms of A will release 6 electrons and 3 atoms of B will need six electrons to complete their octet.

In a neutral molecule, the sum of the bonding valance electrons must be equal. So the products of the negative element and its charges and the positive element and its charge must be equal.

If we have a 3 valance electrons , the ‘A’ charge will be either +3 or -5 for a full octet and valance electron in ‘B’ atoms will mostly result in acquisition of additional electrons (2) for an octet and relative charge of -2.

Balancing the two,

3 × A = -2 × B

To be equal, A = 2 and B = 3

10. Which of the following statements is not correct ?
    (a) Double bond is shorter than a single bond
    (b) Sigma bond is weaker than a pi bond
    (c) Double bond is stronger than a single bond
    (d) Covalent bond is stronger than hydrogen bond Hi, If you feel any difficulty in understanding, Must watch the complete Lecture from the same Teacher, then start the assessment..

Full Lenth Lecture on CHEMICAL BONDING

11. which type of sigma bond is stronger?

(a) p-p
(b) s-p
(c) s-s
(d) All are equal

12. Using VSEPR theory, predict the species which has square pyramidal shape
    (a) SnCl2
    (b) CCl4
    (c) SO3
    (d) BrF5
13. Among the following the maximum covalent character is shown by the compound.
    (a) MgCl2
    (b) FeCl2
    (c) SnCl2
    (d) AlCl3

Explanation

We know that, extent of polarisation ∝ covalent character in ionic bond. Fajans rule states that the polarising power of cation increases, with increase in magnitude of positive charge on the cation Therefore, polarising power ∝ charge of cation.

The polarising power of cation increases with the decrease in the size of a cation. Therefore, polarising (power) ∝ (1)/ (size of cation)
Here the AlCl3 is satisfying the above two conditions i.e., Al is in +3 oxidation state and also has small size. So it has more covalent character.

14. The most polar bond is
    (a) C – F
    (b) C – O
    (c) C – Br
    (d) C – S

Explanation

Polar character is proportional to electronegativity difference between the atoms. In the given molecules, carbon is common. As fluorine has high electronegativity compared to other atoms, C−F bond is most polar.

15. The Trigonal Planer shape is shown by which of the species

(a) BF4-
(b) SiH4
(c) :CH3-
(d) CH3+

Hi, If you feel any difficulty in understanding, Must watch the complete Lecture from the same Teacher, then start the assessment..

Full Lenth Lecture on CHEMICAL BONDING

16. The hybrid state of sulphur in SO2 molecule is :
    (a) sp²
    (b) sp³
    (c) sp
    (d) None of these
17. The number of nodal planes present in s × s antibonding orbitals is
    (a) 1
    (b) 2
    (c) 0
    (d) 3

Explanation

In an antibonding molecular orbital, most of the electron density is located away from the space between the nuclei, as a result of which there is a nodal plane (i.e, a plane at which the electron density is zero) between the nuclei.

18. Which of the following statements is incorrect ?
    (a) The formation of ionic compounds depend upon the ease of formation of the positive and negative ions from the respective neutral atoms.
    (b) Formation of ionic compounds depend upon arrangement of the positive and negative ions in the solid.
    (c) Formation of positive ion involves addition of electron(s) while that of negative ion involves removal of electron(s).
    (d) None of these

Explanation

Ionic bond, also called electrovalent bond, type of linkage formed from the electrostatic attraction between oppositely charged ions in a chemical compound. Such a bond forms when the valence (outermost) electrons of one atom are transferred permanently to another atom.

However, to maintain charge neutrality, strict ratios between anions and cations are observed so that ionic compounds, in general, obey the rules of stoichiometry despite not being molecular compounds.

19. In the formation of a molecule which of the following take part in chemical combination?
    (a) cation
    (b) anion
    (c) valence electron
    (d) inner shell electron
20. In which of the following molecules octet rule is not followed?
    (a) NH3
    (b) CH4
    (c) CO2
    (d) NO

Explanation

Octet rule means when the total number of eight. If the compound is made up of two elements then add the valence electrons, if the total electron is either less than eight or greater than eight then the Octet rule is not followed. Which means octet rule is not followed.draw lewis structures and count electrons.

Some example which not follow octet rule are given in the figure on the next slide.

Hi, If you feel any difficulty in understanding, Must watch the complete Lecture from the same Teacher, then start the assessment..

Full Lenth Lecture on CHEMICAL BONDING

21. According to Molecular orbital theory(MOT) _.

A. oxygen and nitrogen molecules are paramagnetic
B. oxygen molecule is paramagnetic while nitrogen molecule is diamagnetic
C. oxygen and nitrogen both have paramagnetic and diamagnetic
D. None of these

22. Bond Order of O2, F2, N2 respectively are __
    a) +1, +2, +3
    b) +2, +3, +1
    c) +2, +1, +3
    d) +3, +2, +1
23. Which of the following molecule does not exist due to its zero bond order?
    a) H2+
    b) He2+
    c) He2
    d) H2–

Explanation

Molecular orbital electronic configuration of He2 molecule = (σ1s)2 (σ*1s)2. Bond order = 0,

so He2 molecule does not exist.

Hi, If you feel any difficulty in understanding, Must watch the complete Lecture from the same Teacher, then start the assessment..

Full Lenth Lecture on CHEMICAL BONDING

Thank You