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Organic Chemistry: Some Basic Principles and Techniques — Chemistry STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceChemistryOrganic Chemistry: Some Basic Principles and Techniques4 Marks
Question

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
✓

Answer

  1. (a) $\sigma$
  2. (d) All the above
  3. (c) Sublimation
  4. b) no bond resonance
  5.  b) stationery

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A less obvious example of electron transfer is realised when hydrogen combines with oxygen to form water by the reaction:

$2\text{H}_2(\text{g}) + \text{O}_2 (\text{g}) → 2\text{H}_2\text{O} (\text{l})$

Though not simple in its approach, yet we can visualise the H atom as going from a Neutral (zero) state in H2 to a positive state in H2O, the O atom goes from a zero state in O2 To a dinegative state in H2O. It is assumed that There is an electron transfer from H to O and Consequently H2 is oxidised and O2 is reduced. However, as we shall see later, the charge Transfer is only partial and is perhaps better Described as an electron shift rather than a Complete loss of electron by H and gain by O. Two examples of this class Of the reactions are:

$\text{H}_2 (\text{s}) + \text{Cl}_2(\text{g}) → 2\text{HCl} (\text{g})$ And,

$\text{CH}_4 (\text{g}) + 4\text{Cl}_2 (\text{g}) → \text{CCl}_4(\text{l}) + 4\text{HCl (g)}$

In order to keep track of electron shifts in Chemical reactions involving formation of Covalent compounds, a more practical method Of using oxidation number has been Developed. In this method, it is always Assumed that there is a complete transfer of Electron from a less electronegative atom to a More electonegative atom. For example, we Rewrite equations to show Charge on each of the atoms forming part of The reaction:

It may be emphasised that the assumption Of electron transfer is made for book-keeping Purpose only and it will become obvious at a Later stage in this unit that it leads to the simple Description of redox reactions. Oxidation number denotes the Oxidation state of an element in a Compound ascertained according to a set Of rules formulated on the basis that electron pair in a covalent bond belongs Entirely to more electronegative element.  It is not always possible to remember or Make out easily in a compound/ ion, which Element is more electronegative than the other. Therefore, a set of rules has been formulated To determine the oxidation number of an Element in a compound/ ion. We may at this stage, state the rules for the Calculation of oxidation number. These rules are:

1.) In elements, in the free or the uncombined State, each atom bears an oxidation Number of zero. Evidently each atom in H2, O2, Cl2, O3, P4, S8, Na, Mg, Al has the Oxidation number zero.

2.) For ions composed of only one atom, the Oxidation number is equal to the charge On the ion. Thus Na+ Ion has an oxidation Number of +1, Mg2+ ion +2 , Fe3+ ion, +3, Cl –  Ion, –1, O2 – ion, –2; and so on. In their Compounds all alkali metals have Oxidation number of +1, and all alkaline Earth metals have an oxidation number of +2. Aluminium is regarded to have an Oxidation number of +3 in all its Compounds.

3.) The oxidation number of oxygen in most Compounds is –2. However, we come across Two kinds of exceptions here. One arises In the case of peroxides and superoxides, The compounds of oxygen in which oxygen Atoms are directly linked to each other. While in peroxides (e.g., H2 O2, Na2 O2), each Oxygen atom is assigned an oxidation Number of –1, in superoxides (e.g., K O2, Rb O2) each oxygen atom is assigned an Oxidation number of –(½). The second Exception appears rarely, i.e. when oxygen Is bonded to fluorine. In such compounds e.g., oxygen difluoride (OF2) and dioxygen difluoride (O2F2), the oxygen is assigned an oxidation number of +2 and +1, respectively. The number assigned to oxygen will depend upon the bonding state of oxygen but this number would now be a positive figure only.

4.) The oxidation number of hydrogen is +1, Except when it is bonded to metals in binary Compounds (that is compounds containing Two elements). For example, in LiH, NaH, And Ca H2, its oxidation number is –1.

5.) In all its compounds, fluorine has an Oxidation number of –1. Other halogens (Cl, Br, and I) also have an oxidation number Of –1, when they occur as halide ions in Their compounds. Chlorine, bromine and Iodine when combined with oxygen, for Example in oxoacids and oxoanions, have Positive oxidation numbers.

6.) The algebraic sum of the oxidation number Of all the atoms in a compound must be Zero. In polyatomic ion, the algebraic sum Of all the oxidation numbers of atoms of The ion must equal the charge on the ion. Thus, the sum of oxidation number of three Oxygen atoms and one carbon atom in the Carbonate ion, (CO3) 2– must equal –2.

A term that is often used interchangeably With the oxidation number is the oxidation State. Thus in CO2, the oxidation state of Carbon is +4, that is also its oxidation number And similarly the oxidation state as well as Oxidation number of oxygen is – 2. This implies That the oxidation number denotes the Oxidation state of an element in a compound.

The oxidation number/state of a metal in a Compound is sometimes presented according To the notation given by German chemist, Alfred Stock. It is popularly known as Stock Notation. According to this, the oxidation Number is expressed by putting a Roman Numeral representing the oxidation number In parenthesis after the symbol of the metal in The molecular formula. Thus aurous chloride And auric chloride are written as Au(I)Cl and Au(III)Cl3. Similarly, stannous chloride and Stannic chloride are written as Sn(II) Cl2 and Sn(IV)Cl4. This change in oxidation number Implies change in oxidation state, which in Turn helps to identify whether the species is Present in oxidised form or reduced form. Thus, Hg2(I) Cl2 is the reduced form of Hg(II) Cl2.

  1. H atom goes from a … state in H2 to a positive state in H2O in water formation.
  1. Neutral
  2. Positive
  3. Negative
  4. All the above
  1. In oxidation number method, there is a complete transfer of electron from a …. electronegative atom to a … electonegative atom.
  1. more, less
  2. less, more
  3. non, more
  4. non, less
  1. Oxidation number of Mg2 + ion is:
  1. -2
  2. -1
  3. +2
  4. +1
  1. In Na2O2 each oxygen atom is assigned an oxidation number of …
  1. +1
  2. -2
  3. +2
  4. -1
  1. The algebraic sum of the oxidation number of all the atoms in a compound must be…
  1. 0
  2. 1
  3. 2
  4. -2

 

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Read the passage given below and answer the following questions from (i) to (v).

The covalent bond may be classified into twotypes depending upon the types ofoverlapping:(i) Sigma(σ) bond, and (ii) pi($\pi$) bond

  1. Sigma(σ) bond: This type of covalent bondis formed by the end to end (head-on)overlap of bonding orbitals along theinternuclear axis. This is called as headon overlap or axial overlap. This can beformed by any one of the following typesof combinations of atomic orbitals.

s-s overlapping: In this case, there isoverlap of two half filled s-orbitals alongthe internuclear axis.

s-p overlapping: This type of overlapoccurs between half filled s-orbitals of oneatom and half filled p-orbitals of anotheratom.

p–p overlapping: This type of overlaptakes place between half filled p-orbitalsof the two approaching atoms.

  1. pi($\pi$) bond: In the formation of $\pi$ bondthe atomic orbitals overlap in such a waythat their axes remain parallel to each otherand perpendicular to the internuclear axis.The orbitals formed due to sidewiseoverlapping consists of two saucer type charged clouds above and below the planeof the participating atoms.
Basically the strength of a bond depends uponthe extent of overlapping. In case of sigma bond,the overlapping of orbitals takes place to alarger extent. Hence, it is stronger as comparedto the pi bond where the extent of overlappingoccurs to a smaller extent. Further, it isimportant to note that in the formation ofmultiple bonds between two atoms of amolecule, pi bond(s) is formed in addition to asigma bond. In order to explain the characteristicgeometrical shapes of polyatomic moleculeslike CH4,NH3 and H2O etc., Pauling introducedthe concept of hybridisation. According to himthe atomic orbitals combine to form new set ofequivalent orbitals known as hybrid orbitals.Unlike pure orbitals, the hybrid orbitals areused in bond formation. The phenomenon isknown as hybridisation which can be definedas the process of intermixing of the orbitals ofslightly different energies so as to redistributetheir energies, resulting in the formation of newset of orbitals of equivalent energies and shape.For example when one 2s and three 2p-orbitalsof carbon hybridise, there is the formation offour new sp3 hybrid orbitals. Salient features of hybridisation: The mainfeatures of hybridisation are as under:

  1. The number of hybrid orbitals is equal tothe number of the atomic orbitals that gethybridised.
  2. The hybridised orbitals are alwaysequivalent in energy and shape.
  3. The hybrid orbitals are more effective informing stable bonds than the pure atomicorbitals.
  4. These hybrid orbitals are directed in spacein some preferred direction to haveminimum repulsion between electronpairs and thus a stable arrangement.Therefore, the type of hybridisationindicates the geometry of the molecules. Important conditions for hybridisation
  5. The orbitals present in the valence shell of the atom are hybridised.
  6. The orbitals undergoing hybridisation should have almost equal energy.
  7. Promotion of electron is not essential condition prior to hybridisation.
  8. It is not necessary that only half filled orbitals participate in hybridisation.
some cases, even filled orbitals of valence shell take part in hybridisation.

There are various types of hybridisationinvolving s, p and d orbitals. The differenttypes of hybridisation are as under:

  1. sp hybridisation: This type ofhybridisation involves the mixing of one s andone p orbital resulting in the formation of twoequivalent sp hybrid orbitals. The suitableorbitals for sp hybridisation are s and pz, ifthe hybrid orbitals are to lie along the z-axis. Example of molecule having sphybridisationBeCl2: The ground state electronicconfiguration of Be is 1s22s2. In the exited stateone of the 2s-electrons is promoted to vacant 2p orbital to account for its bivalency.One 2s and one 2p-orbital gets hybridised toform two sp hybridised orbitals.
  2. sp2 hybridisation : In this hybridisationthere is involvement of one s and twop-orbitals in order to form three equivalent sp2hybridised orbitals. For example, in BCl3molecule, the ground state electronicconfiguration of central boron atom is1s22s22p1. In the excited state, one of the 2selectrons is promoted to vacant 2p orbital as a result boron has three unpaired electrons.These three orbitals (one 2s and two 2p)hybridise to form three sp2 hybrid orbitals.
  3. sp3 hybridisation: This type ofhybridisation can be explained by taking theexample of CH4 molecule in which there ismixing of one s-orbital and three p-orbitals ofthe valence shell to form four sp3 hybrid orbitalof equivalent energies and shape. There is 25%s-character and 75% p-character in each sp3hybrid orbital. The four sp3 hybrid orbitals soformed are directed towards the four cornersof the tetrahedron. The angle between sp3hybrid orbital is 109.5°.
  1. ....ntroduced the concept of hybridisation.
  1. Pauling
  2. Lewis
  3. Nyholm
  4. Gillespie
  1. Which of the following is an example of sp3 hybridization?
  1. BeCl2
  2. Ch4
  3. BCl3
  4. C2H4
  1. The angle between sp3 hybrid orbital is ….
  1. 5°
  2. 9°
  3. 109.5°
  4. 120°
  1. A sigma bond is formed by the overlapping of …
  1. s−s,
  2. s−p
  3. p−p
  4. All the above
  1. When one 2s and three 2p-orbitals of carbon hybridise, there is the formation of four new …  hybrid orbitals.
  1. sp3
  2. sp2
  3. sp
  4. None of above
→
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)
→
In order to explain the characteristic geometrical shapes of polyatomic molecules, Pauling introduced the concept of hybridisation. The orbitals undergoing hybridisation should have nearly the same energy. There are various type of hybridisations involving s, p and d-type of orbitals. The type of hybridisation gives the characteristic shape of the molecule or ion.

1. Why all the orbitals in a set of hybridised orbitals have the same shape and energy?
2. Out of $XeF _2$ and $SF _2$ which molecule has the same shape as $NO _2^{+}$ion?
3. Out of $XeF _4$ and $XeF _2$ which molecule doesn't have the same type of hybridisation as P (Phosphorus) has in $PF _5$ ?
OR
Unsaturated compounds undergo additional reactions. Why?
→

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 (CH4), ethene (C2H4), ethyne (C2H2) are explained in terms of the use of sp3, sp2 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 H2C=CH2 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 CH2 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 (C2H6), ethene (C2H4), ethyne (C2H2) and methanol (CH3OH) 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, CH3CH2CH2CH2CH2CH2CH2CH3 can be further condensed to CH3(CH2)6CH3. 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 (–CH3) 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. –
→

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 Hm 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 CO2  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°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 CO2  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. Hk
  2. Hm
  3. Hl
  4. Hn
  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 CO2or ‘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. $\triangle\text{w}\text{H}^{\phi}$

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Read the passage given below and answer the following questions from (i) to (vi).
The atomic theory of matter was first proposed on afirm scientific basis by JohnDalton, a British schoolteacher in 1808. His theory, called Dalton’s atomictheory, regarded the atom as the ultimate particle ofmatter Dalton’s atomic theory was able to explainthe law of conservation of mass, law of constantcomposition and law of multiple proportion verysuccessfully. However, it failed to explain the results ofmany experiments.In mid 1850s many scientists mainlyFaraday began to study electrical dischargein partially evacuated tubes, known ascathode ray discharge tubes.Electrical discharge carried out in the modifiedcathode ray tube led to the discovery of canalrays carrying positively charged particles. Thecharacteristics of these positively chargedparticles are listed below.
  1. Unlike cathode rays, mass of positivelycharged particles depends upon thenature of gas present in the cathode raytube. These are simply the positivelycharged gaseous ions.
  2. The charge to mass ratio of the particlesdepends on the gas from which theseoriginate.
  3. Some of the positively charged particlescarry a multiple of the fundamental unitof electrical charge.
  4. The behaviour of these particles in themagnetic or electrical field is opposite tothat observed for electron or cathoderays.
The smallest and lightest positive ion wasobtained from hydrogen and was called
proton. This positively charged particle wascharacterised in 1919. Later, a need was feltfor the presence of electrically neutral particleas one of the constituent of atom. Theseparticles were discovered by Chadwick (1932)by bombarding a thin sheet of beryllium byα-particles. When electrically neutral particleshaving a mass slightly greater than that ofprotons were emitted. He named theseparticles as neutrons.J. J. Thomson, in 1898, proposed that an atom possesses a spherical shape (radiusapproximately 10–10 m) in which the positivecharge is uniformly distributed. The electronsare embedded into it in such a manner as togive the most stable electrostatic arrangementMany different names are given tothis model, for example, plum pudding, raisinpudding or watermelon. This model can be visualised as a pudding or watermelon ofpositive charge with plums or seeds (electrons)embedded into it. An important feature of thismodel is that the mass of the atom is assumed to be uniformly distributed over theatom.Rutherford and his students (Hans Geiger andErnest Marsden) bombarded very thin gold foilwith α–particles. Rutherford’s famous α–particle scattering experiment.The observations of Scattering experiment are as follows-:
  1. most of the α–particles passed throughthe gold foil undeflected.
  2. a small fraction of the α–particles wasdeflected by small angles.
  3.  a very few α–particles (∼1 in 20,000)bounced back, that is, were deflected bynearly 180°.
On the basis of observations andconclusions from this experiment, Rutherford proposed the nuclearmodel of atom. According to this model:
  1. The positive charge and most of the massof the atom was densely concentrated inextremely small region. This very smallportion of the atom was called nucleusby Rutherford.
  2. The nucleus is surrounded by electronsthat move around the nucleus with a veryhigh speed in circular paths called orbits.Thus, Rutherford’s model of atomresembles the solar system in which thenucleus plays the role of sun and theelectrons that of revolving planets.
  3. Electrons and the nucleus are held together by electrostatic forces of attraction.
  1. The atomic theory of matter was first proposed on afirm scientific basis by:
  1. John Dalton
  2. Ernest Rutherford
  3. J.Thomson
  4. Henry Moseley
  1. The cathode rays start from … and move towards the….
  1. Anode, Cathode
  2. Centre, Anode
  3. Cathod, Anode
  4. Cathod, Centre
  1. Negativelycharged particles in atoms, called…
  1. Protons
  2. Electrons
  3. Neutron
  4. Positron
  1. The smallest and lightest positive ion wasobtained from …. and was called proton.
  1. Oxygen
  2. Nitrogen
  3. Carbon
  4. Hydrogen
  1. Electrically neutral particles having a mass slightly greater than that of protons, these particles termed as:
  1. Protons
  2. Electrons
  3. Neutron
  4. Positron
  1. J.J. Thomson’s atomic model is also named as:
  1. Plum pudding
  2. Raisin pudding
  3. Watermelon
  4. All the above
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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 (CH4) 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 C2H6. This hydrocarbon with molecular formula C2H6 is known as ethane. Thus it can consider C2H6 as derived from CH4 by replacing one hydrogen atom by -CH3 group. Go on constructing alkanes by doing this theoretical exercise i.e., replacing hydrogen atom by –CH3 group. The next molecules will be C3H8, C4H10 …

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 Cn H2 n + 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 $\sigma$ 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
  1. LPG stands for ….
  1. Liquid Pressure Gas
  2. Liquefied Petroleum Gas
  3. Liquid Platinum Gas
  4. Liquid Processed Gas
  1. Difference in properties is due to difference in their structures, they are known as …. isomers.
  1. Functional
  2. Positional
  3. Structural
  4. Chain
  1. Structural isomers which differ in chain of carbon atoms are known as…  isomers.
  1. Functional
  2. Positional
  3. Structural
  4. Chain
  1. CNG Stands for ..
  1. Compressed Natural Gas
  2. Condensed Natural Gas
  3. Compressed Neutral Gas
  4. Compressed Neutral Gallium
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Read the passage given below and answer the following questions from 1 to 5.
After having some idea about the terms atomsand molecules, it is appropriate here tounderstand what do we mean by atomic andmolecular masses.One atomicmass unit is defined as a mass exactly equal to one-twelfth of the mass of one carbon – 12 atom.Molecular mass is the sum of atomic masses of the elements present in a molecule. It is obtained by multiplying the atomic mass of each element by the number of its atoms and adding them together. Some substances, such as sodium chloride, do not contain discrete molecules as their constituent units. In such compounds, positive (sodium ion) and negative (chloride ion) entities are arranged in a three dimensional structure. The mole, symbol mol, is the SI unit of amount of substance. One mole contains exactly 6.02214076× 1023 elementary entities. This number is the fixed numerical value of the Avogadro constant, NA , when expressed in the unit mol–1 and is called the Avogadro number. The amount of substance, symbol n, of a system is a measure of the number of specified elementary entities. An elementary entity may be an atom, a molecule, an ion, an electron, any other particle or specified group of particles. It may be emphasised that the mole of a substance always contains the same number of entities, no matter what the substance may be. In order to determine this number precisely, the mass of a carbon–12 atom was determined by a mass spectrometer and found to be equal to 1.992648 × 10–23 g. Knowing that one mole of carbon weighs 12 g, the number of atoms in it is equal to:
        12 g /mol C-12       .
1.992648 × 1023  g / C- 12 atom. = 6.0221367 × 1023 atoms/mol.
The mass of one mole of a substance in grams is called its molar mass. the molar mass in grams is numerically equal to atomic molecular/formula mass in u.An empirical formula represents the simplestwhole number ratio of various atoms present ina compound, whereas, the molecular formulashows the exact number of different types ofatoms present in a molecule of a compound. If the mass per cent of variouselements present in a compound is known, its empiricalformula can be determined. Molecular formulacan further be obtained if the molar mass isknown.Many a time, reactions are carried out with the Amounts of reactants that are different than The amounts as required by a balanced chemical reaction. In such situations, one Reactant is in more amount than the amount required by balanced chemical reaction. The reactant which is present in the least amount Many a time, reactions are carried out with the
amounts of reactants that are different than the amounts as required by a balanced chemical reaction. In such situations, one reactant is in more amount than the amount required by balanced chemical reaction. The reactant which is present in the least amount gets consumed after sometime and after that further reaction does not take place whatever be the amount of the other reactant. Hence, the reactant, which gets consumed first, limits the amount of product formed and is, therefore, called the limiting reagent.
  1. One atomic mass unit (amu) is defined as a mass exactly equal to one-twelfth of the mass of one …atom.
  1. Hydrogen – 1
  2. Carbon – 12
  3. Oxygen -12
  4. Chlorine – 35
  1. The mass of one mole of a substance in grams is called its..
  1. Atomic mass
  2. Molecular Weight
  3. Molecular mass
  4. Molar mass.
  1. … is the sum of atomic massesof the elements present in a molecule.
  1. Atomic mass
  2. Molecular Weight
  3. Molecular mass
  4. Molar mass.
  1. One mole contains exactly …elementary entities.
  1. 02214076 × 1021
  2. 02214076 × 1022
  3. 6.02214076 × 1023
  4. 02214076 × 1024
  1. For which of the following compound , formula mass is preferred instead of molecular mass?
  1. NaCl
  2. C2H6
  3. N2
  4. H2O​
→
The idea of oxidation number has been invariably applied to define oxidation, reduction, oxidising agent (oxidant), reducing agent (reductant) and the redox reaction. To summarise, we may say that:
Oxidation: An increase in the oxidation number of the element in the given substance.
Reduction: A decrease in the oxidation number of the element in the given substance.
Oxidising agent: A reagent which can increase the oxidation number of an element in a given substance. These reagents are called as oxidants also.
Reducing agent: A reagent which lowers the oxidation number of an element in a given substance. These reagents are also called as reductants.
Redox reactions: Reactions which involve change in oxidation number of the interacting species.
Types of Redox Reactions
1.) Combination reactions -A combination reaction may be denoted in the manner:
A + B → C
Either A and B or both A and B must be in the elemental form for such a reaction to be a redox reaction. All combustion reactions, which make use of elemental dioxygen, as well as other reactions involving elements other than dioxygen, are redox reactions. Some important examples of this category are:

2.) Decomposition reactions- Decomposition reactions are the opposite of combination reactions. Precisely, a decomposition reaction leads to the breakdown of a compound into two or more components at least one of which must be in the elemental state.
Examples of this class of reactions are:

It may carefully be noted that there is no change in the oxidation number of hydrogen in methane under combination reactions and that of potassium in potassium chlorate in reaction. This may also be noted here that all decomposition reactions are not redox reactions. For example, decomposition of calcium carbonate is not a redox reaction.
3.) Displacement reactions- In a displacement reaction, an ion (or an atom) in a compound is replaced by an ion (or an atom) of another element. It may be denoted as:
X + YZ → XZ + Y
Displacement reactions fit into two categories: metal displacement and non-metal displacement.
(a) Metal displacement: A metal in a compound can be displaced by another metal in the uncombined state. Metal displacement reactions find many applications in metallurgical processes in which pure metals are obtained from their compounds in ores.
(b) Non-metal displacement: The non-metal displacement redox reactions include hydrogen displacement and a rarely occurring reaction involving oxygen displacement. All alkali metals and some alkaline earth metals (Ca, Sr, and Ba) which are very good reductants, will displace hydrogen from cold water. Many metals, including those which do not react with cold water, are capable of displacing hydrogen from acids. Dihydrogen from acids may even be produced by such metals which do not react with steam. Cadmium and tin are the examples of such metals.
4.) Disproportionation reactions – Disproportionation reactions are a special type of redox reactions. In a disproportionation reaction an element in one oxidation state is simultaneously oxidised and reduced. One of the reacting substances in a disproportionation reaction always contains an element that can exist in at least three oxidation states. The element in the form of reacting substance is in the intermediate oxidation state; and both higher and lower oxidation states of that element are formed in the reaction. The decomposition of hydrogen peroxide is a familiar example of the reaction, where oxygen experiences disproportionation.

Here the oxygen of peroxide, which is present in –1 state, is converted to zero oxidation state in O2 and decreases to –2 oxidation state in H2O.
  1. In … an ion (or an atom) in a compound is replaced by an ion (or an atom) of another element.
  1. displacement reaction
  2. decomposition reaction
  3. disproportionation reaction
  4. combination reaction
  1. leads to the breakdown of a compound into two or more components at least one of which must be in the elemental state.
  1. displacement reaction
  2. decomposition reaction
  3. disproportionation reaction
  4. combination reaction
  1. In …. an element in one oxidation state is simultaneously oxidised and reduced.
  1. displacement reaction
  2. decomposition reaction
  3. disproportionation reaction
  4. combination reaction
  1. Reactions which involve change in oxidation number of the interacting species…
  1. Exothermic reaction
  2. Endothermic reaction
  3. Neutralization reaction
  4. Redox reaction
  1. One of the reacting substances in a disproportionation reaction always contains an element that can exist in at least … oxidation states.
  1. 1
  2. 2
  3. 3
  4. 4
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