Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.	Explain electrogain enthalpy value of elements of 17^(th)group.
Answer» Solution : Electron gain enthalpy value of `17^(th)` group elements are very NEGATIVE. `{:("Element",F,Cl,Br,I,At),(Delta_(eg) H " kJ mol "^(-1),-328,-349,-325,-295,-279):}` Value of electron gain enthalpy of Cl is maximum negative. Remaining value of `e^(-)` gain enthalpy is decreases when atomic number increases. General electronic configuration of halogen elements are `ns^(2)np^(5)`they can acceft 1 electron and becomes stable LIKE noble gases `ns^(2)np^(6)`. So, its value of electron gain enthalpy is negative. NOTE: Electron gain enthalpy value of noble gases are more negative than elements of the group of 16, 17 which are situated right hand side of the noble gas.

2.	Explain effects of acid rain on Taj Mahal.
Answer» SOLUTION :A LARGE numbers of industris and power plants are around near the area of Taj Mahal. Which uses poor quality of coal, kerosene and firewood because of this the air around the Taj Mahal contains fairly HIGH LEVEL of sulphur and nitrogen OXIDES. This resulting in acid rain which reacts with marble of Taj Mahal. `CaCO_3 + H_2 SO_4 to CaSO_4 + H_2 O + CO_2` Thus, this wonderful monument that has attracted people from around the world. As a result, the monument is being slowly disfigured and the marble is getting discoloured and lustreless.

3.	Explain effect of resulting solution on addition of 0.05 M acetate ion to 0.05 M acetic acid solution .
Answer» Solution : As the commonis `[CH_3COO^-]` INCREASES the equilibrium more LEFT side and `[CH_3COOH]`increases. Which is more than `(0.52-x)`. As `[H^+]`decreases and `[H^+] lt x`. PH increases.

4.	Explain effect of temperature on Maxwell Boltzman velocity distribution.
Answer» Solution :Relation between molecular speed and number of molecular with increase in TEMPERATURE : Comparison of Low temperature `T_(1)` and high temperature GRAPH as under. According to Maxwell - Boltzman DISTRIBUTION curve at high temperature `T_(2)` curve is flat and wide that two Temperature curve. (i) Most probable speed of molecules increases with temperature and curve becomes more wide at high temperature `(T_(2))`. (ii) Wide curve indicated that number of molecules having more speed increases with temperature. (iii) Value of Average speed `(u_(AV))` increases at higher temperature `(T_(2))` than `(T_(1))` and most PROBABLY speed is low so, `u_(av)` and `u_(mp)` value is low at high temperature. `u_(av)` and `u_(mp)` value is high at low temperature. Number of molecules at high temperature `(n_(2))gt` number of molecules at lower temperature `(n_(1))`.

5.	Explain effect of temperature on equilibrium by suitable experiment.
Answer» Solution :Effect of temperature on equilibrium can be demonstrated by taking `NO_2` gas (brown in colour) which dimerises into `N_2O_4`gas (colourless). `{:(2NO_(2(g))HARR, N_2O_(4(g)) , , DeltaH-57.2 "kJ mol"^(-1)),("brown"" ""colour gas","colourless gas",):}` Method of experiment : `NO_2` gas prepared by ADDITION of Cu turnings to conc. `HNO_3` is collected in two 5 mL TEST tubes (ensuring same intensity of colour of gas in each tube) and stopper sealed with araldite. Take three 250 mL beakers 1, 2 and 3. Take three 250 ml beakers 1, 2 and 3 containing freezing mixture, water at room temperature and hot water (363K) respectively. Both the test tubes are placed in BEAKER 2 for 8-10 minutes. After this one is placed in beaker 1 and the other in beaker 3. The effect of temperature on direction of reaction is depicted very well in this experiment. This is shown in following figure. Observation: (i) At low temperature in beaker 1, the forward reaction of formation of `N_2O_4` is preferred, as reaction is exothermic, and thus, intensity of brown colour due to `NO_2` decreases. (ii) While in beaker 3, high temperature favours the reverse reaction of formation of `NO_2` and thus, the brown colour intensifies. Deduction In exothermic reaction, the temperature decreases so, forward reaction and if the temperature increases the reverse reaction occurs.

6.	Explain effect of temperature change on chemical equilibrium by example.
Answer» Solution :When a change in TEMPERATURE of equilibrium reaction occurs the value of equilibrium constant `K_c` is changed. In general the temperature dependence of the equilibrium constant depends on the sing of `DeltaH` for the reaction. The RATE of reaction is also changed due to the changes of temperature. Exothermic reaction: The equilibrium constant for an exothermic reaction (negative `DeltaH`) decreases as the temperature increases : REVERSE reaction occurs, `Q_c` increases, products decrease. e.g. Production of ammonia is exothermic reaction. `N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)) , DeltaH =-92.38 "kJ mol"^(-1)` According to Le-Chatelier.s principle, RAISING the temperature shifts the equilibrium to left and decreases the equilibrium concentration of ammonia. In other WORDS, low temperature is favourable for high yield of ammonia, but practically very low temperatures slow down the reaction and thus a catalyst is used. Endothermic reaction : The equilibrium constant for an endothermic reaction (positive `DeltaH`) increases as the temperature increases. Forward reaction will be occurs, `Q_c` decreases and products increases that types of equilibrium established. e.g, following reaction is endothermic. `{:("Reaction:",[Co(H_2O)_6]_((aq))^(2+)+,4Cl_((aq))^(-) hArr, 6H_2O_((l))+,[CoCl_4]_((aq))^(2-)),("Colour :","Pink","Colourless","Colourless","Blue"):}` At room temperature, the equilibrium mixture is blue due to `[CoCl_4]^(2-)` . When cooledin a freezing mixture, the colour of the mixture turns pink due to `[Co(H_2O)_6]^(3+)`

7.	Explain effect of pressure change on equilibrium system by suitable examples.
Answer» Solution :Effect of pressure change on equilibrium : If only gaseous components are there in reaction than only pressure effect on equilibrium. If the moles of gaseous product and reactant are same than the effect of pressure is not observed. If the moles of product and reactant different than pressure AFFECT equilibrium and new equilibrium is established. Pressure increase : (more moles) `to`(less moles of gas) In this way `Q_c ltK_c` reaction moves in forward. `underset"Total 4 mol"(CO_((g)) + 3H_2) hArr underset"Total 2 mol"(CO_(4(g)) + H_2O_((g)))` `therefore` In forward direction mole decreases `4 to 2`. If volume decrease pressure increases : At constant temperature if volume made half than pressure will be double. (p(V) = nRT) Explanation of Le-Chatelier.s principle of pressure effect As total pressure increases partial pressure of product and reactant also increases. partial pressure p `prop`concentration c `prop` moles of gas As per Le-Chatelier.s principle on increasing pressure reaction moves in such direction so that the effect that change is nullify. So forward reaction take place. "On increasing pressure reaction moves in such direction where moles of gases are less." Explain the effect of pressure by reaction quotient `Q_c`: In this METHANATION reaction molar concentration of `[H_2O], [CH_4], [CO]` and `[H_2]`. At equilibrium if total volume made half than the pressure become double. So concentration also become double. So `Q_c` Will be, `Q_c=([CH_(4(g))][H_2O_((g))])/([CO_((g))][H_(2(g))]^3)="2 mol"/"4 mol"` Numerator `lt` denominatorso `Q_c lt K_c . Q_c` decrease for foward reaction. Forward reaction take place and new equilibrium is attain. Derivation By decreasing total volume or by increasing total pressure the reaction moves in such direction where moles of gas decreases and new equilibrium is attained.

8.	Explain effect of mass of molecule from the basis of Maxwell - Boltzman speed distribution.
Answer» Solution :At fixed temperature speed of heavy MOLECULES less than speed of light molecules. For example, speed of `N_(2)` gas is more than speed of `Cl_(2)` gas. (Mass of `N_(2)= 28 MU`) and (Mass of `Cl_(2)=71 mu`) Speed of `N_(2)gt` Speed of `Cl_(2)` .... (consdtant T) `u_(MP) N_(2)gt u_(mp)Cl_(2)` .......(constant T) where, `u_(mp)=` Most probable speed `= (("maximum speed of"),("maximum molecules"))` Value if most Probable speed of `N_(2)` is more than value of most probable speed of `u_(2)`. Speed distribution CURVE of `N_(2)` and `Cl_(2)` is given in below figure.

9.	Explain : Effect of Inert Gas addition on equilibrium.
Answer» SOLUTION :If the volume is kept constant and an INERT gas such as ARGON is added which does not take part in the reaction, the EQUILIBRIUM remains undisturbed. It is because the addition of an inert gas at constant volume does not change the PARTIAL pressure or the molar concentrations of the substance involved in the reaction. The reaction quotient changes only if the added gas is a reactant or product involved in the reaction.

10.	Explain effect of concentration on equilibrium by suitable experiments. underset"Yellow"(Fe_((aq))^(3+))+ underset"Colourless"(SCbarN_((aq))) hArr underset"Deep red"([Fe(SCN)]_((aq))^(2+)) Explain effect by added, (i) Oxalic acid (H_2C_2O_4) (ii) HgCl_2 and (iii) Potassium thiocynate (KSCN) in equilibrium reaction.
Answer» Solution :Experiment (Step-1): A reddish colour appears on adding two drops of 0.002 M potassium thiocynate solution to 1 mL of 0.2 M IRON(III) nitrate solution due to the formation of `[Fe(SCN)]^(2+)`. The intensity of the red colour becomes constant on attaining equilibrium. `underset"Yellow"(Fe_((AQ))^(3+)) + underset"Colourless"(SCbarN_((aq))) hArr underset"Deep red"([Fe(SCN)]_((aq))^(2+))` and At equilibrium `K_c=([Fe(SCN)_((aq))^(2+)])/([Fe_((aq))^(3+)]+[SCbarN_((aq))])` This equilibrium can be shifted in either FORWARD or reverse directions depending on our choice of adding a reactant or a product. Experiment (Step-2): Adding of oxalic acid : Oxalic acid `[H_2C_2O_4]` reacts with `Fe^(3+)` ions to form the stable complex ion `[Fe(C_2O_4)_3]^(3-)` , thus decreasing the concentration of free `Fe^(3+)` `Fe^(3+) + 3C_2O_4^(2-) to [Fe(C_2O_4)]^(3-)` Such `Fe^(3+)` ion is USED and there is disturbance in equilibrium, `[Fe^(3+)]`decreases. In ACCORDANCE with the Le-Chatelier.s principle, the concentration stress of removed `Fe^(3+)` is relieved by dissociation of `[Fe(SCN)]^(2+)` to replenish the `Fe^(3+)` ions. Because the concentration of `[Fe(SCN)]^(2+)` decreases, the intensity of red colour decreases. Experiment (Step-3) : `HgCl_2` is added in equilibrium of Step-1 : Addition of `HgCl_2` also decreases red colour because `Hg^(2+)` reacts with `(SCN^-)` ions to form stable complex ion `[Hg(SCN)_4]^(2-)` and `HgCl_(2(aq)) hArr Hg_((aq))^(2+) + 2Cl_((aq))^(-)` `Hg_((aq))^(2+) + 4SCN^(-) to [Hg(SCN)_4]_((aq))^(2-)` Thus by addition of `HgCl_2SCN^-` in equilibrium decreases so the amount of product decreases. So the reaction moves in forward reaction and intensity of red colour decreases. Experiment (Step-4): If KSCN is added to solution than `[SCN^-]` increases so it consume in reaction and reaction proceed in reverse direction. `Fe_((aq))^(3+) +SCN_((aq))^(-) to [Fe(SCN)]^(2-)` and `SCN^-`decreases and new equilibrium is established. At new `[Fe(SCN)]^(2-)` equilibrium concentration of red colour is more.

11.	Explain : Effect of concentration equilibrium occurs in two drops of 0.2 M potassium thiocynate (KSCN) added in 1 mL. 0.2 M Iron (III) Nitrate solution.
Answer» Solution :EXPERIMENT (Step-1): A reddish colour appears on adding two drops of 0.002 M potassium thiocynate solution to 1 mL of 0.2 M iron(III) nitrate solution due to the formation of `[Fe(SCN)]^(2+)`. The intensity of the red colour becomes constant on ATTAINING equilibrium. `underset"Yellow"(Fe_((aq))^(3+)) + underset"Colourless"(SCbarN_((aq))) hArr underset"Deep red"([Fe(SCN)]_((aq))^(2+))` and At equilibrium `K_c=([Fe(SCN)_((aq))^(2+)])/([Fe_((aq))^(3+)]+[SCbarN_((aq))])` This equilibrium can be shifted in either forward or reverse directions depending on our choice of adding a reactant or a product. Experiment (Step-2): Adding of oxalic acid : Oxalic acid `[H_2C_2O_4]` reacts with `Fe^(3+)` ions to form the STABLE complex ion `[Fe(C_2O_4)_3]^(3-)` , THUS decreasing the concentration of free `Fe^(3+)` `Fe^(3+) + 3C_2O_4^(2-) to [Fe(C_2O_4)]^(3-)` Such `Fe^(3+)` ion is used and there is disturbance in equilibrium, `[Fe^(3+)]`decreases. In accordance with the Le-Chatelier.s principle, the concentration stress of removed `Fe^(3+)` is relieved by dissociation of `[Fe(SCN)]^(2+)` to replenish the `Fe^(3+)` ions. Because the concentration of `[Fe(SCN)]^(2+)` decreases, the intensity of red colour decreases. Experiment (Step-3) : `HgCl_2` is added in equilibrium of Step-1 : Addition of `HgCl_2` also decreases red colour because `Hg^(2+)` reacts with `(SCN^-)` ions to form stable complex ion `[Hg(SCN)_4]^(2-)` and `HgCl_(2(aq)) hArr Hg_((aq))^(2+) + 2Cl_((aq))^(-)` `Hg_((aq))^(2+) + 4SCN^(-) to [Hg(SCN)_4]_((aq))^(2-)` Thus by addition of `HgCl_2SCN^-` in equilibrium decreases so the amount of product decreases. So the reaction moves in forward reaction and intensity of red colour decreases. Experiment (Step-4): If KSCN is added to solution than `[SCN^-]` increases so it consume in reaction and reaction proceed in reverse direction. `Fe_((aq))^(3+) +SCN_((aq))^(-) to [Fe(SCN)]^(2-)` and `SCN^-`decreases and new equilibrium is established. At new `[Fe(SCN)]^(2-)` equilibrium concentration of red colour is more.

12.	Explain effect of high pressure and low temperature according to Boyle.s law.
Answer» Solution :High PRESSURE : At high pressure it shows deviation and at this SITUATION `p PROP (1)/(V)` straight line is not observed. Low TEMPERATURE : At low temperature Boyle.s LAW does not follow completely.

13.	Explain effect of catalyst on chemical equilibrium by example.
Answer» Solution :In chemical reaction to increase or decrease the rate of reaction suitable catalyst is used. Catalyst does not affect the equilibrium. The value of K does not change by catalyst. Catalyst does not affect the COMPONENT of reaction mixture. The catalyst is not seen in equilibrium reaction or in expression of equilibrium constant. Effect of catalyst: (i) A catalyst increases the rate of the chemical reaction by making available a new low energy pathway for the conversion of reactants to products. (ii) It increases the rate of forward and reverse reactions that pass through the same transition state and does not affect equilibrium, (iii) Catalyst lowers the activation energy for the forward and reverse reactions by exactly the same amount. e.g.-1: Let us consider the formation of `NH_3` from `N_2` and `H_2` which is highly exothermic reaction and proceeds with decrease in total number of moles formed as COMPARED to the reactants. `N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g))` Equilibrium constant decreases with increase in temperature. At low temperature rate decreases and it takes long time to reach at equilibrium. German chemist, Fritz Haber discovered that a catalyst consisting of iron catalyse the reaction to OCCUR at a satisfactory rate at temperatures `(500^@ C)`, where the equilibrium concentration of `NH_3` is reasonably favourable. SINCE the number of moles formed in the reaction is less than those of reactants, the yield of `NH_3` can be improved by increasing the pressure. Optimum conditions of temperature and pressure for the synthesis of `NH_3` using catalyst are around `500^@C` and 200 atm. e.g.-2In manufacture of sulphuric acid by contact process, `SO_(2(g)) + O_(2(g)) hArr 2SO_(3(g)) K_c=1.7xx10^26` Though the value of `K_c` is suggestive of reaction going to completion, but practically the OXIDATION of `SO_2` to `SO_3` is very slow. Thus, platinum or divanadium penta-oxide `(V_2O_5)` is used as catalyst to increase the rate of the reaction.

14.	Explain effect of Boyle.s law.
Answer» Solution :Fixed amount of any gas if pressure becomes double than its VOLUME become HALF. When pressure of gas increases volume decreases. Nature of gas is very COMPRESSIBLE if pressure increases gas compressible more. Gases are highly compressible because when a GIVEN mass of a gas is compressed the same number of molecules occupy a smaller space. If increases pressure, gas becomes thick and density increases.

15.	Explain effect of catalyst of Le-Chatelier's principle.
Answer» SOLUTION :A catalyst ALTERS the rates of both forward and backward reactions to the same EXTENT. HENCE the equilibrium is not affected and the equilibrium CONSTANT does not change.

16.	Explain E_2 reaction mechanism with a suitable example.
Answer» Solution :(i) `E_2` REACTION stands BIMOLECULAR elimination reaction of second order. The rate of `E_2` reaction depends on the concentration of ALKYL halide and the base. (ii) Primary alkyl halide undergoes this reaction in the presence of alcoholic KOH. (iii) It is a one step process in which the abstraction of the proton from the ß carbon atom and expulsion of halide from the a carbon atom occur simultaneously. (iv) The mechanism that FOLLOWS is shown below:

17.	Explain dynamic nature of chemical system in laboratory
Answer» SOLUTION :Activity: TAKE two 100 mL measuring cylinders (marked as 1 and 2) and two glass tubes each of 30 cm length and same diameter and also marked 1 and 2. Fill cylinder-1 with coloured water like `KMnO_4` (potassium permanganate) solution and keep second cylinder (number-2) empty. Put one tube in cylinder-1 and second in cylinder-2. The initial stage figure is as under. Putting hollow tube-1 in cylinder-1 the hollow tube will fill with the coloured water. Close the UPPER tip and take it out from the cylinder and transfer the coloured water which is in it into cylinder-2. Repeat the process again similarly put the hollow tube in cylinder 2 and transfer the coloured water from it to cylinder-1. Transfer the coloured water from cylinder-1 to cylinder-2 do this process till the level of coloured water in both the cylinder becomes constant. If we continue inter transferring coloured solution between the solution by the hollow tube there will not be any further change in thelevels of coloured water in two cylinder. Forward REACTION: (coloured water in cylinder-1 is decrease ) `to` (Add water in empty cylinder-2) Reverse reaction: (Decrease water in cylinder-2 ) `to` Add water in cylinder-1) EQUILIBRIUM : (Coloured water in cylinder-1) `hArr` (Coloured water of cylinder -2)

18.	Explain Dumas method of estimation of nitrogen.
Answer» Solution :Dumas method Principle: This method is based on the fact that nitrogeneous compound when heated with cupric oxide in an atmosphere of `CO_(2)` yields free nitrogen. `C_(x)H_(y)N_(z)+(2x+(1)/(2)) CuO rarr x CO_(2)+(y)/(2)H_(2)O+(z)/(2)N_(2)+(2x+(y)/(2))Cu` Traces of nitrogen are reduced to elemental nitrogen by passing over heated copper spiral. Description of the apparatus, `CO_(2)` Generator: `CO_(2)` needed in this process is prepared by heating magnetite or sodium bicarbonate contained in a hard glass tube (or) by the action of dil. HCl on MARBLE in a kipps apparatus. The gas is passed through the combustion tube after dried by bubbling through cone. `H_(2)SO_(4)` . Combustion tube: The combustion tube is heated in a furnace is charged with (a) A roll of oxidised copper gauze to prevent the BACK diffusion of products of combustion and to heat the organic substance mixed with CuO by radiation (b) a weighed amount of organic substance mixed with excess of CuO (c) a layer of CuO packed in about `2 // 3` length of the tube and KEPT in position by loose asbestos PLUG on either side and (d) a reduced copper spiral which reduces any oxides of nitrogen formed during combustion of nitrogen. Schiff.s nitromete: The nitrogen gas obtained by the decomposition of the substance in the combustion tube is mixed with considerable excess of `CO_(2)` . It is estimated by passing nitro mater when `CO_(2)` is absorbed by KOH and the nitrogen gas gets collected in the upper part of the graduated tube. Calculation: Weight of the substance taken = Wg Volume of nitrogen= `V_(1)` L Room temperature = `T_(1)` K Atmospheric pressure =P mm Hg Aqueous tension at room temperature =`P^(.)` mm of Hg `:. "Pressure of dry nitrogen"=P-P^(.)= P_(1)^(.) mm Hg` `P_(0)` , `V_(0)` and `T_(0)` be the pressure, volume and temperature respectively of dry nitrogen at S.T.P. Then, `(P_(0)V_(0))/(T_(0))=(P_(1)V_(1))/(T_(1))` `:. V_(0)=(P_(1)V_(1))/(T_(1)) xx (T_(0))/(P_(0))` `:. V_(0)=(P_(1)V_(1))/(T_(1)) xx (273 K)/(760 mm Hg)` 22.4 L of `N_(2)` at STP weigh 28 g of `N_(2)` `:. V_(0)L "of" `N_(2)` "at STP weigh" (28)/(22.4) xx V_(0)` W g of organic compound contain `(28)/(22.4) xx V_(0) "g of" N_(2)` `:. "100 g of organic contain" (28)/(22.4) xx (V_(0))/(w) xx 100="% of Nitrogen"`

19.	Explain dry-cleaning of clothes or explain grease stains can be removed by petrol.
Answer» Solution :Petrol and grease both are hydrocarbon. Both non-polar. Petrol posses less carbon so it is volatile at low temperature. (ii) Grease is a mixture of higher alkanes, so it is STICKY and non-polar in nature. Both are hydrophobic in nature, both possess similar non-polar property, so based on like dissolve like it gives solubility. So stains of grease becomes SOLUBLE in petrol, REMOVE STAIN in from cloth and so it becomes clean.

20.	Explain distillation under reduced pressure
Answer» Solution :(i) This method is used to purify liquids having very high boiling points (ii) those, which decompoe at or below their boiling points. Such liquids are made to boil at a temperature lower than their normal boiling points by reducing the pressure on their SURFACE A LIQUID boils at a temperature at which its vapour pressure is equal to the external pressure. The pressure is REDUCED with the help of a water pump or VACUUM pump Use: Glycerol can be separated from spent-lye in soap industry by using this technique

21.	Explain disproportionation reaction of chlorine and a conc alkali.
Answer» Solution :`3Cl_2(G)+6OH_((aq))^(-)to3ClO_((aq))^(-)+3Cl_((aq))^(-)+3H_2O(l)` In this reaction , elemental chlorine (ZERO oxidation state) is oxidised to hypochlorite ION (+1 oxidation state) and is reduced to chloride ion (-1 oxidation state). In the above reaction, the hypochlorite ion so formed acts as a bleaching agent.

22.	Explain disproportionation redox reaction of hydrogen peroxide.
Answer» SOLUTION :In this reaction, oxygen present in `H_2O_2` is in -1 oxidation state and gets oxidised to dioxygen (zero oxidation state) and is REDUCED to water (-2 oxidation state) `2H_2O_2(AQ)TO2 H_2O(I)+O_2(g)`

23.	Explain distance of Acid and Base on the base of primary properties.
Answer» SOLUTION :

24.	Explain discovery ofelectron.
Answer» SOLUTION :In1830MichaelFarady showedthat ifelectricityis passedthrougha SOLUTIONOF anelectrolytewhich resultedin the liberationanddepositionof matterat the ELECTRODES andsuggestedthe particulatenatureofelectricity In1850mainlyFaradydiscovercathoderayspasselectricityat anodeto cathodein evacuatedtubeat highpressurethesewerecallednegativechargeelectron.

25.	Explain directional properties of bonds by VB theory.
Answer» Solution :The covalent bond is formed by overlapping of atomic orbitals. The molecule of hydrogen is formed due to the overlap of ls orbitals of two H atoms. In CASE of polyatomic molecules like `CH_(4), NH_(3), H_(2)`O the geometry of the molecules is also IMPORTANT in addition to the bond formation. ln `CH_(4)` the shape is tetrahedral & H - C - H bond angle is `109.5^(@), NH_(3)` gas PYRAMIDAL shape. To explain directional poor HYBRIDIZATION of atomic orbitals REQUIRE.

26.	Explain dipole moment of triatomic molecule (AB_(2)).OR Write a·bout dipole moment of AB_(2) Linear molecule (BeF_(2)) and AB_(2) angular molecule (H_(2)O).
Answer» SOLUTION :TWO types in triatomic MOLECULES (1) Linear `AB_(2)` molecules (2) Angular `AB_(2)` molecules.

27.	Explain direction of reaction by reaction quotient - Q_(c).
Answer» Solution :Predicting the direction of a reaction The equilibrium constant, `K_(c)` helps us in predicting the direction in which a reversible reaction will PROCEED at any state. A useful approach for predicting the direction of a reaction is to USE the concept of 'reaction quotient' `Q_(c)`. The reaction quotient `Q_(c)` has the same mathematical form as the equilibrium constant `K_(c)` expression. `Q_(c)` is the ratio of actual concentrations of products to reactants in the reaction mixture at any point of time of measurement, instead of equilibrium concentrations. Thus, for a reversible reaction `A+BhArrC+D`, Reaction quotient `Q_(c)=([C]_(i)[D]_(i))/([A]_(i)[B]_(i))` whereas, equilibrium constant `K_(c)=([C]_(eq)[D]_(eq))/([A]_(eq)[B]_(eq))`, where subscript 'i' represents the initial or concentrations at any instant of measurement before the attainment of equilibrium and `eq` represents the equilibrium concentrations. The value of the reaction quotient is a measure of the progress of the reaction. The value of `K_(c)` will be small at the initial stages of the reaction. Its value increases with time until at equilibrium it becomes equal to `K_(c)`. i.e., `Q_(c)=K_(c)` Thus, THREE conditions arise: 1. If `Q_(c)=K_(c)`, the reaction is at equilibrium. 2. If at any time `Q_(c)ltK_(c)`, the reaction will procced in the forward direction until equilibrium is reached. 3. If any time, `Q_(c)gtK_(c)`, the reaction will proceed in the backward direction (reverse reaction) until equilibrium is ACHIEVED.

28.	Explain Dipole - induced dipole forces.
Answer» Solution :Dipole - induced dipole FORCES : This type of attractive forces OPERATE between the polar molecules having permanent dipole and the molecules lacking permanent dipole. ex. `HCL - H_(2), CO_(2)` (non Polar) in air `O_(2)-H_(2)O`. Formation of Dipole - Dipole forces : Permanent dipole of the polar molecule induces dipole on the electrically neutral molecule by deforming its electronic cloud. When permanent polar and non polar molecule come near to each other then dipole - induced dipole forces becomes reactive. Permanent dipole of the polar molecule induces dipole on the electrically neutral molecule by deforming its electronic cloud. So, polarity developed in non polar molecule. There are attraction forces present between polar molecule (AB) and induced dipole molecule `(X_(2))`. Which is interaction forces between dipole molecule and non - polar molecule. Characteristics : Thus an induced dipole is developed in the other molecule. In this case also interaction energy is proportional to `1//r^(6)` where r is the distance between two molecules. Induced dipole moment depends upon the dipole moment present in the permanent dipole and the polarisability of the electrically neutral molecule. High polarisability increases the strength of attractive interactions. In this case also CUMULATIVE effect of dispersion forces and dipole - induced dipole interactions exists.

29.	Explain Dipole - Dipole forces snd its characteristics ?
Answer» Solution :Dipole - Dipole forces : Dipole - dipole forces act between the molecules possessing permanent dipole. e.g. `HCl, HF, CO, NO,NH_(3)` forms dipole dipole forces. Ends of the dipoles possess ..Partial charges.. and these charges are shown by Greek letter delta `(delta)`. Partial charges are always less than the unit electronic charge `(1.6xx10^(-19)C)`. Formation of Dipole - Dipole forces : Neighbouring POLAR molecules interact with each other. e.g. : The polar molecules interact with neighbouring molecules. Figure shows electron cloud distribution in the dipole of hydrogen chloride and Fighure shows dipole - dipole interaction between TWO HCl molecules. Characteristics : This interaction is stronger than the London forces but is weaker than ion - ion interaction because only partial charges are involved. The ATTRACTIVE force decreases with the increase of distance between the dipoles. As in the above case here also, the interaction energy is inversely proportional to distance between polar molecules. Dipole - dipole interaction energy between stationary polar molecules (as in solids) is proportional to `1//r^(3)` and that between rotating polar molecules is proportional to `1//r^(6)`, where r is the distance between polar molecules. Note : Besides dipole - dipole interaction, polar molecules can interact by London forces also. Thus cumulative effect is that the total of intermolecular forces in polar molecules increase. `(("Total attraction"),("forces in polar"),("molecules"))=(("Dipole - Dipole"),("attraction"))+(("London"),("attraction"),("forces"))`

30.	Explain differential extraction
Answer» Solution :When an ORGANIC compound is present in an aqueous medium, it is SEPARATED by shaking it with an organic SOLVENT in which it is more solubl than in water.The organic solvent and the aqueous solution should be immiscible with each other so that they form two distinct layers which can be separated by separatory funnel. The organic solvent is later removed by distillation or by evaporation to get back the compound. Differential extraction is carried out in a separatory funnel as shown in FIG If the organic compound is less soluble in the organic solvent, a very large quantity of solvent would be required to extract even a very small quantity of the compound. The technique of continuous extraction is employed in such cases. In this technique same solvent is repeatedly used for extraction of the compound. Extraction means, the compound in its aqueous solution is DISSOLVE in organic solvent in which the solubility of compound is more

31.	Explain difference between ionic product and solubilities product . Explain the reaction with precipitation of sparingly soluble salt.
Answer» Solution :`K_(sp)` : At definitiontemperature the PRODUCE of concentration of ions of sparingly soluble salt in its saturated solution is called solubility product `K_(sp)` . e.g., `BaSO_(4(s)) hArr Ba_((aq))^(2+) + SO_(4(aq))^(2-)` `K_(sp)=[Ba_(aq)^(2+)][SO_((4))^(2-)]`....(Eq.-i) `Q_(sp)` : When two solutions of Sparingly soluble salts mix together than the product of concentration of ions of that two salts is called ionic product `Q_(sp)`. e.g. 0.1 M `Ba(NO_3)_2` mix with 0.05 M `H_2SO_4` in mixture : `Ba(NO_3)_2 to Ba^(2+) + 2NO_3^(-)` `H_2SO_4 to SO_4^(2-) + 2H^(+)` In this mixture the product of concentration of `Ba^(2+)` and `SO_4^(2-)` is called ionic product `Q_(sp) BaSO_4`. `Q_(sp)(BaSO_4)=[Ba^(2+)][SO_4^(2-)]`....(Eq.-ii) Reaction between `K_(sp)` and `Q_(sp)` : If `Q_(sp)=K_(sp)` EQUILIBRIUM is established. If `Q_(sp) lt K_(sp)` precipitation not take place If `Q_(sp) gt K_(sp)` precipitation will take place. By COMPARING the value of `Q_(sp)` and `K_(sp)` precipitation will take place or not can be predicted .

32.	Explain diamond.
Answer» SOLUTION :It has a crystalline lattice. In diamond each carbon atom undergoes `sp^3` HYBRIDISATION and linked to four other carbon ATOMS by using hybridised orbitals in tetrahedral fashion. The C-C bond length is 154 pm. The structure EXTENDS in space and produces a rigid three dimensional network of carbon atoms. In this structure directional covalent BONDS are present throughout the lattice. It is very difficult to break extended covalent bonding and therefore, diamond is a hardest substance on the earth. Uses: It is used as an abrasive for sharpening hard tools, in making dyes and in the manufacture of tungsten filaments for electric light bulbs.

33.	Explain diagonal relationship
Answer» Solution :(i) On moving diagonally across the periodic table, the second and the third period elements show certain similarities. (II) Even though the similarity is not same as we SEE in a group, it is quite pronounced in the following PAIR of elements The similarity in PROPERTIES existing between the diagonally placed elements is CALLED "diagonal relationship

34.	Explain the diagonal relationship
Answer» SOLUTION :On MOVING diagonally across the periodic table, the second and third PERIOD ELEMENTS show certain similarities. It is quite pronounced in the following pair of elements . The SIMILARITY in properties existing between the diagonally placed elements is called diagonal relationship .

35.	Explain delocalization of electron in CH_(3) overset(+)(C )H_(2) and CH_(3)CH= CH_(2)
Answer» Solution :In `CH_(3) OVERSET(+)(C )H_(2)` the `sigma` orbital of `C-H` BOND of `CH_(3)` and THREE vacant 2p orbitals of `overset(+)(C )H_(2)` are parallel and ELECTRON are delocalized. In `CH_(3) CH= CH_(2)` orbital of C-H `sigma`-bond and two 2p orbital of `pi` bond of `CH= CH_(2)` all these three orbital allin parallel and p electron delocalised

36.	Explain dehydrohalogenation of dihalides.
Answer» Solution :Alkynes are prepared by dehydrohalogenation of VICINAL dihalides using alcoholic solution of potassium HYDROXIDE. `underset(((1,2-diromoethane)))underset("Ethylene DIBROMIDE")underset(CH_(2)-Br)underset(\|)(CH_(2) - Br) + 2KOH_(a//c) overset(" "DELTA" ")(rarr) underset("ethyene")(CH equiv) CH + 2KBr + 2H_(2)O`

37.	Explain dehydrohalogenation of a alkyl halide.
Answer» Solution :When alkyl halides areheated with alcoholic potash dehydrohalogenation TAKES place giving alkenes. In this reaction, halogen is removed from the `alpha` - carbon while HYDROGEN is removed from the `beta`-carbon. Hence it is `beta`-elimination reaction. `CH_(3) - underset("1 - Bromopropane(Propyl BROMIDE)")(CH_(2) - CH_(2) - Br) + KOH_(alc) overset(" " DELTA "")(rarr) CH_(3) - underset("Propene")(CH) = CH_(2) + KBr + H_(2)O`

38.	Explain degree of dissociation.
Answer» SOLUTION :The fraction of number of moles dissociated at equilibrium to the total number of moles present initially is called DEGREE of DISSOCIATION (or) degree of IONISATION, represented by `alpha`. `thereforealpha=("Number of moles dissociated at equilibrium")/("Total number of moles originally present")`

39.	Explain decarboxylation with its examples.
Answer» SOLUTION :* Sodium salts of CARBOXYLIC acids on heating with soda lime (mixture of sodium hydroxide and calcium atom less than the carboxylic acid. This process of elimination of carbon dioxide from a caroxylic acid is known as DECARBOXYLATION. `underset("Acid")(RCOOH)underset(+3 NaOH, -H_(2)O,-Na_(2)CO_(3))overset("Soda lime", Delta)rarr underset("Hydrocarbon")(RH)` * Soda lime is a mixture of sodium hydroxide (NaOH) and calcium oxide (CaO). NaOH of soda lime reacts with acid and forms sodium salt of acid. Example -1 : Methane from acetic acid. `underset("Sodium ethanoate")(CH_(3)COO Na^(1))+NaOH underset(Delta)overset(CaO)rarr underset("Methane")(CH_(4))+Na_(2)CO_(3)` Here, formation of sodium ethanote is from acetic acid. Example -2 : Ethane from propanoic acid. `underset("Propanoic acid")(CH_(3)CH_(2)COOH)underset((NaOH//CaO)Delta)overset("Soda lime")rarr underset("Sodium ethanoate")([C_(2)H_(5)COONa^(+)])overset(Delta)rarr underset("Ethane")(C_(2)H_(6))` NOTE : In decarboxylation process, `CO_(2)` is liberated and we get product of 1 carbon less. These reaction is useful when we get 1 carbon less in out product.

40.	Explain decomposition of potassium chlorate.
Answer» Solution :Here OXYGEN is oxidised and CHLORINE ( in `KCIO_3`) is REDUCED. There is no charge in the oxidation number of POTASSIUM. `2KCIO_3(g) to2KCI(g)+3O_2(g)`.

41.	Explain Davisson and Germer experiment.
Answer» Solution :(i) The wave NATURE of electrom was experimentally confirmed by Davisson and Germer. (ii)They allowed the accelerated beam of electrons to FALL on a nickel CRYSTAL and recorded the diffraction pattern. (III)The resultant diffraction pattern is similar to the X-ray diffraction pattern. (iv)The finding of wave nature of lectron LEADS to the development of various experimental techniques such as electron microscope,low energy electron diffraction etc.

42.	Explain Davison and Germer experiment.
Answer» Solution :(i) The WAVE NATURE of ELECTRON was experimentally CONFIRMED by Davison and Germer. (ii) They allowed the accelerated beam of electrons to fall on a nickel crystal and RECORDED the diffraction pattern. (iii) The resultant diffraction pattern is similar to the X-ray diffraction pattern. (iv) The finding of wave nature of electron leads to the development of various

43.	ExplainDalton s atomic theory
Answer» Solution :Theatomictheoryof MATTER was firstproposedby JohnDalton in 1808 . Atomas theultimateparticleof matterAccordingto Daltonatomisindivisibleparticles Thistheorywas ableto explain (i) THELAW ofconservationof mass(II) the LAW of constantcomposition and (iii)the lawof multipleproportionverysuccessfully. Iffailedto explainthe resultsof manyexperimentsfor exampleit was KNOWS thatwithsilkor furgenerateelectricity.

44.	Explain d-block elements of periodic table.
Answer» Solution :"Elements of group 3 to 12 are known as d-block elements." General electronic configuration of these elements are `(n -1) d^(1-10) ns^(0-2)`. These elements are known as d-block elements because their last electron is filled in d-orbital. Characteristic of d-block elements : 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) `d^(10) ns^(2)`do not show most of the properties of transition elements. 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".

45.	Explaind orbital .
Answer» SOLUTION :d orbital: dorbitalexistin energyleveln=3,45…but in n=1 ,2 orbitald orbitalis notexistso 1AND2 dnotexist notof dorbitalsno of orbits = (2L + 1) for dorbitall=2so no of dorbitalsare five(5) sod subshellhas fiveorbitals No of d orbitals :fordorbitalsl=2and magneticquantumnumber`m_(1) = - 2 ,-1,0, +1` SHAPEOF d orbitals : fourdorbitalshave sameshapewithfourlobesSo`d_(xy)d_(yz)` and `d_(zx)`havelobesin theirxy, yz , zxplanewhichfor`d_(x)^(2) ` the lobesare on the x and yaxes. Shapeof `d_(x )`orbitalisdifferentformotherinwhichtwolobesaround2 axis . Energyof d orbital : In anyonesubshelltheenergy of all fived orbitalsis sameandequivalenttheirshapeand sizealsosameit theprincipalquantumnumberincreaseandsizeand energyalsoincreaseandsizeenergyalsoincrease3d `lt 4d lt 5d` Radialnodeand ANGULARNODE : wherethe lobescombinethereelectrondensityis zero in dorbitalsangularnodeis l andradialnode is(n-1-1) for 3dorbitalnode(n-1) =3-2-1=0 Angularnode 1=2 For 3dorbitaltotalnodes=n- 1=3-1=2 nodes.

46.	Explain crystal structure of sodium chloride (NaCl).
Answer» Solution :CRYSTAL STRUCTURE of sodium chloride (NaCl) (rock salt) : The crystal LATTICE structure of NaCl is in three steps. This process is exothermic reaction. (i) Ionization enthalpy `(Delta_(i)H)` : `Na_((g)) rarr Na_((g))^(+) + e^(-) ...Delta_(i) H = 495.8 kJ mol^(-1)` (ii) Electron gain enthalpy `(Delta_(eg)H)` : `(Cl_((g)) + e^(-) rarr Cl_((g))^(-) ... Delta_(eg) H = - 348.7 kJ "mol"^(-1))/(because "(i) + (ii) enthalphy= 147.1 kJ mol"^(-1))` The sum of the two step is endothermic means energy is REQUIRED. (iii) Lattice enthalpy `(Delta_(L)H)`: `Na_((g))^(+) + Cl_((g))^(-) rarr NaCl_(s) .... Delta_(L) H = - 788 kJ " mol"^(-1)` `therefore` (i) + (ii) + (iii) enthalpy = - 640.9 kJ `"mol"^(-1)` Therefore, the energy released in the process is more than the energy absorbed in formation of NaCl.

47.	Explain Covalent radius and ionic radius with an examples.
Answer» Solution :Covalent Radius: "When two atoms are bound together by a single bond in a covalent molecule distance between two atoms from centre is known as covalent radius." e.g. : Bond LENGTH of `Cl - Cl " in " Cl_(2)`molecule is 198 pm. So, its atomic radius is `(198)/2 = 98` pm Metallic Radius : "HALF the internuclear distance separating the METAL cores in the metallic crystal is known as metalic radius. EX.: The distance between two adjacent copper atoms in solid copper is 256 pm hence metalic radius of copper is assigned a value of 128 pm.

48.	Explain covalent radius and ven der Waal's radius.
Answer» Solution :Covalent radius : Each atom of the bonded pair contributes to the bond length. (see figure) Covalent radius of atom : In case of covalent bond the contribution from each atom is called the covalent radius of that atom. Covalent radius : The covalent radius is measured approximately so the radius of an atom.s core which is in contact with the core of an adjacent atom in a bonded situation. The covalent radius is HALF of the distance between two similar atoms joined by a covalent bond in the same molecule. e.g. : The distance between two atom of `Cl_(2)` molecule is 1.98 pm. The covalent radius value of Cl is 1.98 + 2 = 0.99 pm. Bond length of bond between two different atom: AB is covalent molecule `therefore ` Covalent radii of A atom = `r_(A)` `therefore`Covalent radii of B atom = `r_(B)` . `therefore ` Bond length of AB is R = `r_(A) + r_(B)` This given is following figure. van der Waals radii : The van der Waals radius represents the overall SIZE of the atom which incudes its valence shell in a nonbonded situation. The van der Waals radius is half of the distance between two similar atoms in separate molecules in a solid. The convent and van der Waals radius in chlorine molecule are in following figure Where, `r_(c) ` = Covalent radii = `("Distance between centre of two Cl ")/(2) ` = `(198 pm)/(2) ` = 99 pm `r_(vdw)`= van der Waals radius `= ("Dist. betn. the int. nuclear axis of two" Cl_(2))/(2) ` = `(360 pm)/(2)` = 180 pm Remember : Covalent radius is the one half of distance between two atoms of same molecule while van der Wall radius is the one haff of distance between two different molecule Table - l : Some bond length (In pm) Orderofbondlength: `O - H lt C - H lt N - O lt C - O ` = ` C - N lt C - C ` e.g. (i) N- O `gt N = O` (ii) C - O `gt ` C = O (iii) C - N `gt` C = N `gt` C = N (iv) `C - C gt C = C gt C = C ` Table - 2 :Halogen (F, Cl, Br, I join with same atom than the order of Bond length ) Af `lt ` ACl `lt `ABr `lt ` Al . Note-I : In period moving from left to right covalent radius DECREASES. Except F other Halogen Cl, Br, I has max radius. Note-2 : In GROUP as we move down the covalent radius increase. Note: Understand above table, separate and use in deciding order.

49.	Explain covalent in acetylene.
Answer» Solution :`H+C:"":C+HtoH-C-=C-H` Two CARBON and two hydrogen atoms FORM three COVALENT BONDS to give a molecule of acetylene.

50.	Explain covalent bond in carbonadioxide.
Answer» Solution :ONE carbon and TWO oxygen ATOMS, form two covalent bond to given a holecule of carbondioxide. `UNDERSET(..)overset(..)O:+underset(..)overset(..)C+:underset(..)O:tounderset(..)overset(..)O=C=underset(..)overset(..)O`