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 controlled oxidation of methane to methanol and formaldehyde.

Answer»

Solution :Alkanes on heating with limited amounts of air or oxygen at HIGH pressure and in the PRESENCE of SUITABLE catalysts GIVE a number of oxidation products .
`2underset("Methane")(CH_(4)) + O_(2) underset(100 atm)overset(CU//250^(@)C)(rarr) underset("Methyl alcohol(Methanol)")(2 CH_(3) OH)`
`underset("Methane")(CH_(4)) + O_(2) underset(" "Delta" ")overset(MO_(2)O_(3)//275^(@)C)(rarr) underset("Fomaldehyde(Methanol)")(HCHO + H_(2) O)`
Discussion
2.

Explain Contribution of Newland in periodic table.

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Solution :JOHN Alexander Newland in 1865 profundad the Law of octaves.
Law : "EVERY eight ELEMENT had properties similar to the first element."
The relationship was just like every eight note that resembles the first in octaves of MUSIC.

For his work, was later awarded Davy Medal in 1887 by the ROYAL Society, London.
Discussion
3.

Explain conformers of newman projection with example.

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Solution :(a) Newan projection :
(i) In this projection, the molecule is viewed at the C-C bond head on.
(ii) The carbon atom NEARER to the eye is represented by a POINT. Three hydrogen atoms ATTACHED to the front carbon atom are shown by three lines drawn at an angle of `120^(@)` to each other.
(ii) The ear carbon atom (the carbon atom away from the eye) is represented by a CYCLE and the three hydrogen atoms are shown attached to it by the shorter lines drawn at an angle of `120^(@)` to each other. At the end of all lines H is written.
(b) Example : Eclipsed ethane and staggered ethane is as FOLLOWS :
Discussion
4.

Explain concentration of HA, CH_3COO^(-) , H^(+),Na^(+) when 0.05 M sodium acetate is added in 0.05 M acetic acid.

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Solution :`{:(CH_3COOH hArr , CH_3COO^(-)+,H^(+)),((0.05-x)M,x M, x M),(CH_3COONa to , CH_3COO^(-) + , Na^+),((0.05),0.05 M,0.05 M):}`
Thus, in MIX solution
`[CH_3COOH]`=INITIAL concentration is 0.05 M is less equal to x
`x lt lt 0.05 [CH_3COOH] APPROX` 0.05
in solution `CH_3COO^(-)`=0.05 + x`approx` 0.05 M
in solution `[H^+]`= x M and `[Na^+]`=0.05 M
Due to common ion effect of `CH_3COO^-`
ionization of `CH_3COOH` is less, `[H^+]`=x decrease and pH INCREASE .
`[CH_3COO^-]` increase than pH increase.
Discussion
5.

Explain concentration of pure water : Equilibrium of pure water is on left side.

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Solution :Calculation of concentration of pure water :Density of pure water = 1.0 g mL = 1000 g `L^(-1)`
Concentration of water =Water of mol / litre
`=("Weight of 1 L water"/"Molecular mass")=1/"1 Litre"`
`=(1000g)/(18 g "mol"^(-1) XX L^(-1))`
`= 55.55 "mol L"^(-1)` ...(Eq.-iv)
The equilibrium DISSOCIATION of water is towards the reactants. The ions of dissociated water is
`[H^+][OH^-]=1.0xx10^(-7)` M
`therefore` Concentration of pure water = 55.55 M
`therefore` The ratio of dissociated water and undissociated water is,
`="[Concentration of ions]"/"[Concentration of water]"`
`=(1xx10^(-7)M)/(55.55 M)`
`=1.8xx10^(-9) = ~ 2xx10^(-9)` ....(Eq. -v)
Thus, for this very SMALL value the equilibrium lies mainly towards (left side) undissociated water.
Discussion
6.

Explain compressibility factor (Z).

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Solution :COMPRESSIBILITY factor (Z) : Deviation between real gas and idea gas is known as compressibility factor (Z).
`Z=(pV)/(nRT) ""`…..(Eq. -i)
where, Z = 1
Z = Compressibility factor of ideal gas deviation behaviour,
`Z =(V_("real"))/(V_("ideal"))=("real molar volume")/("ideal molar volume")` .......(constant T)
For `Z = 1, Z gt 1, Z lt 1` is give in points(i), (ii), (iii)
(i) Z = 1 OR 1 deviation factor :
For ideal gas Z = 1 at all temperature and pressure , because pV = nRT.
If Z = 1 then gas is ideal gas and `Z to p` graph is parallel line.
At very low pressures all gases shown have Z = 1 and behave as an ideal gas.
Ideal gas law, follow at Z = 1 at Boyle temperature of Boyle point at real pressure.
(ii) `Z gt 1` OR Positive deviation :
At high pressure all real gases shows `Z gt 1` they are less compressible than ideal gas. If `Z gt 1` than permanently gas.
`N_(2), H_(2)O_(2)` shows positive deviation, `Z to p` graph SHOW positive deviation.
According to diagram for `H_(2)` gas at high pressure `Z gt 1`. When pressure increases value of Z becomes more positive.
Ideal gas shows positive deviation of higher than Boyle.s point and value of a are higher than are then intermolecular attraction force are weak.
(iii) `Z lt 1` OR negativedeviation :
At intermediate pressure, most gasses have `Z lt 1`.
For `O_(2), CH_(4), CO_(2)` gas shows negative deviation, `Z to p` graph shows negative deviation.
According to diagram, `Z lt 1` at lower pressure. Initially pressure increases than Z decreases, after that z cross ideal gas line.
At lower temperature then Boyle.s point value of Z decreases at increases pressure, which reacted at minimum value and pressure is continuously increases.
(iv) Graph between `Z to p` for some gases :

(v) RELATION between Z real gas and ideal molar volume :
`Z = (pV_("real"))/(nRT) ""`.....(Eq. -i)
If gas is an ideal,
`V_("real")=(nRT)/(p) ""`......(Eq. -iii)
Value of `(nRT)/(p)` is put in (iv) then we get equation (iii).
`Z=(V_("real"))/(V_("ideal")) ""`....(Eq. -v)
`therefore Z =("Real molar volume")/("Ideal molar volume")`......(constant T)
..Compressibility factor is ratio of real molar volume and ideal molar volume...
Discussion
7.

Explain common in effect for solubility of salts

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Solution :The solubility of sparingly soluble SALT is decrease in presence of common ion. Common ion increase meanspositive ion or negative ion any one ions increase more in solution.
e.g. In saturated solution of AgCl if NaCl is added `Cl^(-1)` ion is common and if `AgNO_3` is added `Ag^+` ion is common so CONCENTRATION of that ions increases. So precipitation take place and solubility of sparingly soluble salt decreases.
eg.-1 : In saturated solution of NaCl is HCl gas is passed than concentration of `Cl^-` increases and solid NaCl increases so solubility of NaCl decreases. In the FIRST example solid AgCl increases and SOLUBILITIES of AgCl decreases.
e.g.-2 : In the solution of sparingly soluble salt AgCl,when NaCl is added , the concentration of common ion `Cl^-` increases or by adding `AgNO_3`, concentration of `Ag^+` increases. Due to increase of `Cl^(-1)` or `Ag^+` , solid AgCl increases and solubility of AgCl decreases.
Discussion
8.

Explain common ion effect taking place in CH_(3)COOH+CH_(3)COONaandNH_(4)OH+NH_(4)Cl.

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Solution :(a) `CH_(3)COOHhArrCH_(3)COO^(-)+H^(+)`
`CH_(3)COONatoCH_(3)COO^(-)+NA^(+)`
Dissociation of acetic acid can be suppressed by adding a strong electrolyte `CH_(3)COONa` containing common ion `CH_(3)COO^(-)`.
(B) `NH_(4)OHhArrNH_(4)^(+)+OH^(-)`
`NH_(4)CL toNH_(4)^(+)+Cl^(-)`
Dissociation of `NH_(4)OH` is suppressed by adding a strong electrolyte `NH_(4)Cl` containing `NH_(4)^(+)` ion as common ion.
Discussion
9.

Explain commercial production of Dihydrogen.

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Solution :(i) ELECTROLYSIS of acidified water using platinum electrodes gives hydrogen.
`2H_2O_((l)) underset"Traces of acid/base"overset"Electrolysis"to 2H_(2(g)) + O_(2(g))`
(ii) High purity (`gt`99.95 %) dihydrogen is obtained by electrolysing warm aqueous barium hydroxide solution between NICKEL electrodes.
(iii) It is obtained as a byproduct in the manufacture of sodium hydroxide and chlorine by the electrolysis of brine solution. During electrolysis, the reactions that take place are :
at anode : `2Cl_((aq))^(-) to Cl_(2(g)) + 2e^(-)`
at cathode : `2H_2O_((l)) + 2e^(-) to H_(2(g)) + 2OH_((aq))^(-)`
The OVERALL reaction is :
`{:(2Na_((aq))^(+)+ 2Cl_((aq))^(-) + 2H_2O_((l))),(""darr),(Cl_(2(g)) + H_(2(g)) + 2Na_((aq))^(+) + 2OH_((aq))^(-)):}`
(iv) Reaction of steam on hydrocarbons or coke at high temperatures in the PRESENCE of catalyst yields hydrogen.
`C_n H_(2n+2) + nH_2O underset"Ni"overset"1270 K"to nCO+(2n+1)H_2`
e.g. , `CH_(4(g)) + H_2O_((g)) underset"Ni"overset"1270 K"to ubrace(CO_((g))+3H_(2(g)))_("water gas ")`
The mixture of CO and `H_2` is called water gas. As this mixture of CO and `H_2` is used for the synthesis of methanol and a number of hydrocarbons, it is also called synthesis gas or .syngas..
Nowadays .syngas. is produced from sewage, saw-dust, scrap wood, newspapers etc.
The process of producing .syngas. from COAL is called .coal gasification..
`C_((s))+H_2O_((g)) overset"1270 K"to CO_((g)) + H_(2(g))`
The production of dihydrogen can be increased by reacting carbon monoxide of syngas mixtures with steam in the presence of iron chromate as catalyst.
`CO_((g))+ H_2O_((g)) underset"catalyst"overset"673 K"to CO_(2(g)) + H_(2(g))`
This is called water-gas shift reaction. Carbon dioxide is removed by scrubbing with sodium arsenite solution.
Presently - 77% of the industrial dihydrogen is produced from petrochemicals, 18% from coal, 4% from electrolysis of aqueous solutions and 1% from other sources.
Discussion
10.

Explain classification of matter based on chornical properties. ( Macrosopic)

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Solution :At the macrosopic or bulk level, matter can be classified as mixture or pure SUBSTANCES. These can be furthor, sub-divided as shown in figure.

Many of the substances present around us are mixture. For example, sugar solution in water, air, TEA etc.
A mixture contains two or more substances present in it which are called its components.
Pure substances have CHARACTERISTICS different from the mixtures. They have fixed composition, where as mixtures may contain the components,in any ratio and their composition is variable. Example Copper, silver, gold, water, glucose etc.
Glucose contains CARBON, hydrogen and oxygen in a fixed ratio and THUS, like all other pure substances has a fixed composition.
The constituents of pure substances cannot be separated by simple physical method.
Discussion
11.

Explain classification of elements based on Newland 's law ofOctaves.

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Solution :Newland's LAW of Octaves :
When elements are arrangedin increasing order of their ATOMIC weights every eighth element RESEMBLES its properties with the first one just like the eight note of a musical table

Linitations : FAILED for heavier elements beyond calcium.
Discussion
12.

Explain chemical reaction of ozonolysis of propene.

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SOLUTION :
Discussion
13.

Explain chemical properties of Hydrogen.

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Solution : The chemical behaviour of dihydrogen is determined, to a large EXTENT, by bond dissociation enthalpy.
The H-H bond dissociation enthalpy is the highest for a single bond between two atoms of any element.
The dissociation of dihydrogen into its atoms is only ~ 0.081% around 2000K which increases to 95.5% at 5000K.
It is relatively inert at room temperature due to the high H-H bond enthalpy. Thus, the atomic hydrogen is produced at a high temperature in an electric arc or under ultraviolet radiations.
Since its orbital is incomplete with `1s^1` electronic configuration, it does combine with almost all the elements.
It accomplishes reactions by (i) loss of the only electron to give `H^+`. (ii) gain of an electron to form `H^(+)`, and (iii) sharing electrons to form a single covalent bond.
Reaction with halogens: It REACTS with halogens, `X_2` to give hydrogen HALIDES, HX,
`H_(2(g)) + X_(2(g)) to 2HX_((g))`[X=F , Cl, Br,I]
While the reaction with fluorine occurs even in the dark, with iodine it requires a catalyst.
Reaction with dioxygen : It reacts with dioxygen to form water. The reaction is HIGHLY exothermic.
`2H_(2(g)) +O_(2(g)) underset"heating"overset"Cytalyst"to 2H_2O_((l)) DeltaH^(ө)=-285.9 "kJ mol"^(-1)`
Reaction with dinitrogen : With dinitrogen it forms ammonia.
`3H_(2(g)) + N_(2(g)) underset"Fe"overset"673K, 200atm"to 2NH_(3(g)) DeltaH^(ө)=-92.6 "kJ mol"^(-1)`
This is the method for the manufacture of ammonia by the Haber process.
Reactions with metals : With many metals it combines at a high temperature to yield the corresponding hydrides.
`H_(2(g)) + 2M_((g)) to 2MH_((g))`, where M is an alkali metal
Reactions with metal ions and metal oxides : It reduces some metal ions in aqueous solution and oxides of metals (less active than iron) into corresponding metals.
`H_(2(g)) + Pd_((aq))^(2+) to Pd_((s)) + 2H_((aq))^(+)`
`yH_(2(g)) + M_x O_(y(s)) to xM_((s)) + yH_2O_((l))`
Reactions with organic compounds : It reacts with many organic compounds in the presence of catalysts to give useful hydrogenated products of COMMERCIAL importance .
For example :
(i) Hydrogenation of vegetable oils using nickel as catalyst gives edible fats (margarine and vanaspati ghee)
(ii) Hydroformylation of olefins yields aldehydes which further undergo reduction to give alcohols.
`H_2+CO+RCH=CH_2 to RCH_2CH_2CHO`
`H_2+RCH_2CH_2CHO to RCH_2CH_2CH_2OH`
Discussion
14.

Explain chemical properties of alkaline earth metals (Group-2) in detail.

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Solution :(i) Reactivity towards air and WATER : Beryllium and magnesium are kinetically inert to oxygen and water because of the formation of an oxide film on their surface.
However, powdered beryllium burns brilliantly on ignition in air to give BEO and `Be_(3)N_(2)`.
Magnesium is more electropositive and burns with dazzling brilliance in air to give Mgo and `Mg_(3)N_(2)`.
Calcium, strontium and barium are readily attacked by air to form the oxide and nitride. They also react with water with increasing vigour even in cold to form hydroxides.
(ii) Reactivity towards the halogens : All the alkaline earth metals combine with halogen at elevated temperatures forming their halides.
`M+X_(2) to MX_(2)(X=F, Cl, Br, I)`
Thermal decomposition of `(NH_(4))_(2), BeF_(4)`is the best route for the preparation of `BeF_(2),and BeCl_(2)` is conveniently made from the oxide.
`BeO+C+Cl_(2)overset(600-800K)hArr BeCl_(2) +CO`
(iii) Reactivity towards hydrogen : All the elements except beryllium combine with hydrogen upon heating to form their hydrides, `MH_(2). BeH_(2)`,however, can be prepared by the reaction of `BeCl_(2)`with `LiAlH_(4)`.
`2BeCl_(2) + LiAlH_(4) to 2BeH_(2) + LiCl + AlCl_(3)`
(iv) Reactivity towards acids : The alkaline earth metals readily react with acids liberating dihydrogen.
`M + 2HCl toMCl_(2) + H_(2)`
(v) Reducing nature : Like alkali metals, the alkaline earth metals are strong reducing agents. This is INDICATED by large negative values of their reduction potentials. However their reducing power is less than those of their corresponding alkali metals.
Beryllium has less negative value COMPARED to other alkaline earth metals. However, its reducing nature is due to large hydration energy associated with the small size of`Be^(2+)`ion and relatively large value of the atomization enthalpy of the metal.
(vi) Solutions in liquid ammonia: Like alkali metals, the alkaline earth metals dissolve in liquid ammonia to give deep blue black solutions forming ammoniated ions. From these solutions, the ammoniates, `[M(NH_(3))_(6)]^(2+)`can be recovered.
`M_((s))+(x+y) NH_(3) to [M(NH_(3))_(x)]^(2+)+2[e(NH_(3))_(y)]^(-)`
Discussion
15.

Explain chemical properties of alkali metal elements (Group I).

Answer»

Solution : The alkali metals are highly reactive due to their large size and low ionization enthalpy. The reactivity of thcsc mctals increases down the group.
Reactivity towards air : The alkali metals tarnish in dry air due to the formation of their OXIDES which in turn react with moisture to form hydroxides.
They burn vigorously in oxygen forming oxides. Lithium forms monoxide, sodium forms peroxide, the other metals form superoxides.
The superoxide `(O_(2)^(-))` ion is stable only in the presence of large cations such as K, Rb, Cs.
`4Li + O_(2) to 2Li_(2)O` (oxide)
`2Na + O_(2) to Na_(2)O_(2)` (peroxide)
`M + O_(2) to MO_(2)` (superoxide) (M = K, Rb, Cs)
In all these oxides the oxidation state of the alkali metal is +1. Lithium shows EXCEPTIONAL behaviour in reacting directly with nitrogen of air to form the nitride, `Li_(3)N` as well.
Because of their high reactivity towards air and water, alkali metals are normally kept in kerosene oil.
Reactivity towards water: The alkali metals react with water to form hydroxide and dihydrogen.
`2M + 2H_(2)O to 2M^(+) + 2OH^(-) + H_(2)`(M = an alkali metal)
It may be noted that although lithium has most negative `E^("ϴ")`VALUE, its reaction with water is less vigorous than that of sodium which has the least negative `E^("ϴ")`value among the alkali metals.
This BEHAVIOR of lithium is attributed to its small size and very high hydration energy. Other metals of the group react explosively with water.
They also react with PROTON donors such as alcohol, gaseous ammonia and alkynes.
Reactivity towards dihydrogen : The alkali metals react with dihydrogen at about 673K (lithium at 1073K) to form hydrides. All the alkali metal hydrides are ionic solids with high melting points.
`2M + H_(2) to 2M^(+)H^(-)`
Reactivity towards halogens : The alkali metals readily react vigorously with halogens to form ionic halides, `M^(+)X^(-)`. However, lithium halides are somewhat covalent.
It is because of the high polarisation capability of lithium ion (The distortion of electron cloud of the anion by the cation is called polarisation).
The `Li^(+)` ion is very small in size and has high tendency to distort electron cloud around the negative halide ion. Since anion with large size can be easily distorted, among halides, lithium iodide is the most covalent in nature.
Reducing nature : The alkali metals are strong reducing agents, lithium being the most and sodium the least powerful. The standard electrode potential `(E^("ϴ"))`which measures the reducing power represents the overall change :
`M_((s)) + M_((g))` (sublimation enthalpy)
` M_((g)) + M_((g))^(+) + e^(-)` (ionization enthalpy)
`M_((g))^(+) + H_(2)O toM_((aq))^(+)` (hydration enthalpy)
With the small size of its ion, lithium has the highest hydration enthalpy which accounts for its high negative`E^("ϴ")`value and its high reducing power.
Discussion
16.

Explain chemical equilibrium by giving example of general reaction.

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Solution :Analogous to the physical systems chemical reactions also attain a state of EQUILIBRIUM. These reactions can OCCUR both in forward and backward directions. When the rates of the forward and reverse reactions become equal, the concentrations of the reactants and the products remain constant. This is the STAGE of chemical equilibrium.
Chemical equilibrium is .dynamic. in nature, it consists of a forward reaction in which the reactants give products and reverse reaction in which products gives the original reactants.
Let we take a normal reversible reaction...
A+B `hArr` C+D ...(i)
At the beginning the solubility of A and B is more and with passage of time, there is accumulation of the products C and D and depletion of the reactants A and B. (See Figure)

So, this leads to a decrease in the rate of forward reaction and an increase in he rate of the reverse reaction. After some time, the two reactions occur at the same rate ,and the system reaches a state of equilibrium .
We can get equilibrium STARTING the reaction from any of the direction. We can also get equilibrium starting from C and D.
Discussion
17.

Explain chemical equilibrium and free energy change.

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Solution :The value of `K_c` GIVES the proportion of PRODUCT and reactant at equilibrium, but the value `K_c` does not change with rate of reaction. The value of `K_c` can.t explain the time when the equilibrium is attain, it can be explain by thermodynamics. By use of Gibbs force energy spontaneity of reaction can be PREDICTED.
(i) `DeltaG` is negative (`DeltaG lt 0`) : `DeltaG` is negative, then the reaction is spontaneous. Reaction PROCEED in forward direction and product will form from reactant equilibrium energy decreases.
(ii) `Delta`G is positive (`DeltaG gt 0`) : `DeltaG` is positive, then reaction is considered non-spontaneous. So, forward reaction will not take PLACE, So, product convert in to reactant.
(iii) `DeltaG = 0`: If `DeltaG = 0` than the reaction is in equilibrium. The reaction will not be continue in any direction. There will be not much energy left for this process.
Discussion
18.

Explain changes of value of electron gain enthalpy of periods and groups.

Answer»

Solution :Less irregularity observed in electron gain enthalpy compared to ionization enthalpy.
When we go left to right in period electron gain enthalpy increases with atomic number.
Because in period we go left to right effective nuclear charge increase so `e^(-)` easily add in small atom and added e is NEAREST to positively charge NUCLEUS.

Electron gain enthalpy in GROUP : We go top to bottom value of electron gain enthalpy becomes less negative.`[S(-200), Se(-195), " Te"(-190), " Po" (-174)]" and " [Cl (-349), " Br" (-325), I (-295) , " At" (-270)]`. Atomic volume increases and added electron becomes away from the centre.
Note: Electron gain enthalpy of oxygen is less negative than sulphur. Electron gain enthalpy fluorine is more than chlorine element.
Because when `e^(-)` is enter in to oxygen and fluorine added `e^(-)` is more in low energy level (n = 2) and SUFFERS significant repulsion from the other electron present in this level. For the n = 3 quantum level (S or CI): The added electron occupies a LARGER region of space and the electron repulsion is much less.
Discussion
19.

Explain characteristic of pi bonds

Answer»

Solution :In a `pi` (pi) bond formation parallel orientation of the two p-orbitals on adjacent atoms is necessary for a proper SIDEWAYS overlap eg. For the formation of `pi` bond in `H_(2)C= CH_(2)`.
All the atoms must be in the same plane. The p orbitals are mutually parallel. Both the p orbitals are perpendicular to the pane of the molecule.
EFFECT on rotation on C-C bond: In `CH_(2)= CH_(2)`, the rotation of one `CH_(2)` fragment with respect to other interferes with maximum overlap of p orbitals and , THEREFORE, such rotation about carbon-carbon double bond (C = C) is restricted.
Availability of electron by `pi` bond and reactive CENTRES of reactions: The electron charge cloud of the `pi` bond is located above and below the plane of bonding atoms. This results in the electrons being easily available to the attacking reagents. In general `pi` bonds provide the most reactive centres in the molecules containing multiple bonds
Discussion
20.

Explain change of energy with graph when internal ethane of C-C rotation after 120^(@).

Answer»

SOLUTION :
Discussion
21.

Explain change in enthalpy related to reaction.

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Solution :In a CHEMICAL reaction, reactants are converted into products and is represented by,
Reactants `to` Products
The enthalpy CHANGE accompanying a reaction is called the reaction enthalpy. The enthalpy change of a chemical reaction, is given by the symbol `Delta_(r) H`.
`Delta_(r) H = ("Sum of enthalpies of products")-(" Sum of enthalpies of reactants" )`
`=sum_(i) a_(i) H_("products") - sum_(i) b_(i) H_("reactants")`
Here symbol `sum` is used for summation and `a_(i) and b_(i)` are the stoichiometric coefficients of the products and reactants respectively in the balanced chemical equation. For example, for the reaction
`CH_(4(g)) + 2O_(2(g)) to CO_(2(g)) + 2H_(2) O_((L))`
`Delta_(r) H = sum_(i) a_(i) H_("products") - sum_(i) b_(i) H_("reactants")`
`= [H_(m) ( CO_(2), g) + 2H_(m) (H_(2) O , l)]- [H_(m) (CH_(4) , g) + 2H_(m) (O_(2) , g) ]`
where, `H_(m)` is the MOLAR enthalpy
Discussion
22.

Explain change in entropy of a system during a reversible process Delta S = (q_(rev))/(T)

Answer»

Solution :As entropy change is the difference in the ENTROPIES, when the system passes from initial state to the FINAL state,
`Delta S = S_(F) - S_(1)`
where, `S_(F)` = entropy of the final state and `S_(1)` = entropy of the initial state.
The absolute value of entropy of a system cannot be measured, but the entropy change `(Delta S)` for a process can be calculated using thermodynamic DATA. Entropy change of a process is the ratio of quantity of heat absorbed to the absolute temperature when the process is carried out reversible and isothermally. If `q_("reversible")` is the heat absorbed at temperature T, then
`Delta S = (q_("reversible"))/(T) = (DeltaH)/(T)`
Eq. (1) is a mathematical form of stating the second law of THERMODYNAMICS.
Entropy change is used to predict the spontaneity of a process. It is found in an isolated system, a process proceeds spontaneously in the DIRECTION where entropy increases. In other words, a process is spontaneous when `Delta S` is positive. More positive the `Delta S`, higher will be the spontaneity.
Discussion
23.

Explain Carius method and its principal for estimation of halogen elements

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Solution :Procedure: A known mass of an organic compound is heated with fuming nitric acid in presence of silver nitrate contained in a hard glass TUBE known as caring tube in a furance. Principle: Carbon adn hydrogen present in organic compound are oxidised to carbon dioxide and water. The halogen present FORMS the corresponding silver halide (Ag X). It is filtered, dried and weighed. `AgNO_(3) + X underset(Delta)overset(HNO_(3))rarr AgX_((s))` mass of organic compound = mg
1 mol of AgX CONTAINS 1 mol of X.
`therefore` mass of halogen in `m_(1)` g of AgX is, mass of halogen `= ("atomic mass of " xx X m_(1))/("molecular mass of AgX")`
`therefore` of halogen `= (("atomic mass of X") (m_(1)) (100))/(("molecular mass of AgX")(m))`
Discussion
24.

Explain Carbylamine reaction.(or) Give the characteristic test for primary amine.

Answer»

Solution : Chloroform REACTS with aliphatic or AROMATIC PRIMARY AMINES and alcoholic caustic potash to give foul smelling alkyl isocyanide (carbylamine).
`underset("Methyl amine") (CH_(3) NH_(2)) + underset("Chloroform") (CHCl_(3)) + 3 KOH to underset("Methyl isocyanide") (CH_(3)NC) + 3 KCl + 3 H_(2)O`
Discussion
25.

Explain calculation of concentration at the value of equilibrium constant.

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Solution :In case of a problem in which we know the initial concentrations but do not know any of the equilibrium concentration the following three STEPS shall be as 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.
(i) The initial concentration. (ii) The change in concentration on going to equilibrium. (ii) State the equilibrium concentration with reference to the beginning concentration and the changes occurring during the reaction.
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 concen trations in to 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 concentration from the calculated value of x.
Step-5: Check your result by using concentration and calculated `K_c`
Discussion
26.

Explainby givingresonsof stabilityof completelyfilled andhalffilledsubshell

Answer»

Solution :Symmetricaldistributionof electron: Thecompletelyfille orhalffilledsubshellhaveand arethereforerelatively smalltheelectronsare morestrongly attractedby THENUCLEUS.

Exchangeenergy : Thestabilizingeffect ariseswhenevertwo ormoreelectrons with thesameand theenergyreleaseddue tothisexchange incalledexchangeenergy .
if `d^(5)` electronconfigurationall fiveelectron havetakeplaceand moreenergyreleaseand OS`d^(5)`ismorestableconfiguration .
Sohalffilledand completelyfilled`3d^(5) ` and `3d^(10)` posses morestability
(i) Relativelysmallshielding
(ii)Smaller coulombicrepulsion energyand
(III) largerexchangeenergy.
Discussion
27.

Explain by examples: (a) Complete (b) Condensed and (c ) Bond-line structural formulas

Answer»

Solution :(a) Complete structure formula (dash):
(i) The COVALENT bond represent by dash (-) or small line
(ii) In structure, every dash (-) means one bonding electron pair or two bonding electron.
(iii) A single dash (-) represents a single bond.
(iv) Double bond is represent by double dash (=).
(v) Triple bond is represent by triple dash `(-=)`.
(vi) Lone pairs of electrons on heteroatoms (e.g., O, N, S, X etc) may or may not be shown.
(vii) Every two atoms are represent by dash (-) and obtain complete structure.

Methanol `(CH_(3)OH)`
`H- underset(underset(H)(|))overset(overset(H)(|))(C )- O - H or H - underset(underset(H)(|))overset(overset(H)(|))(C )- underset(..)overset(..)(O) - H`
Condensed structure formula: These structural FORMULAS can be further abbreviated by omitting some or all of the dashes REPRESENTING covalent bonds and by indicating the number of identical groups attached to an atom by a SUBSCRIPT. The resulting EXPRESSION of the compound is called condensed structure formula.
(i) In structure all single bond indicating dash are not written.
(ii) These formula can be condensed by enclosing the repetitive structural unit within a bracket and place integer as subscript indicating the no. of times the structural unit gets repeated.
(iii) The multiple bond present between C and hetero-atom also not written.
e.g `CH_(3) - overset(overset(O)(||))(C )- CH_(3)` is written instead of `CH_(3)COCH_(3)`
`{:("Ethane","Ethene","Ethyne","Methanol"),(e.g CH_(3)CH_(3),H_(2)C= CH_(2),HC -= CH,CH_(3)OH):}`
e.g The condensed structure of `CH_(3) CH_(2) CH_(2) CH_(2)CH_(2)CH_(2)CH_(2) CH_(3)` is written as a `CH_(3)(CH_(2))_(6)CH_(3)`
(c ) Bond line structural formula: In this method, only lines are used and carbon and hydrogen atoms are not shown.
(i) The lines representing carbon-carbon bonds are drawn in a (zig-zag) fashion.
(ii) The only atoms specifically written are O, X, N , S, P etc.
(iii) The terminals denote methyl `(-CH_(3))` groups.
(iv) The line junctions denote carbon atoms bonded to appropriate number of hydrogens required to satisfy the valency of the carbon atoms.
Discussion
28.

Explain briefly the time independent schrodinger wave equation?

Answer»

Solution :Erwin Schrodinger expressed the wave nature of electron in terms of a differential EQUATION. This equation determines the change of wave FUNCTION in space depending on the field of FORCE in which the electron moves. The time independent Schrodinger equation can be expressed as,
`hatH PSI = EPsi`
Where `hatH` is called Hamiltonian operator, `Psi` is the wave function and is a function of position co-ordinates of the particle and is denoted as `Psi(x, y,z) E` is the energy of the system
`hatH = [(-h^2)/(8pi^2m)((del^2)/(del x^2) + (del^2)/(del y^2) + (del^2)/(del z^2)) + V ]`
Can be written as,
`[(-h^2)/(8pi^2m)((del^2Psi)/(del x^2) + (del^2Psi)/(del y^2) + (del^2Psi)/(del z^2)) + VPsi ] = EPsi`
Multiply by `-(8 pi^2.m)/(h^2)` and rearranging
`(del^2 Psi)/(del x^2) + (del^2Psi)/(dely^2) + (del^2Psi)/(delz^2) + (8pi^2m)/(h^2) (E - V) Psi = 0`
The above schrodinger wave equation does not contain time as a variable and is refferred to as time independent Schrodinger wave equation.
Discussion
29.

Explain briefly the time independent Schrodinger wave equation. S

Answer»

Solution :ERWIN SCHRODINGER EXPRESSED the WAVE nature of ELECTRON in terms of a differential equation. This equation determines the change of wave function in space dependingon the field of force in which the electron moves. The time independent Schrodinger equation can be expressedas ,
`hat(H) Psi=E Psi`
Where `hat(H)`is called Hamiltonian operator, `Psi ` is the wave funcation and is a function of position co-ordinates ofthe particle and is denoted as `Psi`(X, Y, Z)E is the energy of the system
`hat(H) = [(-h^(2))/(8pi^(2) m)((del^(2))/(delx^(2))+(del^(2))/(dely^(2))+(del^(2))/(delz^(2)))+ V ]`
can be written as,
`[(-h^(2))/(8pi^(2)m)((del^(2)Psi)/(delx^(2)) +(del^(2)Psi)/(delx^(2)) +(del^(2)Psi)/(delz^(2)))+V Psi ] = E Psi`
Multiply by ` - (8 pi^(2) m)/(h^(2)) `and rearranging
`(del^(2)Psi)/(delx^(2)) + (del^(2)Psi)/(dely^(2))+(del^(2)Psi)/(delz^(2))+(8 pi ^(2) m)/(h^(2)) (E - V) Psi = 0 `
The above schrodingerwave equation does not containtime as a variable and is referred to as time independent Schrodinger wave equation .
Discussion
30.

Explain briefly the effect of temperature on Le-Chatelier's principle.

Answer»

SOLUTION :The forward reaction is exothermic and hence the backward reaction is endothermic.
If the temperature is increased at equilibrium, the system TRIES to decrease the temperature by absorbing HEAT. The backward reaction which is endothermic GETS favoured. Equilibrium is shifted to the left SIDE.
If the temperature is decreased, the system tries to increase the temperature by liberating heat. The forward reaction (exothermic) gets fovoured. Equilibrium is shifted to the right side.
Discussion
31.

Explain Boyle.s law and derive its equation and plot isothermal graph.

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Solution :On the BASIS of his EXPERIMENTS Robert BOYLE reaches to the conclusion which is KNOWN as Boyle.s law.
Discussion
32.

Explain borax bead test.

Answer»

Solution :Borax bead test : On heating, borax FIRST loses water molecules and SWELLS up. On further heating it turns into a transparent liquid, which SOLIDIFIES in to glass like material known as borax bead.
`Na_2B_4O_7 . 10H_2O oversetDeltato underset"Sodium metaborate"(Na_2B_4O_7) oversetDeltato underset"Boric anhydride"(2NaBO_2 + B_2O_3)`
Sodium 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(BO_2)_2` bead is formed.
Discussion
33.

Explain bond formation in BCl_(3). Explain it is symmetrical trigonal molecute.

Answer»

Solution :In BORON in ground state electron configuration is B (Z = 5 ) : [He] `2s^(2)2p^(1) ` in exited state its configuration will be B* [He] `2s^(1) 2p_(y)^(1) 2p_(z)^(1)`. One electron of 2s enter in 2p ORBITAL. In exited state.
B has three half filled 2s, `2p_(y), 2p_(z)` orbital these orbitals overlap and FORM three `SP^(2)` hybrid orbitals.
Such three sp hybrid orbitals arrange in planar trigonal shape at `120^(@)` angle
These three `sp^(2)` hybrid orbitals overap with three half filled 3p orbital of chlorine and form three B - Cl o bond.
So `BCl_(3)` has trigonal planer shape & Cl - B - Cl `120^(@)` angle as follow:
Discussion
34.

Explain bond dissociation enthalpy.

Answer»

Solution :It is defined as the changed in enthalpy when one mole of covalent bonds of a gaseous covalent compound is broken to FORM PRODUCTS in the gas phase.
`H_(2(G)) rarr 2H_((g)) "" Delta_(H - H) H^(@) = 435 kJ MOL^(-1)`.
Discussion
35.

Explain : BF_3 is Lewis acid but not Bronsted Lawry acid.

Answer»

Solution :In `BF_3`, proton is not PRESENT so it is not become proton DONOR THEREFOR not B.
`therefore` It is LEWIS acid
`F_3B+ : NH_3 to F_3B : NH_3`
In `BF_3`, proton is not present So, it not become proton donor therefore not ACT as a Bronsted Lawry Acid.
Discussion
36.

Explain biological importance of sodium and potassium.

Answer»

Solution : A typical 70 kg MAN contains about 90 g of Na and 170 g of K compared with only 5g of iron and 0.06 g of copper. Sodium ions are found primarily on the outside of cells, being located in blood plasma and in the interstitial fluid which surrounds the cells.
These ions PARTICIPATE in the transmission of nerve signals, in regulating the flow of water across cell membranes and in the transport of sugars and amino ACIDS into cells.
Sodium and potassium, although so similar chemically, differ quantitatively in their ability to penetrate cell membranes, in their transport mechanisms and in their efficiency to activate enzymes. Thus, potassium ions are the most abundant cations within cell fluids, where they activate many enzymes, participate in the oxidation of glucose to produce ATP and, with sodium, are responsible for the transmission of nerve signals.
There is a very considerable variation in the concentration of sodium and potassium ions found on the opposite sides of cell membranes. As a typical example, in blood plasma, sodium is present to the extent of `143"m MOL L"^(-1)`, whereas the potassium level is only `5"m mol L"^(-1)` within the red blood cells.
These concentrations change to `10"m mol L"^(-1)(Na^(+))`and `105"m mol L"^(-1)(K^(+))`. These ionic gradients demonstrate that a discriminatory MECHANISM, called the sodium-potassium pump, operates across the cell membranes which consumes more than one-third of the ATP used by a resting animal and about 15 kg per 24h in a resting human.
Discussion
37.

Explain-Balz-Schiemann reaction.

Answer»

SOLUTION : Fluorobenzene is prepared by treating benzene diazonium chloride with fiuoro boric acid. This reaction PRODUCES diazonium fluoroborate which on heating produces fluorobenzene. This reaction is called Balz-Schiemann reaction.
`C_(6) H_(5) N_(2) Cl + HBF_(4) overset(-HCL) (to) C_(6) H_(5) overset(+)(N_(2)) BF_(4)^(-) overset(Delta)(to) underset("Fluoro benzene") (C_(6) H_(4) F) + BF_(3) + N_(2) uarr`
Discussion
38.

Explain balancing of redox reaction by oxidation number method.

Answer»

Solution :Oxidation number method can be explain by following rules.
Step-1 : Write the correct formula for each reactant and product.
Step-2 : Identify atoms which undergo change in oxidation number in the reaction by assigning the oxidation number to all elements in the reaction.
Step-3 : Calculate the increase or decrease in the oxidation number per atom and for the entire molecule/ion in which it occurs. If these are not equal then multiply by SUITABLE coefficients so that these become equal.
Step-4 : Ascertain the involvement of ions if the reaction is taking place in water, ADD `H^(+)orOH^(-)` ions to the expression on the appropriate side so that the total ionic charges of reactants and products are equal. If the reaction is carried out in acidic solution, USE `H^(+)` ions in the equation, if in basic solution, use `OH^(-)` ions.
Step-5: Make the numbers of hydrogen atoms in the expression on the two sides equal by adding water `(H_(2)O)` molecules to the reactants or products. Now, also CHECK the number of oxygen atoms. If there are the same number of oxygen atoms in the reactants and products, the equation then represents the balanced redox reaction.
Discussion
39.

ExplainBalmerseriesand gives itsequation.

Answer»

Solution :Balmershowedin 1885on the basisofexperimentalobservationsthat ifspectrallinesthen thevisiblelines of THEHYDROGENSPECTRUM obeythe followingformula :
`VEC( V) = 109677 (1)/(2^(2)) - (1)/( N^(2))cm^(-1) `
Theseriesof linedescribedby thisformulaare called the Balmerseries. TheyBalmerseriesofhydrogenspectrumwhichapear i n the visibleregionof theelectromagneticspectrum
Discussion
40.

Explain balancing of redox reaction by half reaction method with suitable example.

Answer»

Solution :Step-1 : Produce unbalanced equation for the reaction in ionic form :
`Fe_((aq))^(+2)+Cr_(2)O_(7(aq))^(-2)toFe_((aq))^(+3)+Cr_((aq))^(+3)`
Step-2 : Separate the equation into half reactions.
OXIDATION half : `Fe_((aq))^(+2)toFe_((aq))^(+3)`
Reduction half : `Cr_(2)O_(7(aq))^(-2)toCr_((aq))^(+3)`
Step-3 : Balance the atoms other than O and H in each half reaction individually. Here the oxidation half reaction is already balanced with respect to Fe atoms. For the reduction half reaction, we multiply the `Cr^(3+)` by 2 to balance Cr atoms.
`Cr_(2)O_(7(aq))^(-2)to2Cr_((aq))^(+3)`
Step-4 : For reactions occurring in acidic medium, add `H_(2)O` to balance O atoms and `H^(+)` to balance H atoms.
`Cr_(2)O_(7(aq))^(-2)+14H_((aq))^(+)to2Cr_((aq))^(+3)+7H_(2)O_((l))`
Step-5 : Add electrons to one side of the half Oxidation half reaction to balance the charges. If need be, make the number of electrons equal in the two half reactions by multiplying one or both half reactions by APPROPRIATE number.
`OHR:Fe_((aq))^(+2)toFe_((aq))^(+3)+e^(-)`
`RHR:Cr_(2)O_(7(aq))^(-2)+14H_((aq))^(+)+6^(-)to2Cr_((aq))^(+3)+7H_(2)O_((l))`
To equalise the number of electrons in both the half reaction, we add two water half reactions, we multiply the oxidation half reaction by 6 and write as : `6Fe_((aq))^(+2)to6Fe_((aq))^(+3)+6e^(-)`
Step-6: We add the two half reactions to achieve the OVERALL reaction and cancel the electrons on each side.
`6Fe_((aq))^(+2)+Cr_(2)O_(7(aq))^(-2)+14H_((aq))^(+)to6Fe_((aq))^(+3)2Cr_((aq))^(+3)+7H_(2)O_((l))`
Discussion
41.

Explain Avogadro Law with figure.

Answer»



Solution :In 1811, Avogadro proposed that equal volumes of gases at the same temperature and PRESSURE should contain equal NUMBER of molecules.
Avogadro made a distinction between atoms and molecules which is quite understandable in the present times.
If we consider again the reaction of hydrogen and oxygen to produce water, we see that two volumes of hydrogen combine with one volume of oxygen to give two volumes of water without LEAVING any unreacted oxygen.
Note thatin the figure below each box contains equal number of molecules. In fact, Avogadro could explain the above result by considering the molecules to be polyatomic. If hydrogen and oxygen were CONSIDERED as diatomic.
Dalton and others believed at that time that atom of the same kind cannot combine and molecules of oxygen or hydrogencontaining two atoms did not exist.
Discussion
42.

Explain : Avogadro.s Law.

Answer»

SOLUTION :In 1811 Italian SCIENTIST Amedeo AVOGADRO tried to combine conclusions of Dalton.s atomic theory and Gay Lussac.s law of combining volumes which is now known as Avogadro law.
Avogadro.s Law : It states that equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.
Mathematically Formula : According to Avogadro.s Law as the temperature and pressure remain constant, the volume DEPENDS upon number of molecules of the gas or in other words amount of the gas.
`V prop n` (constant T and p).....(Eq. -i)
where, n = The number of moles of the gas
`THEREFORE V=k_(4)n ""`.....(Eq. -ii)
Molar volume at STP = 22.7 L
Atoms in one mole `= 6.022xx10^(23)`
The relation of volume and density of gas can be obtained by `M = k_(4)d`.
Discussion
43.

Explain attraction forces of two molecules.

Answer»

Solution :When two MOLECULES COME near EFFECT of repulsion between electron cloud and center of two molecular occurs.
When distance between molecule decreases then magnitude of repulsive FORCES increases.
That.s why compressibility of solid and LIQUID is not easy.
Discussion
44.

ExplainAufbau principlewith example

Answer»

Solution :The wordaufbauin Germanmeansbuildingup L the building up of orbitals meansthe fillingup oforbitalswithelectrons <BR> Principle :in thegroundstateof the atomtheorbitlasare filled inorderof theirincreasingenergies.
Orderof orbitalsare filled :1s , 2s,2p3s, 3p , 4s ,3d,4p5s , 4D , 5p , 6s , 4f , 5d , 6p , 7S
Firstlowerenergyorbitalscompletelyfilledthanafterotherorbitalsarefilled.The capacityoforbitalsforelectronis asfollows

Ex : Be (z=2)electronconfiguration`1s^(2) 2s^(2)P (z=9)` electronconfiguration `1s^(2)2s^(2) 2P^(5)`
Br (z=35)arrangementof electronas under
`1s^(2) 2s^(2) 2p^(6) 3s^(2)4s^(2) 3d^(10)4P^(5)` etc
Discussion
45.

Explain Assumptions (postulates) of the kinetic molecular theory.

Answer»

Solution :Assumption of kinetic molecular theory is related to atoms and molecules.
We can not see molecules or atoms.
So, it is called assumption of kinetic is mictoscopic model of gas.
Gas is very compressible and point masses :Gases consist of large number of IDENTICAL particles (atoms or molecules) that are so small and so far apart on the average that the actual volume of the molecules is negligible in comparison to the empty space between them. They are considered as .point masses.. This assumption explains the great compressibility of gases.
Gases expand and occupy space : There is no force of attraction between the particles of a gas at ordinary temperature and pressure. The support for this assumption comes from the fact that gases expand and occupy all the space available to them.
MOLECULE of stable gas having fixed shape :Particles of a gas are always in constant and random motion. If the particles were at rest and occupied fixed positions, then a gas would have had a fixed shape which is not observed.
Gases have its own pressure : Particles of a gas move in all possible directions in straight lines. During their random motion, they collide with each other and with the walls of the container. Pressure is exerted by the gas as a result of collision of the particles with the walls of the container.
Gases colid each other but their individual energies may changes but the sum of their energies remains constant : Collisions of gas molecules are perfectly elastic. This means that total energy of molecules before and after the collision remains same. There may be exchange of energy between colliding molecules, their individual energies may change, but the sum of their energies remains constant.
It there were loss of kinetic energy, the motion of molecules will stop and gases will settle down. This is contrary to what is actually observed.
The distribution of speeds remains constant at a particular temperature : At any particular time, different particles in the gas have different speeds and hance different kinetic energies. This assumption is REASONABLE because as the particles collide, we expect their speed to change. Even if initial speed of all the particles was same, the molecular collisions will disrupt this uniformity. Consequenctly the particles must have different speeds, which go on changing constantly. It is possible to show that though the individual speeds are changing, the distribution of speeds remains constant at a particular temperature.
On heating the ga, kinetic energy, collision and pressure increases : If a molecule has VARIABLE speed, then it must have a variable kinetic energy. Under these circum - stances, we can talk only about average kinetic energy. In kinetic theory it is assumed that average kinetic energy of the gas molecule is directly proportional to the absolute temperature. It is seen that on heating a gas, at constasnt volume, the pressure increases. On heating the gas, kinetic energy of the particles increases and these strike the walls of the container more frequently thus exerting more pressure.
Uses : Kinetic theory of gases ALLOWS us to derive theoretically, all the gas laws studied in the previous sections. Calculations and predictions based on kinetic theory of gases agree very well with the experimental observations and thus establish the correctness of this model.
Discussion
46.

Explain atomic radius of elements of boron family .

Answer»

Solution :On moving down the group , for each SUCCESSIVE member one extra shell of electrons is added and therefore atomic radius is expected to increase.
However, a deviation can be seen . Atomic radius of Ga is less than that of Al. This can beunderstood from the variationin the inner core of the electronic configuration.
The presence of additional 10 d-electrons offer only POOR screening EFFECT for the outer electrons from the increased nuclear charge in gallium. Consequently , the atomic radius of gallium (135 pm) is less than that of aluminium (143 pm). However , regular PERIODICITY is observed in case of ionic radius.
Discussion
47.

Explain as there is increase in molecular weight of alkane there is increase in boiling point and melting point.

Answer»

Solution :As MOLECULAR weight incrase of ALKANE, there is increase in a boiling point and melting point. Exampes are shown in table:

* In alkane ATOMS there are C-C and C-H bond, all these are covalent bond. Due to less difference in electronagitivity of carbon (2.0) and hydrogen 2.1 most of the alkanes are non polar. So WEAK van-der walls intermolecular attraction force is present in most of alkanes atom.
* As moelcular weight increases, molecular size and SURFACE are also increases. As there is increase in molecular weight there is increase in internal molecular .vam der Waal.s. forces increases, there is increase in a boiling point and freezing point.
Discussion
48.

Explain as to why haloarenes are much less reactive than' haloalkanes towards nucleophilic substitution reactions?

Answer»

Solution :Haloarenes are much LESS reactive than haloalkanes towards nucleophilic substitution reactions due to the following reasons:
(i) RESONANCE effect: In haloarenes the electron pair on the halogen atom is in conjugation with the T-electrons of the ring and the following resonating structures are possible.

C-CI bond ACQUIRES a partial double bornd character due to resonance. As a result. the bond cleavage in haloarenes is difficult than in case of haloalkanesand therefore they are less reactive towards nucleophilic substitution reactions.
(ii) The C-Cl bond length in haloalkanes is 177 pm while in haloarenes it is 169 pm. Since it is difficult to break shorter bond than a longer bond. Therefore, haloarenes are less reactive than haloalkanes towards nucleophilic substitutión reactions.
Discussion
49.

Explainarrangement of orbitalwithincreasingenergyon thebasisof (n+1)rule .

Answer»

Solution :Energy ORDER :
` 1 s lt2s lt2p lt3s LT 3p lt4s lt 3dlt 4p`
thisorderis BASEDON (n+1)(n+l)is samethan NIS moreenergyis more
Discussion
50.

Explain Aqueous tension.

Answer»

SOLUTION :Pressure due to saturated water vapour at a given temperature is CALLED aqueous tension (f)
`:. P_("TOTAL") f+p`, where `p_(x)` is pressure of dry gas pressure of drygas, `p_(x)=(P_("total")-f)`
`:.` Aqueous tension (f) increases on INCREASING temperature.
Discussion