Tuesday, 21 April 2020

1st year Chemistry Chapter 6 Short Questions Notes

1st year Chemistry Chapter 6 Short Questions Notes
Looking for the 11th class Chemistry Chapter 6 Chemical bond notes? Here we have shared the 1st year Chemistry Chapter 6 Short Questions Notes.

01. Bond distance is the compromise distance between two atoms. Explain.
 When two atoms come to close to each other, they have attraction as well as repulsion between them. These bonded atoms stay at the distance, where the attraction is maximum. This distance is also called compromise distance or bond length.
 02. The distinction between a co-ordinate covalent bond and a covalent bond vanishes after bond formation. Explain with the help of an example.
 The coordinate covalent bond is formed when the shared pair of the electron is donated by one of the bonded atoms. This bond is represented by the arrow (→). Arrowhead is from donor atom towards acceptor atom. After the formation of a bond, difference vanishes and it behaves just like covalent.
 
03. The bonds angles of H2O and NH3 are not 109.5° like that of CH4 although O and N atoms are sp3 hybridized. Explain.
 In CH4, NH3 and H2O the central atoms C, N, and O are SP3 hybridized. The bond angle in NH3 and H2O are 107.5o and 104.5o respectively. The decrease in bond angle is because one lone pair of N and two lone pairs of O repel the bonds pair. NH3 has one lone pair so the angle is less reduced while H2O has two lone pairs so the angle is more reduced.

 04. Π bonds are more diffused than σ bond. Or σ bond is stronger than Π bond. Explain.
 In the sigma bond, the electron density is between two nuclei or at the bond axis. In the case of the π bond, the electron density is above and below the nuclear axis. In π bond electron cloud is expanded at more regions than sigma bond, so π bond or π electrons are more diffused than sigma.

 05. The abnormality in bond length and bond strength in HI is less prominent than that of HCl. Explain.
 In the case of HI, the calculated bond energy is 291kJ/mol while experimental bond energy is 299kJ/mol. The difference between the two energies is 8kJ/mol. In the case of HCl, the observed bond energy is 432kJ/mol while the calculated bond energy is 336kJ/mol. The difference between the two energies is 96kJ/mol. Greater energy difference in HCl is due to the greater ionic character of HCl.

 06. The dipole moment of CO2 and CS2 is zero but that of SO2 is 1.61D. Explain.
 CO2 and SO2 have a linear structure. In linear structure dipole moment of individual bond cancel the effect of each other. So the net dipole moment is zero. But cancelation does not happen in SO2 because its structure is angular.
 
07. The melting and boiling points of electrovalent compounds (ionic) are very high as compared to covalent compounds. Why?
 In ionic compound strong ionic forces are present. A large amount of energy is required to break these forces of attraction. In covalent compounds, forces among particles are week so less amount of energy is required to break these forces. That is why ionic compounds have high melting and boiling points.
 
09. Why π − bonds are weaker than σ − bond.
 σ bond is formed by head-to-head overlapping of orbitals and electron density is maximum between the two nuclei on the bond axis. π-bonds are formed by sideway overlapping of orbitals and electrons density is present above and below the bond axis. That’s why π-bonds are weaker than a σ bond.
 
11. Why MOT is superior then VBT?
 MOT theory explains the paramagnetic behavior of the O2 molecule. Also, it explains the non-existence of the He2 molecule.

 12. Why the energy of ABMO is greater than BMO?
 In ABMO the electron density is low on the nuclear axis so nucleus-nucleus repulsion is more and the system is unstable and of high energy.
 In BMO the electrons density is maximum on the nuclear axis between the nuclei. So, nucleus-nucleus repulsion is minimum and the system is stable and of low energy.

13. Why the cationic radius is always smaller than its parent atom?
 The positive ion is always smaller than its parent atom because of two reasons: 
a) Because of the loss of an electron, there is a possibility that a shell will also be lost.
 b) After the loss of an electron, the balance of electron and proton is disturbed. Now there are more protons for fewer electrons. So, there will be a high amount of attraction of protons for electrons and hence size will decrease.

14. Why the anionic radius is always greater than its parent atom?
 The negative ion is always greater than its parent atom because of two reasons: a) The incoming electron may be repelled by the already present electrons in the outermost shell and this repulsion can result in the increase of electronic cloud.
 b) After the gain of the electron, the balance of electron and proton is disturbed. Now there are more electrons for fewer protons. So, the force of attraction between the nucleus and electrons is decreased which results in an increase in size.

15. Why the second ionization energy of magnesium is higher than the first one?
 It is difficult to remove the second electron from an ion as compared to an atom. Because nuclear charge attracts the remaining electrons strongly. So second ionization energy is greater than first ionization energy.

 16. Why the electron affinity of fluorine is less than chlorine?
 Actually, Fluorine has a very small size and seven electrons in 2s and 2p orbitals producing a thick electronic cloud. This thick cloud repels the incoming electron.

 17. Why there is no bond in chemistry which is 100% ionic?
 A 100% ionic bond is formed only when the electron is completely transferred from one atom to the other. It can happen only when the donor atom has zero electronegativity value. However, no atom has zero electronegativity value. So, no bond is 100% ionic. The bond between CsF is 92% ionic which is the highest ionic character. And NaCl is 72% ionic.

 18. What is meant by paramagnetic and diamagnetic substances?
 Any substance which has an unpair of electrons and attracted to strong magnetic field is called a paramagnetic substance.
 Any substance which has a pair of electrons and does not attract towards a strong magnetic field is called a diamagnetic substance.

 19. Why Ionic bonds are non-directional in nature?
 The ionic compounds involve electrostatic lines of forces between oppositely charged ions. Therefore, such bonds are non-rigid and non-directional. Because of this ionic bonds are non-directional in nature.
 
 20. Limitations of the Lewis Model: It doesn’t tell about the following points: i. How electrons avoid each other in spite of their repulsion?
 ii. Shapes of molecules 
iii. The geometry of molecules
 iv. Bond polarity
 v. Bond length
 vi. Various energy transition

 21. Postulates of VBT:
 i. A bond between two atoms is formed by the overlap of half-filled atomic orbitals of atoms.
 ii. The overlapping orbitals retain their identity.
 iii. The number of bonds formed by an atom is equal to the number of half-filled orbitals.
 iv. During the overlapping of orbitals, energy is released.
 VSEPR-Theory
 AB2- Type with two bond pairs: In such molecules, the two bonding electron pairs arranged themselves around the central atom in such a way that to minimize repulsion.
 Example:
 MgCl2, CaCl2, BeCl2
 Angle:
 180°
 Structure:
 Linear
 
AB3- Type with three bond pairs: In such molecules, the three bonding electron pairs arranged themselves around the central atom in such a way that to minimize repulsion.
 Example:
 BF3,BH3, AlCl3
 Angle:
 120°
 Structure:
 Triangular.
 
AB3- Type with two bond pairs and one lone pair:
In such molecules, the two bond pairs and one lone pair arranged themselves around the central atom
in such a way to minimize repulsion.
Example:
Sn̈Cl2
Angle:
Less than 120°
Structure:
Distorted Triangular
Note: Lone pair exert more repulsion on bond pair due to which angle becomes less than 120°.

AB3- Type with two multiple bond pairs and one lone pair:
In such molecules, the two multiple bond pairs and one lone pair arranged themselves around
the central atom in such a way to minimize repulsion.
Example:
S̈O2
Angle:
Less than 120.
Structure:
Distorted Triangular
Note: Lone pair exert more repulsion on bond pair due to which angle becomes less than 120°. And According to the VSEPR theory double bonds behave like a single bond.

AB3- Type with three multiple bond pairs: 
In such molecules, the three multiple bond pairs arranged themselves around the central atom in such a way that to minimize repulsion.
 Example:
 SO3
 Angle:
 120°
 Structure:
 Triangular
 Note: According to VSEPR theory multiple bonds treats like a single bond.

 AB4- Type with four bond pairs:
 In such molecules, the four bond pairs arranged themselves around the central atom in such a way that to minimize repulsion.
 Example:
 CH4, SiH4, CCl4, SnCl4
 Angle:
 109.5°
 Structure:
 Tetrahedral.
 AB4 - Type with one lone pair and three bond pairs: 
A molecule having three bond pairs and one lone pair around the central atom has triangular pyramidal geometry instead of tetrahedral.
 Reason:
 Due to repulsions between lone pair and bond pairs, the geometry will be pyramidal.
 Bond Angle:
 The bond angle reduces to 107.5°
 Example:
 P̈H3, As̈H3, Sb̈H3, N̈H3
 Structure:
 Pyramidal
 AB4 Type with two lone pairs and two bond pairs: 
In this case presence of two lone pairs introduces three types of repulsions i.e. L.p-L.p, L.p-B.p and B.p-B.p. The participation of the lone pair and bond pair in determining the geometry of the H2O molecule. Two corners of the tetrahedron is occupied by each of the two lone pairs and remaining by bond pair.
 Structure:
 Bent or Angular geometry.
 Bond Angle: The angle is further reduced to 104.5° 

Q: Why the angle further reduced in the NF3 molecule to about 102°?
 It is an AB4 type with one lone pair and three bond pairs. The angle of NF3molecule is expected to be the same as NH3molecule i.e 107.5° but in actual it is 102°.
 Reason: The electronegativity of fluorine is greater than Nitrogen its means that in NF3 molecule the bonds become polar and the δ + charge appears on the nitrogen atom. And the δ − charge appears on each fluorine atoms. The δ + charge on nitrogen atom pulls the lone pair towards the nucleus which in turn exerts a stronger repulsion over bond pairs. So, angle further shrink to 102°  
ClF3: It is an AB5 type molecule with two lone pairs and three bond pairs. The electronic geometry would be Trigonal Bipyramidal but in actual it is “T-Shaped” because of the presence of two lone pairs that exert more repulsions.

“Bond Length”

 Definition:
 The average distance between the nuclei of two atoms forming a covalent bond is called Bond Length.
 Determination of Bond Length:
 The bond length can be measured by X-rays or Spectroscopic Techniques.
 Covalent Radii and Bond Length:
 The covalent radii for different elements are almost addictive in nature.
 For Example:
 The atomic radius of Carbon is 77 pm which is half of the C−C bond length i.e. 154 pm The atomic radius of Chlorine is 99 pm which is half of the Cl−Cl bond length i.e. 198 pm.
 So, the bond length of C−Cl will be 77 + 99 = 176 pm.
 Factors Affecting Bond Length
 1) Electronegative Difference:
 When the E.N difference is increase then bond length becomes shortened.
 For Example:
 The si-F bond length in SiF4 is 154-159 pm.
 The addition of their covalent radii Si−F = 117 + 64 = 181 pm.
 The calculated value is greater than observed value.
 Reason:
 The ionic character result in shortening of bond length due to force of attraction between the polar ends.
 2) Hybridization:
 It also explain the shortening of bonds length. As the s-character increase in the hybrid orbitals, the shorter will be bond distance.
 For Example:
 C−C bond length in ethane is 154 pm = sp3 - Hybridization (s-character = 25%).
 C = C bond length in ethane is 133 pm = sp2 - Hybridization (s-character = 33%).
 C≡ C bond length in ethane is 120 pm = sp - Hybridization (s-character = 50%).
 Order: C−C > C = C > C≡ C
 3) Atomic Radius:
 The bond length increase with increase of atomic radius and vice versa.
 

Trends in the Periodic Table

 Along the Periods:
 i. Atomic number increases regularly.
 ii. The nuclear charge will increase.
 iii. Shell will remain the same.
 iv. Shielding effect will also remain same.
 v. The atomic size will decrease.
 For Example:
 C − C Bond length is greater than N − N Along with the Groups:
 i. Atomic number increases irregularly.
 ii. The nuclear charge will increase.
 iii. Shell will also increase.
 iv. Shielding effect will also increase.
 v. The atomic size will increase.
 For Example:
 Si − Si Bond length is greater than C − C “Dipole Moment”
 Dipole:
 Two equal but opposite charges separated by a small distance from a dipole.
 Dipole Moment:
 The product of the electric charge (q) and the distance between the opposite charges (r).
 Mathematical form:
 Dipole moment = μ = q × r
 μ = magnitude of charges × distance between them Representation:
 It is represented by an arrowhead.
 Positive end Negative end
 Note:
 It is a vector quantity that has magnitude as well as direction.
 Units:
 Its unit is “mC”.
 Or it is expressed in Debye units (D).
 Relationship between Debye and mC: 1D = 3.336 × 10−30 mC
 Examples of Dipole Moment
 i. For Diatomic Molecules:
 Their dipole moment is directed from electropositive end s to electronegative ends.

Applications of Dipole Moment
 It is used to calculate:
 i. Percentage of ionic character of a bond.
 ii. Angles between the bonds or geometry of molecules.
 (Explanation from the textbook)
 
 

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