Once an organic compound is extracted from a natural source or sy…

Once an organic compound is extracted from a natural source or synthesised in the laboratory, it is essential to purify it. Various methods used for the purification of organic compounds are based on the nature of the compound and the impurity present in it. Finally, the purity of a compound is ascertained by determining its melting or boiling point. This is one of the most commonly used techniques for the purification of solid organic compounds. In crystallisation Impurities, which impart colour to the solution are removed by adsorbing over activated charcoal. In distillation Liquids having different boiling points vaporise at different temperatures. The vapours are cooled and the liquids so formed are collected separately. Steam Distillation is applied to separate substances which are steam volatile and are immiscible with water. Distillation under reduced pressure: This method is used to purify liquids having very high boiling points.

i. Which method can be used to separate two compounds with different solubilities in a solvent?
ii. Distillation method is used to separate which type of substance?
iii. Which technique is used to separate aniline from aniline water mixture?
OR
Why chloroform and aniline are easily separated by the technique of distillation?

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Read the passage given below and answer the following questions from (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.

A spontaneous process is an … process.

Irreversible
Reversible
Partially irreversible
Partially reversible

$\triangle\text{S}_\text{systeam}+\triangle\text{S}_\text{surrounding}$

< 0
> 0
= 0
None of above

When a system is in equilibrium, the entropy is maximum, and the change in entropy, $\triangle{\text{S}}.....0.$

<
>
=
None of above

… does not discriminate between reversible and irreversible process:

$\triangle\text{H}$
$\triangle\text{S}$
$\triangle\text{G}$
$\triangle\text{U}$

$\text{T}\triangle\text{S}=\ ...$

Kg
J
M
lit

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Read the passage given below and answer the following questions from (i) to (v).
When covalent bond is formed betweentwo similar atoms, for example in $H _2, O _2, Cl _2, N_2 Or F _2$, the shared pair of electrons is equally Attracted by the two atoms. As a result electronPair is situated exactly between the twoldentical nuclei. The bond so formed is calledNonpolar covalent bond. As a result of polarisation, the moleculePossesses the dipole moment which can be defined as the productof the magnitude of the charge and theDistance between the centres of positive andNegative charge. It is usually designated by aGreek letter ' $\mu$ '. Mathematically, it is expressedAs follows :Dipole moment $(\mu)=$ charge $( Q ) \times$ distance ofSeparationDipole moment is usually expressed inDebye units (D). The conversion factor is $1 D =3.33564 \times 10^{-30} C$ mWhere C is coulomb and m is meter. Just as all the covalent bonds haveSome partial ionic character, the ionicBonds also have partial covalentCharacter. The partial covalent character of ionic bonds was discussed by Fajans in terms of the following rules:
- The smaller the size of the cation and theLarger the size of the anion, the greater theCovalent character of an ionic bond.
- The greater the charge on the cation, theGreater the covalent character of the ionic bond.
- For cations of the same size and charge, The one, with electronic configuration( $n -1) d ^0 n s ^0$, typical of transition metals, isMore polarising than the one with a nobleGas configuration, ns2 np6, typical of alkali and alkaline earth metal cations.

Sidgwick and Powell in 1940, proposed a simple theoryBased on the repulsive interactions of theElectron pairs in the valence shell of the atoms.It was further developed and redefined byNyholm and Gillespie (1957).The main postulates of VSEPR theory areAs follows:
- The shape of a molecule depends uponThe number of valence shell electron pairs(bonded or nonbonded) around the centralAtom.
- Pairs of electrons in the valence shell repelone another since their electron clouds arenegatively charged.
- These pairs of electrons tend to occupySuch positions in space that minimiseRepulsion and thus maximise distanceBetween them.
- The valence shell is taken as a sphere withThe electron pairs localising on theSpherical surface at maximum distanceFrom one another.
- A multiple bond is treated as if it is a singleElectron pair and the two or three electronPairs of a multiple bond are treated as aSingle super pair.
- Where two or more resonance structuresCan represent a molecule, the VSEPRModel is applicable to any such structure.
The arrangement of electron pairs and the atoms around the central atom can be : linear,Trigonal planar, tetrahedral, trigonal-Bipyramidal and octahedral. Valence bond theory was introduced byHeitler and London (1927) and developedFurther by Pauling and others. A discussionOf the valence bond theory is based on the knowledge of atomic orbitals, electronicConfigurations of elements.partialmerging of atomic orbitals is called overlappingof atomic orbitals which results in the pairingof electrons. The extent of overlap decides thestrength of a covalent bond. according toorbital overlap concept, the formation of acovalent bond between two atoms results bypairing of electrons present in the valence shellhaving opposite spins. When orbitals of two atoms come close to formbond, their overlap may be positive, negativeor zero depending upon the sign anddirection of orientation of amplitude of orbitalwave function in space. Positive andnegative sign on boundary surface diagramsin the show the sign (phase) of orbitalwave function and are not related to charge.Orbitals forming bond should have same sign(phase) and orientation in space. This is calledpositive overlap. The criterion of overlap, as the main factorfor the formation of covalent bonds appliesuniformly to the homonuclear/heteronucleardiatomic molecules and polyatomic molecules.

Dipole moment is usually expressed in….

Debye
Centimeter
Columbs
Ergs

1D = .....

$33564\times 10^{–28}Cm$
$3.3564\times 10^{–30}Cm$
$33564\times 10^{–32}Cm$
$33564\times 10^{–34}Cm$

Valence bond theory was introduced by ….

Pauling and lewis
Nyholm and Gillespie
Heitler and London
Sidgwick and Powell

Pair is situated exactly between the two Identical nuclei the bond so formed is called …. covalent bond.

Unipolar
Bipolar
Polar
Nonpolar

Pairs of electrons in the valence shell … one another since their electron clouds are negatively charged.

Attract
Repel
Both a) & b)
None if above

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Read the passage given below and answer the following questions from (i) to (iii).
Le Chatelier’s principle is also known as the equilibrium law, used to predict the effect of change on a system at chemical equilibrium. This principle states that equilibrium adjusts the forward and backward reactions in such a way as to accept the change affecting the equilibrium condition. When factor-like concentration, pressure, temperature, inert gas that affect equilibrium are changed, the equilibrium will shift in that direction where the effects caused by these changes are nullified. This principle is also used to manipulate reversible reactions in order to obtain suitable outcomes.

Which one of the following conditions will favour the maximum formation of the product in the reaction?

$\text{A}_{2(\text{g})}+\text{B}_{2(\text{g})}\rightleftharpoons\text{X}_{2(\text{g})},\triangle_\text{r}\text{H}=-\text{XkJ}?$

Low temperature and high pressure.
Low temperature and low pressure.
High temperature and high pressure.
High temperature and low pressure.

For the reversible reaction,

$\text{N}_{2(\text{g})}+3\text{H}_{2(\text{g})}\rightleftharpoons2\text{NH}_{3(\text{g})}+\text{heat}$

The equilibrium shifts in forwarding direction

By increasing the concentration of $NH_3(g)$
By decreasing the pressure.
By decreasing the concentrations of $N_2(g)$ and $H_2(g)$
By increasing pressure and decreasing temperature.

In which one of the following equilibria will the point of equilibrium shift to left when the pressure of the system is increased?

$\text{H}_{2(\text{g})}+\text{l}_{2(\text{g})}\rightleftharpoons2\text{HI}_{(\text{g})}$
$2\text{NH}_{3(\text{g})}\rightleftharpoons\text{N}_{2(\text{g})}+3\text{H}_{2(\text{g})}$
$\text{C}_{(\text{s})}+\text{O}_{2(\text{g})}\rightleftharpoons\text{CO}_{2{\text{g}}}$
$2\text{H}_{2(\text{g})}+\text{O}_{2(\text{g})}\rightleftharpoons2\text{H}_2\text{O}_{(\text{g})}$

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

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

0.952g
3.816g
3.782g
8.64g

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

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

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

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

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

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

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

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

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Read the passage given below and answer the following questions from (i) to (v).
First complete data on pressure-volume-Temperature relations of a substance in bothGaseous and liquid state was obtained byThomas Andrews on Carbon dioxide. He plottedlsotherms of carbon dioxide at variousTemperatures.
Later on it was found That real gases behave in the same manner asCarbon dioxide. Andrews noticed that at highTemperatures isotherms look like that of anldeal gas and the gas cannot be liquified even atVery high pressure. As the temperature isLowered, shape of the curve changes and dataShow considerable deviation from idealBehaviour. At $30.98^{\circ} C$ carbon dioxide remainsGas upto 73 atmospheric pressure. At 73 atmospheric pressure, liquidCarbon dioxide appears for the first time. TheTemperature $30.98^{\circ} C$ is called criticalTemperature (TC) of carbon dioxide. This is theHighest temperature at which liquid carbonDioxide is observed. Above this temperature itls gas. Volume of one mole of the gas at criticalTemperature is called critical volume $\left( V _{ c }\right)$ andPressure at this temperature is called criticalPressure (pc). The critical temperature, pressureand volume are called critical constants. A gasBelow the critical temperature can be liquifiedBy applying pressure, and is called vapour ofThe substance. Carbon dioxide gas below itsCritical temperature is called carbon dioxideVapour.
Intermolecular forces are stronger in liquidState than in gaseous state. Molecules in liquidsAre so close that there is very little empty space between them and under normal conditionsLiquids are denser than gases.Molecules of liquids are held together byAttractive intermolecular forces. Liquids haveDefinite volume because molecules do notSeparate from each other. However, moleculesOf liquids can move past one another freely, Therefore, liquids can flow, can be poured andCan assume the shape of the container in whichThese are stored. If an evacuated container is partially filled withA liquid, a portion of liquid evaporates to fillthe remaining volume of the container withVapour. Initially the liquid evaporates andPressure exerted by vapours on the walls of The container (vapour pressure) increases. AfterSome time it becomes constant, an equilibriumls established between liquid phase andVapour phase. Vapour pressure at this stagels known as equilibrium vapour pressure orSaturated vapour pressure.. Since process ofVapourisation is temperature dependent; the Temperature must be mentioned whilereporting the vapour pressure of a liquid.
When a liquid is heated in an open vessel,The liquid vapourises from the surface. At theTemperature at which vapour pressure of theLiquid becomes equal to the external pressure,Vapourisation can occur throughout the bulkOf the liquid and vapours expand freely intoThe surroundings. The condition of freeVapourisation throughout the liquid is calledBoiling. The temperature at which vapourPressure of liquid is equal to the externalPressure is called boiling temperature at thatPressure. At 1 atm pressure boilingTemperature is called normal boiling point.If pressure is 1 bar then the boiling point isCalled standard boiling point of the liquid. Standard boiling point of the liquid is slightlyLower than the normal boiling point because 1 bar pressure is slightly less than 1 atmPressure . The normal boiling point of water is $100^{\circ} C (373 K)$, its standard boiling point is $99.6^{\circ} C (372.6 K)$.At high altitudes atmospheric pressure isLow. Therefore liquids at high altitudes boil atLower temperatures in comparison to that atSea level. Since water boils at low temperatureOn hills, the pressure cooker is used forCooking food. In hospitals surgical instrumentsAre sterilized in autoclaves in which boilingPoint of water is increased by increasing thePressure above the atmospheric pressure byUsing a weight covering the vent.Boiling does not occur when liquid isHeated in a closed vessel. On heatingContinuously vapour pressure increases.
AtFirst a clear boundary is visible between liquidAnd vapour phase because liquid is more denseThan vapour. As the temperature increases more and more molecules go to vapour phaseAnd density of vapours rises. At the same timeLiquid becomes less dense. It expands becauseMolecules move apart. When density of liquidAnd vapours becomes the same; the clearBoundary between liquid and vapoursDisappears. This temperature is called critical Temperature.

First complete data on Pressure-Volume-Temperature relations of a substance in both Gaseous and liquid state was obtained by:

Thomas Andrews
Fritz London
Robert Boyle
Joseph Lewis Gay Lussac

Critical Temperature (TC) of carbon dioxide is.....

$24^\circ C$
$30.8^\circ C$
$56^\circ C$
$29^\circ C$

The condition of free Vapourisation throughout the liquid is called …

Evaporation
Melting
Boiling
None of above

Standard boiling point of Water is....

$100^\circ C$
$3^\circ C$
$105^\circ C$
$99.6^\circ C$

Boundary between liquid and vapours Disappears,This temperature is called …

Critical temperature
Absolute temperature
Normal temperature
Boiling temperature

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Read the passage given below and answer the following questions from 1 to 5.
Oxidation state and trends in chemical Reactivity Due to small size of boron, the sum of its first Three ionization enthalpies is very high. This Prevents it to form +3 ions and forces it to form Only covalent compounds. But as we move from B to Al, the sum of the first three ionisation Enthalpies of Al considerably decreases, and Is therefore able to form $Al^{3+}$ ions. In fact, Aluminium is a highly electropositive metal. However, down the group, due to poor Shielding effect of intervening d and f orbitals, The increased effective nuclear charge holds ns Electrons tightly (responsible for inert pair Effect) and thereby, restricting their Participation in bonding. As a result of this, Only p-orbital electron may be involved in Bonding. In fact in Ga, In and Tl, both +1 and +3 oxidation states are observed. The relative Stability of +1 oxidation state progressively Increases for heavier elements: A l< Ga < In< Tl. In Thallium +1 oxidation state is predominant whereas the +3 oxidation state is highly Oxidising in character. The compounds in +1 oxidation state, as expected from energy Considerations, are more ionic than those in +3 oxidation state.
Important trends and anomalous properties of boron – certain important trends can be observed in the chemical behaviour of group 13 elements. The tri-chlorides, bromides and iodides of all these elements being covalent in nature are hydrolysed in water. Species like tetrahedral $[M(OH)_4]^–$ and octahedral $[M(H_2O)6]^{3+}$, except in boron, exist in aqueous medium. The monomeric trihalides, being electron deficient, are strong Lewis acids. Boron trifluoride easily reacts with Lewis bases such as $NH_3$ to complete octet around boron. It is due to the absence of d orbitals that the maximum covalence of B is 4. Since the d orbitals are available with Al and other elements, the maximum covalence can be expected beyond 4. Most of the other metal halides (e.g., $AlCl_3$) are dimerised through halogen bridging (e.g., $Al2Cl_6$). The metal species completes its octet by accepting electrons from halogen in these halogen bridged molecules.
i) Reactivity towards air Boron is unreactive in crystalline form. Aluminium forms a very thin oxide layer on The surface which protects the metal from Further attack. Amorphous boron and Aluminium metal on heating in air form $B_2O_3$ And $Al_2O_3$ respectively. With dinitrogen at high Temperature they form nitrides. The nature of these oxides varies down the Group. Boron trioxide is acidic and reacts with Basic (metallic) oxides forming metal borates. Aluminium and gallium oxides are amphoteric And those of indium and thallium are basic in Their properties.
ii) Reactivity towards acids and alkalies Boron does not react with acids and alkalies Even at moderate temperature; but aluminium Dissolves in mineral acids and aqueous alkalies And thus shows amphoteric character. Aluminium dissolves in dilute HCl and Liberates dihydrogen.
$2Al(s) + 6HCl (aq) \rightarrow 2Al_3^+ (aq) + 6Cl^– (aq) + 3H_2 (g)$
However, concentrated nitric acid renders Aluminium passive by forming a protective Oxide layer on the surface. Aluminium also reacts with aqueous alkali And liberates dihydrogen.
$2Al (s) + 2NaOH(aq) + 6H_2O(l) \rightarrow 2 Na+ [Al(OH)_4]^– (aq) + 3H_2(g)$
Sodium Tetrahydroxoaluminate(III).
iii) Reactivity towards halogens These elements react with halogens to form Trihalides (except TlI3). $2E(s) + 3 X_2 (g) \rightarrow 2EX_3 (s) (X = F, Cl, Br, I)$
Borax- It is the most important compound of boron. It is a white crystalline solid of formula $Na_2B_4O_7⋅10H_2O$. In fact it contains the Tetranuclear units and correct Formula; therefore, is $Na2 [B4O5 (OH) 4].8H2O$. Borax dissolves in water to give an alkaline Solution.
$Na_2B_4O7 + 7H_2O \rightarrow 2NaOH + 4H_3BO_3$
On heating, borax first loses water Molecules and swells up. On further heating it Turns into a transparent liquid, which solidifies Into glass like material known as borax Bead. $Na_2B_4O_7.10H_2O \rightarrow Na^2B_4O_7\rightarrow 2NaBO_2+ B2O_3$
Metaborate Boric Anhydride The metaborates of many transition metals Have characteristic colours and, therefore, Borax bead test can be used to identify them In the laboratory. For example, when borax is Heated in a Bunsen burner flame with CoO on A loop of platinum wire, a blue coloured Co(BO2) 2 bead is formed.
Orthoboric acid, $H_3BO_3$ is a white crystalline Solid, with soapy touch. It is sparingly soluble In water but highly soluble in hot water. It can Be prepared by acidifying an aqueous solution Of borax.
$Na_2B_4O_7 + 2HCl + 5H_2O \rightarrow 2NaCl + 4B(OH)_3$
It is also formed by the hydrolysis (reaction With water or dilute acid) of most boron Compounds (halides, hydrides, etc.). It has a layer structure in which planar $BO_3$ units are Joined by hydrogen.

Boron is … in crystalline form.

unreactive
highly reactive
less reactive
only (a) or (c)

Orthoboric acid is …

Amorphous
Crystalline
Polyamorphous
None of above

Aluminium and gallium oxides are … in their properties.

acidic
Basic
amphoteric
None of above

Indium and thallium are … in their properties.

acidic
Alkali
amphoteric
basic

Aluminium is a highly … metal.

electronegative
Neutral
electropositive
None of above

→

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

Case Study Questions Class 11 Chemistry – Equilibrium

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

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

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

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

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

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

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

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

Zero
Positive
Negative
None of above

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

Zero
Positive
Negative
None of above

→

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:

For isothermal irreversible change

$\text{q}=-\text{w}=\text{nRTIn}\frac{\text{V}_\text{f}}{\text{V}_{\text{i}}}$
$=2.303\ \text{nRT}\log\frac{\text{V}_\text{f}}{\text{V}_{\text{i}}}$

For isothermal reversible change
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

For adiabatic change, q = 0 then …

$\triangle\text{U}=\text{w}_{\text{ad}}$
$\triangle\text{U}=\text{q}+\text{w}$
$\triangle\text{U}=\text{w}-\text{q}$
$\triangle\text{U}=\text{w}_{\text{rev}}$

The technique for measure energy changes associated with chemical or physical processes by an experimental technique called …

Colourimetry
Calorimetry
Titration
Photometry

A property whose value depends on the quantity or size of matter present in the system is known as …

Extensive
Intensive
Reversible
Irreversible

If there is no work done …

V = 0
V = 1
V = 2
V = 3

In an endothermic reaction, heat is absorbed, qP is … and $\triangle\text{rΗ}$ will be …

Positive, Positive
Negative, Negative
Positive, Negative
Negative, Positive

→

Read the passage given below and answer the following questions from 1 to 5.
The dipositive oxidation state $(M^{2+})$ is the Predominant valence of Group 2 elements. The Alkaline earth metals form compounds which Are predominantly ionic but less ionic than the Corresponding compounds of alkali metals. This is due to increased nuclear charge and Smaller size. The oxides and other compounds Of beryllium and magnesium are more covalent Than those formed by the heavier and large Sized members (Ca, Sr, Ba). The general Characteristics of some of the compounds of Alkali earth metals are described below.
Oxides and Hydroxides: The alkaline Earth metals burn in oxygen to form the Monoxide, MO which, except for BeO, have Rock-salt structure. The BeO is essentially Covalent in nature. The enthalpies of formation Of these oxides are quite high and consequently They are very stable to heat. BeO is amphoteric While oxides of other elements are ionic in Nature. All these oxides except BeO are basic In nature and react with water to form sparingly Soluble hydroxides.
$MO + H2O \rightarrow M(OH)_2$
The solubility, thermal stability and the Basic character of these hydroxides increase With increasing atomic number from $Mg(OH)_2 To Ba(OH)_2$.The alkaline earth metal Hydroxides are, however, less basic and less Stable than alkali metal hydroxides. Beryllium Hydroxide is amphoteric in nature as it reacts With acid and alkali both.
$Be(OH)_2 + 2OH^– \rightarrow [Be(OH)4]^{2–}$
Beryllate ion
$Be(OH)_2 + 2HCl + 2H_2O \rightarrow [Be(OH)_4]Cl_2$
Halides: Except for beryllium halides, all Other halides of alkaline earth metals are ionic In nature. Beryllium halides are essentially Covalent and soluble in organic solvents. Beryllium chloride has a chain structure in the Solid state as shown below: In the vapour phase $BeCl_2$ tends to form a Chloro-bridged dimer which dissociates into the Linear monomer at high temperatures of the Order of 1200 K. The tendency to form halide Hydrates gradually decreases down the group. The dehydration Of hydrated chlorides, bromides and iodides Of Ca, Sr and Ba can be achieved on heating; However, the corresponding hydrated halides Of Be and Mg on heating suffer hydrolysis. The Fluorides are relatively less soluble than the Chlorides owing to their high lattice energies.
Salts of Oxoacids: The alkaline earth Metals also form salts of oxoacids. Some of These are :
Carbonates: Carbonates of alkaline earth Metals are insoluble in water and can be Precipitated by addition of a sodium or Ammonium carbonate solution to a solution Of a soluble salt of these metals. The solubility Of carbonates in water decreases as the atomic Number of the metal ion increases. All the Carbonates decompose on heating to give Carbon dioxide and the oxide. Beryllium Carbonate is unstable and can be kept only in The atmosphere of $CO_2$. The thermal stability Increases with increasing cationic size.
Sulphates: The sulphates of the alkaline earth Metals are all white solids and stable to heat. $BeSO_4,$ and $MgSO_4$ are readily soluble in water; The solubility decreases from $CaSO_4$ to $ BaSO_4$. The greater hydration enthalpies of $Be_2+$ and $Mg_2+$ ions overcome the lattice enthalpy factor And therefore their sulphates are soluble in Water.
Nitrates: The nitrates are made by dissolution Of the carbonates in dilute nitric acid. Magnesium nitrate crystallises with six Molecules of water, whereas barium nitrate Crystallises as the anhydrous salt. This again Shows a decreasing tendency to form hydrates With increasing size and decreasing hydration Enthalpy. All of them decompose on heating to Give the oxide like lithium nitrate.
$2\text{M}(\text{NO}_3)_2\rightarrow2\text{MO}+4\text{NO}_2+\text{O}_2$
(M = Be. Mg. Ca. Sr. Ba)
Beryllium, the first member of the Group 2 Metals, shows anomalous behaviour as Compared to magnesium and rest of the Members. Further, it shows diagonal Relationship to aluminium which is discussed Subsequently.
i) Beryllium has exceptionally small atomic And ionic sizes and thus does not compare Well with other members of the group. Because of high ionisation enthalpy and Small size it forms compounds which are Largely covalent and get easily hydrolysed.
ii) Beryllium does not exhibit coordination Number more than four as in its valence Shell there are only four orbitals. The Remaining members of the group can have A coordination number of six by making Use of d-orbitals.
iii) The oxide and hydroxide of beryllium, Unlike the hydroxides of other elements in The group, are amphoteric in nature.
Diagonal Relationship between Beryllium and Aluminium-The ionic radius of 4 is estimated to be 31 pm; the charge/radius ratio is nearly the Same as that of the $Al^{3+}$ ion. Hence beryllium Resembles aluminium in some ways. Some of The similarities are:
i) Like aluminium, beryllium is not readily Attacked by acids because of the presence Of an oxide film on the surface of the metal.
ii) Beryllium hydroxide dissolves in excess of Alkali to give a beryllate ion, $[Be(OH)_4]^{2–}$ just As aluminium hydroxide gives aluminate Ion, $[Al(OH)_4]^–.$
iii) The chlorides of both beryllium and Aluminium have $Cl^–$ Bridged chloride Structure in vapour phase. Both the Chlorides are soluble in organic solvents And are strong Lewis acids. They are used As Friedel Craft catalysts.
iv) Beryllium and aluminium ions have strong Tendency to form complexes, $BeF_4^{2–}, AlF_6^{3–}.$

The dipositive oxidation state $(M2^+)$ is the Predominant valence of … elements.

Group 2
Group 1
Group 17
Group 18

… the first member of the Group 2 metals.

Magnesium
Beryllium
Barium
Radium

Except for beryllium halides, all other halides of alkaline earth metals are ionic in nature.

Magnesium halides
beryllium halides
Calcium halides
Radium halides

The ionic radius of $Be^{2+}$ is estimated to be … pm.

310
500
50
31

Beryllium carbonate can be kept only in the atmosphere of …

$N_2$
$H_2$
$CO_2$
$O_2$

→

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