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Structure of Atom — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryStructure of Atom4 Marks
Question
Read the passage given below and answer the following questions from (i) to (v).
The first concreteexplanation for the phenomenon of the blackbody radiation was given byMax Planck in 1900.An ideal body, which emits and absorbs radiations of allfrequencies uniformly, is called a black bodyand the radiation emitted by such a body is called black body radiation. Max Planck arrived at a satisfactory relationshipbymaking an assumption that absorption andemmission of radiation arises from oscillatori.e., atoms in the wall of black body.He suggested that atoms andmolecules could emit or absorb energy onlyin discrete quantities and not in a continuousmanner. He gave the name quantum to thesmallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation. The energy (E) of aquantum of radiation is proportionalto its frequency (ν) and is expressed byequation .
$E = hυ.$
The proportionality constant, ‘h’ is knownas Planck’s constant and has the value6.$626\times 10^{–34}$ Js.In 1887, H. Hertz performed a very interestingexperiment in which electrons (or electriccurrent) were ejected when certain metals (forexample potassium, rubidium, caesium etc.)were exposed to a beam of light. The phenomenon is calledPhotoelectric effect. The results observed inthis experiment were:
  1. The electrons are ejected from the metalsurface as soon as the beam of light strikesthe surface, i.e., there is no time lagbetween the striking of light beam and theejection of electrons from the metal surface.
  2. The number of electrons ejected is proportional to the intensity or brightness of light.
  3. For each metal, there is a characteristicminimum frequency,ν0(also known asthreshold frequency) below which photoelectric effect is not observed. At afrequency $ν >ν_0$, the ejected electrons comeout with certain kinetic energy. The kineticenergies of these electrons increase withthe increase of frequency of the light used.
The particle nature of light posed a dilemmafor scientists. Theonly way to resolve the dilemma was to acceptthe idea that light possesses both particle andwave-like properties, i.e., light has dualbehaviour. Depending on the experiment, wefind that light behaves either as a wave or as astream of particles. Whenever radiationinteracts with matter, it displays particle likeproperties in contrast to the wavelike properties (interference and diffraction), whichit exhibits when it propagates. This conceptwas totally alien to the way the scientiststhought about matter and radiation and it tookthem a long time to become convincedof itsvalidity.
The study of emission or absorption spectra is referred to as spectroscopy.The emission spectra of atoms inthe gas phase, on the other hand, do not showa continuous spread of wavelength from redto violet, rather they emit light only at specificwavelengths with dark spaces between them.Such spectra are called line spectra or atomicspectra.The Swedishspectroscopist, Johannes Rydberg, noted that
all series of lines in the hydrogen spectrumcould be described by the following expression:
$\bar{\text{v}}=109,677\big(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\big)\text{cm}^{-1}$
The value $109,677 cm^{–1}$​​​​​​​ is called theRydberg constant for hydrogen. The first fiveseries of lines that correspond to $n_1= 1, 2, 3,4, 5$ are known as Lyman, Balmer, Paschen,Bracket and Pfund series, respectively.Neils Bohr (1913) was the first to explainquantitatively the general features of thestructure of hydrogen atom and its spectrum.He used Planck’s concept of quantisation ofenergy. Though the theory is not the modernquantum mechanics, it can still be used to rationalize many points in the atomic structureand spectra. Bohr’s model for hydrogen atomis based on the following postulates:
  1. The electron in the hydrogen atom canmove around the nucleus in a circular pathof fixed radius and energy. These paths arecalled orbits, stationary states or allowedenergy states. These orbits are arrangedconcentrically around the nucleus.
  2. The energy of an electron in the orbit doesnot change with time. However, theelectron will move from a lower stationarystate to a higher stationary state whenrequired amount of energy is absorbedby the electron or energy is emitted when electron moves from higher stationarystate to lower stationary state. The energychange does not takeplace in a continuous manner.
  3. The frequency of radiation absorbed oremitted when transition occurs between two stationary states that differ in energyby $\triangle\text{E},$ is given by:
$\text{v}=\frac{\triangle\text{E}}{\text{h}}=\frac{\text{E}_2-\text{E}_1}{\text{h}}$

Where E1 and E2 are the energies of the lower and higher allowed energy statesrespectively. This expression is commonly known as Bohr’s frequency rule.
  1. The angular momentum of an electron isquantised. In a given stationary state itcan be expressed as in equation
$\text{m}_{\text{e}}\text{vr}=\text{n}.\frac{\text{h}}{2\pi}\text{n}=1,2,3.....$
  1. The first concrete explanation for the phenomenon of the black body radiation was given by ….in 1900.
  1. Max Planck
  2. De Broglie
  3. Albert Einstein,
  4. Niels Bohr
  1. Which of the following equation is Planck’s equation?
  1. $E= mc^2​​​​​​​$
  2. $E = hυ$
  3. $E= hc^2​​​​​​​$
  4. $E= vc^2.$
  1. What is nature of light?
  1. Wave
  2. Particle
  3. Wave and Particle
  4. None of above
  1. The value …. is called theRydberg constant for hydrogen.
  1. $109,674cm^{–1}​​​​​​​$
  2. $109,675cm^{–1}​​​​​​​$
  3. $109,676cm^{–1}​​​​​​​$
  4. $109,677cm^{–1}$​​​​​​​
  1. …was the first to explain quantitatively the general features of the structure of hydrogen atom and its spectrum.
  1. Max Planck
  2. De Broglie
  3. Albert Einstein,
  4. Niels Bohr
✓

Answer

  1. (a) Max Planck
  1. (b) E = hυ
  1. (c) Wave and Particle
  1. (d) $109,677cm^{–1}$
  1. (d) Niels Bohr

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Read the passage given below and answer the following questions from 1 to 5.
Alkenes are unsaturated hydrocarbons containing at least one double bond. What should be the general formula of alkenes? If there is one double bond between two carbon atoms in alkenes, they must possess two hydrogen atoms less than alkanes. Hence, general formula for alkenes is $C_nH_{2n}$. Alkenes are also known as olefins (oil forming) since the first member, ethylene or ethene $(C_2H_4)$ was found to form an oily liquid on reaction with chlorine.
Structure of Double Bond Carbon-carbon double bond in alkenes consists of one strong sigma $(\sigma)$ bond (bond enthalpy about $397\ kJ\ mol^{–1)}$ due to head-on overlapping of $sp^2$ hybridised orbitals and one weak pi $\pi$ bond (bond enthalpy about $284\ kJ\ mol^{–1})$ obtained by lateral or sideways overlapping of the two 2p orbitals of the two carbon atoms. The double bond is shorter in bond length (134 pm) than the C–C single bond (154 pm). You have already read that the pi $(\pi)$ bond is a weaker bond due to poor sideways overlapping between the two 2p orbitals. Thus, the presence of the pi $(\pi)$bond makes alkenes behave as sources of loosely held mobile electrons. Therefore, alkenes are easily attacked by reagents or compounds which are in search of electrons. Such reagents are called electrophilic reagents. The presence of weaker$(\pi)$-bond makes alkenes unstable molecules in comparison to alkanes and thus, alkenes can be changed into single bond compounds by combining with the electrophilic reagents. Strength of the double bond (bond enthalpy, $681\ kJ\ mol^{–1}$) is greater than that of a carbon-carbon single bond in ethane (bond enthalpy, $348\ kJ\ mol^{–1}$). Orbital diagrams of ethene molecule are shown in Figure.

Geometrical isomerism: Doubly bonded Carbon atoms have to satisfy the remaining two Valences by joining with two atoms or groups. If the two atoms or groups attached to each Carbon atom are different, they can be Represented by YX C = C XY like structure. YX C = C XY can be represented in space in the Following two ways:

In (a), the two identical atoms i.e., both the X or both the Y lie on the same side of the Double bond but in (b) the two X or two Y lie Across the double bond or on the opposite Sides of the double bond. This results in Different geometry of (a) and (b) i.e. disposition Of atoms or groups in space in the two Arrangements is different. Therefore, they are Stereoisomers. They would have the same Geometry if atoms or groups around C = C bond Can be rotated but rotation around C = C bond Is not free. It is restricted. For understanding This concept, take two pieces of strong Cardboards and join them with the help of two Nails. Hold one cardboard in your one hand And try to rotate the other. Can you really rotate The other cardboard ? The answer is no. The Rotation is restricted. This illustrates that the Restricted rotation of atoms or groups around The doubly bonded carbon atoms gives rise to Different geometries of such compounds. The Stereoisomers of this type are called Geometrical isomers. The isomer of the type (a), in which two identical atoms or groups lie On the same side of the double bond is called Cis isomer and the other isomer of the type (b), in which identical atoms or groups lie on The opposite sides of the double bond is called Trans isomer. Thus cis and trans isomers Have the same structure but have different Configuration (arrangement of atoms or groups In space). Due to different arrangement of Atoms or groups in space, these isomers differ In their properties like melting point, boiling Point, dipole moment, solubility etc. Geometrical or cis-trans isomers of but-2-ene Are represented below:

Cis form of alkene is found to be more polar Than the trans form. For example, dipole Moment of cis - but - 2-ene is 0.33 Debye, Whereas, dipole moment of the trans form Is almost zero or it can be said that trans - but - 2 -ene is non-polar. This can be understood by drawing geometries of the two forms as given below from which it is clear that in the trans - but - 2 -ene, the two methyl groups are in opposite directions, Therefore, dipole moments of $C - CH_3$ bonds cancel, thus making the trans form non-polar.

In the case of solids, it is observed that the trans isomer has higher melting point than the cis form. Geometrical or cis-trans isomerism is also shown by alkenes of the types XYC = CXZ and XYC = CZW
Preparation – From alkynes: Alkynes on partial reduction with calculated amount of dihydrogen in the presence of palladised charcoal partially deactivated with poisons like sulphur
compounds or quinoline give alkenes. Partially deactivated palladised charcoal is known as Lindlar’s catalyst. Alkenes thus obtained are having cis geometry. However, alkynes on reduction with sodium in liquid ammonia form trans alkenes.

From alkyl halides: Alkyl halides (R-X) on heating with alcoholic potash (potassium hydroxide dissolved in alcohol, say, ethanol) eliminate one molecule of halogen acid to form alkenes. This reaction is known as dehydrohalogenation i.e., removal of halogen acid. This is example of $\beta-$elimination reaction, since hydrogen atom is eliminated from the $\beta$ carbon atom (carbon atom next to the carbon to which halogen is attached).

Nature of halogen atom and the alkyl group determine rate of the reaction. It is observed that for halogens, the rate is: iodine > bromine > chlorine, while for alkyl groups it is: tert > secondary > primary.
Physical properties Alkenes as a class resemble alkanes in physical properties, except in types of isomerism and difference in polar nature. The first three members are gases, the next fourteen are liquids and the higher ones are solids. Ethene is a colourless gas with a faint sweet smell. All other alkenes are colourless and odourless, insoluble in water but fairly soluble in non- polar solvents like benzene, petroleum ether. They show a regular increase in boiling point with increase in size i.e., every $–CH_2$ group added increases boiling point by 20–30 K. Like alkanes, straight chain alkenes have higher boiling point than isomeric branched chain compounds.
  1. The first three members of alkenes are …?
  1. Gases
  2. Liquids
  3. Solids
  4. None of above
  1. General formula for alkenes is ….?
  1. $C_nH_{2n+1}$
  2. $C_nH_{2n}$
  3. $C_nH_{2n-1}$
  4. $C_nH_{2n+2}$
  1. The colour of ethene gas is …?
  1. Red
  2. White
  3. Pale Green
  4. None of above
  1. The bond length of carbon carbon double bond is … pm ?
  1. 154
  2. 143
  3. 134
  4. 120
  1. Alkenes are also knows as …?
  1. Olefines
  2. Paraffines
  3. Oleofines
  4. Paracetofines
→
The s-Block Elements The elements of Group 1 (alkali metals) and Group 2 (alkaline earth metals) which have ns1and ns2 outermost electronic configuration belong to the s-Block Elements. They are all reactive metals with low ionization enthalpies. They lose the outermost electron(s) readily to form 1+ ion (in the case of alkali metals) or 2+ ion (in the case of alkaline earth metals). The metallic character and the reactivity increase as we go down the group. Because of high reactivity they are never found pure in nature. The compounds of the s-block elements, with the exception of those of lithium and beryllium are predominantly ionic. The p-Block Elements comprise those belonging to Group 13 to 18 and these together with the s-Block Elements are called the Representative Elements or Main Group Elements. The outermost electronic configuration varies from ns2np1 to ns2np6 in each period. At the end of each period is a noble gas element with a closed valence shell ns2np6 configuration. All the orbitals in the valence shell of the noble gases are completely filled by electrons and it is very difficult to alter this stable arrangement by the addition or removal of electrons. The noble gases thus exhibit very low chemical reactivity. Preceding the noble gas family are two chemically important groups of non-metals. They are the halogens (Group 17) and the chalcogens (Group 16).The non-metallic character increases as we move from left to right across a period and metallic character increases as we go down the group. These are the elements of Group 3 to 12 in the centre of the Periodic Table. These are characterised by the filling of inner d orbitals by electrons and are therefore referred to as d-Block Elements. These elements have the general outer electronic configuration (n-1)d1-10ns0-2 . They are all metals. They mostly form coloured ions, exhibit variable valence (oxidation states), paramagnetism and oftenly used as catalysts. However, Zn, Cd and Hg which have the electronic configuration, (n-1) d10ns2 do not show most of the properties of transition elements. In a way, transition metals form a bridge between the chemically active metals of s-block elements and the less active elements of Groups 13 and 14 and thus take their familiar name “Transition Elements”.The two rows of elements at the bottom of the Periodic Table, called the Lanthanoids, Ce(Z = 58) – Lu(Z = 71) and Actinoids, Th(Z = 90) – Lr (Z = 103) are characterised by the outer electronic configuration (n-2)f 1-14 (n-1)d 0–1ns2 . The last electron added to each element is filled in f- orbital. These two series of elements are hence called the Inner- Transition Elements (f-Block Elements). They are all metals. Within each series, the properties of the elements are quite similar. The chemistry of the early actinoids is more complicated than the corresponding lanthanoids, due to the large number of oxidation states possible for these actinoid elements. Actinoid elements are radioactive. Many of the actinoid elements have been made only in nanogram quantities or even less by nuclear reactions and their chemistry is not fully studied. The elements after uranium are called Transuranium Elements. The elements can be divided into Metals and Non-Metals. In contrast, non-metals are located at the top right hand side of the Periodic Table. The elements become more metallic as we go down a group; the non- metallic character increases as one goes from left to right across the Periodic Table. Periodic Table show properties that are characteristic of both metals and non- metals. These elements are called Semi-metals or Metalloids.
  1. Alkali metal and alkaline earth metal belongs to ..
  1. S – block
  2. P – block
  3. D – block
  4. F – block
  1. The metallic character and the reactivity … as we go down the group.
  1. Decreases
  2. Increases
  3. Remains Constant
  4. None of Above
  1. Group … Elements known as chalcogens.
  1. 12
  2. 14
  3. 16
  4. 18
  1. Elements Ce(Z = 58) to Lu(Z = 71) are known as:
  1. Halogens
  2. Chalcogens
  3. Actinoids
  4. Lanthenoids
  1. The elements after uranium are called … Elements.
  1. Halogens
  2. Chalcogens
  3. Actinoids
  4. Transuranium
→
When anions and cations approach each other, the valence shell of anions are pulled towards the cation nucleus and thus, the shape of the anion is deformed. The phenomenon of deformation of anion by a cation is known as polarization and the ability of the cation to polarize the anion is called as polarizing power of cation. Due to polarization, sharing of electrons occurs between two ions to some extent and the bond shows some covalent character.
The magnitude of polarization depends upon a number of factors.

1. Out of $AlCl _3$ and $AlI _3$ which halides show maximum polarization? (1)
2. Out of $AlCl _3$ and $CaCl _2$ which one is more covalent in nature? (1)
3. The non-aqueous solvent like ether is added to the mixture of $LiCl , NaCl$ and KCl . Which will be extracted into the ether? (2)
OR
Out of $CaF _2$ and $CaI _2$ which one has a minimum melting point? (2)
→
Read the passage given below and answer the following questions from (i) to (v).$\frac{1}{2}\text{N}_2(\text{g})+\frac{3}{2}\text{H}_2(\text{g})=\text{NH}_3(\text{g});\triangle_\text{r}\text{H}^\ominus=-46.1\text{kJ}\text{mol}^{-1}$
$\frac{1}{2}\text{H}_2(\text{g})+\frac{1}{2}\text{Cl}_2(\text{g})=\text{HCl}(\text{g});\triangle_\text{r}\text{H}^\ominus=-92.32\text{kJ}\text{mol}^{-1}$
$\text{H}_2(\text{g})+\frac{1}{2}\text{O}_2(\text{g})\rightarrow\text{H}_2\text{O}(\text{l});\triangle_\text{r}\text{H}^\ominus=-285.8\text{kJ}\text{mol}^{-1}$
A spontaneous process is an irreversible process and may only be reversed by some external agency. If we examine the phenomenon like flow of Water down hill or fall of a stone on to the Ground, we find that there is a net decrease in Potential energy in the direction of change. By Analogy, we may be tempted to state that a Chemical reaction is spontaneous in a given Direction, because decrease in energy has Taken place, as in the case of exothermic Reactions. For example: The decrease in enthalpy in passing from Reactants to products may be shown for any Exothermic reaction on an enthalpy diagram. Thus, the postulate that driving force for a Chemical reaction may be due to decrease in Energy sounds ‘reasonable’ as the basis of Evidence so far ! Now let us examine the following reactions:$\frac{1}{2}\text{N}_2(\text{g})+\text{O}_2(\text{g})\rightarrow\text{NO}_2(\text{g});\triangle_\text{r}\text{H}^\ominus=+33.2\text{kJ}\text{mol}^{-1}$
$\text{C}(\text{graphite,s})+2\text{s}(\text{l})\rightarrow\text{CS}_2(\text{l});\triangle_\text{r}\text{H}^\ominus=+128.5\text{kJ}\text{mol}^{-1}$
Therefore, it becomes obvious that while decrease in enthalpy may be a contributory factor for spontaneity, but it is not true for all cases . Whenever heat is added to the system, it increases molecular motions causing increased randomness in the system. Thus heat (q) has randomising influence on the system. Temperature is the measure of average chaotic motion of particles in the system. The entropy change is inversely proportional to the temperature. $\triangle\text{S}$ is related with q and T for a reversible reaction as:$\triangle\text{S}=\frac{\text{q}_\text{rev}}{\text{T}}$
The total entropy change $(\triangle\text{S}_\text{total})$ for the system and surrounding of a spontaneous process is given by $\triangle\text{S}_\text{total}=\triangle\text{S}_\text{system}+\triangle\text{S}_\text{surr}.0$ When a system is in equilibrium, the entropy is maximum, and the change in entropy,$\triangle\text{S}=0.$ We can say that entropy for a spontaneous process increases till it reaches maximum and at equilibrium the change in entropy is zero. Since entropy is a state property, we can calculate the change in entropy of a reversible process by$\triangle\text{S}_{\text{sys}}=\frac{\text{q}_\text{rev,rew}}{\text{T}}$
G = H – TS Gibbs function, G is an extensive property and a state function. The change in Gibbs energy for the system, $\triangle\text{G}_{\text{sys}}$can be written as$\therefore\triangle\text{G}_{\text{sys}}=\triangle\text{H}_{\text{sys}}-\text{T}\triangle\text{S}_{\text{sys}}-\text{S}_\text{sys}\triangle\text{T}$
At constant temperature, $\triangle\text{T}=0$$\triangle\text{G}_{\text{sys}}=\triangle\text{H}_{\text{sys}}-\text{T}\triangle\text{S}_{\text{sys}}$
Usually the subscript ‘system’ is dropped and we simply write this equation as$\triangle\text{G}=\triangle\text{H}–\text{T}\triangle\text{S}$
Thus, Gibbs energy change = enthalpy change – temperature × entropy change, and is referred to as the Gibbs equation, one of the most important equations in chemistry. Here, we have considered both terms together for spontaneity: energy (in terms of $\triangle\text{H}$) and entropy ($\triangle\text{s},$ a measure of disorder) as indicated earlier. Dimensionally if we analyse, we find that $\triangle\text{G}$ has units of energy because, both $\triangle\text{H}$ and the $\text{T}\triangle\text{S}$ are energy terms, since $\text{T}\triangle\text{S}=(\text{K}) (\text{J/K}) = \text{J}.$ Now let us consider how $\triangle\text{G}$ is related to reaction spontaneity. We know, $\triangle\text{S}_\text{systeam}+\triangle\text{S}_\text{surrounding}$ If the system is in thermal equilibrium with The surrounding, then the temperature of the Surrounding is same as that of the system. Also, increase in enthalpy of the surrounding Is equal to decrease in the enthalpy of the System. Therefore, entropy change of Surroundings,$\triangle\text{S}_\text{surr}=\frac{\triangle\text{H}_\text{surr}}{\text{T}}-\frac{\triangle\text{H}_\text{sys}}{\text{T}}$
$\triangle\text{S}_\text{total}=\text{S}_\text{sys}+\Big(-\frac{\triangle\text{H}_\text{sys}}{\text{T}}\Big)$
Rearrangine the above equation:$\text{T}\triangle\text{S}_{\text{total}}=\text{T}\triangle\text{S}_{\text{sys}}-\triangle\text{H}_{\text{sys}}$
For spontaneous process,$\triangle\text{S}_\text{total}>0,$ so
$\text{T}\triangle\text{S}_{\text{sys}}-\triangle\text{H}_{\text{sys}}>\text{O}$
$\Rightarrow-(\triangle\text{H}_{\text{sys}}-\text{T}\triangle\text{S}_{\text{sys}})$
Using equation, the above equation can Be written as:$-\triangle\text{G}>\text{O}$
$\triangle\text{G}=\triangle\text{H}-\text{T}\triangle\text{S},0$
$\triangle\text{H}_\text{sys}$
Is the enthalpy change of a reaction, $\text{T}\triangle\text{S}_\text{sys}$ Is the energy which is not available to Do useful work. So $\triangle\text{G}$ is the net energy Available to do useful work and is thus a Measure of the ‘free energy. For this reason, it Is also known as the free energy of the reaction. $\triangle\text{G}$ gives a criteria for spontaneity at Constant pressure and temperature. If $\triangle\text{G}$ is negative (< 0), the process is b) If $\triangle\text{G}$ is positive (> 0), the process is non Entropy and Second Law of Thermodynamics – For an isolated system the change in energy remains constant. Therefore, increase in entropy in such systems is the natural direction of a spontaneous change. This, in fact is the second law of thermodynamics. Like first law of thermodynamics, second law can also be stated in several ways. The second law of thermodynamics explains why spontaneous exothermic reactions are so common. In exothermic reactions heat released by the reaction increases the disorder of the surroundings and overall entropy change is positive which makes the reaction spontaneous. Absolute Entropy and Third Law of Thermodynamics Molecules of a substance may move in a straight line in any direction, they may spin like a top and the bonds in the molecules may stretch and compress. These motions of the molecule are called translational, rotational and vibrational motion respectively. When temperature of the system rises, these motions become more vigorous and entropy increases. On the other hand when temperature is lowered, the entropy decreases. The entropy of any pure crystalline substance approaches zero as the temperature approaches absolute zero. This is called third law of thermodynamics. This is so because there is perfect order in a crystal at absolute zero. The statement is confined to pure crystalline solids because theoretical arguments and practical evidences have shown that entropy of solutions and super cooled liquids is not zero at 0K. The importance of the third law lies in the fact that it permits the calculation of absolute values of entropy of pure substance from thermal data alone. For a pure substance, this can be done by summing q T rev increments from 0K to 298K. Standard entropies can be used to calculate standard entropy changes by a Hess’s law type of calculation.$\text{A}+\text{B}\rightleftharpoons\text{C}+\text{D};$ is $\triangle_\text{r}\text{G}=0$
A knowledge of the sign and Magnitude of the free energy change of a Chemical reaction allows: Prediction of the spontaneity of the Chemical reaction. Prediction of the useful work that could Be extracted from it. So far we have considered free energy Changes in irreversible reactions. Let us now Examine the free energy changes in reversible Reactions.‘Reversible’ under strict thermodynamic Sense is a special way of carrying out a Process such that system is at all times in Perfect equilibrium with its surroundings. When applied to a chemical reaction, the Term ‘reversible’ indicates that a given Reaction can proceed in either direction Simultaneously, so that a dynamic Equilibrium is set up. This means that the Reactions in both the directions should proceed with a decrease in free energy, which seems impossible. It is possible only if at equilibrium the free energy of the system is minimum. If it is not, the system would spontaneously change to configuration of lower free energy. So, the criterion for equilibrium Gibbs energy for a reaction in which all reactants and products are in standard state, $\triangle_\text{r}\text{G}=0$ is related to the equilibrium constant of the reaction as follows:$0=\triangle_\text{r}\text{G}^{\ominus}+\text{RT}\text{ln}\text{K}$
or $\triangle_\text{r}\text{G}^{\ominus}=-\text{RT}\text{ln}\text{K}$ or $\triangle_\text{r}\text{G}^{\ominus}=-2.303\text{RT}\log\text{K}$ We also know that$\triangle_\text{r}\text{G}^{\ominus}=\triangle_\text{r}\text{H}^{\ominus}-\text{T}\triangle_\text{r}\text{S}^{\ominus}-\text{RT}\text{ln}\text{K}$
For strongly endothermic reactions, the value of $\triangle_\text{r}\text{H}^\phi$ may be large and positive. In such a case, value of K will be much smaller than 1 and the reaction is unlikely to form much product. In case of exothermic reactions, $\triangle_\text{r}\text{H}^\phi$ is large and negative, and $\triangle_\text{r}\text{G}^\phi$ is likely to be large and negative too. In such cases, K will be much larger than 1. We may expect strongly exothermic reactions to have a large K, and hence can go to near completion. $\triangle_\text{r}\text{G}^\phi$ also depends upon $\triangle_\text{r}\text{S}^\phi,$ if the changes in the entropy of reaction is also taken into account, the value of K or extent of chemical reaction will also be affected, depending upon whether $\triangle_\text{r}\text{S}^\phi$ Is positive or Negative, It is possible to obtain an estimate of $\triangle{\text{G}}^0$ From the measurement of $\triangle{\text{H}}^0$ And $\triangle{\text{S}}^0,$ And then calculate K at any temperature For economic yields of the products. If K is measured directly in the Laboratory, value of $\triangle{\text{G}}^0$ At any other Temperature can be calculated.
  1. A spontaneous process is an … process.
  1. Irreversible
  2. Reversible
  3. Partially irreversible
  4. Partially reversible
  1. $\triangle\text{S}_\text{systeam}+\triangle\text{S}_\text{surrounding}$
  1. < 0
  2. > 0
  3. = 0
  4. None of above
  1. When a system is in equilibrium, the entropy is maximum, and the change in entropy, $\triangle{\text{S}}.....0.$
  1. <
  2. >
  3. =
  4. None of above
  1. … does not discriminate between reversible and irreversible process:
  1. $\triangle\text{H}$
  2. $\triangle\text{S}$
  3. $\triangle\text{G}$
  4. $\triangle\text{U}$
  1. $\text{T}\triangle\text{S}=\ ...$
  1. Kg
  2. J
  3. M
  4. lit
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Read the passage given below and answer the following questions from 1 to 5.
Important compounds of calcium are calcium Oxide, calcium hydroxide, calcium sulphate, Calcium carbonate and cement. These are Industrially important compounds. The large Scale preparation of these compounds and Their uses are described below. Calcium Oxide or Quick Lime CaO – It is prepared on a commercial scale by Heating limestone $(CaCO_3)$ in a rotary kiln at 1070-1270 K.
$\text{CaCO}_3\overset{\text{heat}}{\rightleftharpoons}\text{CaO} + \text{CO}_2$
The carbon dioxide is removed as soon as It is produced to enable the reaction to proceed To completion. Calcium oxide is a white amorphous solid. It has a melting point of 2870 K. On exposure To atmosphere, it absorbs moisture and carbon Dioxide.
$\text{CaO}+\text{H}_2\text{O}\rightarrow\text{Ca}{\text{OH}}_2$
$\text{CaO}+\text{CO}_2\rightarrow\text{CaCO}_3$
The addition of limited amount of water Breaks the lump of lime. This process is called Slaking of lime. Quick lime slaked with soda Gives solid sodalime. Being a basic oxide, it Combines with acidic oxides at high Temperature.
$\text{CaO}+\text{SiO}_2\rightarrow\text{CaSiO}_3$
$6\text{CaO}+\text{P}_4\text{O}_{10}\rightarrow2\text{Ca}_3(\text{PO}_4)_2$
Uses: It is an important primary material for Manufacturing cement and is the cheapest Form of alkali. It is used in the manufacture of sodium Carbonate from caustic soda. It is employed in the purification of sugar And in the manufacture of dye stuffs.
Calcium Carbonate – $CaCO_3$ occurs in nature in several Forms like limestone, chalk, marble etc. It can Be prepared by passing carbon dioxide Through slaked lime or by the addition of Sodium carbonate to calcium chloride.
$\text{Ca}\text{(OH)}_2+\text{CO}_2\rightarrow\text{CaCO}_3+\text{H}_2\text{O}$
$\text{CaCl}_2+\text{Na}_2\text{CO}_3+\text{CaCO}_3+2\text{NaCl}$
excess of carbon dioxide should be Avoided since this leads to the formation of Water soluble calcium hydrogencarbonate. Calcium carbonate is a white fluffy powder. It is almost insoluble in water. When heated To 1200 K, it decomposes to evolve carbon Dioxide. It is used as a building material in the form of Marble and in the manufacture of quick lime. Calcium carbonate along with magnesium Carbonate is used as a flux in the extraction of Metals such as iron. Specially precipitated CaCO3 Is extensively used in the manufacture Of high quality paper. It is also used as an Antacid, mild abrasive in tooth paste, a Constituent of chewing gum, and a filler in Cosmetics.
Calcium Sulphate (Plaster of Paris), $\text{CaSO}_4·\frac{1}{2}\text{H}_2\text{O}^ –$ It is a hemihydrate of calcium sulphate. It is Obtained when gypsum, $CaSO_4·2H_2O$, is Heated to 393 K.
$2(\text{CaSO}_4.2\text{H}_2)\rightarrow2\text{(CaSO}_4)\text{H}_2\text{O}+3\text{H}_2\text{O}$
above 393 K, no water of crystallisation is left and anhydrous calcium sulphate, $CaSO_4$ is Formed. This is known as ‘dead burnt plaster’. It has a remarkable property of setting with Water. On mixing with an adequate quantity Of water it forms a plastic mass that gets into a Hard solid in 5 to 15 minutes.
Uses: The largest use of Plaster of Paris is in the Building industry as well as plasters. It is used For immoblising the affected part of organ where There is a bone fracture or sprain. It is also Employed in dentistry, in ornamental work and For making casts of statues and busts.
Cement: Cement is an important building Material. it was first introduced in England in 1824 by Joseph Aspdin. It is also called Portland cement because it resembles with the Natural limestone quarried in the Isle of Portland, England. Cement is a product obtained by Combining a material rich in lime, $CaO$ with Other material such as clay which contains Silica, $SiO_2$ along with the oxides of Aluminium, iron and magnesium. The average Composition of Portland cement is : CaO, 50-60%; $SiO_2, 20-25\%; Al2O_3, 5-10\%; MgO, 2-3\%; Fe2O_3, 1-2\%$ and $SO_3, 1-2\%$. For a good Quality cement, the ratio of silica $(SiO_2)$ to Alumina $(Al2O_3)$ should be between 2.5 and 4 And the ratio of lime (CaO) to the total of the Oxides of silicon $(SiO_2)$ aluminium $(Al2O_3)$ And iron $(Fe2O_3)$ should be as close as possible To 2. The raw materials for the manufacture of Cement are limestone and clay. When clay and Lime are strongly heated together they fuse and React to form ‘cement clinker’. This clinker is Mixed with 2-3% by weight of gypsum $(CaSO_4·2H_2O)$ to form cement. Thus important Ingredients present in Portland cement are Dicalcium silicate $(Ca_2SiO4)$ 26%, tricalcium silicate $(Ca_3SiO_5)$ 51% and tricalcium Aluminate $(Ca_3Al_2O_6)$ 11%.
Setting of Cement: When mixed with water, The setting of cement takes place to give a hard Mass. This is due to the hydration of the Molecules of the constituents and their Rearrangement. The purpose of adding Gypsum is only to slow down the process of Setting of the cement so that it gets sufficiently Hardened.
Uses: Cement has become a commodity of National necessity for any country next to iron And steel. It is used in concrete and reinforced Concrete, in plastering and in the construction Of bridges, dams and buildings.
Biological importance of magnesium and calcium - An adult body contains about 25g of Mg and 1200g of Ca compared with only 5g of iron And 0.06 g of copper. The daily requirement In the human body has been estimated to be 200 – 300 mg. All enzymes that utilise ATP in phosphate Transfer require magnesium as the cofactor. The main pigment for the absorption of light In plants is chlorophyll which contains Magnesium. About 99% of body calcium is Present in bones and teeth. It also plays Important roles in neuromuscular function, Interneuronal transmission, cell membrane Integrity and blood coagulation. The calcium Concentration in plasma is regulated at about $100\ mgL^{–1}.$ It is maintained by two hormones: Calcitonin and parathyroid hormone. Do you Know that bone is not an inert and unchanging Substance but is continuously being Solubilised and redeposited to the extent of 400mg per day in man? All this calcium Passes through the plasma.
  1. Quick Lime is prepared on a commercial scale by heating … in a rotary kiln at 1070-1270 K.
  1. $CaCO_3$
  2. $Ca_3Al_2O_6$
  3. $CaSO_42H_2O$
  4. $CaO$
  1. An adult body contains about … of Ca
  1. 600g
  2. 1200g
  3. 1800g
  4. 2400g
  1. The calcium Concentration in plasma is regulated at about …
  1. 10
  2. 50
  3. 100
  4. 500
  1. It was first introduced in England in 1824 by ….
  1. Edgar Dobbs
  2. Egor Cheliev
  3. James Parker
  4. Joseph Aspdin
  1. Molecular Formula of plaster of paris is …
  1. $CaSO_42H_2O$
  2. $\text{CaSO}_4·\frac{1}{2}\text{H}_2\text{O}^ –$
  3. $Ca_2SiO_4$
  4. $Ca_3Al_2O_6$
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Read the passage given below and answer the following questions from $(i)$ to $(v).$
Predicting the Direction of the Reaction- The equilibrium constant helps in predicting the direction in which a given reaction will proceed at any stage. For this purpose, we calculate the reaction quotient $Q$. The reaction quotient, $Q (Qc$ with molar concentrations and $QP$ with partial pressures$)$ is defined in the same way as the equilibrium constant Kc except that the concentrations in $Qc$ are not necessarily equilibrium values. For a general reaction:
$\text{a}\text{A}+\text{b}\text{B}\rightleftharpoons\text{c}\text{C}+\text{d}\text{D}$
$\text{Q}\text{c}=\frac{[\text{C}]^\text{c}[\text{D}]^\text{d}}{[\text{A}]^\text{a}[\text{B}]^\text{b}}$
Then,
If $Qc > Kc,$ the reaction will proceed in the direction of reactants (reverse reaction).
If $Qc < Kc,$ the reaction will proceed in the direction of the products (forward reaction).
If $Qc = Kc,$ the reaction mixture is already at equilibrium. Consider the gaseous reaction of $H_2$ with $I_2$ ,
$\text{H}_{2(\text{g})}+\text{l}_{2(\text{g})}\rightleftharpoons2\text{Hl}_{(\text{g})};\text{kc}=57.0\text{at}700\text{k}.$
Suppose we have molar concentrations $[H_2 ]t =0.10M, [I_2 ]t = 0.20 M$ and $[HI]t = 0.40 M.$ (the subscript t on the concentration symbols means that the concentrations were measured at some arbitrary time t, not necessarily at equilibrium). Thus, the reaction quotient, Qc at this stage of the reaction is given by,
$\text{Qc}=\frac{[\text{Hl}]\text{t}^2}{[\text{H}]^2]_\text{t}[\text{l}_2]_\text{t}}=\frac{(0.40)_2}{(0.10)\times(0.20)}=8.0$
Now, in this case, $Qc (8.0)$ does not equal Kc $(57.0)$, so the mixture of $H2 _{(g)}, I2 _{(g)} $ and $HI_{(g)} $ is not at equilibrium; that is, more $H2 _{(g)} $ and $I 2 _{(g)} $ will react to form more $HI_{(g)} $ and their concentrations will decrease till $Qc = Kc.$ The reaction quotient, Qc is useful in predicting the direction of reaction by comparing the values of Qc and Kc.Thus, we can make the following generalisations concerning the direction of the reaction
If $Qc < Kc,$ net reaction goes from left to right
If $Qc > Kc,$ net reaction goes from right to left.
If $Qc = Kc,$ no net reaction occurs.
Calculating Equilibrium Concentrations In case of a problem in which we know the initial concentrations but do not know any of the equilibrium concentrations, the following three steps shall be followed:
Step 1) Write the balanced equation for the reaction.
Step 2) Under the balanced equation, make a table that lists for each substance involved in the reaction: $(a)$ the initial concentration, $(b)$ the change in concentration on going to equilibrium, and $(c)$ the equilibrium concentration. In constructing the table, define $x$ as the concentration (mol/L) of one of the substances that reacts on going to equilibrium, then use the stoichiometry of the reaction to determine the concentrations of the other substances in terms of $x.$
Step 3) Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x. If you are to solve a quadratic equation choose the mathematical solution that makes chemical sense.
Step 4) Calculate the equilibrium concentrations from the calculated value of $x.$
Step 5) Check your results by substituting them into the equilibrium equation.
Relationship between equilibrium constant K, reaction quotient Q and gibbs energy G The value of Kc for a reaction does not depend on the rate of the reaction. However, it is directly related to the thermodynamics of the reaction and in particular, to the change in Gibbs energy, $\triangle\text{G}.$ If,
$\triangle\text{G}$ is negative, then the reaction is spontaneous and proceeds in the forward direction.
$\triangle\text{G}$ is positive, then reaction is considered non-spontaneous. Instead, as reverse reaction would have a negative $\triangle\text{G},$ the products of the forward reaction shall be converted to the reactants.
$\triangle\text{G}$ is 0, reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction. A mathematical expression of this thermodynamic view of equilibrium can be described by the following equation:
$\triangle\text{G}=\triangle\text{G}^\phi+\text{RT}\text{lnQ}$
where, $\triangle\text{G}^\phi$ is standard Gibbs energy. At equilibrium, when $\triangle\text{G}=0$ and $Q = Kc,$ the equation becomes,
$\triangle\text{G}=\text{G}^\phi+\text{RT}\text{lnk}=0$
$\triangle\text{G}^\phi=-\text{RT}\text{lnk}$
$\text{Ink}=\frac{-\triangle\text{G}^\phi}{\text{RT}}$
Taking antilog of both sides, we get,
$\text{K}=\text{e}-\frac{\triangle\text{G}0}{\text{RT}}$
Hence, using the equation, the reaction spontaneity can be interpreted in terms of the value of $\triangle\text{G}^\phi.$
If $\triangle\text{G}^\phi>0$ then $\frac{-\triangle\text{G}^\phi}{\text{RT}}$ is positive, and $>1$, making $K > 1$, which implies a spontaneous reaction or the reaction which proceeds in the forward direction to such an extent that the products are present predominantly.
If $\triangle\text{G}^\phi>0,$ then $\frac{-\triangle\text{G}^\phi}{\text{RT}}$ is negative, and $< 1$, that is, $K < 1$, which implies a non-spontaneous reaction or a reaction which proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.
Factors affecting equilibria One of the principal goals of chemical synthesis is to maximise the conversion of the reactants to products while minimizing the expenditure of energy. This implies maximum yield of products at mild temperature and pressure conditions. If it does not happen, then the experimental conditions need to be adjusted. For example, in the Haber process for the synthesis of ammonia from $N_2$ and $H_2,$ the choice of experimental conditions is of real economic importance. Annual world production of ammonia is about hundred million tones, primarily for use as fertilizers. Equilibrium constant, Kc is independent of initial concentrations. But if a system at equilibrium is subjected to a change in the concentration of one or more of the reacting substances, then the system is no longer at equilibrium; and net reaction takes place in some direction until the system returns to equilibrium once again. Similarly, a change in temperature or pressure of the system may also alter the equilibrium. In order to decide what course the reaction adopts and make a qualitative prediction about the effect of a change in conditions on equilibrium we use Le Chatelier’s principle. It states that a change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change. This is applicable to all physical and chemical equilibria.
Effect of Concentration Change In general, when equilibrium is disturbed by the addition/removal of any reactant/ products, Le Chatelier’s principle predicts that:
The concentration stress of an added reactant/product is relieved by net reaction in the direction that consumes the added substance.
The concentration stress of a removed reactant/product is relieved by net reaction in the direction that replenishes the removed substance. or in other words, “When the concentration of any of the reactants or products in a reaction at equilibrium is changed, the composition of the equilibrium mixture changes so as to minimize the effect of concentration changes”. Let us take the reaction,
$\text{H}_{2(\text{g})}+\text{l}_{2(\text{g})}\rightleftharpoons2\text{Hl}_{(\text{g})}$
If $H_2$ is added to the reaction mixture at equilibrium, then the equilibrium of the reaction is disturbed. In order to restore it, the reaction proceeds in a direction wherein $H_2$ is consumed, i.e., more of $H_2$ and $I_2$ react to form HI and finally the equilibrium shifts in right (forward) direction. This is in accordance with the Le Chatelier’s principle which implies that in case of addition of a reactant/product, a new equilibrium will be set up in which the concentration of the reactant/product should be less than what it was after the addition but more than what it was in the original mixture. The same point can be explained in terms of the reaction quotient, $Qc,$​​​​​​​​​​​​​​
Case Study Questions Class 11 Chemistry – Equilibrium
$\text{Qc}=\frac{[\text{HI}]^2}{[\text{H}]_2[\text{I}]_2}$
Addition of hydrogen at equilibrium results in value of $Qc$ being less than $Kc$. Thus, in order to attain equilibrium again reaction moves in the forward direction. Similarly, we can say that removal of a product also boosts the forward reaction and increases the concentration of the products and this has great commercial application in cases of reactions, where the product is a gas or a volatile substance. In case of manufacture of ammonia, ammonia is liquified and removed from the reaction mixture so that reaction keeps moving in forward direction. Similarly, in the large scale production of $CaO$ (used as important building material) from $CaCO_3$, constant removal of $CO_2$ from the kiln drives the reaction to completion. It should be remembered that continuous removal of a product maintains Qc at a value less than Kc and reaction continues to move in the forward direction.
  1. If … the reaction will proceed in the direction of reactants (reverse reaction).
  1. $Qc > Kc$
  2. $Qc < Kc$
  3. $Qc = Kc$
  4. None of above
  1. If … the reaction will proceed in the direction of the products (forward reaction).
  1. $Qc > Kc$
  2. $Qc < Kc$
  3. $Qc = Kc$
  4. None of above
  1. If … the reaction mixture is already at equilibrium. Consider the gaseous reaction.
  1. $Qc > Kc$
  2. $Qc < Kc$
  3. $Qc = Kc$
  4. All of above
  1. If $\triangle\text{G}$ is …. then the reaction is spontaneous and proceeds in the forward direction.
  1. Zero
  2. Positive
  3. Negative
  4. None of above
  1. $\triangle\text{G}$ is … reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction.
  1. Zero
  2. Positive
  3. Negative
  4. None of above
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The molecular orbital theory is based on the principle of a linear combination of atomic orbitals. According to this approach when atomic orbitals of the atoms come closer, they undergo constructive interference as well as destructive interference giving molecular orbitals, i.e., two atomic orbitals overlap to form two molecular orbitals, one of which lies at a lower energy level (bonding molecular orbital). Each molecular orbital can hold one or two electrons in accordance with Pauli's exclusion principle and Hund's rule of maximum multiplicity. For molecules up to $N _2$, the order of filling of orbitals is:
Image
Bond order $=\frac{1}{2}$ [bonding electrons - antibonding electrons]
Bond order gives the following information:
i. If bond order is greater than zero, the molecule/ion exists otherwise not.
ii. Higher the bond order, higher is the bond dissociation energy.
iii. Higher the bond order, greater is the bond stability.
iv. Higher the bond order, shorter is the bond length.

1. Arrange the following negative stabilities of $CN , CN ^{+}$and $CN ^{-}$in increasing order of bond.
2. The molecular orbital theory is preferred over valence bond theory. Why?
3. Ethyne is acidic in nature in comparison to ethene and ethane. Why is it so?
OR
Bonding molecular orbital is lowered by a greater amount of energy than the amount by which antibonding molecular orbital is raised. Is this statement correct?
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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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The orbital wave function or $\psi$ for an electronin an atom has no physical meaning. It issimply a mathematical function of thecoordinates of the electron. However, fordifferent orbitals the plots of correspondingwave functions as a function of r (the distancefrom the nucleus) are different. According to the German physicist, MaxBorn, the square of the wave function(i.e.,$\psi^2$) at a point gives the probability densityof the electron at that point. Boundary surface diagrams of constantprobability density for different orbitals give afairly good representation of the shapes of theorbitals. In this representation, a boundarysurface or contour surface is drawn in spacefor an orbital on which the value of probabilitydensity $\mid\psi\mid2$ is constant. In principle manysuch boundary surfaces may be possible.However, for a given orbital, only thatboundary surface diagram of constantprobability density* is taken to be goodrepresentation of the shape of the orbital whichencloses a region or volume in which theprobability of finding the electron is very high,say, 90%.
In hydrogen atom, electron has the same energy when it is in the2s orbital as when it is present in 2p orbital.The orbitals having the same energy are calleddegenerate. The 1s orbital in a hydrogenatom, as said earlier, corresponds to the moststable condition and is called the ground stateand an electron residing in this orbital is moststrongly held by the nucleus.
An electron inthe 2s, 2p or higher orbitals in a hydrogen atomis in excited state.The filling of electrons into the orbitals ofdifferent atoms takes place according to theaufbau principle which is based on the Pauli’sexclusion principle, the Hund’s rule ofmaximum multiplicity and the relativeenergies of the orbitals. Theaufbausprinciple states : In the ground state of theatoms, the orbitals are filled in order oftheir increasing energies. In other words,electrons first occupy the lowest energy orbitalavailable to them and enter into higher energyorbitals only after the lower energy orbitals arefilled.The number of electrons to be filled in variousorbitals is restricted by the exclusion principle,given by the Austrian scientist Wolfgang Pauli(1926). According to this principle : No twoelectrons in an atom can have the sameset of four quantum numbers. Pauliexclusion principle can also be stated as : “Onlytwo electrons may exist in the same orbitaland these electrons must have oppositespin.” This means that the two electrons canhave the same value of three quantum numbersn, l and $m_l$, but must have the opposite spinquantum number.Hund’s Rule of Maximum Multiplicity rule deals with the filling of electrons into the orbitals belonging to the same subshell. It states : pairing ofelectrons in the orbitals belonging to thesame subshell (p, d or f) does not take placeuntil each orbital belonging to thatsubshell has got one electron each i.e., itis singly occupied.
The distribution of electrons into orbitals of anatom is called its electronic configuration.If one keeps in mind the basic rules whichgovern the filling of different atomic orbitals,the electronic configurations of different atomscan be written very easily.The electronic configuration of differentatoms can be represented in two ways. Forexample :
  1. $s^a p^bd^c$​​​​​​​…… notation
  2. Orbital diagram
  1. …at a point gives the probability density of the electron at that point.
  1. $\psi\times2$
  2. $-\psi^2$
  3. $\psi$
  4. $\psi^2$
  1. Only …. electrons may exist in the same orbital and these electrons must have opposite spin.
  1. One
  2. Two
  3. Three
  4. Four
  1. …deals with the filling of electrons into the orbitals belonging to the same subshell.
  1. Hund’s Rule of Maximum Multiplicity rule
  2. Pauli’s exclusion principle
  3. Aufbau principle
  4. Werner Heisenberg
  1. Electrons first occupy the …. energy orbital available to them and enter into … energy orbitals.
  1. Lowest, Higher
  2. Higher, Lowest
  3. Middle, Higher
  4. Higher, Middle
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A process or change is said to be reversible, if a change is brought out in such a way that the process could, at any moment, be reversed by an infinitesimal change. A reversible process proceeds infinitely slowly by a series of equilibrium states such that system and the surroundings are always in near equilibrium with each other. Processes other than reversible processes are known as irreversible processes.
Isothermal and free expansion of an ideal gas For isothermal (T = constant) expansion of an ideal gas into vacuum; w = 0 since pex = 0. Also, Joule determined experimentally that q = 0; therefore, $\triangle\text{U}=0, \triangle =+\text{Uqw}$ can be expressed for isothermal irreversible and reversible changes as follows:
  1. For isothermal irreversible change
$\text{q}=-\text{w}=\text{nRTIn}\frac{\text{V}_\text{f}}{\text{V}_{\text{i}}}$
$=2.303\ \text{nRT}\log\frac{\text{V}_\text{f}}{\text{V}_{\text{i}}}$
  1. For isothermal reversible change
  2. For adiabatic change, q = 0
$\triangle\text{U}=\text{w}_{\text{ad}}$
In thermodynamics, a distinction is made between extensive properties and intensive properties. An extensive property is a property whose value depends on the quantity or size of matter present in the system. For example, mass, volume, internal energy, enthalpy, heat capacity, etc. are extensive properties. Those properties which do not depend on the quantity or size of matter present are known as intensive properties. For example temperature, density, pressure etc. are intensive properties. A molar property, χm, is the value of an extensive property χ of the system for 1 mol of the substance. If n is the amount of matter, χ χ m = n is independent of the amount of matter. Other examples are molar volume .
Measurement of $\triangle\text{U}$ and $\triangle{Η}$: Calorimetry We can measure energy changes associated with chemical or physical processes by an experimental technique called calorimetry. In calorimetry, the process is carried out in a vessel called calorimeter, which is immersed in a known volume of a liquid. Knowing the heat capacity of the liquid in which calorimeter is immersed and the heat capacity of calorimeter, it is possible to determine the heat evolved in the process by measuring temperature changes. Measurements are made under two different conditions: i) at constant volume, qV ii) at constant pressure, qP
Explain the determination of DeltaU of a reaction calorimetrically.
∆U Measurements For chemical reactions, heat absorbed at constant volume, is measured in a bomb calorimeter. Here, a steel vessel (the bomb) is immersed in a water bath. The whole device is called calorimeter. The steel vessel is immersed in water bath to ensure that no heat is lost to the surroundings. A combustible substance is burnt in pure dioxygen supplied in the steel bomb. Heat evolved during the reaction is transferred to the water around the bomb and its temperature is monitored. Since the bomb calorimeter is sealed, its volume does not change i.e., the energy changes associated with reactions are measured at constant volume. Under these conditions, no work is done as the reaction is carried out at constant volume in the bomb calorimeter. Even for reactions involving gases, there is no work done as $\triangle\text{v}=0$ Temperature change of the calorimeter produced by the completed reaction is then converted to qV, by using the known heat capacity of the calorimeter with the help of equation
b)$\triangle{Η}$ Measurements Measurement of heat change at constant pressure (generally under atmospheric pressure) can be done in a calorimeter shown in Figure. We know that $\triangle{Η}=\text{qp}$ (at constant p) and, therefore, heat absorbed or evolved, qP at constant pressure is also called the heat of reaction or enthalpy of reaction, $\triangle\text{rΗ}$ In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, qP will be negative and ∆rH will also be negative. Similarly in an endothermic reaction, heat is absorbed, qP is positive and $\triangle\text{rΗ}$ will be positive.
THERMODYNAMICS - NCERT Class 11 Chemistry
  1. For adiabatic change, q = 0 then …
  1. $\triangle\text{U}=\text{w}_{\text{ad}}$
  2. $\triangle\text{U}=\text{q}+\text{w}$
  3. $\triangle\text{U}=\text{w}-\text{q}$
  4. $\triangle\text{U}=\text{w}_{\text{rev}}$
  1. The technique for measure energy changes associated with chemical or physical processes by an experimental technique called …
  1. Colourimetry
  2. Calorimetry
  3. Titration
  4. Photometry
  1. A property whose value depends on the quantity or size of matter present in the system is known as …
  1. Extensive
  2. Intensive
  3. Reversible
  4. Irreversible
  1. If there is no work done …
  1. V = 0
  2. V = 1
  3. V = 2
  4. V = 3
  1. In an endothermic reaction, heat is absorbed, qP is … and $\triangle\text{rΗ}$ will be …
  1. Positive, Positive
  2. Negative, Negative
  3. Positive, Negative
  4. Negative, Positive
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