how to solve chemical bonding questions

Chemical Bonding – Common Questions

Here’s a compilation of various types of chemical bonding questions that keep repeating themselves over and over again! Please make sure you use the correct key words/phrases when answering the questions!

Example Questions 1.???????? Explain why sodium chloride has a very high melting point.

Ans: Sodium chloride is an ionic compound.?A lot of heat energy is needed to overcome the strong electrostatic forces of attraction between the oppositely ? charged ions (Na + and Cl – ), so sodium chloride has a high melting point.

2.???????? Explain why sodium chloride will conduct electricity when molten (aqueous).

Ans: When molten (aqueous), the ions are free to move throughout the liquid (solution). The mobile ions carry their charges with them as they move throughout the liquid (solution), thus allowing sodium chloride to conduct electricity.

3.???????? Explain why sodium chloride does not conduct electricity when solid.

Ans: When solid, the ions (Na + and Cl – ) are held in fixed positions by strong ionic bonds.? The ions are not able to move in the solid, so there are no mobile ions to carry the current, so sodium chloride does not conduct electricity when solid.

4. ? ? ? ? Explain why carbon dioxide has a low boiling point.

Ans: Carbon dioxide is a simple covalent compound. Little heat energy is needed to overcome the weak intermolecular forces of attraction (Van Der Waals forces of attraction) between the carbon dioxide molecules during boiling, so carbon dioxide has a low boiling point.

5.???????? Explain why carbon dioxide does not conduct electricity in any state.

Ans: Carbon dioxide is a covalent compound. There are no free moving ions and all the valence electrons are used for chemical bonding.?There are no free / mobile charged particles (ions and electrons) to conduct electricity.

Recent Posts

• ‘O’ Level Mole Revision Guide…
• [Qualitative analysis] Cher… please help!
• Copper… and its many different colours
• 3 colours that you need to know…
• 3 things you need to know about Acids

Class 10 Chemistry Chapter 2 Chemical Bonding Important Questions

Here, you'll discover significant inquiries pertaining to Chapter 2: Chemical Bonding for ICSE Class 10 Chemistry. These inquiries are carefully designed to aid students in preparing for the ICSE Class 10 Chemistry Examination in 2023–24. Engaging with different question formats allows students to address uncertainties, improve their exam preparedness, boost their self-assurance, and polish their ability to solve problems.

Introduction

What is chemical bonding , class 10 chemistry chapter 2 chemical bonding important questions and answers, icse class 10 chemistry chapter wise important questions, frequently asked questions.

In Class 10 Chemical Bonding, you will delve into the Electron Dot Structures of Electrovalent compounds like NaCl, MgCl 2 , and CaO. You'll also explore the distinctive properties of electrovalent compounds, including their physical states, melting and boiling points, conductivity (both heat and electricity), dissociation in solutions, and when in a molten state, which will be connected to electrolysis. Moreover, the chapter covers the Electron Dot Structures of covalent molecules based on the concepts of duplet and octet of electrons. Examples include hydrogen, oxygen, chlorine, nitrogen, ammonia, carbon tetrachloride, and methane. It will also discuss Polar Covalent compounds, which are determined by differences in electronegativity. Examples provided are HCl, NH 3 , and H 2 O, along with their respective structures. Furthermore, you'll learn about the characteristic properties of covalent compounds, such as their physical states, melting and boiling points, conductivity (heat and electricity), and ionisation in solution. A comparison between Electrovalent and Covalent compounds will be made. The chapter will define Coordinate Bonding and explain the lone pair effect observed in the oxygen atom of the water molecule and the nitrogen atom of the ammonia molecule. These explanations will help illustrate the formation of H3O + and OH - ions in water and the NH4+ ion. The concept of a lone pair will be clarified, and the formation of hydronium ions and ammonium ions will be elucidated using electron dot diagrams. If you're looking for chemical bonding class 10 ICSE important questions or questions on chemical bonding class 10, make sure to refer to the Oswal textbooks or oswal.io for comprehensive practice.

In ICSE Class 10 Chapter 2, "Chemical Bonding," the concept of chemical bonding comes to the forefront. This fundamental process involves the establishment of chemical connections among two or more atoms, molecules, or ions, ultimately culminating in the formation of a chemical compound. These chemical bonds play a pivotal role in maintaining the unity of the constituent elements within the resulting compound. The driving force behind this cohesion, which brings various components, such as atoms and ions, together and stabilises them by reducing the overall energy, is aptly termed chemical bonding. As we delve deeper into this topic, it becomes evident that the strength of these chemical bonds among constituents significantly influences the stability of the resulting compound. Stronger bonds contribute to greater stability, ensuring the compound's durability. Conversely, weak chemical bonding between constituents results in diminished stability, rendering the compound susceptible to further reactions aimed at producing more stable compounds with stronger bonds. In their pursuit of stability, atoms strive to minimise their energy levels. In interactions between different forms of matter, forces come into play. When these forces are attractive, they lead to an energy reduction. Conversely, repulsive forces drive an increase in energy. The attractive force responsible for binding two atoms together is recognized as a chemical bond. Thus, the study of chemical bonding forms the cornerstone of understanding the intricate relationships between atoms, molecules, and ions, influencing the stability and reactivity of compounds. For questions on chemical bonding class 10 ICSE important questions, you should refer to your class materials and textbooks for a comprehensive set of practice questions and exercises.

Q1. In the given figure how many pairs of bonds are contained in the Hydrogen molecule Electron dot structures of Non-polar covalent compound ?

(a) Four (b) Five (c) Six (d) One

Ans . (d) One

Explanation: In the given figure, one pair of bonds contains in the hydrogen molecule Electron dot structures of Non-polar covalent compound. The electronegativity difference is zero (diatomic molecule) and type of bonding is Non-polar covalent single bond.

Q2. A polar covalent bond will be formed in which one of these pair of atoms:

(a) hf (b) h 2 (c) cl 2 (d) o 2.

Ans . (a) HF Explanation: HF is a polar molecule in which the H-F bond is a polar covalent bond due to unequal sharing of electrons between more electronegative F and less electronegative H atoms.

Q3. There are three elements E, F, G with atomic numbers 19, 8 and 17 respectively. (i) Classify the elements as metals and non-metals. (ii) Give the molecular formula of the compound formed between E and G and state the type of chemical bond in this compound.

Explanation: (i) 19E is a metal.8F and 17G are non-metals. (ii) Molecular formula - EG ‍ Type of bond - Ionic bond

Q4. How is a coordinate bond formed?

Explanation: The sharing of an electron pair from a single atom results in the formation of coordinate bonds, a sort of alternative covalent link. The same atom is responsible for both shared electron contributions.

Q5. Write the important characteristics of covalent compounds.

Explanation: The characteristics of covalent compounds 1. Covalent compounds can be found in the form of gases, liquids, or soft solids. 2. Covalent compounds typically have low melting and boiling points. 3. Insoluble in water, covalent compounds dissolve in organic solvents. 4. In a solid, molten, or liquid form, they don't conduct electricity.

The exploration of "Chemical Bonding" in ICSE Class 10 Chemistry has provided a fundamental understanding of the forces that bind atoms, molecules, and ions together to form chemical compounds. The study of chemical bonds, both strong and weak, is paramount in comprehending the stability and reactivity of these compounds.Throughout this chapter, we've delved into the intricacies of attractive and repulsive forces that govern the behaviour of matter at the atomic and molecular levels. For those seeking to excel in this critical area of chemistry, additional practice resources can be invaluable. Oswal.io offers a comprehensive collection of questions and study materials tailored to enhance your learning experience.

Q1 : Why do atoms react, and how?

Ans: Atoms achieve stability when they possess eight electrons in their outermost orbit, which eliminates their inclination to engage in chemical reactions. On the other hand, atoms with fewer than eight electrons tend to interact with other atoms to attain an octet of electrons in their outermost orbit, thereby achieving stability. Atoms with a slight surplus of more than eight electrons may relinquish some to atoms deficient in electrons. Atoms that are unable to either lose or gain electrons may engage in electron sharing to attain an octet configuration. In cases where molecules still lack an octet configuration even after the reaction, they have the option to accept lone pairs of electrons from other atoms or molecules.

Q2: Name the forces that keep reacting atoms together.

Ans:  In metals, there is an overlapping of outer orbitals of atoms, causing the electrons within them to be shared among multiple atoms rather than being associated with a specific atom. This shared electron behaviour is what binds all the atoms together, a phenomenon known as metallic bonding. When atoms undergo a process of losing and gaining electrons, they transform into ions and are held together by the electrostatic forces of attraction, which is referred to as an ionic bond. In cases where atoms equally contribute and share electrons, these shared electrons become the binding force that holds the atoms together, forming a covalent bond.In situations involving electron-deficient or molecules containing free lone pairs, they can satisfy the electron-deficient atom's octet requirement. The shared electrons serve as a bridge between the electron-rich atom and the electron-deficient atom, creating a coordinate bond.

Q3 : What are hybridised orbitals? What are their uses?

Ans: Sub-orbitals with reasonably comparable energy levels can combine to create a fresh set of orbitals with the same count as the contributing orbitals. These newly formed orbitals are known as hybridised orbitals. They serve a valuable purpose in elucidating the resemblances observed in the bond length, bond angles, structure, shape, and magnetic characteristics of molecules, with these properties being proportional to the contributing orbitals' numbers.

Q4 : sp 3 and dsp 2 are four hybridised orbitals. But one is a tetrahedral shape, and the other is square planar. Why?

Ans:  Sp 3 orbitals originate from the s-subshell, where electrons are evenly spread around the nucleus, and the p-subshell, with electrons distributed along three perpendicular axes. Consequently, hybridised orbitals possess electron distribution in three-dimensional space, aligning with tetrahedral directions. In the case of dsp 2 hybridization, all participating orbitals share the same plane for their electron distribution. Consequently, the resultant hybridised orbitals also lie within the same plane, leading to the formation of a square planar geometry.

Q5 : The oxygen molecule is paramagnetic. Is there an explanation?

Ans: An oxygen atom forms an oxygen molecule by sharing two electrons with another oxygen atom. Oxygen molecules display paramagnetism, indicating the presence of unpaired electrons. To explain this phenomenon, a molecular orbital theory has been introduced. According to this theory, atoms relinquish their individual orbitals and instead create an equivalent number of orbitals that encompass the entire molecule, hence the term "molecular orbital." The process of filling these orbitals in ascending energy levels results in the presence of unpaired electrons, which accounts for the paramagnetic characteristics observed in oxygen molecules.

Chapter Wise  Important Questions for ICSE Board Class 10 Chemistry

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Phone: (+91) 7895987722 ‍ Mail: [email protected]

• Privacy policy
• Terms & Conditions
• Oswal Publishers
• Books with Oswal.io
• Download solutions

Need Any Help?

Feel free to connect with us for any query

Head Office

Ground Floor, 1 / 12 Sahitya Kunj, M.G Road Agra – 282002,Uttar Pradesh (India) +91-562-25277711/2/3/4 or +91-7455077222 [email protected]

Corporate Office

Noida One, B 702, B7, Sector 62, Noida, Uttar Pradesh 201309 +91-745 507 7222 [email protected]

Copyright 2022 OSWAL PUBLISHERS Simplifying Exams

Lets Connect

• chemical-bonding-quiz

Chemical Bonding Quiz

Chemical bonding - practice problems with solutions.

• PRO Courses Guides New Tech Help Pro Expert Videos About wikiHow Pro Upgrade Sign In
• EDIT Edit this Article
• EXPLORE Tech Help Pro About Us Random Article Quizzes Request a New Article Community Dashboard This Or That Game Popular Categories Arts and Entertainment Artwork Books Movies Computers and Electronics Computers Phone Skills Technology Hacks Health Men's Health Mental Health Women's Health Relationships Dating Love Relationship Issues Hobbies and Crafts Crafts Drawing Games Education & Communication Communication Skills Personal Development Studying Personal Care and Style Fashion Hair Care Personal Hygiene Youth Personal Care School Stuff Dating All Categories Arts and Entertainment Finance and Business Home and Garden Relationship Quizzes Cars & Other Vehicles Food and Entertaining Personal Care and Style Sports and Fitness Computers and Electronics Health Pets and Animals Travel Education & Communication Hobbies and Crafts Philosophy and Religion Work World Family Life Holidays and Traditions Relationships Youth
• Browse Articles
• Learn Something New
• Quizzes Hot
• This Or That Game
• Train Your Brain
• Explore More
• Support wikiHow
• About wikiHow
• Log in / Sign up
• Education and Communications

How to Balance Chemical Equations

Last Updated: October 13, 2023 Fact Checked

This article was co-authored by Bess Ruff, MA . Bess Ruff is a Geography PhD student at Florida State University. She received her MA in Environmental Science and Management from the University of California, Santa Barbara in 2016. She has conducted survey work for marine spatial planning projects in the Caribbean and provided research support as a graduate fellow for the Sustainable Fisheries Group. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 4,945,834 times.

Taking a dive into the world of chemical equations? These problems can seem tricky at a glance, but they’re easy to figure out once you learn the basic steps and rules to balancing them. Not to worry; we’ll walk you through exactly how to figure out just about any problem, no matter how many atoms and molecules you're working with. Dealing with especially complex equations? We’ve got you covered there, too—scroll to section 2 for a handy tutorial on solving trickier equations with an algebraic balance.

Doing a Traditional Balance

• C 3 H 8 + O 2 --> H 2 O + CO 2
• This reaction occurs when propane (C 3 H 8 ) is burned in the presence of oxygen to produce water and carbon dioxide.

• For example, you have 3 oxygen atoms on the right side, but that total results from addition.
• Left side: 3 carbon (C3), 8 hydrogen (H8) and 2 oxygen (O2).
• Right side: 1 carbon (C), 2 hydrogen (H2) and 3 oxygen (O + O2).

• You'll need to recount your atoms before balancing the hydrogen and oxygen, as you'll likely need to use coefficients to balance the other atoms in the equation.

• C 3 H 8 + O 2 --> H 2 O + 3 CO 2
• The coefficient 3 in front of carbon on the right side indicates 3 carbon atoms just as the subscript 3 on the left side indicates 3 carbon atoms.
• In a chemical equation, you can change coefficients, but you must never alter the subscripts.

• C 3 H 8 + O 2 --> 4 H 2 O + 3CO 2
• On the right side, you now added a 4 as the coefficient because the subscript showed that you already had 2 hydrogen atoms.
• When you multiply the coefficient 4 times by the subscript 2, you end up with 8.

• Add a coefficient of 5 to the oxygen molecule on the left side of the equation. You now have 10 oxygen atoms on each side.

• The carbon, hydrogen, and oxygen atoms are balanced. Your equation is complete.
• The other 6 atoms of oxygen come from 3CO 2 .(3x2=6 atoms of oxygen+ the other 4=10)

Completing an Algebraic Balance

This method, also known as Bottomley's method, is especially useful for more complex reactions, although it does take a bit longer.

• PCl 5 + H 2 O --> H 3 PO 4 + HCl

• a PCl 5 + b H 2 O --> c H 3 PO 4 + d HCl

• On the left side, there are 2 b atoms of hydrogen (2 for every molecule of H 2 O), while on the right side, there are 3 c + d atoms of hydrogen (3 for every molecule of H 3 PO 4 and 1 for every molecule of HCl). Since the number of atoms of hydrogen has to be equal on both sides, 2 b must be equal to 3 c + d .
• Cl: 5 a = d
• H: 2 b =3 c + d

• To quickly do this, take one variable and assign a value to it. Let's make a = 1. Then start solving the system of equations to get the following values:
• Since P: a = c, we know that c = 1.
• Since Cl: 5a = d, we know that d = 5
• 2b = 3(1) + 5
• If the value you assigned returns fractional values, just multiply all values by the least common multiple (LCM) of the denominators to get rid of the fractions. If there is only one fraction, multiply all values by that values denominator.
• If the value you assigned returns coefficients that have a greatest common factor (GCF), simplify the chemical equation by dividing each value by the GCF.

Community Q&A

• Remember to simplify! If all of your coefficients can be divided by the same number, do so to get the simplest result. Thanks Helpful 46 Not Helpful 11
• If you're stuck, you can type the equation into the online balancer to balance it. Just remember that you won't have access to an online balancer when you're taking an exam, so don't become dependent on it. Thanks Helpful 55 Not Helpful 24
• To get rid of fractions, multiply the entire equation (both the left and right sides) by the number in the denominator of your fraction. Thanks Helpful 28 Not Helpful 11

• During the balancing process, you may use fractions to assist you, but the equation is not balanced as long as there are still coefficients using fractions. You never make half of a molecule or half of an atom in a chemical reaction. Thanks Helpful 27 Not Helpful 12

You Might Also Like

• ↑ https://www.khanacademy.org/science/ap-biology/chemistry-of-life/elements-of-life/a/matter-elements-atoms-article
• ↑ https://www.mathsisfun.com/algebra/add-subtract-balance.html
• ↑ https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/balancing-chemical-equations/v/balancing-chemical-equations-introduction
• ↑ https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Balancing_Redox_reactions
• ↑ https://www.mathsisfun.com/algebra/completing-square.html

About This Article

To balance a chemical equation, first write out your given formula with the reactants on the left of the arrow and the products on the right. For example, your equation should look something like "H2 + O2 → H2O." Count the number of atoms in each element on each side of the equation and list them under that side. For the equation H2 + O2 → H2O, there are 2 hydrogen atoms being added to 2 oxygen atoms on the left, so you would write "H=2" and "O=2" under the left side. There are 2 hydrogen atoms and 1 oxygen atom on the right, so you would write "H=2" and "O=1" under the right side. Since the number of atoms in each element isn't identical on both sides, the equation is not balanced. To balance the equation, you'll need to add coefficients to change the number of atoms on one side to match the other. For the equation H2 + O2 → H2O, you would add the coefficient 2 before H2O on the right side so that there are 2 oxygen atoms on each side of the equation, like H2 + O2 → 2H2O. However, subscripts can't be changed and are always multiplied by the coefficient, which means there are now 4 hydrogen atoms on the right side of the equation and only 2 hydrogen atoms on the left side. To balance this, add the coefficient 2 before H2 on the left side of the equation so there are 4 hydrogen atoms on each side, like 2H2 + O2 → 2H2O. Now the number of atoms in each element is the same on both sides of the equation, so the equation is balanced. Remember that if there's no coefficient in front of an element, it's assumed that the coefficient is 1. To learn how to balance chemical equations algebraically, scroll down! Did this summary help you? Yes No

• Send fan mail to authors

Reader Success Stories

Mar 11, 2022

Did this article help you?

E. M. Brown

Dec 3, 2016

Mar 23, 2022

Theresa Hill

Nov 21, 2017

Featured Articles

Trending Articles

Watch Articles

• Terms of Use
• Privacy Policy
• Do Not Sell or Share My Info
• Not Selling Info

wikiHow Tech Help Pro:

Develop the tech skills you need for work and life

• Chemistry Concept Questions and Answers

Bond Energy Questions

Bond Energy is also known as the average Bond Enthalpy and gives a measure of the strength of a chemical bond. The larger the bond energy the more will be the stability of the molecules.

Bond Energy Chemistry Questions with Solutions

Q1: How can we determine the strength of a bond by the radius of an atom?

Answer: Any of the radius i.e. metallic radius, ionic radius or covalent radius can be used to determine the strength of bond formed by the given atoms.

For example: Boron (B) has a covalent radius of 83 pm. It makes a B-B bond of bond length 175 pm in B 2 Cl 4 . This indicates that the B-B bond length is even higher than the radius of an atom of B. Thus, this is a weak bond. However, in the case of Rhenium (Re), the metallic radius is 137.5 pm. The Re-Re bond length in Re 2 Cl 8 is 224 pm. This indicates that the Re-Re is a very strong quadruple bond.

Q2. What is the difference between a primary and secondary bond? Give 2 examples of each.

Answer: A Primary bond is formed by the mutual sharing or complete transfer of electrons. This type of bond is stronger than the secondary bond. For example: O 2 , NaCl.

A secondary bond is formed either due to permanent dipole or due to certain imbalances in the symmetrical charge distribution within the atom- creating a dipole. These dipoles induce a charge imbalance in their immediate neighbouring atoms. For example: N 2 , H 2 O.

Q3. How is the melting temperature of solids related to the bond energy?

Answer: The melting temperature is associated with the breaking of bonds of solid substances. The higher the melting temperature, the stronger the bond. The stronger the bond, the higher the bond energy. Hence, the melting temperature of solids has a direct relationship with the bond energy.

Q4. What is the enthalpy change for the given reaction?

CH 3 CH 3 + Cl 2 → CH 3 CH 2 Cl + HCl

Given the enthalpies:

Answer: The total enthalpy for a reaction is calculated by the formula:

ΔH o = ⅀H bonds broken – ⅀H bonds formed

ΔH o = (413 + 239) kJ mol -1 – (339 + 427) kJ mol -1

ΔH o = -114 kJ mol -1

Q5. Choose the correct statement.

• A (-) ΔH reaction indicates that energy is released.
• A (+) ΔH reaction indicates that energy is released.
• A (-) ΔH reaction indicates that energy is absorbed.
• A (-) ΔH reaction indicates that the reaction is endothermic.

Answer: (a.)

Explanation: In an exothermic reaction, the total energy change in reactants and products comes out to be negative because the products’ energy is lesser than the reactants energy. Hence, (-) ΔH reaction indicates the release of energy.

Q6. Pick the false statement.

• Single bonds are more easily broken than double bonds.
• Double bonds are more easily broken than triple bonds.
• Triple bonds are more easily broken than double bonds.
• The bonds formed in the endothermic reactions are weaker than the bonds broken.

Answer: (c.)

Explanation: Triple bonds are stronger than double bonds. Hence, triple bonds cannot be more easily broken than double bonds.

Q7. What is the enthalpy change in the following decomposition reaction of HCl?

2HCl → H 2 + Cl 2

The average bond enthalpies (kj mol -1 ) for the concerned bonds are:

H-Cl = 431 kj mol -1

Cl-Cl = 242 kj mol -1

H-H = 436 kj mol -1

ΔH o = ⅀H Bonds broken – ⅀H Bonds formed

ΔH o = 2(431) kj mol -1 – (242 + 436) kj mol -1

ΔH o = +185 kJ mol -1

Hence, the reaction is endothermic.

Q8. Why does FeCl 3 have a greater covalent character than FeCl 2 ?

Answer: This is because the charge on Fe in FeCl 3 is greater i.e. +3 than in FeCl 2 i.e. +2. Due to this reason, the size of the Fe +3 ion is smaller than Fe +2 . Hence, Fe +3 ion creates more polarisation and hence, more neutralization of charge takes place. Therefore, the ionic character of FeCl 3 decreases and the covalent character increases.

Q9. The combustion of 1 g Graphite produces 20.7 kJ heat. Calculate its molar enthalpy change.

Answer: Molar enthalpy change is calculated as:

Molar enthalpy change for combustion of 1 g Graphite = Enthalpy of Combustion of 1 g Graphite x Molar Mass

Enthalpy of Combustion of 1 g Graphite = -20.7 kJ g -1 (energy is released)

Since graphite (C) is made from carbon, its molar mass is 12 g mol -1

ΔH combustion = -20.7 kJ g -1 x 12 g mol -1

ΔH combustion =-2.48 x 10 2 kJ mol -1

This reaction is exothermic in nature.

Q10. Predict the spontaneity of the reaction:

N 2 O 4 (g) ⇌ 2NO 2 (g); Given K P = 0.98 and T = 298 K.

Answer: From the relationship between standard free energy change and the equilibrium constant, we have:

Δ r G o = -RT lnK P

Since K P = 0.98, Δ r G o comes out to be negative. This indicates that the reaction is spontaneous.

Q11. Why is the molar enthalpy of vaporisation of acetone less than that of water?

Answer: In this, the inter-molecular forces come into play. Since the water molecules have strong H-bonding, water has higher molar enthalpy of vaporisation.

Q12. Comment on the bond energies of 4 C-H bonds in CH 4 .

Answer: Bond energy is the amount of energy required to break 1 mole of bonds present in between the gaseous molecules. Since the bond energies of 1 st , 2 nd , 3 rd and 4 th C-H bonds in CH 4 are not equal. An average value of the bond energies are taken.

Q13. Predict the enthalpy change for the isomerization of acetonitrile into methyl isocyanide.

Answer: The isomerization of acetonitrile into methyl isocyanide can be written into equation as:

H 3 C-CN → H 3 C-NC

In this, a C-C bond is broken and a new C-N bond is formed.

Bond dissociation energy for C-C = 347 kJ mol -1

Bond dissociation energy for C-N = 305 kJ mol -1

ΔH o = (347 -305) kJ mol -1

ΔH o = 42 kJ mol -1

The reaction is endothermic.

Q14. Name 3 ions that are isoelectronic to Neon.

Answer: Neon has 10 electrons. The ions isoelectronic to Neon are Na + , Mg 2+ , and F – .

Q15. Out of NaCl and KCl, which will have more exothermic lattice energy?

Answer: The anions in both salts are the same. Both the cations have the same charge but differ in size. Since, Na + is smaller than K + , NaCl has more exothermic lattice energy.

Practise Questions on Bond Energy

Q1. Which of the following alkali metals has the least melting point?

Q2. Why does Calcium have an abnormally higher boiling point than magnesium?

Q3. Name the energy associated when 1 mole of an ionic compound is formed from gaseous ions.

Q4. In Spite of the large electronegativity gap between Be and O in the BeO molecule, BeO is covalent. Why?

Q5. What are the decomposition products of Group II nitrates?

Click the PDF to check the answers for Practice Questions. Download PDF

Recommended Videos

Types of bonds.

Energy Change in Bond formation

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

• Share Share

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons

Margin Size

• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

5: Covalent Bonding & Simple Molecular Compounds: Questions

• Last updated
• Save as PDF
• Page ID 44590

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)