Loading [MathJax]/extensions/TeX/AMSsymbols.js
Skip to main content

Chemical Bonding

Question. Explain with reason why are𝛑 bonds more diffused than 𝛔 bond. 

Answer. As in 𝛔 bond electron density is maximum between the nuclei and symmetrical about line joining the two nuclei. On the other hand in 𝛑 bond electron density lies above and below the nuclei. So, 𝛑 bond is more diffused (soft) than 𝛔 bond. 

Question. Define bond energy or enthalpy. Explain the various parameters which determine its strength. 

Answer. Bond energy is the average amount of energy required to break one mole of a specific bond in gaseous substance into gaseous atoms. It is also called bond enthalpy which is measured under standard conditions i.e 298 K and 1 atm pressure. 

Factors: Following are factors of bond energy.

1. The electronegativity difference of bonded atoms. Greater the electronegativity difference, greater is bond enthalpy. 

2. Sizes of the atoms: Larger the size of atoms, smaller is bond enthalpy. 

3. Bond length: longer the bond length, weaker is bond enthalpy. 

Question. How do you compare the bond strengths of (i) Polar and non polar molecules (ii) 𝛔 and 𝛑 bond 

Answer. Polar molecule and Non-polar molecule: The molecule which contains atoms of different elements is called polar molecule The molecule which contains atoms of same element is called non polar molecule. In non polar molecule, the bonding electron pair is equally shard between the atoms having same electronegativities. Due to even (equal) distribution of electrons, bonded atoms remain electrically neutral. Whereas in a polar molecule, the bonding pair is not equally shard between atoms having different electronegativities. Due to uneven (unequal) distribution of electrons, one end of molecule is partially positive charge and an other partially negative charge. So polar molecule has an additional strength due to this ionic character. So polar molecule is more stronger then non polar molecule.

Sigma Bond (𝛔-bond) and Pi-bond( 𝛑 bond): In 𝛔 bond, electron density is maximum between the nuclei and symmetrical about line joining the two nuclei. So 𝛔 electrons are tightly held between the nuclei. On the other hand in 𝛑 bond electron density lies above and below the nuclei. So, 𝛑 bond is more diffused (soft) than 𝛔 bond. That is why 𝛑 bond is weaker than 𝛔 bond. 

Question. Calculate the bond energy of H-Br.The bond energy of H-H is 436 Kjmol-1 and Br-Br is 193 kJmol-1. 

Answer. Bond energy of H-Br = ?

 Bond energy of one mole H-H = 436 kJmol-1 

Bond energy of one molecule of H-H bond = (436 )/(6.023 x 10^23 ) kJ/molecule = 72.42 x 10^-23 kJ/molecule 

Bond energy of one atom of H-H bond= (72.42 x 10^-23 )/2kJ/atom = 36.21 x 10-23 Kj /atom 

Bond energy of one mole of Br-Br bond = 193 kJmol-1 

Bond energy of Br-Br = (193 )/(6.023 x 10^23 ) kJ/molecule = 32.04 x 10-23 kJ/molecule

 Bond energy of Br-Br = (72.42 x 10^(-23 ))/2kJ/atom = 16.02 x 10-23 Kj /atom

Bond energy of H-Br = 36.21 x 10^-23 Kj /atom of H + 16.02 x 10-23 Kj /atom of Br

                                   = 52.23 x 10-23 Kj / molecule of H-Br

                                    = 52.23 x 10-23 x 6.023x1023 kJ /mole  

                                    = 3.14.5 kJ/mol 

Dipole Moment 

Question. Define dipole moment. Give its various units. Find the relationship between Debye and mC. Answer.

Definition. The product of electric charge (q) and distance (r)between the positive and negative centers.                  

 Mathematically,

                          μ = q x r 

Units: mC and Debye Relationship between Debye and mC: 1D =3.336 X 10^-30 mC 

Question.  The bond length of H-Br is 1.4 x 10^-10 m. Its observed dipole moment is 0.79 D.

 Find the percentage ionic character of the bond. Unit positive charge = 1.6022 x 10^-19 and 1 D =3.336 X 10^(-30) mC 

Solution. % age ionic character = (μ obs/(μ ionic ) x 100 ) μobs=0.79 D

                μ ionic = q X r = 1.6022 x 10^-19 x 1.4 x 10^-10 

                            = 2.243 x 10-29 m C 

               μ ionic   =(2.243 X 10^(-29) mC )/(3.336 X 10^(-30) mC)

                              = 6.723 D 

             Since 1D  =3.336 X 10^(-30) mC 

           % age ionic character  = μ obs./ μ ionic

                                                = ((0.79)/(6.732 ) x 100 ) 

                                                = 11.7%

 Question. Dipole moment of CO2 is zero but that of CO is 0.12D. Why?

 Answer. Dipole moment of CO2 is zero but that of CO is 0.12 D. Reason is that CO2 has a linear structure, where the dipoles are equal and opposite, cancel each other effect. Whereas CO has single dipole moment directed from carbon and oxygen. μ = 0 D μ = 1.61 D 

Question. The dipole moment of CO2 is zero and that of H2O is 1.85 D. Why? 

Answer. Dipole moment of CO2 is zero but that of H2O is 1.85 D. Reason is that CO2 has a linear structure, where the dipoles are equal and opposite, cancel each other effect. Whereas dipole moment of water cancel the linear structure of molecule. A linear H2O molecule could have zero dipole moment. Calculations show that water has a bent or angular structure with bond angle of 104.5O. μ = 0 D μ = 1.49 D 

Question. NH3 and BF3 are both tetra atomic molecules, but the dipole moment of BF3 is zero, while that of NH3 is 1.49D. Justify it. 

Answer. Dipole moment of BF3 is zero but that of NH3 is 1.49 D. Reason is that BF3 has a triangular planer structure, where the dipoles are equal and opposite, cancel each other effect. Whereas dipole moment of NH3 cancel the triangular planer structure of molecule. A triangular planer NH3 molecule could have zero dipole moment. Calculations show that NH3 has a pyramidal structure, with net dipole moment directed towards the direction of lone pair. μ = 1.49 D μ = 0 D 

Question. How the %age of ionic character of the polar bond be determined? 

Answer. For this purpose we require to know actual dipole moment μobs of molecule and bond length, which can be determine experimentally. The dipole moment μionic. of 100% ionic compound is calculated by multiplying full charge of electron or proton with bond length. % age ionic character can determined by following formula. % age ionic character = □((μobs )/(μionic ) x 100 ) 

Question. Why the dipole moment of CH4 is zero? 

Answer. Reason is that CH4 has a regular tetrahedral structure, where the dipoles are equal and opposite, cancel each other effect. 

Question. Why many of ionic compounds are not soluble in water? 

Answer. Although, most of ionic compounds are soluble in water but still there are many of ionic compounds are insoluble in water. Reason is that ionic compounds are only soluble in water when their hydration energy is greater than lattice energy. So the ionic compounds which are not soluble in water have lower hydration energy than lattice energy. 

Question. Why covalent compounds dissolve easily in non-polar solvent? 

Answer. Covalent compounds are only soluble in non polar solvent when attractive forces of solvent molecules with covalent compounds are enough to overcome the intermolecular forces of attraction. 

Question. Ionic compounds do not show isomerism and covalent compounds show isomerism. Why? Answer. As ionic compound involve just electrostatic forces between oppositely charged ions. Therefore, such bonds are non rigid and non directional. Due to this ionic compounds do not show isomerism. On the other hand, covalent compounds involve overlapping of atomic orbitals . therefore such bonds are rigid and directional. Due to this covalent compounds show isomerism. 


Comments

Popular posts from this blog

Class 9 Chemistry Notes for Federal Board

Class 9 Chemistry Notes for Federal Board Introduction Class 9 Chemistry notes for the Federal Board have been written according to the National Curriculum 2006. It develops various skills such as questioning, observations, predicting, measuring, classifying, designing experiments, analyzing and interpreting the data. among students.  We have made these notes very simple.  After reading these notes, students can easily understand all topics. Students can  answer all types of questions including  essay type, short and multiple choice, self assessments and think tank. We have also solved exercises according to the Federal board examination pattern.  Students, after going through 9th class chemistry notes can get maximum marks in their exams. They can lay down a strong foundation of scientific knowledge among students. In future, notes for class 9 will be available in pdf format. Students will be able to note pdf free download  very soon. Students will ha...

Class 10 Chemistry Notes for Federal Board

Class 10 Chemistry Notes for Federal Board Introduction Class 10 Chemistry Notes for Federal Board have been written according to  the National Curriculum 2006.  These notes will make students able to learn topics by their own. 10th class notes for all chapters are very and easy to understand. After studying these notes, students will become confident to answer all type of questions. Chemistry notes for class 10  provide solution to all kinds of  questions including essay type, short and multiple choice, self assessments and think tank. These notes also contain all past papers of reduced syllabus  as well as complete syllabus. Soon, students will have access to 10th class chemistry notes in Urdu pdf download form. We are also working hard to provide all chemistry notes in pdf format and student will be able to download these notes. We also trying to upload video lectures about class 10 chemistry. Don't forget, you can test your chemistry by using online test. Fu...

Theories of Covalent Bonding and Shapes of Molecules: MCQs

  chemicalAlgebra 1. The bond energies of F 2 , Cl 2 , Br 2 and I 2 are 37, 58, 46 and 36 Kcal, respectively. The strongest bond is formed in A) F 2 B) Cl 2 C) Br 2 D) I 2 ... Answer is B) Description: The bond energy is proportional to bond length. So, the bond strength of Cl 2 is higher. show Answer 2. The percentage of ionic character of bond between two atoms is calculated from the difference between their A) Ionic B) Electronegativities C) Electron affinities D) Ionization energies ... Answer is B) Description: The percentage of ionic character depends upon the difference of E.N. difference between the bonded atom. Greater the E. N. difference, greater the ionic character and hence greater the bond energy show Answer 3. The geometry of PF 5 molecule is A) Planer B) Square planer C) Trigonal pyramidal D) Tetrahedral ... Answer is C) Description: ...

Chemistry

T he Federal Board of Intermediate & Secondary Education (FBISE) is an autonomous body that conducts the examinations in the institutions affiliated with it in Pakistan as well as abroad.  Punjab government has ten boards of Intermediate & Secondary Education including Bahawalpur Board, D.G. Khan Board, Faisalabad Board, Gujranwala Board, Lahore Board, Multan Board, Rawalpindi Board and Sargodha Board and each board conducts exams independently.   Federal Board generally follows the international norms and  standards. Federal Board paper pattern is quite tricky and similar to O-Level and A Level formats. These questions need conceptual approach to be solved. Punjab board’s Paper pattern is comparatively easy up to some extent.  We are currently focusing on the subject of Chemistry for all classes affiliated with Federal Board ( IX ,  X ,  XI ,  XII )  and Punjab Board (  XI   XII ). We intend to extend it to all subjects i...

Basic Definitions| Matter, Substance, Element, Compound. Mixture, Homogeneous mixture and Heterogeneous mixture Examples (for Grade - 9 students)

Basic Definitions| Matter, Substance, Element, Compound. Mixture, Homogeneous mixture and Heterogeneous mixture Examples (for Grade - 9 students) Question: Define following terms:  Matter Substance Element Compound Mixture Homogeneous mixture Heterogeneous mixture Answer Matter:   Matter can be anything that occupies space (volume) and has mass. Example, book, table, walls, chairs, plants, animal, etc.  Substance:   Substance is a matter that has a particular set of characteristics that differ from the characteristics of another kind of matter. For example, oxygen, water, carbon monoxide, carbon dioxide, urea, glucose, common salt etc are different substances.  Element:   A substance that cannot be converted to other simpler substances is known as an element. or A substance which has all atoms of same atomic number is called element. For example substances like oxygen, hydrogen, iron, copper, aluminum etc. are elements.  Compound:  ...

Experimental Techniques in Chemistry

Analytical Chemistry: The branch of chemistry that deals with the quantitative and qualitative analysis of substances is called Analytical Chemistry. Question 2.1: What is difference between quantitative and quantitative analysis? What are various steps involved in quantitative analysis? Answer:   Qualitative analysis: The analysis in which elements are identified is called qualitative analysis.  Quantitative analysis: The analysis in the form of which relative amounts of elements are determined is called quantitative analysis.  There are four main steps involved in quantitative analysis. 1. Obtaining a sample for analysis 2. Separation of desired constituent 3. Measurement and calculations of results 4. Drawing conclusions from analysis Question 2.2: Define filtration. How do we select the filter media? Answer:   The process of separating solids or suspended particles from liquids by using a filter medium is called filtration. Question...

Alkyl halides

The compounds in which one hydrogen atom of alkane is replaced by a halogen atom is called alkyl halide. Alternatively, The halogen derivatives of alkanes with general formula R-X are Halide called alkyl halide. For example CH4  ( Alkane )            CH3-X  ( Alkyl halide )             X =   F, Cl, Br and I  Table Of Contents Types Mono, di, tri or poly haloalkane depending upon the number of halogen atom. For example C lassification of Alkyl Halides Alkyl halides can be classified as primary,  secondary and Tertiary alkyl halides. 1.  Primary Alkyl Halides The alkyl halides in which a halogen atom is bonded with primary carbon is called primary alkyl halide.  Primary carbon is the carbon that is attached  one or no carbon atom.    The Secondary Alkyl Halides The alkyl halides in which a halogen atom is bonded with secondary carbon ...

CHAPTER # 15 ORGANIC COMPOUNDS

  ORGANIC COMPOUNDS Modern definition. C ompounds of carbon, hydrogen and their derivatives, are called organic compounds. For example, CH 4  , CH 3 Cl etc. Contents Sources of organic Compounds:  Following are the main sources of organic compounds: i. Fossil fuels or Fossil remains.  ii. Plants and Natural product  iii. Partial &Total Synthesis iv.  Fermentation and Biotechnology Fossil fuels or Fossil remains:   The main sources of fossil fuels or fossil remains are:  a. coal b. petroleum c. gas a. Coal:  Coal is one of the organic compounds. Coal major source of into coke, and coal law" on pyrolysis Com can be converted destructive distillation. More than 200 organic compounds" have been directly isolated from coal tar, which are used. in ...drug. dyes, other. preparation of perfumes.. photographic developers and. b. Petroleum: Petroleum Sticky Liquid ground in that is block Thick seeps out of the s...

Shielding Effect (Grade 9 Students) and its trend in Periodic Table.

Question: What is shielding  effect? Give its trends in the periodic table. Answer . Shielding Effect  The decrease in force of attraction between nucleus  and valence electrons due to presence of inner shell electrons is called shielding effect.  It is also known as screening effect. It increases with increase in inner shell electrons. For example, Mg has greater shielding effect than Be. This is because Mg has eight more inner-shell electrons than Be. Trend in Periodic Table In period As we move from left to right in a period, shielding effect remains constant. Reason is that the number of inner shell electrons remains constant along the period. In group As we move from top to bottom in a group, shielding effect increases. This is due to fact that the number of inner shell increases and hence, inner shell electrons also increase. For example, Na has greater shielding effect than Li because of eight more inner shell electrons than Li...

Class 9 Electrochemistry MCQs and Answers with explanation

1. In which of the following changes, the nitrogen atom is reduced? a) N 2 to NO b) N 2 to NO 2 c) N 2 to NH 3 d) N 2 to HNO 3 ... Answer is b) Explanation: Oxidation number of an elemnet in free state is zero. So Oxidation No. of N 2 = 0 Calculate oxidation No. of nitrogen in each compound. Option a) Let, oxidation No. of N in NO is X. Sum of oxidation numbers of all atoms present in a neutral compound is zero. Oxidation No. of N + (Oxidation No. of O ) = 0 X-2 = 0 X = +2 X = +2 Oxidation No. of N in NO 2 is +2 Option b) Let, oxidation state of N in NO 2 is X. Sum of oxidation numbers of all atoms present in a neutral compound is zero. Oxidation No. of N + 2(Oxidation No. of O ) = 0 X + 2(-2) = 0 X -4 = 0 X = +4 Oxidation No. of N in NO 2 is +4 Option c) Let, oxidation state of N in NH 3 is X. Sum of oxidation numbers of all atoms present in a neutral compound is zero. Oxidation No. of N + 3(Oxidation No. of H )...

For All Topics Click here

Show more

Contact Form

Name

Email *

Message *