Words and Expressions Class 10 Solutions Unit 1 A Letter to God

Words and Expressions Class 10 Solutions Unit 1 A Letter to God

Let’s Begin:

Question 1.
You have read about Lencho in ‘A Letter to God’, First Flight, textbook in English for Class X. How has the story of Lencho moved you ? What do you think about Lencho’s faith in God and his attitude towards the officials in the post office ? Do you think he is right in asking God for support and then blaming the people at the post office ? Discuss his actions with your friends. Why does he do so ?
Answer:
Lencho was a poor farmer. He was quite hardworking too, an ox of a man. Due to severe rain and hailstorm his ripe corn crops were badly damaged and ruined. He got worried and anxious at this grave loss as that meant that his family would have to go hungry that year. Lencho had a firm faith in the Almighty and believed that only God could help him out of this bad situation. With great hopes he decided to write a letter to God to compensate him for the loss that he had suffered. It was naivety and sheer innocence on part of Lencho to ask and expect monetary and financial help from God. Lencho was a sincere, responsible and hardworking fellow.

He had devoted his utmost and sincerest efforts in sowing his crop fields. He could not imagine and believe all his efforts going waste in the wink of an eye. As a responsible fellow he was really concerned about the well being of his family and couldn’t bear to see them embroiled in any kind of difficulty. It was out of care, compassion towards his loved ones that he tried to do something unusual and out of the ordinary i.e. writing a letter to an invisible force that he thought is omni present. I feel great sympathy for the poor fellow as he had to suffer the loss for reasons that were out of his control. However, I also feel that it was quite inappropriate and unreasonable on part of Lencho to blame the post-office staff for the inadequate amount of money that he had received. The post office employees did nothing wrong but tried to help the poor fellow by doing their best. Lencho should have thanked God and the post-office staff for receiving such great assistance at the time of need.

Question 2.
The summary of the story of Lencho is given below. The sequence of the story is jumbled up.
Read the story again and rearrange the jumbled sentences to make a coherent story.

Write the correct order in numbers in the column below Jumbled sentences
(i) But it rained heavily and hailstorm thrashed his corn field and all the corn was destroyed.
(ii) He wrote ‘To God’ on the envelope and posted it.
(iii) Lencho predicted it will rain and it did. He and his wife were happy.
(iv) Lencho felt desperate but believed that God would help him.
(v) Their corn field was longing for water.
(vi) The postman and the postmaster laughed at Lencho’s letter but were surprised by the faith of the man.
(vii) He wrote a letter to God asking for a hundred pesos to sow his field again and to survive until the next crop came up.
(viii) After some days, Lencho went to the post office to receive the money as the postmaster watched him.
(ix) He immediately went to the counter and asked for a pen and paper, wrote a few lines and posted the letter.
(x) Lencho, his wife, and their children lived in a lonely house in a valley.
(xi) When the postmaster opened the letter, it said “Oh! Thanks for sending the money. But I have received only seventy pesos. Next time you please send the money directly to me. The people at the post office are a bunch of crooks. They have taken the thirty pesos.
(xii) He opened the letter and was happy to find the money. He counted and found it was only seventy pesos.
(xiii) The postmaster collected about 70 pesos from his colleagues and sent it to Lencho.

Answer:
(i) 4
(ii) 7
(iii) 3
(iv) 5
(v) 2
(vi) 8
(vii) 6
(viii) 10
(ix) 12
(x) 1
(xi) 13
(xii) 11
(xiii) 9.

Create a graphic description based on the given summary of ‘A Letter to God’.
A. Lencho in his corn field.
Words and Expressions Class 10 Solutions Unit 1 A Letter to God 1
B. Lencho writing letter to God asking for compensation.
Words and Expressions Class 10 Solutions Unit 1 A Letter to God 2
C. The Post-master surprised to read the letter and the other staff also astonished.
Words and Expressions Class 10 Solutions Unit 1 A Letter to God 3
D. Post office employees helping Lencho by providing monetary help that he had asked for.
Words and Expressions Class 10 Solutions Unit 1 A Letter to God 4
E. Lencho surprised on receiving a sum less than what he had asked for; decides to write another letter to God complaining about the same.
Words and Expressions Class 10 Solutions Unit 1 A Letter to God 4a

Reading Comprehension:

Text – I
Given below is a story of the tailor who has been known for his skills in cheating and stealing. Read in groups of four or in pairs. You may discuss the events and incidences of the story with your friends and write them in a sequential manner.

The Trader and the Tailor:
There was a tailor who was known for beating others in the art of being light-fingered and thievery. A trader swore that even with a hundred attempts, the tailor would not be able to take a coil of thread from him without his knowledge.

The trader was told that many others, more intelligent than him, had been beaten by the tailor. They said he should not imagine himself so great, for his ego would only give him worse troubles in competition with the tailor. Still, the trader became more competitive and made a wager that the tailor would not be able to rob him of anything.

He wagered an Arab horse with those who taunted him. If the tailor failed to steal, then they would have to give him a horse instead. That night, the trader lay awake troubled by the situation and could not sleep a wink.

In the morning, he put a piece of satin cloth under his arm and went to the bazaar and entered the shop and saluted the cunning rogue warmly. The tailor sprang up from his seat and welcomed the trader, inquiring about his health with cordiality exceeding even that of the trader, planting in his heart great feelings of affection for him. When the trader heard these songs of sweetness from the tailor, he flung down the piece of silk saying, “Cut this into a coat for me and make it wide below my navel so as not to hamper my legs and tight about it to show off my figure.”

The tailor answered, “O! kindly man, I will do you a hundred services,” and accepted the order. He measured the satin and inspected the working surface and all the time chatted away to the trader in idle gossip, about other amirs and of bounties and gifts he had received from them, and about misers and their mean ways, and made the trader laugh with hysterics. During this beguiling talk, he was snipping away with his scissors rapidly, cutting as fast as his lips moved.

The trader was laughing, the tailor was cutting, the trader closed his eyes in joy, the tailor cut extra pieces, tucking them under his thighs, hidden away from all but God. From his delight at the tailor’s tales, the trader’s former boast went out of his mind. What satin? what boast? What wager? The trader was drunk on the jokes told by the tailor.

Then the tailor told such an incredible story that the trader fell over on his back with laughter. The tailor swiftly stitched a swatch of the satin to the hem of his underpants while the trader was paying no attention at all, greedily sucking with every guffaw at the jests the tailor told. The tailor continued to tell funnier and funnier tales and jokes until the trader was completely within his power.

With his eyes shut and his reason vanished, the bewildered and boastful trader was drunk with joy. And the tailor continued to cut, filching yet more of the beautiful cloth, with nothing to stop him now. And yet the trader begged for more.

You who becomes the slave of the jest, no story is more laughable than you yourself, think on this at the edge of your grave. How long will you listen to the lies of this world that leave your mind and spirit unhinged ? The Universal Tailor will cut and stitch the hems of a hundred travellers, silly as children.

Eventually the tailor became bored with the whole story and told the trader that he better leave before another tale was told, “For if I tell another one, the coat will be too tight for you, and you will stop laughing and weep tears of blood.”
by Rumi (Abridged)

Did you enjoy the story of ‘The Trader and the Tailor* ? The story is summarised below. Read the story again with your friend and complete the paragraph using short sentences with the given hints.

Once there was a tailor who was known for ___________ (cheat and thievery). He could cheat anyone with his cunning way of deceiving people. A trader challenged that the tailor would not be able to cheat him. He ___________ (go) to the tailor with ___________ (satin). The tailor ___________ (welcome). The trader (please) by the tailor’s praise. He told the tailor to make a coat for him. The tailor ___________ (praise) the trader (maximum). The trader ___________ (joy). The tailor ___________(recount) funny tales and the trader forgot ___________ (bet). The tailor started snipping ___________ and ___________ hide). Eyes of the trader ___________ (tears and joy) and yet the trader kept on asking for ___________ (stories). At last, the tailor told the trader that (tight) and ___________ (weep-blood).
Answer:
beating others in the art of cheating and thievery, went, a piece of satin tucked under his arm, welcomed him with great cordiality, was very pleased, praised, to his maximum, experienced great joy, recounted innumerable, about the bet completely, the cloth rapidly, tucking in the extra pieces under his thighs hiding from the trader, more and more, if he told another story the coat would be too tight for him, he would weep tears of blood.

Vocabulary:

Question 1.
In this story you have come across a word “light-fingered” which means inclined to steal things. Some such words are given below. Use appropriate words and fill in the blanks.
light-headed, light year, light-hearted, light sleeper, light-duty
(a) The ___________ is a unit of distance used to measure distance in space.
(b) The farmer bought a ___________ truck to transport their produce, as it uses less dies.el.
(c) We had a fairly ___________ discussion with our colleagues. .
(d) Rita felt dizzy and ___________ after skydiving.
(e) He is a ___________, a slightest noise wakes him.
Answer:
(a) light year
(b) light-duty
(c) light-hearted
(d) light-headed
(e) light sleeper

Grammar:

Question 1.
You have come across Relative Clauses in the lesson ‘A Letter to God’. Read about the relative clauses again. Complete the sentences given below in the table by adding the most suitable clause from the box and make it a relative clause. The first one has been done for you.
contains meaning of word stitches clothes tell lies won the best actor award this year hit a century is faithful need to fill in the application is a crook help themselves won the nobel prize believe in hard work.
(а) A tailor is a person who stitches clothes.
(b) A dictionary is a book ___________
(c) The book is about a scientist ___________
(d) What is the name of the player ___________
(e) I don’t like people ___________
(f) What is the name of the actor ___________
(g) Students ___________ should meet the principal in her office.
(h) Friends of the trader thought the tailor ___________ and can cheat anyone.
(i) Lencho ___________ thought God would help him.
(j) Those ___________ will succeed.
(k) God helps those ___________
Answer:
(b) which contains meaning of words
(c) who won the nobel prize
(d) who hit a century
(e) who tell lies
(f) who won the best actor award this year
(g) who need to fill in the application
(h) who is a crook
(i) who is faithful
(j) who believe in hard work
(k) who help themselves

Editing:

Question 1.
Suppose Lencho gets a hundred pesos in his envelope. He writes a letter to God expressing his gratitude. However, he has missed a word in each line. Help him with those words so that his letter is complete.
Dear God!
Thank for – Thank you for
(а) sending a hundred pesos help. ___________
(b) I am very for this kind gesture. ___________
(c) You were my only and your ___________
(d) help has my faith in you. ___________
(e) This money will help me until the crop ___________
(f) and I promise to hard in the future. ___________
Sincerely yours,
Lencho
Answer:
(a) sending me a
(b) very grateful for
(c) only hope and
(d) has increased my
(e) crop comes/grows
(f) to work hard.

Listening:

Question 1.
Here is an interesting story. The story has been told in two ways. Story A has all the odd sentences (1,8,5,…) and story B has the even sentences (2,4,6,…). Working in pairs, one of you take up Story A and the other take up Story B. The person with Story A will read out the first sentence and the partner (person with story B) will write in the space provided. Then the person with Story B will read sentence two and the partner will write in the space for sentence two. Likewise complete the whole story. Once the story is completed, both of you will read out to each other to check whether you have listened carefully to your partner and written the complete story.
Story – A
The Race:
1. Every evening the animals of the Machlipatti used to sit under the banyan tree.
2. _____________________
3. He said, “Let’s run a race!” “Who will run a race with you ?
4. _____________________
5. Choo-Choo, the rat, shouted, “No, he’s not the fastest runner.
6. _____________________
7. Everyone laughed at him.
8. _____________________
9. The starting point was the peepul tree and the finishing point was the neem tree.
10. _____________________
11. Kho-Kho and Choo-Choo started running.
12. _____________________
13. “He must be far behind.”
14. _____________________
15. He shouted, “I’ve won the race! I’ve won the race!”
16. _____________________

Story – B
The Race:
1. _____________________
2. One day Kho-Kho, the rabbit, was feeling bored.
3. _____________________
4. No one can run faster than you,” said Bhalu, the bear, eating a mango.
5. _____________________
6. I can run faster than him.”
7. _____________________
8. Then it was decided to have a race between Kho-Kho and Choo-Choo.
9. _____________________
10. Coocku, the cock, crowed to start the race.
11. _____________________
12. After running half the distance, Kho-Kho looked back but Choo-Choo was nowhere to be seen.
13. _____________________
14. When he was near the finishing point, Choo-Choo jumped over Kho-Kho’s shoulder and touched the neem tree first.
15. _____________________
16. Everyone cheered for Choo-Choo.

Note : Do it yourself.

Speaking:

Question 1.
Lencho, the main character in the story, is a farmer. He supports his family through farming. The conflict in the story happens when a hailstorm strikes the field and destroys his farm. All the hard work, for nothing! The entire land is wrecked by catastrophe. There is no one who can help. Lencho’s complete faith in God is amazing during the calamity.
So, one type of conflict dominating in the story is man versus nature. Discuss with your partner the natural calamities that play havoc in the lives of people. Collect photographs and make a visual presentation to share in the class.
Answer:
A natural disaster is a major adverse event resulting from natural processes of the Earth. Examples are : floods, hurricanes, tornadoes, volcanic eruptions, tsunamis and other geologic processes.

A natural disaster can cause loss of life or damage property and typically leaves some economic damage in its wake the severity of which depends on the affected population’s resilience and also on the infrastructure available.

Earthquake : An earthquake is the result of a sudden release of energy in the Earth’s crust that creates seismic waves. At the Earth’s surface, earthquakes manifest themselves by vibration, shaking, and sometimes displacement of the ground. Earthquakes by themselves rarely kill people or wildlife. It is usually the secondary events that they trigger such as building collapse, fires, tsunamis and volcanoes. Many of these could possibly be avoided by better construction, safety systems, early warning and planning.
Words and Expressions Class 10 Solutions Unit 1 A Letter to God 5

Flood : A flood is an overflow of water that ‘submerges’ land. Flooding may result from the volume of water within a body of water, such as a river or lake, which overflows, causing some of the water to escape its boundaries.
Words and Expressions Class 10 Solutions Unit 1 A Letter to God 6
Floods have large social consequences for communities and individuals. As most people are well aware, the immediate impacts of flooding include loss of human life, damage to property, destruction of crops, loss of live stock, and deterioration of health condi tions owing to waterborn diseases.

Question 2.
Try writing a small poem on the topic man versus nature. Read aloud the poem with the required stress and intonation. Given below is a poem written by Norman Littleford, which may help you think and compose.
Man vs Nature:
The heavens roared with thunder
as lightning filled the skies
was this God getting angry
or Nature’s big surprise.
Is Mother Nature telling us
the best way that she can
to stop the interference
and abuse she gets from man.
We marvel at her beauty
each time we look around
then dig up all her treasures
from their natural burial ground.
We forget that Nature gives us
all that keeps us living
we take it all and still want more
but never think of giving.
We build across the countryside
progressing every day
but Mother Nature can’t progress
for man is in the way.
by Norman Lulls ford
Answer:
Man vs Nature:
Mother Nature cradles the human life
giving endless care for humanity to thrive.
Pouring generously from its womb,
the elixir of life.
But man’s never ending torture is rife,
his ever growing greed pushing the mother
to her tomb.
Man is wild, an erroneous child.
By ripping the nature of all its resources,
he is planting the seeds of his own remorses.
Mother Nature is loving and forgiving
and could only curse herself for her faulty upbringing.
She weeps in silent tears,
but doesn’t complain about her wears.
It’s time man gives nature its due respect
and doesn’t mar it for its ulterior prospects
He must realise and stop being naive…
With his selflish acts he is digging his own grave. – Moina

Writing:

Question 1.
The tailor here is a cheat, but Lencho in the story, ‘A Letter to God’, is naive and simple. He has immense faith in God. How do you distinguish between Lencho and the tailor? What makes them different ? Working in pairs, list the actions and qualities of Lencho and the tailor and draw their character sketches. The given cues will help you to draw a character sketch of these characters.

Action: What does the person do ? What quality it reveals?
Lencho lost all his crops. He is helpless. Prays to God ___________________
How does he manage the situation ? ___________________
Does he harm anyone by his action ? ___________________
How do others think of him ? ___________________
What is your judgement of him ? ___________________
He received 70 pesos and he thought 30 pesos had been stolen. ___________________

Now, write the character sketch of Lencho and the tailor.
Answer:
Character sketch of Lencho:
Lencho was a poor farmer. He was hardworking. He was a simpleton and a naive. He had an optimistic attitude towards life. He was a firm believer of God and its omnipresence. He had an ufflinching and unwavering faith in the Almighty. He was a person full of hope. When hi’s crop fields were badly damaged, due to hailstorm, he didn’t lose hope instead demonstrated great patience and determination. He had a fighting attitude. Instead of getting all disheartened and dismounted at his loss he fought bravely with the situation at hand by devising an optimum solution. His writing a letter to God showcased his child like innocence. He was a responsible person and was greatly concerned about the well being of his family as was clearly seen in his earnest efforts to find a solution to his problem.

Character sketch of The tailor:
The tailor in the story ‘The Trader and the Tailor’ was a mean and a dishonest person. He was a trickster and often indulged in thievery and robbing people of one thing or the other. He was quick witted and smart and knew exactly how to exploit the weaknesses of others to his advantage. He invariably got successful in outwitting others with his cleverness and presence of mind. He was greedy and was known as someone who wouldn’t not leave any opportunity to steal from people. The story projects and highlights his great skills of storytelling. He would weave one story after the other in order to keep the trader glued to him so that he could work his way towards stealing without getting noticed or caught.

Question 2.
Writing a portrait or description of a person
Read the following description about two persons, Raavi and Mohini.

Raavi Mohini
A portrait or a picture of a handsome young man in full sleeves shirt. A portrait or a picture of a middle aged lady who is a singer.
Raavi is an engineer in a manufacturing firm. She studied music and became a playback singer.
He likes to make, fix, and repair things. She likes to be with people and participates in social functions.
He loves to watch television and goes to the movies thrice a month. She loves music and reading books.
He is congenial and supportive. She is firm and a lady with a purpose.
He does not like to play, but goes for a morning walk everyday. She does not get time to exercise.

Now, write a description of any one of the above based on the details given. You may follow the process approach to writing. First, you read the given information in the box and make a list of describable (adjectives) points used for each person, in your own words. Secondly, take the help of these points to develop an outline in sentence form. Thirdly, use the outline and prepare the first draft of the description of the person and then edit the draft. Prepare the final copy after making all changes. The table below gives you ideas on how the Process Approach is helpful for improving your writing.
Writing: The Process Approach

The processes What do you do ?
Brainstorming Note down as many ideas as you can about a given topic. Then arrange the ideas in a sequence.
Outlining Create an outline of your writing – which ideas will go first, how will they be supported by evidence, will there be a picture, etc.
Drafting Write the first draft; here the focus is on the content and not the form.
Revising Revise the first draft, improve it. Add or delete ideas and improve the language.
Proofreading Proofread the draft with the help of the teacher; here the focus is on the form and not the content.
Final draft Write the final draft now.

Answer:
Raavi
Raavi is a handsome young man. He is highly educated and a professional human being. He works as an engineer in a manufacturing firm. He is intelligent, creative and resourceful. He is innovative and relishes making his own things, repairing and fixing the items that need mending and repair to a fully operational state. He has a great sense of clothing and style and likes to dress up well on all occasions. He is a movie buff and likes to go for movies whenever he gets time off from work. He loves to watch television. He is a people person and a peace lover. He has a compassionate and sympathetic attitude towards others. Although he is not much into play he is quite health-conscious and a fitness-freak. He is a nature lover and enjoys going for a morning walk everyday to keep his mind and body in a healthy state.

Project:

Question 1.
You may have come across some people in your neighbourhood who might have inspired you with their actions. Talk to them about –

  • their childhood, their education, job, their likes and dislikes
  • their contribution to the enrichment of their village or town, job, country, etc.
  • their interesting features and characteristics

Request for their pictures. Based on the collected information, write their portraits and present it to the class. You may read on ‘How to do project work ?’ given below.

How to do project work in the English language classroom?
In addition to the textbook activities, project work could be used as an effective strategy to promote collaborative ways of language learning where we work together in groups to do an activity. We collect information, ideas; observe the language being used; use language in real life situations; exchange views and debate on ideas; write the ideas into a report; edit our writing and produce the report in a suitable form. This helps in constructing ideas and analysing critical judgments of the peers to arrive at a decision or conclusion.

It would be of much benefit to learners, if projects planned are of an interdisciplinary nature.

Designing and managing projects:
There are many ways students can be initiated to carry out projects.

  • Read books, newspapers, listen to the radio, watch TV, consult websites and encyclopedia, talk to people, observe events and proceedings, interview people, record audio/ video, collect pictures and illustrations, interpret them, and report.
  • Write field notes to remember and organise the data collected.
  • Bring your own ideas to the class and have a class discussion.
  • Use language meaningfully and precisely. If the meaning of learning is ‘meaning making’, the purpose is achieved fully while doing projects.
  • Learners should do the projects in group independently. They should not ask their parents to do these or buy readymade products available in the market.
  • They may seek information or some help from parents or elders.

Planning, writing the draft and editing:
Having collected the required data, information and ideas, plan and organise the data and interpret them to write the draft. It could be posters, a script of a play (enact and direct a play), songs, essays, etc. Editing is an important part of writing the report. Students should be made aware of process approach involved in writing a good piece of work. Then present it to the whole class or to the school during the assembly or on special occasions (in case of play, songs, etc.).
Answer:
Mr. Pratap Singh (Brigadier General) lives in my neighbourhood. He is 65 years of age and is a retired army personnel. He stays at his current residence with his wife who served as an assistant professor (chemistry) in Delhi University. His two sons are married and happily settled abroad. Both his sons are engineering graduates from premier institutes. After completing his schooling from Delhi Public School, Noida, in 1st division, Mr. Pratap Singh joined Delhi university’s reputed college, St. Stephens to pursue a degree course in B.Sc. In his last year’at college he got interested in joining the Indian Army. He appeared for an Entrance test for the same and qualified the test in his first attempt. It was a moment of great pride for him as well as for his family. After completing his graduation he joined the Indian army as lieutinent.

Hailing from an educated and disciplined family he had always lived a very disciplined and strict life. His parents always made sure that he recieved a good upbringing so that he could learn to shoulder responsibilities early in life. As a child he was always encouraged to accomplish all his tasks by himself with guided assistance from his parents. Mr. Pratap Singh is a very jovial and fun loving person. He has a great persona. He appears quite a determined, hardworking and disciplined man. He is always found abounding in positivity. He learnt to have a perfect work-life balance. He is a great husband, father and a very successful Army Officer.

He is very fond of hiking and reading books of different genre. Himself being a modest fellow and a man of integrity he dislikes dishonesty and lack of sympathy in people. I am a great admirer of Mr Singh like many others in our locality. I truly appreciate him for his great contribution in setting up a sports complex in our area so that people can benefit from it in staying fit and healthy. He is a true gentleman and will always be remembered for his greatness.

Words and Expressions Class 10 Solutions

Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4

Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4

Class 11 Chemistry Notes students can refer to the Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4https://www.cbselabs.com/chemical-bonding-molecular-structure-cbse-notes-class-11-chemistry/ Pdf here. They can also access the CBSE Class 11 Chemical Bonding and Molecular Structure Chapter 4 Notes while gearing up for their Board exams.

Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4

• Chemical Bond
The force that holds different atoms in a molecule is called chemical bond.
• Octet Rule
Atoms of different elements take part in chemical combination in order to complete their octet or to attain the noble gas configuration.
• Valence Electrons
It is the outermost shell electron which takes part in chemical combination.
Chemical Bonding Class 11
• Facts Stated by Kossel in Relation to Chemical Bonding

— In the periodic table, the highly electronegative halogens and the highly electro-positive alkali metals are separated by noble gases.
— Formation of an anion and cation by the halogens and alkali metals are formed by gain of electron and loss of electron respectively.
— Both the negative and positive ions acquire the noble gas configuration.
— The negative and positive ions are stabilized by electrostatic attraction Example,
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-1
• Modes of Chemical Combination
— By the transfer of electrons: The chemical bond which formed by the complete transfer of one or more electrons from one atom to another is termed as electrovalent bond or ionic bond.
— By sharing of electrons: The bond which is formed by the equal sharing of electrons between one or two atoms is called covalent bond. In these bonds electrons are contributed by both.
— Co-ordinate bond: When the electrons are contributed by one atom and shared by both, the bond is formed and it is known as dative bond or co-ordinate bond.
Class 11 Chemistry Chapter 4 Notes
• Ionic or Electrovalent Bond

Ionic or Electrovalent bond is formed by the complete transfer of electrons from one atom to another. Generally, it is formed between metals and non-metals. We can say that it is the electrostatic force of attraction which holds the oppositely charged ions together.
The compounds which is formed by ionic or electrovalent bond is known as electrovalent compounds. For Example, ,
(i) NaCl is an electrovalent compound. Formation of NaCl is given below:
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-2
Na+ ion has the configuration of Ne while Cl ion represents the configuration of Ar.
(ii) Formation of magnesium oxide from magnesium and oxygen.
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-3
Electrovalency: Electrovalency is the number of electrons lost or gained during the formation of an ionic bond or electrovalent bond.
Chemical Bonding And Molecular Structure
• Factors Affecting the Formation of Ionic Bond

(i) Ionization enthalpy: As we know that ionization enthalpy of any element is the amount of energy required to remove an electron from outermost shell of an isolated gaseous atom to convert it into cation.
Hence, lesser the ionization enthalpy, easier will be the formation of a cation and have greater chance to form an ionic bond. Due to this reason alkali metals have more tendency to form an ionic bond.
For example, in formation of Na+ ion I.E = 496 kJ/mole
While in case of magnesium, it is 743 kJ/mole. That’s why the formation of positive ion for sodium is easier than that of magnesium.
Therefore, we can conclude that lower the ionization enthalpy, greater the chances of ionic bond formation.
(ii) Electron gain enthalpy (Electron affinities): It is defined as the energy released when an isolated gaseous atom takes up an electron to form anion. Greater the negative electron gain enthalpy, easier will be the formation of anion. Consequently, the probability of formation of ionic bond increases.
For example. Halogens possess high electron affinity. So, the formation of anion is very common in halogens.
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-4
(iii) Lattice energy or enthalpy: It is defined as the amount of energy required to separate 1 mole of ionic compound into separate oppositely charged ions.
Lattice energy of an ionic compound depends upon following factors:
(i) Size of the ions: Smaller the size, greater will be the lattice energy.
(ii) Charge on the ions: Greater the magnitude of charge, greater the interionic attraction and hence higher the lattice energy.
Chapter 4 Chemistry Class 11 Notes
• General Characteristics of ionic Compounds

(i) Physical’State: They generally exist as crystalline solids, known as crystal lattice. Ionic compounds do not exist as single molecules like other gaseous molecules e.g., H2 , N2 , 02 , Cl2 etc.
(ii) Melting and boiling points: Since ionic compounds contain high interionic force between them, they generally have high melting and boiling points.
(iii) Solubility: They are soluble in polar solvents such as water but do not dissolve in organic solvents like benzene, CCl4etc.
(iv) Electrical conductivity: In solid state they are poor conductors of electricity but in molten state or when dissolved in water, they conduct electricity.
(v) Ionic reactions: Ionic compounds produce ions in the solution which gives very fast reaction with oppositely charged ions.
For example,
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-5
• Covalent Bond—Lewis-Langmuir Concept
When the bond is formed between two or more atoms by mutual contribution and sharing of electrons, it is known as covalent bond.
If the combining atoms are same the covalent molecule is known as homoatomic. If they are different, they are known as heteroatomic molecule.
For Example,
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-6
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-7

How to calculate bond order is simple and corresponds to the number of bonds.

Class 11 Chemistry Chemical Bonding And Molecular Structure Notes Pdf
• Lewis Representation of Simple Molecules (the Lewis Structures)

The Lewis dot Structure can be written through the following steps:
(i) Calculate the total number of valence electrons of the combining atoms.
(ii) Each anion means addition of one electron and each cation means removal of one electron. This gives the total number of electrons to be distributed.
(iii) By knowing the chemical symbols of the combining atoms.
(iv) After placing shared pairs of electrons for single bond, the remaining electrons may account for either multiple bonds or as lone pairs. It is to be noted that octet of each atom should be completed.
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-8
Notes Of Chemical Bonding Class 11
• Formal Charge

In polyatomic ions, the net charge is the charge on the ion as a whole and not by particular atom. However, charges can be assigned to individual atoms or ions. These are called formal charges.
It can be expressed as
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-9
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-10

Get the free “Valence Electrons Calculator” widget for your website, blog, WordPress, … calculates the valence electrons of any element.

Class 11th Chemistry Chapter 4 Notes
• Limitations of the Octet Rule

(i) The incomplete octet of the central atoms: In some covalent compounds central atom has less than eight electrons, i.e., it has an incomplete octet. For example,
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-11
Li, Be and B have 1, 2, and 3 valence electrons only.
(ii) Odd-electron molecules: There are certain molecules which have odd number of electrons the octet rule is not applied for all the atoms.
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-12
(iii) The expanded Octet: In many compounds there are more than eight valence electrons around the central atom. It is termed as expanded octet. For Example,
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-13
Class 11 Chapter 4 Chemistry Notes
• Other Drawbacks of Octet Theory

(i) Some noble gases, also combine with oxygen and fluorine to form a number of compounds like XeF2 , XeOF2 etc.
(ii) This theory does not account for the shape of the molecule.
(iii) It does not give any idea about the energy of The molecule and relative stability.
• Bond Length
It is defined as the equilibrium distance between the centres of the nuclei of the two bonded atoms. It is expressed in terms of A. Experimentally, it can be defined by X-ray diffraction or electron diffraction method.
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-14
• Bond Angle
It is defined as -the angle between the lines representing the orbitals containing the bonding – electrons.
It helps us in determining the shape. It can be expressed in degree. Bond angle can be experimentally determined by spectroscopic methods.
Chemical Bonding Notes Pdf
• Bond Enthalpy

It is defined as the amount of energy required to break one mole of bonds of a particular type to separate them into gaseous atoms.
Bond Enthalpy is also known as bond dissociation enthalpy or simple bond enthalpy. Unit of bond enthalpy = kJ mol-1
Greater the bond enthalpy, stronger is the bond. For e.g., the H—H bond enthalpy in hydrogen is 435.8 kJ mol-1.
The magnitude of bond enthalpy is also related to bond multiplicity. Greater the bond multiplicity, more will be the bond enthalpy. For e.g., bond enthalpy of C —C bond is 347 kJ mol-1 while that of C = C bond is 610 kJ mol-1.
In polyatomic molecules, the term mean or average bond enthalpy is used.
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-15
• Bond Order
According to Lewis, in a covalent bond, the bond order is given by the number of bonds between two atoms in a molecule. For example,
Bond order of H2 (H —H) =1
Bond order of 02 (O = O) =2
Bond order of N2 (N = N) =3
Isoelectronic molecules and ions have identical bond orders. For example, F2 and O22- have bond order = 1. N2, CO and NO+ have bond order = 3. With the increase in bond order, bond enthalpy increases and bond length decreases. For example,
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-16
• Resonance Structures
There are many molecules whose behaviour cannot be explained by a single-Lew is structure, Tor example, Lewis structure of Ozone represented as follows:
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-17
Thus, according to the concept of resonance, whenever a single Lewis structure cannot explain all the properties of the molecule, the molecule is then supposed to have many structures with similar energy. Positions of nuclei, bonding and nonbonding pairs of electrons are taken as the canonical structure of the hybrid which describes the molecule accurately. For 03, the two structures shown above are canonical structures and the III structure represents the structure of 03 more accurately. This is also called resonance hybrid.
Some resonating structures of some more molecules and ions are shown as follows:
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-18
Notes Of Chemistry Class 11 Chapter 4
• Polarity of Bonds

Polar and Non-Polar Covalent bonds
Non-Polar Covalent bonds: When the atoms joined by covalent bond are the same like; H2, 02, Cl2, the shared pair of electrons is equally attracted by two atoms and thus the shared electron pair is equidistant to both of them.
Alternatively, we can say that it lies exactly in the centre of the bonding atoms. As a result, no poles are developed and the bond is called as non-polar covalent bond. The corresponding molecules are known as non-polar molecules.
For Example,
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-19
Polar bond: When covalent bonds formed between different atoms of different electronegativity, shared electron pair between two atoms gets displaced towards highly electronegative atoms.
For Example, in HCl molecule, since electronegativity of chlorine is high as compared to hydrogen thus, electron pair is displaced more towards chlorine atom, thus chlorine will acquire a partial negative charge (δ) and hydrogen atom have a partial positive charge (δ+) with the magnitude of charge same as on chlorination. Such covalent bond is called polar covalent bond.
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-20
• Dipole Moment
Due to polarity, polar molecules are also known as dipole molecules and they possess dipole moment. Dipole moment is defined as the product of magnitude of the positive or negative charge and the distance between the charges.
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-21
• Applications of Dipole Moment
(i) For determining the polarity of the molecules.
(ii) In finding the shapes of the molecules.
For example, the molecules with zero dipole moment will be linear or symmetrical. Those molecules which have unsymmetrical shapes will be either bent or angular.
(e.g., NH3with μ = 1.47 D).
(iii) In calculating the percentage ionic character of polar bonds.
• The Valence Shell Electron Pair Repulsion (VSEPR) Theory
Sidgwick and Powell in 1940, proposed a simple theory based on repulsive character of electron pairs in the valence shell of the atoms. It was further developed by Nyholm and Gillespie (1957).
Main Postulates are the following:
(i) The exact shape of molecule depends upon the number of electron pairs (bonded or non bonded) around the central atoms.
(ii) The electron pairs have a tendency to repel each other since they exist around the central atom and the electron clouds are negatively charged.
(iii) Electron pairs try to take such position which can minimize the rupulsion between them.
(iv) The valence shell is taken as a sphere with the electron pairs placed at maximum distance.
(v) A multiple bond is treated as if it is a single electron pair and the electron pairs which constitute the bond as single pairs.
• Valence Bond Theory
Valence bond theory was introduced by Heitler and London (1927) and developed by Pauling and others. It is based on the concept of atomic orbitals and the electronic configuration of the atoms.
Let us consider the formation of hydrogen molecule based on valence-bond theory.
Let two hydrogen atoms A and B having their nuclei N and N and electrons present in them are eA and e.
As these two atoms come closer new attractive and repulsive forces begin to operate.
(i) The nucleus of one atom is attracted towards its own electron and the electron of the other and vice versa.
(ii) Repulsive forces arise between the electrons of two atoms and nuclei of two atoms. Attractive forces tend to bring the two atoms closer whereas repulsive forces tend to push them apart.
• Orbital Overlap Concept
According to orbital overlap concept, covalent bond formed between atoms results in the overlap of orbitals belonging to the atoms having opposite spins of electrons. Formation of hydrogen molecule as a result of overlap of the two atomic orbitals of hydrogen atoms is shown in the figures that follows:
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-22
Stability of a Molecular orbital depends upon the extent of the overlap of the atomic orbitals.
• Types of Orbital Overlap
Depending upon the type of overlapping, the covalent bonds are of two types, known as sigma (σ ) and pi (π) bonds.
(i) Sigma (σ bond): Sigma bond is formed by the end to end (head-on) overlap of bonding orbitals along the internuclear axis.
The axial overlap involving these orbitals is of three types:
• s-s overlapping: In this case, there is overlap of two half-filled s-orbitals along the internuclear axis as shown below:
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-23
• s-p overlapping: This type of overlapping occurs between half-filled s-orbitals of one atom and half filled p-orbitals of another atoms.
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-24
• p-p overlapping: This type of overlapping takes place between half filled p-orbitals of the two approaching atoms.
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-25
(ii) pi (π bond): π bond is formed by the atomic orbitals when they overlap in such a way that their axes remain parallel to each other and perpendicular to the internuclear axis.The orbital formed is due to lateral overlapping or side wise overlapping.
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-26
• Strength of Sigma and pf Bonds
Sigma bond (σ bond) is formed by the axial overlapping of the atomic orbitals while the π-bond is formed by side wise overlapping. Since axial overlapping is greater as compared to side wise. Thus, the sigma bond is said to be stronger bond in comparison to a π-bond.
Distinction between sigma and n bonds
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-27
• Hybridisation
Hybridisation is the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies resulting in the formation of new set of orbitals of equivalent energies and shape.
Salient Features of Hybridisation:
(i) Orbitals with almost equal energy take part in the hybridisation.
(ii) Number of hybrid orbitals produced is equal to the number of atomic orbitals mixed,
(iii) Geometry of a covalent molecule can be indicated by the type of hybridisation.
(iv) The hybrid orbitals are more effective in forming stable bonds than the pure atomic orbitals.
Conditions necessary for hybridisation:
(i) Orbitals of valence shell take part in the hybridisation.
(ii) Orbitals involved in hybridisation should have almost equal energy.
(iii) Promotion of electron is not necessary condition prior to hybridisation.
(iv) In some cases filled orbitals of valence shell also take part in hybridisation.
Types of Hybridisation:
(i) sp hybridisation: When one s and one p-orbital hybridise to form two equivalent orbitals, the orbital is known as sp hybrid orbital, and the type of hybridisation is called sp hybridisation.
Each of the hybrid orbitals formed has 50% s-characer and 50%, p-character. This type of hybridisation is also known as diagonal hybridisation.
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-28
(ii) sp2 hybridisation: In this type, one s and two p-orbitals hybridise to form three equivalent sp2 hybridised orbitals.
All the three hybrid orbitals remain in the same plane making an angle of 120°. Example. A few compounds in which sp2 hybridisation takes place are BF3, BH3, BCl3 carbon compounds containing double bond etc.
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-29
(iii) sp3 hybridisation: In this type, one s and three p-orbitals in the valence shell of an atom get hybridised to form four equivalent hybrid orbitals. There is 25% s-character and 75% p-character in each sp3 hybrid orbital. The four sp3 orbitals are directed towards four corners of the tetrahedron.
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-30
The angle between sp3 hybrid orbitals is 109.5°.
A compound in which sp3 hybridisation occurs is, (CH4). The structures of NH2 and H20 molecules can also be explained with the help of sp3 hybridisation.
• Formation of Molecular Orbitals: Linear Combination of Atomic Orbitals (LCAO)
The formation of molecular orbitals can be explained by the linear combination of atomic orbitals. Combination takes place either by addition or by subtraction of wave function as shown below.
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-31
The molecular orbital formed by addition of atomic orbitals is called bonding molecular orbital while molecular orbital formed by subtraction of atomic orbitals is called antibonding molecular orbital.
Conditions for the combination of atomic orbitals:
(1) The combining atomic orbitals must have almost equal energy.
(2) The combining atomic orbitals must have same symmetry about the molecular axis.
(3) The combining atomic orbitals must overlap to the maximum extent.
• Types of Molecular Orbitals
Sigma (σ) Molecular Orbitals: They are symmetrical around the bond-axis.
pi (π) Molecular Orbitals: They are not symmetrical, because of the presence of positive lobes above and negative lobes below the molecular plane.
• Electronic configuration and Molecular Behaviour
The distribution of electrons among various molecular orbitals is called electronic configuration of the molecule.
• Stability of Molecules
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-32
• Bond Order
Bond order is defined as half of the difference between the number of electrons present in bonding and antibonding molecular orbitals.
Bond order (B.O.) = 1/2 [Nb-Na]
The bond order may be a whole number, a fraction or even zero.
It may also be positive or negative.
Nature of the bond: Integral bond order value for single double and triple bond will be 1, 2 and 3 respectively.
Bond-Length: Bond order is inversely proportional to bond-length. Thus, greater the bond order, smaller will be the bond-length.
Magnetic Nature: If all the molecular orbitals have paired electrons, the substance is diamagnetic. If one or more molecular orbitals have unpaired electrons, it is paramagnetic e.g., 02 molecule.
• Bonding in Some Homonuclear (Diatomic) Molecules
(1) Hydrogen molecule (H2): It is formed by the combination of two hydrogen atoms. Each hydrogen atom has one electron in Is orbital, so, the electronic configuration of hydrogen molecule is
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-33
This indicates that two hydrogen atoms are bonded by a single covalent bond. Bond dissociation energy of hydrogen has been found = 438 kJ/mole. Bond-Length = 74 pm
No unpaired electron is present therefore,, it is diamagnetic.
(2) Helium molecule (He2): Each helium atom contains 2 electrons, thus in He2 molecule there would be 4 electrons.
The electrons will be accommodated in σ1s and σ*1s molecular orbitals:
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-34
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-35
• Hydrogen Bonding
When highly electronegative elements like nitrogen, oxygen, flourine are attached to hydrogen to form covalent bond, the electrons of the covalent bond are shifted towards the more electronegative atom. Thus, partial positive charge develops on hydrogen atom which forms a bond with the other electronegative atom. This bond is known as hydrogen bond and it is weaker than the covalent bond. For example, in HF molecule, hydrogen bond exists between hydrogen atom of one molecule and fluorine atom of another molecule.
It can be depicted as
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-36
• Types of H-Bonds
(i) Intermolecular hydrogen bond (ii) Intramolecular hydrogen bond.
(i) Intermolecular hydrogen bond: It is formed between two different molecules of the same or different compounds. For Example, in HF molecules, water molecules etc.
(ii) Intramolecular hydrogen bond: In this type, hydrogen atom is in between the two highly electronegative F, N, O atoms present within the same molecule. For example, in o-nitrophenol, the hydrogen is in between the two oxygen atoms.
Chemical Bonding and Molecular Structure Class 11 Notes Chemistry Chapter 4 img-37

Class 11 Chemistry Notes

Hydrocarbons Class 11 Notes Chemistry Chapter 13

Hydrocarbons Class 11 Notes Chemistry Chapter 13

Class 11 Chemistry Notes students can refer to the Hydrocarbons Class 11 Notes Chemistry Chapter 13 https://www.cbselabs.com/hydrocarbons-cbse-notes-class-11-chemistry/ Pdf here. They can also access the CBSE Class 11 Hydrocarbons Chapter 13 Notes while gearing up for their Board exams.

Hydrocarbons Class 11 Notes Chemistry Chapter 13

Hydrocarbons Class 11 Notes Chapter 13

• Hydrocarbon
A compound of carbon and hydrogen is known as hydrocarbon.
• Saturated Hydrocarbon
A hydrocarbon is said to be saturated if it contains only C—C single bonds.
For example: Ethane CH3—CH
• Unsaturated Hydrocarbon
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-1
• Aromatic Hydrocarbon
Benzene and its derivatives are called aromatic compounds.
Example:
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-2
Hydrocarbon Notes Class 11
• Alicyclic Compounds

Cyclic compounds which consist only of carbon atoms are called alicyclic or carboeyclic compounds.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-3
Notes Of Hydrocarbons Class 11
• Heterocyclic Compounds

Cyclic compounds in which the ring atoms are of carbon and some other element (For example, N, S, or O) are called heterocyclic compounds.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-4
Hydrocarbon Class 11 Notes

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• Alkanes
Alkanes are the simplest organic compounds made of carbon and hydrogen only.
They have the general formula CnHC2n+2 (where n = 1, 2, 3, etc.)
The carbon atoms in their molecules are bonded to each other by single covalent bonds. Since the carbon skeleton of alkanes is fully saturated’ with hydrogens, they are also called saturated hydrocarbons. Alkanes contain strong C —C and C —H bonds. Therefore, this class of hydrocarbons are relatively chemically inert. Hence they are sometimes referred to as paraffins (Latin parum affinis = little affinity). First three members of this class can be represented as
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-5
Structure:
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-6
In methane carbon forms single bonds with four hydrogen atoms. All H—G—H bond angles are of 109.5°. Methane has a tetrahedral structure. C—C and C—H bonds are formed by head-on overlapping of sp3 hybrid orbitals of carbon and Is orbitals of hydrogen atoms.
Hydrocarbons Notes
• Nomenclature Guidelines

Use the following step-by-step procedure to write the IUPAC names from the structural formulas. Consider the following structural formula:
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-7
Step 1. Identify the longest chain: In the given example, longest chain has seven carbons. The seven carbon chain is heptane.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-8
Step 2. Number the chain: The chain is numbered from left to right. This gives lowest numbers to the attached alkyl group.
Step 3. Identify the alkyl group: There are two methyl groups at C-2 and C-3, there is one ethyl group of C-4.
Step 4. Write the IUPAC name: In this case the IUPAC name is 4-Ethyl-2,3-dimethyl heptane. Always keep in mind (a) Numbers are separated from each other by commas. (b) Numbers are separated from names by hyphens, (c) Prefixes di, tri are not taken into account in alphabetising substituent names.
• Newman Projections
In this projection, the molecule is viewed at the C—C bond head on.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-9
• Relative Stability of Conformations
In staggered form of ethane there are maximum repulsive forces, minimum energy and maximum stability of molecule. On the other hand, when the staggered form changes in the eclipsed form the electron clouds of the carbon hydrogen bonds come closer to each other resulting in increase in electron cloud repulsions, molecule have to possess more energy and thus has lower stability.
Torsional Angle: Magnitude of torsional strain depends upon the angle of rotation about C—C bond. This angle is also called dihedral angle or torsional angle.

Addition to Alkyne and Alkene Reactions Cheat Sheet, Cheat Sheet for Organic Chemistry.

Chapter 13 Chemistry Class 11 Notes
• Alkenes

Alkenes are hydrocarbons that contain a carbon-carbon double bond (C=C) in their molecule.
They have the general formula
Structure:
Let us consider (H2C=CH2) for illustrating the orbital make up of alkenes.
In ethylene the carbon atoms are sp2 hybridized- They are attached to each other by a a bond and a σ bond.
The a bond results from the overlap of two sp2 hybrid orbitals. The π bond is formed from overlap of the unhybridized p-orbitals. Ethylene is a planar molecule.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-10
Ppints to be noted
(i) The carbon-carbon double bond in alkenes is made up of one σ and one π-bond.
(ii) Alkenes are more reactive than Alkanes. This is due to the availability of n electrons.
Hydrocarbons Class 11 Pdf Notes
• Nomenclature

In IUPAC system
(i) The name of the hydrocarbon is based on the parent alkene having the longest ‘ carbon chain of which double bond is apart.
(ii) This chain is numbered from the end near the double bond and its position is indicated by the number of the carbon atom not which the double bond originates,
(iii) The name of the parent alkene with the position number of the double bond is written first and then the names of other substituents prefixed to it.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-11
(iv) When there are two or three double bonds in a molecule, the ending-one of the corresponding alkane is replaced by-a diene to get the name.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-12
Ch 13 Chemistry Class 11 Notes
• Isomerism

Structural Isomerism: Ethene and propene have no structural isomers, but there are three structures of butenes.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-13
Of these, two are straight chain structures with the difference being in the position of double bond in the molecules.
These are position isomers and third structure is a branched-chain isomer.
Geometrical Isomerism: It is known that a carbon-carbon double bond is made up of one σ bond and one π-bond. The π-bond presents free rotation about the double bond.
This presentation of rotation about the carbon-carbon double bond gives rise to the phenomenon of geometrical isomerism. An alkene having a formula RCH=CHR can have two stereoisomers, depending upon whether the two alkyl groups are on the same or opposite sides of the double bond. If they are on the same side, then it is called cis-isomer. If they are on opposite sides, then it is called trans-isomer.
Due to different arrangement of atoms or groups in space, these isomers differ in their properties like melting point, boiling point, dipole moment, solubility, etc.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-14
Class 11 Chemistry Ch 13 Notes
• Alkynes

Alkynes are characterised by the presence of a triple bond in the molecule.
Their general formula is CnH2n-2.
The first and the most important member of this series of hydrocarbons is acetylene, HC=CH, and hence they are also called the Acetylenes.
Structure: Let us consider ethyne (HC=CH) for illustrating the orbital make up of ethyne. In ethyne, the carbon atoms are sp hybridized. They are attached to each other by a σ-bond and two π-bonds.
The σ -bond results from the overlap of two sp hybrid orbitals. The π bonds are formed from the separate overlap of the two p-orbitals from the two adjacent carbon atoms.
The other sp hybrid orbital of each carbon atom forms a σ bond with another carbon or hydrogen atom. Ethyne is a linear molecule.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-15
Points to be noted:
(i) The carbon-carbon triple bond in alkynes is made up of one σ and two π bonds.
(ii) Like alkenes, alkynes undergo addition reaction. These reactions are due to the availability of more exposed π electrons.
• Nomenclature
IUPAC System: The IUPAC names of alkynes are obtained by dropping the ending-ane of the parent alkane and adding the suffix-yne. Carbon chain including the triple bond is – numbered from the end nearest this bond. The position of the triple bond is indicated by prefixing the number of carbon preceding it to the name of the alkyne.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-16
Preparation:
From calcium carbide: Ethyne is prepared by treating calcium carbide with water. Calcium carbide is prepared as follows:
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-17
From vicinal dihalides: When reacted with vicinal dihalides, alcoholic potassium hydroxide undergo dehydrohalogenation. One molecule of hydrogen halide is eliminated to form alkenyl halide which on treatment with sodamide gives alkyne.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-18
Class 11 Chemistry Chapter Hydrocarbon Notes
• Aromatic Hydrocarbons

These hydrocarbons are also known as ‘arenes’. Most of such compounds were found to contain benzene ring.
Aromatic compounds containing benzene ring are known as benzenoids and those not containing a benzene ring are known as non-benzenoids. Some examples of arenes are given below.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-19
Nomenclature and Isomerism: Benzene and its homologous are generally called by their common names which are accepted by the IUPAC system. The homologous of benzene having a single alkyl group are named as Alkyl benzenes.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-20
Structure of Benzene: By elemental analysis, it is found that molecular formula of benzene is C6H6. This indicates that benzene is a highly unsaturated compound. In 1865, Kekule gave the cyclic planar structure of benzene with six carbons with alternate double and single bonds.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-21
The Kekule structure indicates the possibility of two isomeric 1,2-dibromobenzenes. In one of the isomers, the bromine atoms are attached to the doubly bonded carbon atoms whereas in the other they are attached to the singly bonded carbon.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-22
In fact, only one ortho-dibromobenzene could be prepared.
To overcome this problem Kekule suggested that benzene was a mixture of two forms.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-23
Failure of Kekule’s structure: Kekule structure of benzene failed to explain the unique stability and its preference to substitution reaction than addition reactions.
Resonance Structure of Benzene: The phenomenon in which two or more structures can be written for a substance which involve identical positions of atoms is called resonance. In benzene’s Kekule’s structures (1) and (2) represent the resonance structures. Actual structure – of the molecule is represented by hybrid of the these two structures.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-24
Orbital structure of benzene: All six carbon atoms in benzene are sp2 hybridized. The sp2 hybrid orbitals overlap with each other and with s orbitals of the six hydrogen atoms forming C—C and C—H σ-bonds.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-25
X-Ray diffraction data indicates that benzene is a planar molecule. The data indicates that all the six C—C bond length are of the same order (139 pm) which is intermediate between (C—C) single bond (154 pm) and C—C double bond (133 pm). Thus the presence of pure double bond in benzene gives the idea of reductance of benzene to show addition reaction under normal condition. The is, It explains the unusual behaviour of benzene.
Aromaticity: It is a property of the sp2 hybridized planar rings in which the p orbitals allow cyclic delocalization of π electrons.
Conditions for Aromaticity:
(i) An aromatic compound is cyclic and planar.
(ii) Each atom in an aromatic ring has a p orbital. These p orbitals must be parallel so that a continuous overlap is possible around the ring.
(iii) The cyclic π molecular orbital (electron cloud) formed by overlap of p orbitals must contain (4n + 2) π electrons. Where n = integer (0, 1, 2, 3, etc.). This is known as Huckel rule.
Some Examples of Atomic Compounds are given below:
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-26
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-27
Preparation of Benzene: Benzene is commercially isolated from coaltar. However, there are some synthetic methods which is applied in the laboratory for the preparation of benzene.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-28
Physical Properties of Benzene:
(i) Benzene is a colourless liquid.
(ii) It is’ insoluble in water. It is soluble in alcohol, ether, chloroform etc.
(iii) Benzene itself is a good solvent for many organic and inorganic substances e.g., fat, resins, sulphur and iodine.
(iv) It bums with a luminous, sooty flame in contrast to alkanes and alkenes which usually bum with a bluish flame.

Follow along with the alkyne reactions cheat sheet.
Chemical Properties:
Benzene undergeos following types of chemical reactions.
(i) Electrophillic Substitution Reaction
(ii) Addition Reaction
Electrophillic Substitution Reactions:
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-29
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-30
Benzene on treatment with excess of chlorine in the presence of anhydrous AlCl3 can be chlorinated to hexachlorobenzene (C6Cl6)
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-31
Mechanism of electrophilic substitution reactions:
All electrophilic substitution reactions follow the same three step mechanism.
Setp 1. Formation of an electrophile:
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-32
Step 2. The electrophile attacks the aromatic ring to form a carbonium ion.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-33
Step 3. Loss of proton gives the substitution product.
Activating groups: These group activates the benzene ring for the attack by an electrophile. Example, —OH; —NH2, —NHR, —NHCOCH3, —OCH3 —CH3 —C2H5 etc.
Deactivating groups: Due to deactivating group because of strong —I effect, overall electron density on benzene ring decreases. It makes further substitution difficult.
Metadirecting group: The groups which direct the incoming group to meta position are called meta directing groups. Some examples of meta directing groups are —N02, —CN, —CHO, —COR, —COOH, —COOR, -S03H etc.
Let us consider the example of nitro group: Since Nitro group due to its strong -I effect reduces the electron density in benzene ring. Nitrobenzene is a resonance hybrid of following structures.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-34
Carcinogenicity and Toxicity: Some polynuclear hydrocarbons containing more than two benzene rings fused together become toxic and they are having cancer producing property. They are actually formed due to incomplete combustion of some organic materials like tobacco, coal and petroleum, etc.
Some of the carcinogenic hydrocarbons are given below.
Hydrocarbons Class 11 Notes Chemistry Chapter 13 img-35
• Hydrocarbons: They are compounds of carbon and hydrogen only.
Open Chain saturated compound—Alkane
Unsaturated Compound—Alkenes and Alkynes Aromatic Compound—Benzene and its derivatives Terminal alkynes are weakly acidic in nature.
• Conformation: Spatial arrangements obtained by rotation around sigma bonds.
• Eclipsed Conformation: Less stable because of more repulsion between bond pairs of electrons.
• Staggered: It is more stable since there is less repulsion between bond pairs of electrons.
• Geometrical isomerism: Observed only in compounds containing a double bond.
• Stability of benzene. Is explained on the basis of resonance hybrid.
• Arenes: Take part in electrophilic substitution reaction.
Aromaticity is determined by Huckle’s rule (4n + 2) rule

Class 11 Chemistry Notes

CBSE Sample Papers for Class 12 Chemistry Term 2 Set 8 with Solutions

Students can access the CBSE Sample Papers for Class 12 Chemistry with Solutions and marking scheme Term 2 Set 7 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Chemistry Standard Term 2 Set 8 with Solutions

Max. Marks: 35
Time: 2 Hours

General Instructions:
Read the following instructions carefully:

  • There are 12 questions in this question paper with internal choice.
  • Section A – Q. No. 1 to 3 are very short answer questions carrying 2 marks each.
  • Section B – Q. No. 4 to 11 are short answer questions carrying 3 marks each.
  • Section C- Q. No. 12 is case-based question carrying 5 marks.
  • All questions are compulsory.
  • Use of log tables and calculators is not allowed.

Section – A

Question 1.
If the concentration is expressed in mol L-1 units and time in seconds, what would be the unit of K*
(a) For a zero-order reaction
(b) For a first order reaction
(c) Why molecularity of a reaction cannot be zero?
Answer:
(a) For zero order reaction unit of K is mol L-1 sec-1.
(b) For first order reaction unit of K is sec’-1.
(c) Molecularity is the number of given reactant molecules or atoms that are colliding in the elementary reaction. Thus, a minimum of one reactant molecule, atom or ion is required to initiate a chemical
reaction. Hence, molecularity cannot be zero.

Question 2.
Complete the following equations:
CBSE Sample Papers for Class 12 Chemistry Term 2 Set 8 with Solutions 1
Answer:
CBSE Sample Papers for Class 12 Chemistry Term 2 Set 8 with Solutions 2

Question 3.
Predict the product of the following reactions:
CBSE Sample Papers for Class 12 Chemistry Term 2 Set 8 with Solutions 3
Answer:
CBSE Sample Papers for Class 12 Chemistry Term 2 Set 8 with Solutions 4

Section – B

Question 4.
(a) Name the isomerism shown by the complex [C0(NH3)5(NO2)]Cl2.
(b) What type of bond is present between metal and carbon in metal carbonyls?
(c) [FeF6]3- has five unpaired electrons while [Fe(CN)6]3-has only one unpaired electron. Explain.
OR
The crystal field theory assumes that the interaction between the metal ion and ligand is purely electrostatic. When ligands approach central metal atom/ion the five degenerated orbitals of the central metal atom become differential. In a complex, the central metal atom or ion is surrounded by various atoms or groups of atoms called ligands.
(a) What is the type of bond in metal complex according to crystal field theory?
(b) What is the number of ligands in Cu2 [Fe(CN)6].
(c) Which one is lower energy set between t2g and eg, after splitting of five degenerate orbital or metal atom/ion on complexation in tetrahedral complex?
Answer:
(a) Linkage isomerism occurs in complex compounds with ambidentate ligands like NO2, SCN, CN, S2O3-2.
These ligands have two donor atoms but at a time only one atom is directly linked to central metal atom of the complex. Linkage isomerism is shown by [Co(NH3)5(NO2)] Cl2. The isomers are [Co(NH3)5NO2]Cl and [Co(NH3)5ONO]Cl2.

(b) In a metal carbonyl, the metal-carbon bond possesses both σ and π character. Here M – C π bond is formed by the donation of lone pair of electrons from carbonyl carbon to the vacant orbital of metal. M – C π bond is formed by the back donation of electrons from a filled d-orbital of metal into the vacant antibondirig n orbital of carbon monoxide (Carbonyl ligand).

(c) In [FeF6]3-Fe(III) has got 3d5 systgem. As F is a weak field ligand, according to crystal field theory (CFT) electronic arrangement will be as follows:
CBSE Sample Papers for Class 12 Chemistry Term 2 Set 8 with Solutions 5
Hence, it has 5 unpaired electrons.
In [Fe(CH)6]3- Fe(III) has got 3d5 system. As CN is a strong field ligand, according to CFT electronic
the arrangement will be as follows:
CBSE Sample Papers for Class 12 Chemistry Term 2 Set 8 with Solutions 6
Hence, it has only one unpaired electron.
OR
(a) Ionic bond or electrostatic bond.
(b) Number of ligands = 6.
(c) Lower energy set = eg.

Ionic Strength Calculator is a free online tool that displays the ionic strength for the given concentration and valency.

Question 5.
Following ions are given:
Cr2+, Cu2+, Cu+, Fe2+, Fe3+, Mn3+
Identify the ion which is
(a) A strong reducing agent.
(b) Unstable in aqueous solution.
(c) A strong oxiding agent.
Give suitable reason in each case.
OR
Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:
(a) Electronic configurations
(b) Oxidation states
(c) Ionisation enthalpies.
Answer:
(a) Cr2+’ is a stronger reducing agent because it can lose one of its electron to become more stable Cr3+
in which the t2g level of d-orbital is half filled arid eg level is empty.
(b) Cu+ is unstable in aqueous solution because it disproportionates in water to form Cu2+ and Cu.
(c) Mn3+ is a stronger oxidising agent because it has 4 electrons in its valence shell and when it gains
one electron to form Mn2+, it results in the half-filled (d5) configuration that has extra stability.
OR
(a) In the 1st, 2nd and 3rd transition series, the 3d, 4d and 5d orbitais are respectively filled.
We know that elements in the same vertical column generally have similar electronic configurations.
In the first transition series, two elements show unusual electronic configurations:
Cr (24) = 3d54s1 (Because of extra stability of half filled and fully filled orbitais)
Cu (29) = 3d10 4s1
Similarly, there are exceptions in the second transition series. These are:
Mo(42) = 4d55s1
Te(43) = 4d65s1
Ru(44) = 4d75s1
Rh(45) = 4d85s1
Pd(46) = 4d105s0
Ag(47) = 4d105s1
There are some exceptions in third transition series as well. These are:
W(74) = 5d46s2
Pt(78) = 5d96s1
Au(79) =5d106s1

As a result of these exceptions, it happens many times that the electronic configurations of the elements present in the same group are dissimilar.
(b) In each of the three transition series the number of oxidation states shown by the elements is the maximum in the middle and minimum at the extreme ends.

However, +2 and +3 oxidation states are quite stable for all elements present in the first transition series.
All metals present in the first transition series form stable compounds in the +2 and +3 oxidation states.
The stability of the +2 and +3 oxidation states decreases in the second and the third transition series,
wherein higher oxidation states are more common.

For example, chromate ion (CrO4)2- is strong oxidant while molybdate (MoO4)2- and tungstate (WO4)2-
are stable. For example: WCl6, ReF7, RuO4, etc., are stable but no such compound of 1st transition series exists.

(c) In each of the three transition series, the first ionisation enthalpy increases from left to right. The first ionisation enthalpies of the third transition series are higher than those of the first and second transition series. This occurs due to the poor shielding effect of 4f electrons in the third transition series. The removal of one electron alters the relative energies of 4s and 3d orbitals. Hence, there is marginal and irregular increase in ionisation enthalpies.

Question 6.
(a) How would you account for the fact that the transition elements have high enthalpies of atomisation?
(b) Name the catalyst used in Ostwald’s process for the manufacturing of nitric acid.
(c) Transition metals generally form coloured compounds. Explain why?
Answer:
(a) Transition elements have large number of unpaired electrons in their atoms due to which they have stronger interatomic interactions and hence stronger bonding between atoms, leading to high
atomisation enthalpies.

(b) Platinised asbestos is used as catalyst in the Ostwald’s process for the manufacturing of nitric acid.

(c) Transition metals generally form coloured compounds because they generally possess one or more unpaired electrons. When visible light falls on a transition metal compound or ion, the unpaired electrons present in the lower energy d-orbital get promoted to high energy d-orbitals, called d-d transition, due to the absorption of visible light. Since the energy involved in d-d transition is quantised, only a definite wavelength gets absorbed, remaining wavelengths present in the visible region got transmitted. Therefore, transmitted light shows some colour complementary to the absorbed colour.

Question 7.
Write one difference in each of the following:
(a) Multimolecular colloid and Associated colloid
(b) Coagulation and Peptization
(c) Lyophobic sol and Lyophilic sol
Answer:
(a) Multìmolecular colloids: It consists of aggregates of a large number of atoms or smaller molecules. Its size is in the range of 1-1000rim. Associated colloids: At low concentration, they act as strong electrolytes whereas at high concentrations form aggregates or micelles

(b) Coagulation is the process of aggregating together the colloidal particles so as to change them into large sized particle by the addition of excess of an electrolyte whereas peptisation is the process of converting a fresh precipitate into colloidal particles by adding small amount of suitable electrolyte.

(c) Lyophobic colloidal sols are not hydrated and have weak affinity with the dispersion medium whereas lyophilic colloidal sols are heavily hydrated and have strong affinity with the dispersion medium.

Question 8.
(a) A galvanic cell consists of metallic zinc plate dipped in 0.1 M zinc nitrate solution and a lead plate dipped in 0.02 M lead nitrate solution. Write chemical equation for the electrode reaction and calculate the e.m.f. of the cell at 25°C.
E0Zn2+/Zn = – 0-76 V, E°pb2+/Pb = – 0.13 V
(b) Why does the conductivity of a solution decrease with dilution?
(c) What is the use of salt bridge in an electrochemical cell?
Answer:
(a) The half cell reactions can be written as:
Pb2++ 2e → Pb, E°pb2+/Pb=-0.13V
Zn2+ + 2e → Zn, E°Zn2+ /Zn =-0.76 V
So, E°cell = E°cathode — E°anode
=-0.13V-(-0.76V)
=-0.13+0.76 -0.63V
Now, overall reaction:
Zn+Pb2+ → Zn2+Pb;Ecell=?

Ecell = E°cell – \(\frac{0.0591}{2} \log \left[\frac{\mathrm{Zn}^{+2}}{\mathrm{~Pb}^{+2}}\right]\)
= 0.63 – \(\frac{0.06}{2} \log \left(\frac{0.1}{0.02}\right)\)
=0.63-0.03 log5
=0.63-0.03 x 0.6910
= 0.63 — 0.02 = 0.61V

(b) The conductivity of a solution is the conductance of ions present in a unit volume of the solution. The number of ions (responsible for carrying current) decreases when the solution is diluted. As a result, the conductivity of a solution decreases with dilution.

(c) The purpose of the salt bridge is to act as a source of spectator ions that can migrate Into each of the half cells to preserve electrical neutrality.

Question 9.
(a) An organic compound X has molecular formula C5H10O. It does not reduce Fehling’s solution but
forms a bisulphate compound. It also gives positive iodoform test. What are possible structures of X? Explain your reasoning relating structure.
(b) Write IUPAC name of the following:
CBSE Sample Papers for Class 12 Chemistry Term 2 Set 8 with Solutions 7
OR
Predict the product:
CBSE Sample Papers for Class 12 Chemistry Term 2 Set 8 with Solutions 8
Answer:
(a) As the compound gives adduct with bisulphite, i.e., bisulphate compound, it contains a carbonyl group. But it does not reduce Fehling’s solution so it must be a ketone. Now, as this gives a positive test with iodoform, it is methyl ketone. So, based on the above information the following are the possible structures of the compound:
CBSE Sample Papers for Class 12 Chemistry Term 2 Set 8 with Solutions 9

Question 10.
Answer the following questions:
(a) What is Hinsberg reagent?
(b) Why do primary amines have higher boiling points than tertiary amines?
(c) Why are aliphatic amines stronger bases than aromatic amines?
Answer:
(a) Benzenesulphonyl chloride (C6H5 – SO2Cl) is known as the Hinsberg reagent.

(b) Primary amines are involved in intermolecular association due to H-bonding. While 3° amines do not have intermolecular association due to lack of H-bonding. Thus, Molecular association of primary amine leads to the increase of its effective molar mass and thus, 1° amine has higher boiling point than 3° amine.

(c) Due to the -I effect of the benzene ring, the electrons on the N-atom are involved in resonance and are available as base. Therefore, the electrons on the N-atom in aromatic amines cannot be donated easily. This is the reason that aliphatic amines are stronger bases than aromatic amines.

Question 11.
Predict the product:
CBSE Sample Papers for Class 12 Chemistry Term 2 Set 8 with Solutions 10
Or
(a) Arrange the following in increasing acidic strength:
CBSE Sample Papers for Class 12 Chemistry Term 2 Set 8 with Solutions 11
(b) Distinguish between:
(i) Acetone and formaldehyde
(ii) C6H5COCH3 and C6H5CH2 – CHO
Answer:
CBSE Sample Papers for Class 12 Chemistry Term 2 Set 8 with Solutions 12
CBSE Sample Papers for Class 12 Chemistry Term 2 Set 8 with Solutions 13

Section – C

Question 12.
Read the passage given below and answer the questions that follow:
The term order of a reaction is the sum of exports of concentration terms appearing in the rate equation. It is an experimentally determinable quantity. It may be whole number, fraction, zero on even negative. Knowledge of order does not require knowledge of mechanism. Order of a ‘reaction’ may change with change in experimental condition namely pressure, temperature etc.

A 1st order reaction is a reaction whose rate-determining step (r.d.s) involves only one molecule. Thus the step is A → Product.
A 2nd order reaction may be of two types:
2 A→ Y Product
or A + B → Product.
(a) Give the unit of rate constant of a reaction.
(b) In what condition order and molecularity of a reaction becomes equal.
(c) In which type of reaction order and molecularity are different.
(d) Write expression for rate constant of first-order reaction. Mention unit of rate constant of a first-order reaction.
OR
(d) Show that half-life period of a first-order reaction is independent of initial concentration of the reactant.
Answer:
(a) Unit of rate constant = (Concentration)1-n Time-1
When n order of reaction
(b) In elementary reaction, order and molecularity are same.
(c) In complex multistep reaction, order and molecularity are different.
(d) Rate = K [R]1, unit of k in first order reaction rate.
= \(\frac{\mathrm{mol} \cdot \mathrm{L}^{-1}}{s} \times \frac{1}{\left(\mathrm{~mol}, \mathrm{~L}^{-1}\right)} \) = s-1
OR
(d) From First order reaction:
k= \(\frac{2.303}{t} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]} \)
At t1/2‘[R] = \(\frac{[\mathrm{R}]_{0}}{2}\)
CBSE Sample Papers for Class 12 Chemistry Term 2 Set 8 with Solutions 15
So, in a first-order reaction, t1/2 is independent of initial concentration.

Some Basic Concepts of Chemistry Class 11 Notes Chapter 1

Some Basic Concepts of Chemistry Class 11 Notes Chapter 1

Class 11 Chemistry Notes students can refer to the Some Basic Concepts of Chemistry Class 11 Notes Chapter 1 https://www.cbselabs.com/basic-concepts-chemistry-cbse-notes-class-11-chemistry/ Pdf here. They can also access the CBSE Class 11 Some Basic Concepts of Chemistry Chapter 1 Notes while gearing up for their Board exams.

Some Basic Concepts of Chemistry Class 11 Notes Chapter 1

Class 11 Chemistry Chapter 1 Notes 

• Importance of Chemistry
Chemistry has a direct impact on our life and has wide range of applications in different fields. These are given below:
(A) In Agriculture and Food:
(i) It has provided chemical fertilizers such as urea, calcium phosphate, sodium nitrate, ammonium phosphate etc.
(ii) It has helped to protect the crops from insects and harmful bacteria, by the use ‘ of certain effective insecticides, fungicides and pesticides.
(iii) The use of preservatives has helped to preserve food products like jam, butter, squashes etc. for longer periods.
(B) In Health and Sanitation:
(i) It has provided mankind with a large number of life-saving drugs. Today, dysentery and pneumonia are curable due to discovery of sulpha drugs and penicillin life-saving drugs. Cisplatin and taxol have been found to be very effective for cancer therapy and AZT (Azidothymidine) is used for AIDS victims.
(ii) Disinfectants such as phenol are used to kill the micro-organisms present in drains, toilet, floors etc.
(iii) A low concentration of chlorine i.e., 0.2 to 0.4 parts per million (ppm) is used ’ for sterilization of water to make it fit for drinking purposes.
(C) Saving the Environment:
The rapid industrialisation all over the world has resulted in lot of pollution.
Poisonous gases and chemicals are being constantly released in the atmosphere. They are polluting environment at an alarming rate. Scientists are working day and night to develop substitutes which may cause lower pollution. For example, CNG (Compressed Natural Gas), a substitute of petrol, is very effective in checking pollution caused by automobiles.
(D) Application in Industry:
Chemistry has played an important role in developing many industrially ^ manufactured fertilizers, alkalis, acids, salts, dyes, polymers, drugs, soaps,
detergents, metal alloys and other inorganic and organic chemicals including new materials contribute in a big way to the national economy.
Some Basic Concepts Of Chemistry
• Matter

Anything which has mass and occupies space is called matter.
For example, book, pencil, water, air are composed of matter as we know that they have
mass and they occupy space.
• Classification of Matter
There are two ways of classifying the matter:
(A) Physical classification
(B) Chemical classification
(A) Physical Classification:
Matter can exist in three physical states:
1. Solids 2. Liquids 3. Gases
1. Solids: The particles are held very close to each other in an orderly fashion and there is not much freedom of movement.
Characteristics of solids: Solids have definite volume and definite shape.
2. Liquids: In liquids, the particles are close to each other but can move around. Characteristics of liquids: Liquids have definite volume but not definite shape.
3. Gases: In gases, the particles are far apart as compared to those present in solid or liquid states. Their movement is easy and fast.
Characteristics of Gases: Gases have neither definite volume nor definite shape. They completely occupy the container in which they are placed.
(B) Chemical Classification:
Based upon the composition, matter can be divided into two main types:
1. Pure Substances 2. Mixtures.
1. Pure substances: A pure substance may be defined as a single substance (or matter) which cannot be separated by simple physical methods.
Pure substances can be further classified as (i) Elements (ii) Compounds
(i) Elements: An element consists of only one type of particles. These particles may be atoms or molecules.
For example, sodium, copper, silver, hydrogen, oxygen etc. are some examples of elements. They all contain atoms of one type. However, atoms of different elements are different in nature. Some elements such as sodium . or copper contain single atoms held together as their constituent particles whereas in some others two or more atoms combine to give molecules of the element. Thus, hydrogen, nitrogen and oxygen gases consist of molecules in which two atoms combine to give the respective molecules of the element.
Chemistry Class 11 Chapter 1 Notes
(ii) Compounds:
It may be defined as a pure substance containing two or more elements combined together in a fixed proportion by weight and can be decomposed into these elements by suitable chemical methods. Moreover, the properties of a compound are altogether different from the constituting elements.
The compounds have been classified into two types. These are:
(i) Inorganic Compounds: These are compounds which are obtained from non-living sources such as rocks and minerals. A few
examples are: Common salt, marble, gypsum, washing soda etc.
(ii) Organic Compounds are the compounds which are present in plants and animals. All the organic compounds have been found to contain carbon as their essential constituent. For example, carbohydrates, proteins, oils, fats etc.

The Counting Atomic Calculator is a tool for calculating the atomic number and the mass number based on numbers of atom components.

Some Basic Concepts Of Chemistry Notes
2. Mixtures:
The combination of two or more elements or compounds which are not chemically combined together and may also be present in any proportion, is called mixture. A few examples of mixtures are: milk, sea water, petrol, lime water, paint glass, cement, wood etc.
Types of mixtures: Mixtures are of two types:
(i) Homogeneous mixtures: A mixture is said to be homogeneous if it has a uniform composition throughout and there are no visible boundaries of separation between the constituents.
For example: A mixture of sugar solution in water has the same sugar water composition throughout and all portions have the same sweetness.
(ii) Heterogeneous mixtures: A mixture is said to be heterogeneous if it does not have uniform composition throughout and has visible boundaries of separation between the various constituents. The different constituents of a heterogeneous mixture can be seen even with naked eye.
For example: When iron filings and sulphur powder are mixed together, the mixture formed is heterogeneous. It has greyish-yellow appearance and the two constituents, iron and sulphur, can be easily identified with naked eye.
Some Basic Concepts Of Chemistry Class 11 Notes
• Differences between Compounds and Mixtures

Compounds
1. In a compound, two or more elements are combined chemically.
2. In a compound, the elements are present in the fixed ratio by mass. This ratio cannot change.
3. CompoUnds are always homogeneous i.e., they havethe same composition throughout.
4 In a compound, constituents cannot be separated by physical methods
5. In a compound, the constituents lose their identities i.e., i compound does not show the characteristics of the constituting elements.
Mixtures
1. In a mixture, or more elements or compounds are simply mixed and not combined chemically.
2. In a mixture the constituents are not present in fixed ratio. It can vary
3. Mixtures may be either homogeneous or heterogeneous in nature.
4. Constituents of mixtures can be separated by physical methods.
5, In a mixture, the constituents do not lose their identities i.e., a mixture shows the characteristics of all the constituents .
We have discussed the physical and chemical classification of matter. A flow sheet representation of the same is given below.
Some Basic Concepts of Chemistry Class 11 Notes Chapter 1 img-1
• Properties of Matter and Their Measurements
Physical Properties: Those properties which can be measured or observed without changing the identity or the composition of the substance.
Some examples of physical properties are colour, odour, melting point, boiling point etc. Chemical Properties: It requires a chemical change to occur. The examples of chemical properties are characteristic reactions of different substances. These include acidity, basicity, combustibility etc.
• Units of Measurement
Fundamental Units: The quantities mass, length and time are called fundamental quantities and their units are known as fundamental units.
There are seven basic units of measurement for the quantities: length, mass, time, temperature, amount of substance, electric current and luminous intensity.
Si-System: This system of measurement is the most common system employed throughout the world.
It has given units of all the seven basic quantities listed above.
Some Basic Concepts of Chemistry Class 11 Notes Chapter 1 img-2
Some Basic Concepts Of Chemistry Notes Pdf
• Definitions of Basic SI Units

1. Metre: It is the length of the path travelled by light in vacuum during a time interval of 1/299792458 of a second.
2. Kilogram: It is the unit of mass. It is equal to the mass of the international prototype
of the kilogram. ,
3. Second: It is the duration of 9192631, 770 periods of radiation which correspond to the transition between the two hyper fine levels of the ground state of caesium- 133 atom.
4. Kelvin: It is the unit of thermodynamic temperature and is equal to 1/273.16 of the thermodynamic temperature of the triple point of water.
5. Ampere: The ampere is that constant current which if maintained in two straight parallel conductors of infinite length, of negligible circular cross section and placed, 1 metre apart in vacuum, would produce between these conductors a force equal to 2 x 10-7 N per metre of length.
6. Candela: It may be defined as the luminous intensity in a given direction, from a source which emits monochromatic radiation of frequency 540 x 1012 Hz and that has a radiant intensity in that direction of 1/ 683 watt per steradian.
7. Mole: It is the amount of substance which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon -12. Its symbol is ‘mol’.
Class 11th Chemistry Chapter 1 Notes
• Mass and Weight

Mass: Mass of a substance is the amount of matter present in it.
The mass of a substance is constant.
The mass of a substance can be determined accurately in the laboratory by using an analytical
balance. SI unit of mass is kilogram.
Some Basic Concepts of Chemistry Class 11 Notes Chapter 1 img-3
Weight: It is the force exerted by gravity on an object. Weight of substance may vary from one place to another due to change in gravity.
Volume: Volume means the space occupied by matter. It has the units of (length)3. In SI units, volume is expressed in metre3 (m3). However, a popular unit of measuring volume, particularly in liquids is litre (L) but it is not in SI units or an S.I. unit.
Mathematically,
1L = 1000 mL = 1000 cm3 = 1dm3.
Volume of liquids can be measured by different devices like burette, pipette, cylinder, measuring flask etc. All of them have been calibrated.
Some Basic Concepts of Chemistry Class 11 Notes Chapter 1 img-4
Temperature: There are three scales in which temperature can be measured. These are known as Celsius scale (°C), Fahrenheit scale (°F) and Kelvin scale (K).
Some Basic Concepts of Chemistry Class 11 Notes Chapter 1 img-5
-> Thermometres with Celsius scale are calibrated from 0°C to 100°C.
-> Thermometres with Fahrenheit scale are calibrated from 32°F to 212°F.
-> Kelvin’scale of temperature is S.I. scale and is very common these days.Temperature on this scale is shown by the sign K.
The temperature on two scales are related to each other by the relationship
Some Basic Concepts of Chemistry Class 11 Notes Chapter 1 img-6
Density: Density of a substance is its amount of mass per unit volume. So, SI unit of density can be obtained as follows:
Some Basic Concepts of Chemistry Class 11 Notes Chapter 1 img-7
This unit is quite large and a chemist often expresses density in g cm3 where mass is expressed in gram and volume is expressed in cm3.
• Uncertainty in Measurements
All scientific measurements involve certain degree of error or uncertainty. The errors which arise depend upon two factors.
(i) Skill and accuracy of the worker (ii) Limitations of measuring instruments.

Find the maximum yield of a chemical reaction with our theoretical yield calculator.

Notes Of Class 11 Chemistry Chapter 1
• Scientific Notation

It is an exponential notation in which any number can be represented in the form N x 10n where n is an exponent having positive or negative values and N can vary between 1 to 10. Thus, 232.508 can be written as 2.32508 x 102 in scientific notation.
Now let us see how calculations are carried out with numbers expressed in scientific notation.
(i) Calculation involving multiplication and division
Some Basic Concepts of Chemistry Class 11 Notes Chapter 1 img-8
(ii) Calculation involving addition and subtraction: For these two operations, the first numbers are written in such a way that they have the same exponent. After that, the coefficients are added or subtracted as the case may be. For example,
Some Basic Concepts of Chemistry Class 11 Notes Chapter 1 img-9
• Significant Figures
Significant figures are meaningful digits which are known with certainty. There are certain rules for determining the number of significant figures. These are stated below:
1. All non-zero digits are significant. For example, in 285 cm, there are three significant figures and in 0.25 mL, there are two significant figures.
2. Zeros preceding to first non-zero digit are not significant. Such zeros indicates the position of decimal point.
For example, 0.03 has one significant figure and 0.0052 has two significant figures.
3. Zeros between two non-zero digits are significant. Thus, 2.005 has four significant figures.
4. Zeros at the end or right of a number are significant provided they are on the right side of the decimal point. For example, 0.200 g has three significant figures.
5. Counting numbers of objects. For example, 2 balls or 20 eggs have infinite significant figures as these are exact numbers and can be represented by writing infinite number of zeros after placing a decimal.
i.e., 2 = 2.000000
or 20 = 20.000000
Basic Concepts Of Chemistry Class 11
• Addition and Subtraction of Significant Figures

In addition or subtraction of the numbers having different precisions, the final result should be reported to the same number of decimal places as in the term having the least number of decimal places.
For example, let us carry out the addition of three numbers 3.52, 2.3 and 6.24, having different precisions or different number of decimal places.
Some Basic Concepts of Chemistry Class 11 Notes Chapter 1 img-10
The final result has two decimal places but the answer has to be reported only up to one decimal place, i.e., the answer would be 12.0.
Subtraction of numbers can be done in the same way as the addition.
Some Basic Concepts of Chemistry Class 11 Notes Chapter 1 img-11
The final result has four decimal places. But it has to be reported only up to two decimal places, i.e., the answer would be 11.36.
• Multiplication and Division of Significant Figures
In the multiplication or division, the final result should be reported upto the same number of significant figures as present in the least precise number.
Multiplication of Numbers: 2.2120 x 0.011 = 0.024332
According to the rule the final result = 0.024
Division of Numbers: 4.2211÷3.76 = 1.12263
The correct answer = 1.12
• Dimensional Analysis
Often while calculating, there is a need to convert units from one system to other. The method used to accomplish this is called factor label method or unit factor method or dimensional analysis.
Class11 Chemistry Ch1 Notes
• Laws of Chemical Combinations

The combination of elements to form compounds is governed by the following five basic laws.
(i) Law of Conservation of Mass
(ii) Law of Definite Proportions
(iii) Law of Multiple Proportions
(iv) Law of Gaseous Volume (Gay Lussac’s Law)
(v) Avogadro’s Law
Importance Of Chemistry Class 11
(i) Law of Conservation of Mass

The law was established by a French chemist, A. Lavoisier. The law states:
In all physical and chemical changes, the total mass of the reactants is equal to that of the products.
In other words, matter can neither be created nor destroyed.
The following experiments illustrate the truth of this law.
(a) When matter undergoes a physical change.
Some Basic Concepts of Chemistry Class 11 Notes Chapter 1 img-12
It is found that there is no change in weight though a physical change has taken place.
(b) When matter undergoes a chemical change.
For example, decomposition of mercuric oxide.
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During the above decomposition reaction, matter is neither gained nor lost.
(ii) Law of Definite Proportions
According to this law:
A pure chemical compound always consists of the same elements combined together in a fixed proportion by weight.
For example, Carbon dioxide may be formed in a number of ways i.e.,
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(iii) Law of Multiple Proportions
If two elements combine to form two or more compounds, the weight of one of the elements which combines with a fixed weight of the other in these compounds, bears simple whole number ratio by weight.
For example,
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(iv) Gay Lussac’s Law of Gaseous Volumes
The law states that, under similar conditions of temperature and pressure, whenever gases combine, they do so in volumes which bear simple whole number ratio with each other and also with the gaseous products. The law may be illustrated by the following examples.
(a) Combination between hydrogen and chlorine:
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(b) Combination between nitrogen and hydrogen: The two gases lead to the formation of ammonia gas under suitable conditions. The chemical equation is
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(v) Avogadro’s Law: Avogadro proposed that, equal volumes of gases at the same temperature and pressure should contain equal number of molecules.
For example,
If we consider the reaction of hydrogen and oxygen to produce water, we see that two volumes of hydrogen combine with one volume of oxygen to give two volumes of water without leaving any unreacted oxygen.
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• Dalton’s Atomic Theory
In 1808, Dalton published ‘A New System of Chemical Philosophy’ in which he proposed the following:
1. Matter consists of indivisible atoms.
2. All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass.
3. Compounds are formed when atoms of different elements combine in a fixed ratio.
4. Chemical reactions involve reorganisation of atoms. These are neither created nor destroyed in a chemical reaction.
• Atomic Mass
The atomic mass of an element is the number of times an atom of that element is heavier than an atom of carbon taken as 12. It may be noted that the atomic masses as obtained above are the relative atomic masses and not the actual masses of the atoms.
One atomic mass unit (amu) is equal to l/12th of the mass of an atom of carbon-12 isotope. It is also known as unified mass.
Average Atomic Mass
Most of the elements exist as isotopes which are different atoms of the same element with different mass numbers and the same atomic number. Therefore, the atomic mass of an element must be its average atomic mass and it may be defined as the average relative mass of an atom of an element as compared to the mass of carbon atoms (C-12) taken as 12w.
Molecular Mass
Molecular mass is the sum of atomic masses of the elements present in a molecule. It is obtained by multiplying the atomic mass of each element by number of its atoms and adding them together.
For example,
Molecular mass of methane (CH4)
= 12.011 u + 4 (1.008 u)
= 16.043 u
Formula Mass
Ionic compounds such as NaCl, KNO3, Na2C03 etc. do not consist of molecules i.e., single entities but exist “as ions closely packed together in a three dimensional space as shown in -Fig. 1.5.
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In such cases, the formula is used to calculate the formula mass instead of molecular mass. Thus, formula mass of NaCl = Atomic mass of sodium + atomic mass of chlorine
= 23.0 u + 35.5 u = 58.5 u.
• Mole Concept
It is found that one gram atom of any element contains the same number of atoms and one gram molecule of any substance contains the same number of molecules. This number has been experimentally determined and found to be equal to 6.022137 x 1023 The value is generally called Avogadro’s number or Avogadro’s constant.
It is usually represented by NA:
Avogadro’s Number, NA = 6.022 × 1023
• Percentage Composition
One can check the purity of a given sample by analysing this data. Let us understand by taking the example of water (H20). Since water contains hydrogen and oxygen, the percentage composition of both these elements can be calculated as follows:
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• Empirical Formula
The formula of the compound which gives the simplest whole number ratio of the atoms of yarious elements present in one molecule of the compound.
For example, the formula of hydrogen peroxide is H202. In order to express its empirical formula, we have to take out a common factor 2. The simplest whole number ratio of the atoms is 1:1 and the empirical formula is HO. Similarly, the formula of glucose is C6H1206. In order to get the simplest whole number of the atoms,
Common factor = 6
The ratio is = 1 : 2 : 1 The empirical formula of glucose = CH20
• Molecular Formula
The formula of a compound which gives the actual ratio of the atoms of various elements present in one molecule of the compound.
For example, molecular formula of hydrogen peroxide = H202and Glucose = C6H1206
Molecular formula = n x Empirical formula
Where n is the common factor and also called multiplying factor. The value of n may be 1, 2, 3, 4, 5, 6 etc.
In case n is 1, Molecular formula of a compound = Empirical formula of the compound.
• Stoichiometry and Stoichiometric Calculations
The word ‘stoichiometry’ is derived from two Greek words—Stoicheion (meaning element) and metron (meaning measure). Stoichiometry, thus deals with the calculation of masses (sometimes volume also) of the reactants and the products involved in a chemical reaction. Let us consider the combustion of methane. A balanced equation for this reaction is as given below:
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Limiting Reactant/Reagent
Sometimes, in alchemical equation, the reactants present are not the amount as required according to the balanced equation. The amount of products formed then depends upon the reactant which has reacted completely. This reactant which reacts completely in the reaction is called the limiting reactant or limiting reagent. The reactant which is not consumed completely in the reaction is called excess reactant.
Reactions in Solutions
When the reactions are carried out in solutions, the amount of substance present in its given volume can be expressed in any of the following ways:
1. Mass percent or weight percent (w/w%)
2. Mole fraction
3. Molarity
4. Molality
1. Mass percent: It is obtained by using the following relation:
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2. Mole fraction: It is the ratio of number of moles of a particular component to the total number of moles of the solution. For a solution containing n2 moles of the solute dissolved in n1 moles of the solvent,
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3. Molarity: It is defined as the number of moles of solute in 1 litre of the solution.
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4. Molality: It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by m.
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• All substances contain matter which can exist in three states — solid, liquid or gas.
• Matter can also be classified into elements, compounds and mixtures.
• Element: An element contains particles of only one type which may be atoms or molecules.
• Compounds are formed when atoms of two or more elements combine in a fixed ratio to each other.
• Mixtures: Many of the substances present around us are mixtures.
• Scientific notation: The measurement of quantities in chemistry are spread over a wide rhnge of 10-31to 1023. Hence, a convenient system of expressing the number in scientific notation is used.
• Scientific figures: The uncertainty is taken care of by specifying the number of significant figures in which the observations are reported.
• Dimensional analysis: It helps to express the measured quantities in different systems of units.
• Laws of Chemical Combinations are:
(i) Law of Conservation of Mass
(ii) Law of Definite Proportions
(iii) Law of Multiple Proportions
(iv) Gay Lussac’s Law of Gaseous Volumes
(v) Avogadro’s Law.
• Atomic mass: The atomic mass of an element is expressed relative to 12C isotope of carbon which has an exact value of 12u.
• Average atomic mass: Obtained by taking into account the natural aboundance of different isotopes of that element.
• Molecular mass: The molecular mass of a molecule is obtained by taking sum of atomic masses of different atoms present in a molecule.
• Avogadro number: The number of atoms, molecules or any other particles present in a given system are expressed in terms of Avogadro constant.
= 6.022 x 1023
• Balanced chemical equation: A balanced equation has the same number of atoms of each element on both sides of the equation.
• Stoichiometry: The quantitative study of the reactants required or the products formed is called stoichiometry. Using stoichiometric calculations, the amounts of one or more reactants required to produce a particular amount of product can be determined and vice-versa.

Class 11 Chemistry Notes