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Redox Reactions — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryRedox Reactions4 Marks
Question
Read the passage given below and answer the following questions from 1 to 5. Chemistry deals with varieties of matter and change of one kind of matter into the other. Transformation of matter from one kind into another occurs through the various types of reactions. One important category of such reactions is Redox Reactions. Originally, the term oxidation was used to describe the addition of oxygen to an element or a compound. Because of the presence of dioxygen in the atmosphere (~20%), many elements combine with it and this is the principal reason why they commonly occur on the earth in the form of their oxides. The following reactions represent oxidation processes: $2\text{Mg}(\text{s})\rightarrow2\text{MgO}\text{ (S)}$ $\text{S}(\text{s})+\text{O}_2(\text{g})\rightarrow\text{SO}_2\text{ g}$ the term “oxidation” is defined as the addition of oxygen/electronegative element to a substance or removal of hydrogen/ electropositive element from a substance. In the beginning, reduction was considered as removal of oxygen from a compound. However, the term reduction has been broadened these days to include removal of oxygen/electronegative element from a substance or addition of hydrogen/ electropositive element to a substance. According to the definition given above, the following are the examples of reduction processes: $2\text{HgO}\text{S}\xrightarrow{\triangle}2\text{ Hg}(1)+\text{O}_2(\text{g})$ $2\text{HgCl}_2(\text{aq})+\text{SnCl}_2(\text{aq})\rightarrow\text{Hg}_2\text{Cl}_2(\text{s})+\text{SnCl}_4(\text{aq})$ In reaction simultaneous oxidation of stannous chloride to stannic chloride is also occurring because of the addition of electronegative element chlorine to it. It was soon realised that oxidation and reduction always occur simultaneously (as will be apparent by re-examining all the equations given above), hence, the word “redox” was coined for this class of chemical reactions.The reactions:
$2\text{Na}(\text{s})+\text{Cl2}\text{g}\rightarrow2\text{NaCl}(\text{s})$ $4\text{Na}(\text{s})+\text{O}_2\text{g}\rightarrow2\text{Na}_2\text{o}(\text{s})$ $2\text{Na}(\text{s})+\text{S}\text{(s)}\rightarrow2\text{Na}_2\text{S}(\text{s})$ are redox reactions because in each of these reactions sodium is oxidised due to the addition of either oxygen or more electronegative element to sodium. Simultaneously, chlorine, oxygen and sulphur are reduced because to each of these, the electropositive element sodium has been added. From our knowledge of chemical bonding we also know that sodium chloride, sodium oxide and sodium sulphide are ionic compounds and perhaps better written as $\mathrm{Na}+\mathrm{Cl}-(\mathrm{s}),\left(\mathrm{Na}^{+}\right)_2 \mathrm{O}^2-(\mathrm{s})$, and $\left(\mathrm{Na}^{+}\right)_2 \mathrm{~S}^{2-}(\mathrm{s})$. Development of charges on the species produced suggests us to rewrite the reactions in the following manner: For convenience, each of the above processes can be considered as two separate steps, one involving the loss of electrons and the other the gain of electrons. As an illustration, we may further elaborate one of these, say, the formation of sodium chloride. $2\text{Na}(\text{s})\rightarrow2\text{Na}^+\text{g}+2\bar{\text{e}}$ $\text{Cl}_2\text{g}+2\bar{\text{e}}\rightarrow2\text{C}\bar{\text{I}}\text{ (g)}$ Each of the above steps is called a half reaction, which explicitly shows involvement of electrons. Sum of the half reactions gives the overall reaction $2\text{Na}(\text{s})+\text{Cl}_2\text{(g)}\rightarrow2\text{Na}^+\text{CI}(\text{s})\text{ or } 2\text{NaCI}(\text{s})$ Above Reactions suggest that half reactions that involve loss of electrons are called oxidation reactions. Similarly, the half reactions that involve gain of electrons are called reduction reactions. To summarise, we may mention that Oxidation: Loss of electron(s) by any species. Reduction: Gain of electron(s) by any species. Oxidising agent: Acceptor of electron(s). Reducing agent: Donor of electron(s).
  1. Addition of electronegative element to a substance is known as..
  1. Oxidation
  2. Reduction
  3. Redox reaction
  4. All the above
  1. Removal of electronegative element to a substance is known as ..
  1. Oxidation
  2. Reduction
  3. Redox reaction
  4. All the above
  1. Acceptor of electrons is …
  1. Reducing Agent
  2. Catalytic Agent
  3. Oxidising Agent
  4. None of above
  1. Donor of electrons is…
  1. Organic Agent
  2. Catalytic Agent
  3. Oxidising Agent
  4. Reducing Agent
  1. Oxidation and Reduction occurs simultaneously is known as …
  1. Exothermic reaction
  2. Endothermic reaction
  3. Redox reaction
  4. Neutralization reaction
✓

Answer

  1. (a) Oxidation
  2. (b) Reduction
  3. (c) Oxidising Agent
  4. (d) Reducing Agent
  5. (c) Redox reaction

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Read the passage given below and answer the following questions from (i) to (v).
In a chemical reaction, reactants are converted into products and is represented by, Reactants → Products The enthalpy change accompanying a reaction is called the reaction enthalpy. The enthalpy change of a chemical reaction, is given by the symbol $\triangle\text{rH}.$
$\triangle\text{rH}$ = (sum of enthalpies of products) – (sum of enthalpies of reactants)
$\sum\limits_\text{t}\text{a}_{\text{t}}\text{H}_\text{products}-\sum\limits_\text{t}\text{b}_\text{t}\text{H}_\text{reactants}$
Here symbol ∑ (sigma) is used for summation and ai and bi are the stoichiometric coefficients of the products and reactants respectively in the balanced chemical equation. For example, for the reaction
$\text{CH}_4(\text{g})+2\text{O}_2(\text{g})\rightarrow\text{CO}_2(\text{g})+2\text{H}_2\text{O}(\text{l})$
$\triangle_\text{r}\text{H}=\sum\limits_\text{t}\text{a}_{\text{t}}\text{H}_\text{products}-\sum\limits_\text{t}\text{b}_\text{t}\text{H}_\text{reactants}$
$=[\text{H}_\text{m}(\text{CO}_2,\text{g})+2\text{H}_\text{m}(\text{H}_2\text{O},\text{l})]-[\text{H}_\text{m}(\text{CH}_4,\text{g})+2\text{H}_\text{m}(\text{O}_2,\text{g})]$
where $H_m$ is the molar enthalpy. Enthalpy change is a very useful quantity. Knowledge of this quantity is required when one needs to plan the heating or cooling required to maintain an industrial chemical reaction at constant temperature. It is also required to calculate temperature dependence of equilibrium constant.
Standard Enthalpy of Reactions Enthalpy of a reaction depends on the conditions under which a reaction is carried out. It is, therefore, necessary that we must specify some standard conditions. The standard enthalpy of reaction is the enthalpy change for a reaction when all the participating substances are in their standard states. The standard state of a substance at a specified temperature is its pure form at 1 bar. For example, the standard state of liquid ethanol at 298K is pure liquid ethanol at 1 bar; standard state of solid iron at 500K is pure iron at 1 bar. Usually data are taken at 298K. Standard conditions are denoted by adding the superscript 0 to the symbol $\triangle\text{H},$ e.g., $\triangle\text{H}^\phi$
Enthalpy Changes during Phase Transformations Phase transformations also involve energy changes. Ice, for example, requires heat for melting. Normally this melting takes place at constant pressure (atmospheric pressure) and during phase change, temperature remains constant (at 273K).
$\text{H}_2\text{O}(\text{s})\rightarrow\text{H}_2\text{O}(\text{l});\triangle_{\text{fus}}\text{H}^\phi=6.00\text{kJ}\ \text{mol}^{-1}$
Here $\triangle\text{vap}\text{H}^\phi$ is enthalpy of fusion in standard state. If water freezes, then process is reversed and equal amount of heat is given off to the surroundings. The enthalpy change that accompanies melting of one mole of a solid substance in standard state is called standard enthalpy of fusion or molar enthalpy of fusion, $\triangle\text{fus}\text{H}0.$Melting of a solid is endothermic, so all enthalpies of fusion are positive. Water requires heat for evaporation. At constant temperature of its boiling point Tb and at constant pressure:
$\text{H}_2\text{O}(\text{l})\rightarrow\text{H}_2\text{O}(\text{g});\triangle_{\text{vap}}\text{H}^\phi=+40.79\text{kJ}\ \text{mol}^{-1}$
$\triangle\text{vap}\text{H}^\phi$ is the standard enthalpy of vaporisation. Amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure (1bar) is called its standard enthalpy of vaporization or molar enthalpy of vaporization, $\triangle\text{vap}\text{H}^\phi.$ Sublimation is direct conversion of a solid into its vapour. Solid $CO_2 $or ‘dry ice’ sublimes at 195K with $\triangle\text{sub}\text{H}^\phi=25.2\text{kJ}\text{mol}^{–1};$ naphthalene sublimes slowly and for this $\triangle\text{sub}\text{H}0= 73.0\text{kJ}\text{mol}^{–1}.$ Standard enthalpy of sublimation, $\triangle\text{sub}\text{H}^\phi$ is the change in enthalpy when one mole of a solid substance sublimes at a constant temperature and under standard pressure (1bar). The magnitude of the enthalpy change depends on the strength of the intermolecular interactions in the substance undergoing the phase transfomations. For example, the strong hydrogen bonds between water molecules hold them tightly in liquid phase. For an organic liquid, such as acetone, the intermolecular dipole-dipole interactions are significantly weaker. Thus, it requires less heat to vaporise 1 mol of acetone than it does to vaporize 1mol of water.
Standard Enthalpy of Formation The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (also known as reference states) is called Standard Molar Enthalpy of Formation. Its symbol is $\triangle\text{f}\text{H}^\phi$ where the subscript ‘ f ’ indicates that one mole of the compound in question has been formed in its standard state from its elements in their most stable states of aggregation. The reference state of an element is its most stable state of aggregation at $25^\circ C$ and 1 bar pressure.
Hess’s Law of Constant Heat Summation We know that enthalpy is a state function, therefore the change in enthalpy is independent of the path between initial state (reactants) and final state (products). In other words, enthalpy change for a reaction is the same whether it occurs in one step or in a series of steps. This may be stated as follows in the form of Hess’s Law. If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature. Let us understand the importance of this law with the help of an example. Consider the enthalpy change for the reaction
$\text{C}(\text{graphite,s})+\frac{1}{2}\text{O}_2(\text{g})\rightarrow\text{CO}(\text{g});\triangle_\text{r}\text{H}^{\ominus}=?$
Although CO(g) is the major product, some $CO_2 $gas is always produced in this reaction. Therefore, we cannot measure enthalpy change for the above reaction directly. However, if we can find some other reactions involving related species, it is possible to calculate the enthalpy change for the above reaction. Let us consider the following reactions:
$\text{C}(\text{graphite,s})+\text{O}_2(\text{g}) \rightarrow\text{CO}_2(\text{g});\triangle\text{r}\text{H}^{\phi}=–393.5\text{kJ}\text{mol}^{–1}(\text{i})$
$\text{CO}(\text{g})+\frac{1}{2}\text{O}_2(\text{g})\rightarrow\text{CO}_2(\text{g})\triangle_\text{r}\text{H}^{\phi}=-283.0\text{kJ}\text{mol}^{-1}(\text{ii})$
We can combine the above two reactions in such a way so as to obtain the desired reaction. To get one mole of CO(g) on the right, we reverse equation (ii). In this, heat is absorbed instead of being released, so we change sign of $\triangle\text{r}\text{H}^\phi$ value
$\text{CO}_2(\text{g})\rightarrow\text{CO}(\text{g})+\frac{1}{2}\text{O}_2(\text{g});\triangle\text{r}\text{H}^{\phi}=+283.0\text{kJ}\text{mol}^{-1}...(\text{iii})$
Adding equation (i) and (iii), we get the desired equation,
$\text{C}(\text{graphite,s})+\frac{1}{2}\text{O}_2(\text{g})\rightarrow\text{CO}(\text{g});$
for which $\triangle_\text{r}\text{H}^{\phi}=(-393.5+283.0)=-110.5\text{kJ}\text{mol}^{-1}$
In general, if enthalpy of an overall reaction A → B along one route is $\triangle\text{rH}$ and$​​\triangle\text{rH}_1,\triangle\text{rH}_2,\triangle\text{rH}_3$ representing enthalpies of reactions leading to same product, B along another route, then we have
$\triangle\text{rH}=​​\triangle\text{rH}_1+\triangle\text{rH}_2+\triangle\text{rH}_3$

It can be represented as:
  1. The enthalpy change of a chemical reaction, is given by the symbol …
  1. $\triangle\text{rH}$
  2. $\triangle\text{rG}$
  3. $\triangle\text{rF}$
  4. $\triangle\text{rR}$'
  1. The molar enthalpy is denoted by:
  1. $H_k$
  2. $H_m$
  3. $H_l$
  4. $H_n$
  1. …is enthalpy of fusion in standard state.
  1. $\triangle\text{fus}\text{H}^{\phi}$
  2. $\triangle_\text{r}\text{H}^{\phi}$
  3. $\triangle\text{vap}\text{H}^{\phi}$
  4. $\triangle\text{w}\text{H}^{\phi}$
  1. Solid $CO_2$or ‘dry ice’ sublimes at..
  1. $100K$
  2. $195K$
  3. $150K$
  4. $200K$
  1. … is the standard enthalpy of vaporisation.
  1. $\triangle\text{fus}\text{H}^{\phi}$
  2. $\triangle_\text{r}\text{H}^{\phi}$
  3. $\triangle\text{vap}\text{H}^{\phi}$
  4. $\triangle\text{w}\text{H}^{\phi}$
→
Chromatography is an important technique extensively used to separate mixtures into their components, purify compounds and also test the purity of compounds. Based on the principle involved, chromatography is classified into different categories. Two of these are Adsorption chromatography and Partition chromatography. Two main types of chromatographic techniques are based on the principle of differential adsorption column chromatography, and thin-layer chromatography. Adsorption chromatography is based on the fact that different compounds are adsorbed on an adsorbent to different degrees. Column chromatography involves the separation of a mixture over a column of adsorbent (stationary phase) packed in a glass tube. Thin-layer chromatography (TLC) is another type of adsorption chromatography, which involves the separation of substances of a mixture over a thin layer of an adsorbent coated on a glass plate. Partition chromatography is based on the continuous differential partitioning of components of a mixture between stationary and mobile phases.

1. Which adsorbent is used in adsorption chromatography?
2. How do you visualize colourless compounds after separation in Paper Chromatography?
3. Why paper chromatography is a type of partition chromatography?
OR
Which chromatography is shown in following image?
Image
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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? (1)
2. Does Pyridine, pyrrole, thiophene are all heteroaromatic compounds (1)
3. Difference between heterocyclic and homocyclic compound. (2)
OR
Is tetrahydrofuran is aromatic compounds? (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.
  1. 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.
  1. $\sigma$
  2. $\pi$
  3. $\delta$
  4. $\eta$
  1. Which of the is an example of technique used for purification.
  1. Distillation
  2. Differential extraction
  3. Chromatography
  4. All the above
  1. On heating, some solid substances change from solid to vapour state without passing through liquid state is known as …
  1. Melting
  2. Boiling
  3. Sublimation
  4. Condensation
  1. The hyperconjugation may also be regarded as ….
  1. bonding resonance
  2. no bond resonance
  3. no bond induction
  4. bonding induction
  1. Chromatography paper contains water trapped in it, which acts as the … phase.
  1. mobile
  2. stationery
  3. Secondary
  4. quaternary
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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 $+ 2H_2O \rightarrow 2NaOH + 2Hg + H_2$​​​​​​​
  1. NaOH Sodium hydroxide is generally prepared Commercially by the electrolysis of … in Castner-Kellner cell.
  1. $NaCl$
  2. $Na_2CO_3$
  3. $NaHCO_3$
  4. $NaNH_2$
  1. … is used in the manufacture of soft Soap.
  1. Sodium Hydroxide
  2. Potassium Hydroxide
  3. Aluminium hydroxide
  4. Beryllium hydroxide
  1. … is used in devising Photoelectric cells.
  1. Hydrogen
  2. Lithium
  3. Caesium
  4. Helium
  1. … compounds were Earlier used as anti-knock additives to petrol.
  1. Organomagnesium
  2. Organosilicon
  3. Organochloride
  4. Organolead
  1. The sodium amalgam is treated with water to gives ….
  1. $NaOH$
  2. $Na_2CO_3$
  3. $NaHCO_3$
  4. $NaNH_2$​​​​​​​
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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. Finally, the purity of a compound is ascertained by determining its melting or boiling point. This is one of the most commonly used techniques for the purification of solid organic compounds. In crystallisation Impurities, which impart colour to the solution are removed by adsorbing over activated charcoal. In distillation Liquids having different boiling points vaporise at different temperatures. The vapours are cooled and the liquids so formed are collected separately. Steam Distillation is applied to separate substances which are steam volatile and are immiscible with water. Distillation under reduced pressure: This method is used to purify liquids having very high boiling points.

i. Which method can be used to separate two compounds with different solubilities in a solvent?
ii. Distillation method is used to separate which type of substance?
iii. Which technique is used to separate aniline from aniline water mixture?
OR
Why chloroform and aniline are easily separated by the technique of distillation?
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Read the passage given below and answer the following questions from 1 to 5.
F Wohler synthesised an organic compound, urea from an inorganic compound, ammonium cyanate.
The knowledge of fundamental concepts of molecular structure helps in understanding and predicting the properties of organic compounds. You have already learnt theories of valency and molecular structure. Also, you already know that tetravalence of carbon and the formation of covalent bonds by it are explained in terms of its electronic configuration and the hybridisation of s and p orbitals. It may be recalled that formation and the shapes of molecules like methane $(CH_4)$, ethene $(C_2H_4)$, ethyne $(C_2H_2)$ are explained in terms of the use of $sp^3, sp^2$ and sp hybrid orbitals by carbon atoms in the respective molecules. Hybridisation influences the bond length and bond enthalpy (strength) in compounds. The sp hybrid orbital contains more s character and hence it is closer to its nucleus and forms shorter and stronger bonds than the sp3 hybrid orbital.The sp2 hybrid orbital is intermediate in s character between sp and sp3 and, hence, the length and enthalpy of the bonds it forms, are also intermediate between them. The change in hybridisation affects the electronegativity of carbon. The greater the s character of the hybrid orbitals, the greater is the electronegativity. Thus, a carbon atom having an sp hybrid orbital with 50% s character is more electronegative than that possessing sp2 or sp3 hybridised orbitals. This relative electronegativity is reflected in several physical and chemical properties of the molecules concerned, about which you will learn in later units.
Characteristic Features of π Bonds In a π (pi) bond formation, parallel orientation of the two p orbitals on adjacent atoms is necessary for a proper sideways overlap. Thus, in $H_2C=CH_2$ molecule all the atoms must be in the same plane. The p orbitals are mutually parallel and both the p orbitals are perpendicular to the plane of the molecule. Rotation of one $CH_2$ fragment with respect to other interferes with maximum overlap of p orbitals and, therefore, such rotation about carbon-carbon double bond (C=C) is restricted. The electron charge cloud of the π bond is located above and below the plane of bonding atoms. This results in the electrons being easily available to the attacking reagents. In general, π bonds provide the most reactive centres in the molecules containing multiple bonds.

Structures of organic compounds are represented in several ways. The Lewis structure or dot structure, dash structure, condensed structure and bond line structural formulas are some of the specific types. The Lewis structures, however, can be simplified by representing the two-electron covalent bond by a dash (–). Such a structural formula focuses on the electrons involved in bond formation. A single dash represents a single bond, double dash is used for double bond and a triple dash represents triple bond. Lone- pairs of electrons on heteroatoms (e.g., oxygen, nitrogen, sulphur, halogens etc.) may or may not be shown. Thus, ethane $(C_2H_6)$, ethene $(C_2H_4)$, ethyne $(C_2H_2)$ and methanol $(CH_3OH)$ can be represented by the following structural formulas. Such structural representations are called complete structural formulas.
These structural formulas can be further abbreviated by omitting some or all of the dashes representing covalent bonds and by indicating the number of identical groups attached to an atom by a subscript. The resulting expression of the compound is called a condensed structural formula. Thus, ethane, ethene, ethyne and methanol can be written as:

Similarly, $CH_3CH_2CH_2CH_2CH_2CH_2CH_2CH_3$ can be further condensed to $CH_3(CH_2)_6CH_3$. For further simplification, organic chemists use another way of representing the structures, in which only lines are used. In this bond-line structural representation of organic compounds, carbon and hydrogen atoms are not shown and the lines representing carbon-carbon bonds are drawn in a zig-zag fashion. The only atoms specifically written are oxygen, chlorine, nitrogen etc. The terminals denote methyl $(–CH_3)$ groups (unless indicated otherwise by a functional group), while the line junctions denote carbon atoms bonded to appropriate number of hydrogens required to satisfy the valency of the carbon atoms. Some of the examples are represented as follows: (i) 3-Methyloctane can be represented in various forms as:
  1. … synthesised an organic compound, urea from an inorganic compound, ammonium cyanate.
  1. Wohler
  2. Adams
  3. Roger
  4. William Evans
  1. Dot structure is also known as …
  1. Zig zag structure
  2. Lewis structure
  3. Line structure
  4. Bond line structure
  1. Terminals in zigzig structure denotes … Group.
  1. Bromyl
  2. Propyl
  3. Methyl
  4. Pentyl
  1. Triple dash represents …
  1. Single bond
  2. Double bond
  3. Triple bond
  4. Equivalent bond
  1. Lewis structures representing the two-electron covalent bond by …
  1. .
  2. :
  3. ?
  4. –
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The uncertainty in the experimental or the calculated values is indicated by mentioning the number of significant figures. Significant figures are meaningful digits which are known with certainty plus one which is estimated or uncertain. The uncertainty is indicated by writing the certain digits and the last uncertain digit. there are certain rules for determining the Number of significant figures. These are Stated below:
  • All non-zero digits are significant. For Example in 285cm, there are three Significant figures and in 0.25 mL, there are two significant figures.
  • Zeros preceding to first non-zero digit are not significant. such zero indicates the position of decimal point. thus, 0.03 has one significant figure and 0.0052 has two significant figures.
  • Zeros between two non-zero digits are significant. thus, 2.005 has four Significant figures.
  • 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. But, if otherwise, the terminal zeros are not significant if there is no decimal point.
Precision refers to the closeness of various measurements for the same quantity. However, accuracy is the agreement of a particular valueto the true value of the result.
LAWS OF CHEMICALCOMBINATIONS- The combination of elements to form compounds is governed by the following five basic laws.
  1. Law of Conservation of Mass-This law was put forth by Antoine Lavoisierin 1789. He performed careful experimental studies for combustion reactions and reached to the conclusion that in all physical andchemical changes, there is no net change inmassduring the process. Hence, he reachedto the conclusion that matter can neither becreated nor destroyed. This is called ‘Law ofConservation of Mass’.
  2. Law of Definite Proportions-This law was given by, a French chemist, Joseph Proust. He stated that a given compound always contains exactly the same proportion of elements by weight.
  3. Law of Multiple Proportions-This law was proposed by John Dalton. According to this law, if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers. For example, hydrogen combines with oxygen to form two compounds, namely, water and hydrogen peroxide.
Hydrogen + Oxygen→ Water

2g 16g 18g

Hydrogen + Oxygen → Hydrogen Peroxide

2g 32g 34g

Here, the masses of oxygen (i.e., 16 g and 32 g), which combine with a fixed mass of hydrogen (2g) bear a simple ratio, i.e., 16:32 or 1:2.
  1. Gay Lussac’s Law of Gaseous Volumes-This law was given by Gay Lussac in 1808. Heobserved that when gases combine or are produced in a chemicalreaction they do so in asimple ratio by volume,provided all gases are at the same temperature and pressure.
  2. Avogadro’s Law – In 1811, Avogadro proposed that equal volumes of all gases at the same temperature and pressure should contain equal number of molecules.
In 1808, Dalton published ‘A New System of Chemical Philosophy’, in which he proposed the following :
  1. Matter consists of indivisible atoms.
  2. All 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 reorganisati on of atoms. These are neither created nor destroyed in a chemical reaction.
  1. … refers to the closeness of variousmeasurements for the same quantity.
  1. Accuracy
  2. Reliability
  3. Precision
  4. Uncertainty
  1. Law of Conservation of mass was put forth by ….in 1789.
  1. Joseph Proust
  2. Antoine Lavoisier
  3. Joseph Louis
  4. Gay Lussac
  1. Which of the following number has twosignificant figures.
  1. 0.0052
  2. 052
  3. 52
  4. 0052
  1. … is the agreement of a particular valueto the true value of the result.
  1. Accuracy
  2. Reliability
  3. Precision
  4. Uncertainty
  1. Law of Multiple Proportions proposed by....
  1. Joseph Proust
  2. Antoine Lavoisier
  3. Joseph Louis
  4. John Dalton
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Read the passage given below and answer the following questions from 1 to 5 .
Group 14 elements: the carbon family-Carbon, silicon, germanium, tin lead and Flerovium are the members of group 14. Carbon Is the seventeenth most abundant element by Mass in the earth's crust. It is widely Distributed in nature in free as well as in the Combined state. In elemental state it is available As coal, graphite and diamond; however, in Combined state it is present as metal Carbonates, hydrocarbons and carbon dioxide Gas ( $0.03 \%$ ) in air. One can emphatically say That carbon is the most versatile element in the World. Its combination with other elements Such as dihydrogen, dioxygen, chlorine and Sulphur provides an astonishing array of Materials ranging from living tissues to drugs And plastics.
The valence shell electronic configuration of These elements is $ns ^2 np ^2$. The inner core of the Electronic configuration of elements in this Group also differs.
Covalent Radius There is a considerable increase in covalent Radius from C to Si, thereafter from Si to Pb a Small increase in radius is observed. This is Due to the presence of completely filled d and f Orbitals in heavier members. Ionization Enthalpy The first ionization enthalpy of group 14 Members is higher than the corresponding Members of group 13. The influence of inner Core electrons is visible here also. In general the lonisation enthalpy decreases down the group. Small decrease in $\Delta iH$ from Si to Ge to Sn and Slight increase in $\Delta i H$ from Sn to Pb is the Consequence of poor shielding effect of Intervening $d$ and $f$ orbitals and increase in size Of the atom. Electronegativity Due to small size, the elements of this group Are slightly more electronegative than group 13 elements. The electronegativity values for Elements from Si to Pb are almost the same.
Physical Properties All members of group14 are solids. Carbon and Silicon are non-metals, germanium is a metalloid, Whereas tin and lead are soft metals with low Melting points. Melting points and boiling points Of group 14 elements are much higher than those Of corresponding elements of group 13.
(i) Reactivity towards oxygen All members when heated in oxygen form Oxides. There are mainly two types of oxides, i.e., monoxide and dioxide of formula MO and $MO _2$ respectively. SiO only exists at high Temperature. Oxides in higher oxidation states Of elements are generally more acidic than Those in lower oxidation states. The dioxides $CO _2$, SiO 2 and $GeO _2$ are acidic, whereas $SnO _2$ and $PbO _2$ are amphoteric in nature. Among monoxides, CO is neutral, GeO is Distinctly acidic whereas SnO and PbO are Amphoteric.
(ii) Reactivity towards water Carbon, silicon and germanium are not Affected by water. Tin decomposes steam to Form dioxide and dihydrogen gas.
$Sn+2 H_2 O \rightarrow SnO_2+2 H_2$
Lead is unaffected by water, probably Because of a protective oxide film formation.
(iii) Reactivity towards halogen These elements can form halides of formula $M X_2$ and $M X_4$ (where $\left.X=F, C l, B r, I\right)$. Except Carbon, all other members react directly with Halogen under suitable condition to make Halides. Most of the $MX _4$ are covalent in nature. The central metal atom in these halides Undergoes $sp ^3$ hybridisation and the molecule Is tetrahedral in shape. Exceptions are $SnF _4$ And $PbF _4$, which are ionic in nature.
Carbon atoms have the tendency to link With one another through covalent bonds to Form chains and rings. This property is called Catenation. This is because C-C bonds are Very strong. Down the group the size increases and electronegativity decreases, and, thereby, Tendency to show catenation decreases. This Can be clearly seen from bond enthalpies Values. The order of catenation is $C > Si > Ge \approx Sn$. Lead does not show catenation.
  1. Which of the following is not the member of group 14 ?
  1. boron
  2. silicon
  3. germanium
  4. tin
  1. … does not show catenation.
  1. Carbon
  2. Lead
  3. Silicon
  4. Germanium
  1. Which of following elements are affected by water ?
  1. carbon
  2. silicon
  3. germanium
  4. All the above
  1. The valence shell electronic configuration of Group 14 elements is …
  1. $ns^2np^4$
  2. $ns^2np^5$
  3. $ns^2np^2$
  4. $ns^2np^3$
  1. Half-life of $^{14}C$ is … years.
  1. 6570
  2. 4570
  3. 5770
  4. 1970
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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.
  1. Alkanes contain carbon-carbon … bonds.
  1. sigma $\sigma$
  2. pi bond$\pi$
  3. delta$\delta$
  4. eta $\eta$
  1. C-C single bond is hindered by a small energy barrier of…. $kJ ~mol^{–1}$​​​​​​​
  1. 10 - 200
  2. 1 - 20
  3. 100 - 427
  4. 342 - 786
  1. A decomposition reaction into smaller fragments by the application of heat is called as ….
  1. pyrolysis
  2. cracking
  3. both (a) & (b)
  4. combustion
  1. Which of the following steps are involving in free radical chain mechanism
  1. initiation
  2. propagation
  3. termination
  4. All the above
  1. 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.
  1. pyrolysis
  2. cracking
  3. both (a) & (b)
  4. combustion
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