Read the passage given below and answer the following questions from 1 to 5. The term ‘hydrocarbon’ is self-explanatory which means compounds of carbon and hydrogen only. Hydrocarbons play a key role in our daily life. You must be familiar with the terms ‘LPG’ and ‘CNG’ used as fuels. LPG is the abbreviated form of liquified petroleum gas whereas CNG stands for compressed natural gas. Another term ‘LNG’ (liquified natural gas) is also in news these days. This is also a fuel and is obtained by liquifaction of natural gas. Petrol, diesel and kerosene oil are obtained by the fractional distillation of petroleum found under the earth’s crust. Coal gas is obtained by the destructive distillation of coal. Natural gas is found in upper strata during drilling of oil wells. The gas after compression is known as compressed natural gas. LPG is used as a domestic fuel with the least pollution. Kerosene oil is also used as a domestic fuel but it causes some pollution. Automobiles need fuels like petrol, diesel and CNG. Petrol and CNG operated automobiles cause less pollution. All these fuels contain mixture of hydrocarbons, which are sources of energy. Hydrocarbons are also used for the manufacture of polymers like polythene, polypropene, polystyrene etc. Higher hydrocarbons are used as solvents for paints. They are also used as the starting materials for manufacture of many dyes and drugs. Thus, you can well understand the importance of hydrocarbons in your daily life. In this unit, you will learn more about hydrocarbons. Classification Hydrocarbons are of different types. Depending upon the types of carbon-carbon bonds present, they can be classified into three main categories – (i) saturated (ii) unsaturated and (iii) aromatic hydrocarbons. Saturated hydrocarbons contain carbon-carbon and carbon-hydrogen single bonds. If different carbon atoms are joined together to form open chain of carbon atoms with single bonds, they are termed as alkanes. On the other hand, if carbon atoms form a closed chain or a ring, they are termed as cycloalkanes. Unsaturated hydrocarbons contain carbon-carbon multiple bonds – double bonds, triple bonds or both. Aromatic hydrocarbons are a special type of cyclic compounds. You can construct a large number of models of such molecules of both types (open chain and close chain) keeping in mind that carbon is tetravalent and hydrogen is monovalent. For making models of alkanes, you can use toothpicks for bonds and plasticine balls for atoms. For alkenes, alkynes and aromatic hydrocarbons, spring models can be constructed. Alkanes As already mentioned, alkanes are saturated open chain hydrocarbons containing carbon – carbon single bonds. Methane (CH_4) is the first member of this family. Methane is a gas found in coal mines and marshy places. If replace one hydrogen atom of methane by carbon and join the required number of hydrogens to satisfy the tetravalence of the other carbon atom,it get C_2H_6. This hydrocarbon with molecular formula C_2H_6 is known as ethane. Thus it can consider C_2H_6 as derived from CH_4 by replacing one hydrogen atom by -CH_3 group. Go on constructing alkanes by doing this theoretical exercise i.e., replacing hydrogen atom by –CH_3 group. The next molecules will be C_3H_8, C_4H_ 10 … These hydrocarbons are inert under normal conditions as they do not react with acids, bases and other reagents. Hence, they were earlier known as paraffins (latin: parum, little; affinis, affinity). the general formula for alkanes is C_nH_2n + 2. It represents any particular homologue when n is given appropriate value. methane has a tetrahedral structure, in which carbon atom lies at the centre and the four hydrogen atoms lie at the four corners of a regular tetrahedron. All H-C-H bond angles are of 109.5^ . In alkanes, tetrahedra are joined together in which C-C and C-H bond lengths are 154 pm and 112 pm respectively. We have already read that C–C and C–H bonds are formed by head-on overlapping of sp 3 hybrid orbitals of carbon and 1s orbitals of hydrogen atoms. Difference in properties is due to difference in their structures, they are known as structural isomers. structural isomers which differ in chain of carbon atoms are known as chain isomers. Preparation- Petroleum and natural gas are the main sources of alkanes. However, alkanes can be prepared by following methods: 1) From unsaturated hydrocarbons Dihydrogen gas adds to alkenes and alkynes in the presence of finely divided catalysts like platinum, palladium or nickel to form alkanes. This process is called hydrogenation. These metals adsorb dihydrogen gas on their surfaces and activate the hydrogen – hydrogen bond. Platinum and palladium catalyse the reaction at room temperature but relatively higher temperature and pressure are required with nickel catalysts. 2) From alkyl halides i) Alkyl halides (except fluorides) on reduction with zinc and dilute hydrochloric acid give alkanes. ii) Alkyl halides on treatment with sodium metal in dry ethereal (free from moisture) solution give higher alkanes. This reaction is known as Wurtz reaction and is used for the preparation of higher alkanes containing even number of carbon atoms. 1. Which of the following catalyst is used for hydrogenation… 1. platinum 2. palladium 3. nickel 4. All the above 2. LPG stands for …. 1. Liquid Pressure Gas 2. Liquefied Petroleum Gas 3. Liquid Platinum Gas 4. Liquid Processed Gas 3. Difference in properties is due to difference in their structures, they are known as …. isomers. 1. Functional 2. Positional 3. Structural 4. Chain 4. Structural isomers which differ in chain of carbon atoms are known as… isomers. 1. Functional 2. Positional 3. Structural 4. Chain 5. CNG Stands for .. 1. Compressed Natural Gas 2. Condensed Natural Gas 3. Compressed Neutral Gas 4. Compressed Neutral Gallium

1. (d) All the above 2. (b) Liquefied Petroleum Gas 3. (c) Structural 4. (d) chain 5. (a) compressed natural gas

Read the passage given below and answer the following questions f…

Read the passage given below and answer the following questions from (i) to (v).
The identity of a substance is defined not only by the types of atoms or ions it contains, but by the quantity of each type of atom or ion. The experimental approach required the introduction of a new unit for amount of substances, the mole, which remains indispensable in modern chemical science. The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter. A mole is defined as the amount of substance containing the same number of discrete entities (atoms, molecules, ions, etc.) as the number of atoms in a sample of pure 12 C weighing exactly 12 g . One Latin connotation for the word "mole" is "large mass" or "bulk," which is consistent with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules and so forth. The number of entities composing a mole has been experimentally determined to be $6.02214179 \times 10^{23}$. $6.02214179 \times 10^{23}$, a fundamental constant named Avogadro's number (NA) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of "per mole," a conveniently rounded version being $6.022 \times 10^{23} / mol$. Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole ( $g / mol$ ). The following questions are multiple choice questions. Choose the most appropriate answer:
i. A sample of copper sulphate pentahydrate contains 8.64 g of oxygen. How many grams of Cu is present in the sample?

A sample of copper sulphate pentahydrate contains 8.64g of oxygen. How many grams of Cu is present in the sample?

0.952g
3.816g
3.782g
8.64g

A gas mixture contains 50% helium and $50\%$ methane by volume. What is the percent by \ weight of methane in the mixture?

$19.97\%$
$20.05\%$
$50\%$
$80.03\%$

The mass of oxygen gas which occupies 5.6 litres at STP could be:

Gram atomic mass of oxygen
One fourth of the gram atomic mass of oxygen
Double the gram atomic mass of oxygen
Half of the gram atomic mass of oxygen

What is the mass of one molecule of yellow phosphorus? (Atomic mass of phosphorus = 30)

$1.993 \times 10^{-22}$ mg
$1.993 \times 10^{-19}$ mg
$4.983 \times 10^{-20}$ mg
$4.983 \times 10^{-23}$ mg

The number of moles of oxygen in 1L of air containing $21\%$ oxygen by volume, in standard conditions is:

$0.186$ mol
$0.21$ mol
$2.10$ mol
$0.0093$ mol

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The molecular orbital theory is based on the principle of a linear combination of atomic orbitals. According to this approach when atomic orbitals of the atoms come closer, they undergo constructive interference as well as destructive interference giving molecular orbitals, i.e., two atomic orbitals overlap to form two molecular orbitals, one of which lies at a lower energy level (bonding molecular orbital). Each molecular orbital can hold one or two electrons in accordance with Pauli's exclusion principle and Hund's rule of maximum multiplicity.
For molecules up to $N _2$, the order of filling of orbitals is:

Bond order $=\frac{1}{2}$ [bonding electrons - antibonding electrons]
Bond order gives the following information:
I. If bond order is greater than zero, the molecule/ion exists otherwise not.
II. Higher the bond order, higher is the bond dissociation energy.
III. Higher the bond order, greater is the bond stability.
IV. Higher the bond order, shorter is the bond length.

1. Arrange the following negative stabilities of $CN , CN ^{+}$and $CN ^{-}$in increasing order of bond. (1)
2. The molecular orbital theory is preferred over valence bond theory. Why? (1)
3. Ethyne is acidic in nature in comparison to ethene and ethane. Why is it so? (2)
OR
Bonding molecular orbital is lowered by a greater amount of energy than the amount by which antibonding molecular orbital is raised. Is this statement correct? (2)

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Read the passage given below and answer the following questions from 1 to 5.

Alkanes are generally inert towards acids, bases, oxidising and reducing agents. However, they undergo the following reactions under certain conditions.
1) Substitution reactions One or more hydrogen atoms of alkanes can be replaced by halogens, nitro group and sulphonic acid group. Halogenation takes place either at higher temperature (573-773K) or in the presence of diffused sunlight or ultraviolet light. Lower alkanes do not undergo nitration and sulphonation reactions. These reactions in which hydrogen atoms of alkanes are substituted are known as substitution reactions. As an example, chlorination of methane is given below: Halogenation
$\text{CH}_3-\text{CH}_3+\text{CL}_2\xrightarrow{\text{hv}}\text{CH}_3-\text{CH}_2\text{Cl}+\text{HCl}$

It is found that the rate of reaction of alkanes with halogens is $F_2 > C_{l2} > Br_2 > I_2$. Rate of replacement of hydrogens of alkanes is: $3^\circ > 2^\circ > 1^\circ$ . Fluorination is too violent to be controlled. Iodination is very slow and a reversible reaction. It can be carried out in the presence of oxidizing agents like $HIO_3$ or $HNO_3$.
$\text{CH}_4+\text{I}_2\rightleftharpoons\text{CH}_3\text{I}+\text{HI}$
$\text{HIO}_3+5\text{HI}\rightarrow3\text{I}_2+3\text{H}_2\text{O}$
Halogenation is supposed to proceed via free radical chain mechanism involving three steps namely initiation, propagation and termination.

The General combustion equation for any alkane is:
$\text{C}_\text{n}\text{H}_{2\text{n}+2}+\Bigg(\frac{3\text{n}+1}{2}\Bigg)\text{O}_2\rightarrow\text{nCO}_2+(\text{n}+1)\text{H}_2\text{O}$
Combustion
Alkanes on heating in the presence of air or dioxygen are completely oxidized to carbon dioxide and water with the evolution of large amount of heat.
Due to the evolution of large amount of heat during combustion, alkanes are used as fuels. During incomplete combustion of alkanes with insufficient amount of air or dioxygen, carbon black is formed which is used in the manufacture of ink, printer ink, black pigments and as filters.

Controlled oxidation Alkanes on heating with a regulated supply of dioxygen or air at high pressure and in the presence of suitable catalysts give a variety of oxidation products.
Ordinarily alkanes resist oxidation but alkanes having tertiary H atom can be oxidized to corresponding alcohols by potassium permanganate .
Pyrolysis Higher alkanes on heating to higher temperature decompose into lower alkanes, alkenes etc. Such a decomposition reaction into smaller fragments by the application of heat is called pyrolysis or cracking.

Pyrolysis of alkanes is believed to be a free radical reaction. Preparation of oil gas or petrol gas from kerosene oil or petrol involves the principle of pyrolysis. For example, dodecane, a constituent of kerosene oil on heating to 973K in the presence of platinum, palladium or nickel gives a mixture of heptane and pentene.

Conformations- Alkanes contain carbon-carbon sigma $(\sigma)$bonds. Electron distribution of the sigma molecular orbital is symmetrical around the internuclear axis of the C–C bond which is not disturbed due to rotation about its axis. This permits free rotation about C–C single bond. This rotation results into different spatial arrangements of atoms in space which can change into one another. Such spatial arrangements of atoms which can be converted into one another by rotation around a C-C single bond are called conformations or conformers or rotamers. Alkanes can thus have infinite number of conformations by rotation around C-C single bonds. However, it may be remembered that rotation around a C-C single bond is not completely free. It is hindered by a small energy barrier of $1^{-20}kJ\ mol–1$ due to weak repulsive interaction between the adjacent bonds. Such a type of repulsive interaction is called torsional strain. Conformations of ethane : Ethane molecule $(C_2H_6)$ contains a carbon – carbon single bond with each carbon atom attached to three hydrogen atoms. Considering the ball and stick model of ethane, keep one carbon atom stationary and rotate the other carbon atom around the C-C axis. This rotation results into infinite number of spatial arrangements of hydrogen atoms attached to one carbon atom with respect to the hydrogen atoms attached to the other carbon atom. These are called conformational isomers (conformers). Thus there are infinite number of conformations of ethane. However, there are two extreme cases. One such conformation in which hydrogen atoms attached to two carbons are as closed together as possible is called eclipsed conformation and the other in which hydrogens are as far apart as possible is known as the staggered conformation. Any other intermediate conformation is called a skew conformation. It may be remembered that in all the conformations, the bond angles and the bond lengths remain the same. Eclipsed and the staggered conformations can be represented by Sawhorse and Newman projections.

Alkanes contain carbon-carbon … bonds.

sigma $\sigma$
pi bond$\pi$
delta$\delta$
eta $\eta$

C-C single bond is hindered by a small energy barrier of…. $kJ ~mol^{–1}$

10 - 200
1 - 20
100 - 427
342 - 786

A decomposition reaction into smaller fragments by the application of heat is called as ….

pyrolysis
cracking
both (a) & (b)
combustion

Which of the following steps are involving in free radical chain mechanism

initiation
propagation
termination
All the above

The … reaction in which alkanes on heating in the presence of air or dioxygen are completely oxidized to carbon dioxide and water with the evolution of large amount of heat.

pyrolysis
cracking
both (a) & (b)
combustion

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Read the passage given below and answer the following questions from $1$ to $5$. Lithium metal is used to make useful alloys, for example with lead to make ‘white metal’ Bearings for motor engines, with aluminium to make aircraft parts, and with magnesium to make armour plates. It is used in Thermonuclear reactions. Lithium is also used to make electrochemical cells. Sodium is used To make a Na/Pb alloy needed to make $PbEt_4$ and $PbMe_4$. These organolead compounds were Earlier used as anti-knock additives to petrol, But nowadays vehicles use lead-free petrol. Liquid sodium metal is used as a coolant in Fast breeder nuclear reactors. Potassium has a vital role in biological systems. Potassium Chloride is used as a fertilizer. Potassium Hydroxide is used in the manufacture of soft Soap. It is also used as an excellent absorbent of carbon dioxide. Caesium is used in devising Photoelectric cells. Points of Difference between Lithium and other Alkali Metals –
i) Lithium is much harder. Its m.p. and b.p. are higher than the other alkali metals.
ii) Lithium is least reactive but the strongest Reducing agent among all the alkali metals. On combustion in air it forms mainly Monoxide, $Li_2O$ and the nitride, $Li_3N$ unlike Other alkali metals.
iii) LiCl is deliquescent and crystallises as a Hydrate, LiCl.$2H_2O$ whereas other alkali Metal chlorides do not form hydrates.
iv) Lithium hydrogencarbonate is not Obtained in the solid form while all other Elements form solid hydrogencarbonate.
v) Lithium unlike other alkali metals forms No ethynide on reaction with ethyne.
vi) Lithium nitrate when heated gives lithium Oxide, $Li_2O$, whereas other alkali metal Nitrates decompose to give the Corresponding nitrite. $4\text{LiNO}_3\rightarrow2\text{L}\text{i}_2\text{O}+4\text{NO}_2+\text{O}_2$ $2\text{NaNO}_3\rightarrow2\text{NaNO}_2+\text{O}_2$ vii) LiF and $Li_2O$ are comparatively much less Soluble in water than the corresponding Compounds of other alkali metals. Sodium carbonate is generally prepared by Solvay Process. In this process, advantage is Taken of the low solubility of sodium Hydrogencarbonate whereby it gets Precipitated in the reaction of sodium chloride with ammonium hydrogencarbonate. The Latter is prepared by passing $CO_2$ to a Concentrated solution of sodium chloride Saturated with ammonia, where ammonium Carbonate followed by ammonium Hydrogencarbonate are formed. The equations For the complete process may be written as $2\text{NH}_3+\text{H}_2\text{O}+\text{CO}_2\rightarrow{\text{(NH}_4)_2}\text{CO}_3$ $(\text{NH}_4)_2\text{CO}_3+\text{H}_2\text{O}+\text{CO}_2\rightarrow2\text{NH}_4\text{HCO}_3$ $\text{NH}_4\text{HCO}_3+\text{NaCl}\rightarrow\text{NH}_4\text{Cl}+\text{NaHCO}_3$ Sodium hydrogencarbonate crystal Separates. these are heated to give sodium Carbonate. The most abundant source of sodium chloride is sea water which contains $2.7$ to $2.9 \%$ by Mass of the salt. In tropical countries like India, Common salt is generally obtained by Evaporation of sea water. Approximately $50$ Lakh tons of salt are produced annually in India by solar evaporation. Crude sodium Chloride, generally obtained by crystallisation Of brine solution, contains sodium sulphate, Calcium sulphate, calcium chloride and Magnesium chloride as impurities. Calcium Chloride, $CaCl_2$, and magnesium chloride, $MgCl_2$ are impurities because they are Deliquescent (absorb moisture easily from the Atmosphere). To obtain pure sodium chloride, The crude salt is dissolved in minimum amount Of water and filtered to remove insoluble Impurities. The solution is then saturated with Hydrogen chloride gas. Crystals of pure Sodium chloride separate out. Calcium and Magnesium chloride, being more soluble than Sodium chloride, remain in solution. Sodium Hydroxide (Caustic Soda), NaOH is generally prepared Commercially by the electrolysis of sodium Chloride in Castner-Kellner cell. A brine Solution is electrolysed using a mercury Cathode and a carbon anode. Sodium metal Discharged at the cathode combines with Mercury to form sodium amalgam. Chlorine Gas is evolved at the anode. Cathod: $\text{Na}^++\bar{\text{e}}\xrightarrow{\text{Hg}}\text{Na}-\text{amalgam}$ Anode: $\text{Cl}^-\rightarrow\frac{1}{2}\text{Cl}_2+\text{e}^-$ The amalgam is treated with water to give Sodium hydroxxide and hydrogen gas. $2$ Na - amalgam $+ 2H_2O \rightarrow 2NaOH + 2Hg + H_2$

NaOH Sodium hydroxide is generally prepared Commercially by the electrolysis of … in Castner-Kellner cell.

$NaCl$
$Na_2CO_3$
$NaHCO_3$
$NaNH_2$

… is used in the manufacture of soft Soap.

Sodium Hydroxide
Potassium Hydroxide
Aluminium hydroxide
Beryllium hydroxide

… is used in devising Photoelectric cells.

Hydrogen
Lithium
Caesium
Helium

… compounds were Earlier used as anti-knock additives to petrol.

Organomagnesium
Organosilicon
Organochloride
Organolead

The sodium amalgam is treated with water to gives ….

$NaOH$
$Na_2CO_3$
$NaHCO_3$
$NaNH_2$

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Read the passage given below and answer the following questions from 1 to 5.

Hyperconjugation is a general stabilising interaction. It involves delocalisation of σ electrons of C—H bond of an alkyl group directly attached to an atom of unsaturated system or to an atom with an unshared p orbital. The σ electrons of C—H bond of the alkyl group enter into partial conjugation with the attached unsaturated system or with the unshared p orbital. Hyperconjugation is a permanent effect. To understand hyperconjugation effect, let us take an example of CH 32 + (ethyl cation) in which the positively charged carbon atom has an empty p orbital. One of the C-H bonds of the methyl group can align in the plane of this empty p orbital and the electrons constituting the C-H bond in plane with this p orbital can then be delocalised into the empty p orbital as depicted in Figure.
This type of overlap stabilises the carbocation because electron density from the adjacent σ bond helps in dispersing the positive charge.

In general, greater the number of alkyl groups attached to a positively charged carbon atom, the greater is the hyperconjugation interaction and stabilisation of the cation. Thus, we have the following relative stability of carbocations:

Hyperconjugation is also possible in alkenes and alkylarenes. Delocalisation of electrons by hyperconjugation in the case of alkene can be depicted as in Figure.
There are various ways of looking at the hyperconjugative effect. One of the way is to regard C—H bond as possessing partial ionic character due to resonance.
The hyperconjugation may also be regarded as no bond resonance.

The hyperconjugation may also be regarded as no bond resonance.
Methods of purification of organic compounds Once an organic compound is extracted from a natural source or synthesised in the laboratory, it is essential to purify it. Various methods used for the purification of organic compounds are based on the nature of the compound and the impurity present in it. The common techniques used for purification are as follows :
i) Sublimation
ii) Crystallisation
iii) Distillation
iv) Differential extraction and
v) Chromatography
Finally, the purity of a compound is ascertained by determining its melting or boiling point. Most of the pure compounds have sharp melting points and boiling points. New methods of checking the purity of an organic compound are based on different types of chromatographic and spectroscopic techniques.
Sublimation On heating, some solid substances change from solid to vapour state without passing through liquid state. The purification technique based on the above principle is known as sublimation and is used to separate sublimable compounds from non- sublimable impurities.
Crystallisation This is one of the most commonly used techniques for the purification of solid organic compounds. It is based on the difference in the solubilities of the compound and the impurities in a suitable solvent. The impure compound is dissolved in a solvent in which it is sparingly soluble at room temperature but appreciably soluble at higher temperature. The solution is concentrated to get a nearly saturated solution. On cooling the solution, pure compound crystallises out and is removed by filtration. The filtrate (mother liquor) contains impurities and small quantity of the compound. If the compound is highly soluble in one solvent and very little soluble in another solvent, crystallisation can be satisfactorily carried out in a mixture of these solvents. Impurities, which impart colour to the solution are removed by adsorbing over activated charcoal. Repeated crystallisation becomes necessary for the purification of compounds containing impurities of comparable solubilities.
Distillation This important method is used to separate
i) volatile liquids from nonvolatile impurities and
ii) the liquids having sufficient difference in their boiling points.

Liquids having different boiling points vaporise at different temperatures. The vapours are cooled and the liquids so formed are collected separately. Chloroform (b.p 334 K) and aniline (b.p. 457 K) are easily separated by the technique of distillation (Fig 12.5). The liquid mixture is taken in a round bottom flask and heated carefully. On boiling, the vapours of lower boiling component are formed first. The vapours are condensed by using a condenser and the liquid is collected in a receiver. The vapours of higher boiling component form later and the liquid can be collected separately.
Partition Chromatography: Partition chromatography is based on continuous differential partitioning of components of a mixture between stationary and mobile phases. Paper chromatography is a type of partition chromatography. In paper chromatography, a special quality paper known as chromatography paper is used. Chromatography paper contains water trapped in it, which acts as the stationary phase. A strip of chromatography paper spotted at the base with the solution of the mixture is suspended in a suitable solvent or a mixture of solvents (Fig. 12.13). This solvent acts as the mobile phase. The solvent rises up the paper by capillary action and flows over the spot. The paper selectively retains different components according to their differing partition in the two phases. The paper strip so developed is known as a chromatogram. The spots of the separated coloured compounds are visible at different heights from the position of initial spot on the chromatogram. The spots of the separated colourless compounds may be observed either under ultraviolet light or by the use of an appropriate spray reagent as discussed under thin layer chromatography.

Hyperconjunction involves delocalisation of … electrons of C—H bond of an alkyl group directly attached to an atom of unsaturated system or to an atom with an unshared p orbital.

$\sigma$
$\pi$
$\delta$
$\eta$

Which of the is an example of technique used for purification.

Distillation
Differential extraction
Chromatography
All the above

On heating, some solid substances change from solid to vapour state without passing through liquid state is known as …

Melting
Boiling
Sublimation
Condensation

The hyperconjugation may also be regarded as ….

bonding resonance
no bond resonance
no bond induction
bonding induction

Chromatography paper contains water trapped in it, which acts as the … phase.

mobile
stationery
Secondary
quaternary

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Read the passage given below and answer the following questions from $(i)$ to $(v).$
Predicting the Direction of the Reaction- The equilibrium constant helps in predicting the direction in which a given reaction will proceed at any stage. For this purpose, we calculate the reaction quotient $Q$. The reaction quotient, $Q (Qc$ with molar concentrations and $QP$ with partial pressures$)$ is defined in the same way as the equilibrium constant Kc except that the concentrations in $Qc$ are not necessarily equilibrium values. For a general reaction:
$\text{a}\text{A}+\text{b}\text{B}\rightleftharpoons\text{c}\text{C}+\text{d}\text{D}$
$\text{Q}\text{c}=\frac{[\text{C}]^\text{c}[\text{D}]^\text{d}}{[\text{A}]^\text{a}[\text{B}]^\text{b}}$
Then,
If $Qc > Kc,$ the reaction will proceed in the direction of reactants (reverse reaction).
If $Qc < Kc,$ the reaction will proceed in the direction of the products (forward reaction).
If $Qc = Kc,$ the reaction mixture is already at equilibrium. Consider the gaseous reaction of $H_2$ with $I_2$ ,
$\text{H}_{2(\text{g})}+\text{l}_{2(\text{g})}\rightleftharpoons2\text{Hl}_{(\text{g})};\text{kc}=57.0\text{at}700\text{k}.$
Suppose we have molar concentrations $[H_2 ]t =0.10M, [I_2 ]t = 0.20 M$ and $[HI]t = 0.40 M.$ (the subscript t on the concentration symbols means that the concentrations were measured at some arbitrary time t, not necessarily at equilibrium). Thus, the reaction quotient, Qc at this stage of the reaction is given by,
$\text{Qc}=\frac{[\text{Hl}]\text{t}^2}{[\text{H}]^2]_\text{t}[\text{l}_2]_\text{t}}=\frac{(0.40)_2}{(0.10)\times(0.20)}=8.0$
Now, in this case, $Qc (8.0)$ does not equal Kc $(57.0)$, so the mixture of $H2 _{(g)}, I2 _{(g)} $ and $HI_{(g)} $ is not at equilibrium; that is, more $H2 _{(g)} $ and $I 2 _{(g)} $ will react to form more $HI_{(g)} $ and their concentrations will decrease till $Qc = Kc.$ The reaction quotient, Qc is useful in predicting the direction of reaction by comparing the values of Qc and Kc.Thus, we can make the following generalisations concerning the direction of the reaction
If $Qc < Kc,$ net reaction goes from left to right
If $Qc > Kc,$ net reaction goes from right to left.
If $Qc = Kc,$ no net reaction occurs.
Calculating Equilibrium Concentrations In case of a problem in which we know the initial concentrations but do not know any of the equilibrium concentrations, the following three steps shall be followed:
Step 1) Write the balanced equation for the reaction.
Step 2) Under the balanced equation, make a table that lists for each substance involved in the reaction: $(a)$ the initial concentration, $(b)$ the change in concentration on going to equilibrium, and $(c)$ the equilibrium concentration. In constructing the table, define $x$ as the concentration (mol/L) of one of the substances that reacts on going to equilibrium, then use the stoichiometry of the reaction to determine the concentrations of the other substances in terms of $x.$
Step 3) Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x. If you are to solve a quadratic equation choose the mathematical solution that makes chemical sense.
Step 4) Calculate the equilibrium concentrations from the calculated value of $x.$
Step 5) Check your results by substituting them into the equilibrium equation.
Relationship between equilibrium constant K, reaction quotient Q and gibbs energy G The value of Kc for a reaction does not depend on the rate of the reaction. However, it is directly related to the thermodynamics of the reaction and in particular, to the change in Gibbs energy, $\triangle\text{G}.$ If,
$\triangle\text{G}$ is negative, then the reaction is spontaneous and proceeds in the forward direction.
$\triangle\text{G}$ is positive, then reaction is considered non-spontaneous. Instead, as reverse reaction would have a negative $\triangle\text{G},$ the products of the forward reaction shall be converted to the reactants.
$\triangle\text{G}$ is 0, reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction. A mathematical expression of this thermodynamic view of equilibrium can be described by the following equation:
$\triangle\text{G}=\triangle\text{G}^\phi+\text{RT}\text{lnQ}$
where, $\triangle\text{G}^\phi$ is standard Gibbs energy. At equilibrium, when $\triangle\text{G}=0$ and $Q = Kc,$ the equation becomes,
$\triangle\text{G}=\text{G}^\phi+\text{RT}\text{lnk}=0$
$\triangle\text{G}^\phi=-\text{RT}\text{lnk}$
$\text{Ink}=\frac{-\triangle\text{G}^\phi}{\text{RT}}$
Taking antilog of both sides, we get,
$\text{K}=\text{e}-\frac{\triangle\text{G}0}{\text{RT}}$
Hence, using the equation, the reaction spontaneity can be interpreted in terms of the value of $\triangle\text{G}^\phi.$
If $\triangle\text{G}^\phi>0$ then $\frac{-\triangle\text{G}^\phi}{\text{RT}}$ is positive, and $>1$, making $K > 1$, which implies a spontaneous reaction or the reaction which proceeds in the forward direction to such an extent that the products are present predominantly.
If $\triangle\text{G}^\phi>0,$ then $\frac{-\triangle\text{G}^\phi}{\text{RT}}$ is negative, and $< 1$, that is, $K < 1$, which implies a non-spontaneous reaction or a reaction which proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.
Factors affecting equilibria One of the principal goals of chemical synthesis is to maximise the conversion of the reactants to products while minimizing the expenditure of energy. This implies maximum yield of products at mild temperature and pressure conditions. If it does not happen, then the experimental conditions need to be adjusted. For example, in the Haber process for the synthesis of ammonia from $N_2$ and $H_2,$ the choice of experimental conditions is of real economic importance. Annual world production of ammonia is about hundred million tones, primarily for use as fertilizers. Equilibrium constant, Kc is independent of initial concentrations. But if a system at equilibrium is subjected to a change in the concentration of one or more of the reacting substances, then the system is no longer at equilibrium; and net reaction takes place in some direction until the system returns to equilibrium once again. Similarly, a change in temperature or pressure of the system may also alter the equilibrium. In order to decide what course the reaction adopts and make a qualitative prediction about the effect of a change in conditions on equilibrium we use Le Chatelier’s principle. It states that a change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change. This is applicable to all physical and chemical equilibria.
Effect of Concentration Change In general, when equilibrium is disturbed by the addition/removal of any reactant/ products, Le Chatelier’s principle predicts that:
The concentration stress of an added reactant/product is relieved by net reaction in the direction that consumes the added substance.
The concentration stress of a removed reactant/product is relieved by net reaction in the direction that replenishes the removed substance. or in other words, “When the concentration of any of the reactants or products in a reaction at equilibrium is changed, the composition of the equilibrium mixture changes so as to minimize the effect of concentration changes”. Let us take the reaction,
$\text{H}_{2(\text{g})}+\text{l}_{2(\text{g})}\rightleftharpoons2\text{Hl}_{(\text{g})}$
If $H_2$ is added to the reaction mixture at equilibrium, then the equilibrium of the reaction is disturbed. In order to restore it, the reaction proceeds in a direction wherein $H_2$ is consumed, i.e., more of $H_2$ and $I_2$ react to form HI and finally the equilibrium shifts in right (forward) direction. This is in accordance with the Le Chatelier’s principle which implies that in case of addition of a reactant/product, a new equilibrium will be set up in which the concentration of the reactant/product should be less than what it was after the addition but more than what it was in the original mixture. The same point can be explained in terms of the reaction quotient, $Qc,$

Case Study Questions Class 11 Chemistry – Equilibrium

$\text{Qc}=\frac{[\text{HI}]^2}{[\text{H}]_2[\text{I}]_2}$
Addition of hydrogen at equilibrium results in value of $Qc$ being less than $Kc$. Thus, in order to attain equilibrium again reaction moves in the forward direction. Similarly, we can say that removal of a product also boosts the forward reaction and increases the concentration of the products and this has great commercial application in cases of reactions, where the product is a gas or a volatile substance. In case of manufacture of ammonia, ammonia is liquified and removed from the reaction mixture so that reaction keeps moving in forward direction. Similarly, in the large scale production of $CaO$ (used as important building material) from $CaCO_3$, constant removal of $CO_2$ from the kiln drives the reaction to completion. It should be remembered that continuous removal of a product maintains Qc at a value less than Kc and reaction continues to move in the forward direction.

If … the reaction will proceed in the direction of reactants (reverse reaction).

$Qc > Kc$
$Qc < Kc$
$Qc = Kc$
None of above

If … the reaction will proceed in the direction of the products (forward reaction).

$Qc > Kc$
$Qc < Kc$
$Qc = Kc$
None of above

If … the reaction mixture is already at equilibrium. Consider the gaseous reaction.

$Qc > Kc$
$Qc < Kc$
$Qc = Kc$
All of above

If $\triangle\text{G}$ is …. then the reaction is spontaneous and proceeds in the forward direction.

Zero
Positive
Negative
None of above

$\triangle\text{G}$ is … reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction.

Zero
Positive
Negative
None of above

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IUPAC (International Union of Pure and Applied Chemistry) system of nomenclature. Common names are useful and in many cases indispensable, particularly when the alternative systematic names are lengthy and complicated. A systematic name of an organic compound is generally derived by identifying the parent hydrocarbon and the functional group(s) attached to it. By using prefixes and suffixes, the parent name can be modified to obtain the actual name. In a branched-chain compound, small chains of carbon atoms are attached at one or more carbon atoms of the parent chain. The small carbon chains (branches) are called alkyl groups. An alkyl group is derived from a saturated hydrocarbon by removing a hydrogen atom from carbon. Abbreviations are used for some alkyl groups. For example, methyl is abbreviated as Me, ethyl as Et, propyl as Pr and butyl as Bu.

1. Draw the structure of 3-Ethyl-4,4-dimethylheptane. (1)
2. How is the numbering in branched chain hydrocarbon done?
3. Derive the structure of 2-Chlorohexane. (2)
OR
Why $CH _4$ after becoming- $CH _3$ called a methyl group? (2)

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The existing large number of organic compounds and their ever-increasing numbers has made it necessary to classify them on the basis of their structures. Organic compounds are broadly classified as open-chain compounds which are also called aliphatic compounds. Aliphatic compounds further classified as homocyclic and heterocyclic compounds. Aromatic compounds are special types of compounds. Alicyclic compounds, aromatic compounds may also have heteroatom in the ring. Such compounds are called heterocyclic aromatic compounds. Organic compounds can also be classified on the basis of functional groups, into families or homologous series. The members of a homologous series can be represented by general molecular formula and the successive members differ from each other in a molecular formula by a $- CH _2$ unit.

1. The successive members of a homologous series differ by which mass of amu?
OR
Is tetrahydrofuran is aromatic compounds?
2. Does Pyridine, pyrrole, thiophene are all heteroaromatic compounds
3. Difference between heterocyclic and homocyclic compound.

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Read the passage given below and answer the following questions from 1 to 5.
The s-block elements of the Periodic Table are those in which the last electron enters the outermost s-orbital. as the s-orbital can accommodate only two electrons, two Groups (1 & 2) belong to the s-block of the Periodic Table. Group 1 of the Periodic Table consists of the elements: Lithium, sodium, potassium, rubidium, caesium and Francium. They are collectively known as the alkali metals. These are so called because they form hydroxides on Reaction with water which are strongly alkaline in nature. The elements of Group 2 include beryllium, magnesium, Calcium, strontium, barium and radium. These elements With the exception of beryllium are commonly known as The alkaline earth metals. These are so called because their Oxides and hydroxides are alkaline in nature and these Metal oxides are found in the earth’s crust. The general electronic configuration of s-block elements is [noble gas] ns1 for alkali metals and [noble gas] $ns^2$ for Alkaline earth metals.

All the alkali metals have one valence electron, ns1 outside the noble gas core. The loosely held s-electron in the outermost Valence shell of these elements makes them the Most electropositive metals. They readily lose Electron to give monovalent M+ Ions. The monovalent ions (M+) are smaller than the parent atom. Hence they Are never found in free state in nature.
The alkali metal atoms have the largest sizes In a particular period of the periodic table. With increase in atomic number, the atom becomes Larger. the atomic and ionic Radii of alkali metals increase on moving down the group i.e., they increase in size while going From Li to Cs.
The ionization enthalpies of the alkali metals Are considerably low and decrease down the Group from Li to Cs. this is because the effect of increasing size outweighs the increasing Nuclear charge, and the outermost electron is very well screened from the nuclear charge.
The hydration enthalpies of alkali metal ions Decrease with increase in ionic sizes. $Li^+ > Na^+ > K^+ > Rb^+ > Cs^+ > Li^+$ Has maximum degree of hydration and For this reason lithium salts are mostly Hydrated, e.g., $LiCl·2H_2O$.
All the alkali metals are silvery white, soft and Light metals. Because of the large size, these Elements have low density which increases down The group from Li to Cs. However, potassium is Lighter than sodium. The melting and boiling Points of the alkali metals are low indicating Weak metallic bonding due to the presence of Only a single valence electron in them. The alkali Metals and their salts impart characteristic Colour to an oxidizing flame. This is because the Heat from the flame excites the outermost orbital Electron to a higher energy level. When the excited Electron comes back to the ground state . Alkali metals can therefore, be detected by The respective flame tests and can be Determined by flame photometry or atomic Absorption spectroscopy. These elements when Irradiated with light, the light energy absorbed May be sufficient to make an atom lose electron. This property makes caesium and potassium Useful as electrodes in photoelectric cells.
The alkali metals are highly reactive due to Their large size and low ionization enthalpy. The Reactivity of these metals increases down the Group.
Reactivity towards air: The alkali metals Tarnish in dry air due to the formation of Their oxides which in turn react with Moisture to form hydroxides. They burn Vigorously in oxygen forming oxides. Lithium forms monoxide, sodium forms Peroxide, the other metals form Superoxide. The superoxide $O^{2–}$ Ion is Stable only in the presence of large cations Such as K, Rb, Cs.
Reactivity towards water: The alkali Metals react with water to form hydroxide And dihydrogen.
$2\text{M}+2\text{H}_2\text{O}\rightarrow2\text{M}^++2\text{OH}^-+\text{H}_2$
(M = analkali metal)
Reactivity towards dihydrogen: The Alkali metals react with dihydrogen at About 673K (lithium at 1073K) to form Hydrides. All the alkali metal hydrides are Ionic solids with high melting points.
Reactivity towards halogens: The alkali Metals readily react vigorously with Halogens to form ionic halides, $M^+X^–$ .
Reducing nature: The alkali metals are Strong reducing agents, lithium being the Most and sodium the least powerful.
Solutions in liquid ammonia: The alkali Metals dissolve in liquid ammonia giving Deep blue solutions which are conducting in nature.

The general electronic configuration of s-block elements is … for alkali metals.

[noble gas] $ns^1$
[noble gas] $ns^2$
[noble gas] $ns^1np^1$
[noble gas] $ns^1np^2$

The general electronic configuration of s-block elements is … for alkaline earth metals.

noble gas] $ns^1$
[noble gas] $ns^2$
[noble gas] $ns^1np^1$
[noble gas] $ns^1np^2$

The atomic and ionic Radii of alkali metals … on moving down the group.

constant
decrease
increase
All the above

The hydration enthalpies of alkali metal ions … with … in ionic sizes.

increase, decrease
increase, increase
decrease, decrease
decrease, increase

Which of the following element is strong reducing agent?

Lithium
Sodium
Fluorine
Helium

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Read the passage given below and answer the following questions from 1 to 5.
Alkynes– Like alkenes, alkynes are also unsaturated hydrocarbons. They contain at least one triple bond between two carbon atoms. The number of hydrogen atoms is still less in alkynes as compared to alkenes or alkanes. Their general formula is $C_nH_{2n–2}.$ The first stable member of alkyne series is ethyne which is popularly known as acetylene. Acetylene is used for arc welding purposes in the form of oxyacetylene flame obtained by mixing acetylene with oxygen gas. Alkynes are starting materials for a large number of organic compounds. Hence, it is interesting to study this class of organic compounds.
Structure of Triple Bond Ethyne is the simplest molecule of alkyne series. Structure of ethyne is shown in Figure. Each carbon atom of ethyne has two sp hybridised orbitals. Carbon-carbon sigma $(\sigma)$ bond is obtained by the head-on overlapping of the two sp hybridised orbitals of the two carbon atoms. The remaining sp hybridised orbital of each carbon atom undergoes overlapping along the internuclear axis with the 1s orbital of each of the two hydrogen atoms forming two C-H sigma bonds. H-C-C bond angle is of $180^\circ .$

Each carbon has two unhybridised p orbitals which are perpendicular to each other as well as to the plane of the C C sigma bond. The $2 p$ orbitals of one carbon atom are parallel to the $2 p$ orbitals of the other carbon atom, which undergo lateral or sideways overlapping to form two pi ( $\pi$ ) bonds between two carbon atoms. Thus ethyne molecule consists of one $C - C \sigma$ bond, two $C - H \sigma$ bonds and two $C - C \pi$ bonds. The strength of $C \equiv C$ bond (bond enthalpy $823 kJ mol ^{-1}$ ) is more than those of $C = C$ bond (bond enthalpy $681 kJ mol ^{-1}$ ) and $C - C$ bond (bond enthalpy 348 kJ $mol ^{-1}$ ). The $C \equiv C$ bond length is shorter $(120 pm )$ than those of $C = C (133 pm )$ and $C - C (154 pm )$. Electron cloud between two carbon atoms is cylindrically symmetrical about the internuclear axis. Thus, ethyne is a linear molecule. Preparation - From calcium carbide: On industrial scale, ethyne is prepared by treating calcium carbide with water. Calcium carbide is prepared by heating quick lime with coke. Quick lime can be obtained by heating limestone as shown in the following reactions:

$\text{C}\text{aC}_2+2\text{H}_2\text{O}\rightarrow\text{Ca}(\text{OH})_2+\text{C}_2\text{H}_2$
From vicinal dihalides: Vicinal dihalides on treatment with alcoholic potassium hydroxide undergo dehydrohalogenation. One molecule of hydrogen halide is eliminated to form alkenyl halide which on treatment with sodamide gives alkyne.

Aromatic hydrocarbon- These hydrocarbons are also known as ‘arenes’. Since most of them possess pleasant odour, the class of compounds was named as ‘aromatic compounds’. Most of such compounds were found to contain benzene ring. Benzene ring is highly unsaturated but in a majority of reactions of aromatic compounds, the unsaturation of benzene ring is retained. However, there are examples of aromatic hydrocarbons which do not contain a benzene ring but instead contain other highly unsaturated 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:

Friedel-Crafts alkylation reaction: When benzene is treated with an alkyl halide in the presence of anhydrous aluminium chloride, alkylbenene is formed.
Friedel-Crafts acylation reaction: The reaction of benzene with an acyl halide or acid anhydride in the presence of Lewis acids ($AlCl_3$) yields acyl benzene.

If excess of electrophilic reagent is used, further substitution reaction may take place in which other hydrogen atoms of benzene ring may also be successively replaced by the electrophile. For example, benzene on treatment with excess of chlorine in the presence of anhydrous $AlCl_3$ can be chlorinated to hexachlorobenzene $(C_6Cl_6)$

The general formula of Alkynes is …?

$C_nH_{2n–2}$
$C_nH_{2n–2}$
$C_nH_{2n–2}$
$C_nH_{2n–2}$

Calcium carbide is prepared by heating quick lime with …?

Backing soda
Coke
Carbide
Salt

Vicinal dihalides on treatment with alcoholic potassium hydroxide undergo …?

Dehydrogenation.
Hydrohalogenation.
Dehydrohalogenation
Dehalogenation.

The bond enthalpy of C ≡ C is …?

523 kJ mol
623 kJ mol
723 kJ mol
823 kJ mol

The C ≡ C bond length is …?

1200 pm
620 pm
240 pm
120 pm

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