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.

Peptizing agent is

Answer»

Always an electrolyte
Always a non-electrolyte
Electrolyte or non-electrolyte
A lyophilic colloid

Solution :Process of CONVERTING precipitate into colloidal particles by adding SUITABLE electrolyte CALLED peptization and stabilizing agent (ELECTROLYTES) as peptizing agent.
Discussion
2.

Peptisation is a process of

Answer»

PRECIPITATION of colloidal particles
purification of colloids
DISPERSING precipitate into colloidal SOLUTION
movement of colloidal particles in the electric field.

Solution :PEPTISATION is a process of dispersing precipitate or any other insoluble material with proper particle size, into colloidal solution.
Discussion
3.

Peptisation denotes

Answer»

DIGESTION of food
hydrolysis of proteins
breaking and DISPERSION into COLLOIDAL state
precipitation of a SOLID from colloidate state.

Answer :C
Discussion
4.

Pentene-1 with HCl gives

Answer»

3-chloropentane
2-chloropentane
1,2-dichloropentane
1-chloropentane.

Answer :B
Discussion
5.

Peeling of ozone umbrella, which protects us from DV rays is caused by:

Answer»

PAN
FREONS
`CO_(2)`
NO

Answer :B::D
Discussion
6.

Peqfor NH_(4)COONH_(2(s)) harr 2NH_(3(g))+ CO_(2(g)) at certain temperature is 0.9 atm. Then, partial pressure of Ammonia at equilibrium (in atm)

Answer»

0.9
0.81
0.03
0.6

Solution :`UNDERSET("at eqm")(NH_(4)COONH_(2(s))) harr underset(2P)(2NH_(3))+underset(P)(CO_(2))`
`P_(EQ)=3P=0.9 implies P=0.3`
`:. P_(NH_(3))=2P=0.6`
Discussion
7.

PCl_(5)(s) reacts with H_(2)O(l) accoding to the equation : PCl_(5) (s)+4H_(2)O(l)toH_(3)PO_(4)(aq)+_5HCl(aq) What is DeltaH^(@) for this rection in Kj//mol^(-1)

Answer»

`-722.2`
`-533.2`
`533.2`
722.2

Answer :B
Discussion
8.

PCl_(5),PCl_(3) "and "Cl_2 are at eqilibrium at 500 K and having concentration 1.59 M PCl_(3), 1. 59 M Cl_2 and 1.41 M PCl_5 Calculate K_C for the reaction PCl_5 hArr PCl_3 +Cl_2

Answer»

Solution :The equlibrium CONSTANT `K_C` for the above reaction can be WRITTENAS : `K_(C)=([PCl_3][Cl_2])/([PCl_5])=((1.59)^2)/(1.41)=1.79`
Discussion
9.

PCl_5,PCl_3 and Cl_2 are at equilibrium at 500 K in a closed container and their concentrations are 0.8 xx 10^(-3) "mol L"^(-1), 1.2 xx 10^(-3) "mol L"^(-1) and 1.2 xx 10 "mol L"^(-1), respectively. The value of K_c for the reaction PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g)) will be

Answer»

`1.8xx10^(3) "MOL L"^(-1)`
`1.8xx10^(-3)`
`1.8xx10^(-3)"mol L"^(-1)`
`0.55xx10^4`

Solution :For the REACTION, `PCl_(5) hArr PCl_(3) + Cl_2`
At 500 K in a closed CONTAINER,
`[PCl_5]=0.8 xx 10^(-3) "mol"^(-1)`
`[PCl_5]=1.2xx10^(-3) "mol"^(-1)`
`[Cl_2]=1.2xx10^(-3) "mol"^(-1)`
`K_c=([PCl_3][Cl_2])/([PCl_5])=((1.2xx10^(-3))xx(1.2xx10^(-3)))/((0.8xx10^(-3)))`
`=1.8xx10^(-3)`
Discussion
10.

PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) 'alpha' is the degree of dissociation of PCl_(5) at equilibrium pressure 'P'. Which among the following is the correct expression for degree of dissociation of 'alpha' ?

Answer»

<P>`ALPHA = SQRT((K_(p))/(P+K_(p)))`
`alpha = sqrt((P+K_(p))/(K_(p)))`
`alpha=sqrt(K_(p)P)`
`alpha = sqrt(P//K_(p))`

ANSWER :A
Discussion
11.

PCl_(5), PCl_(3) and Cl_(2) are at equilibrium at 500 K in a closed container and their concentrations are 0.8xx10^(-3)mol L^(-1), 1.2xx10^(-3)mol L^(-1) and 1.2xx10^(-3)mol L^(-1) respectively. The value of K_(c) for the reaction PCl_(5)(g) hArrPCl_(3)(g)+Cl_(2)(g) will be

Answer»

`1.8xx10^(-3) ` mol `L^(_1)`
`1.8xx10^(-3)`
`1.8xx10^(-3) L "mol"^(-1)`
`0.55xx10^(4)`

Solution :`K_(c)=([PCl_(3)][Cl_(2)])/([PCl_(5)]) = ((1.2xx10^(-3))xx(1.2xx10^(-3)))/(0.8xx10^(-3))=1.8xx10^(-3)`.
Discussion
12.

PCl_(5), PCl_(3) and Cl_(2) are at equilibrium at 500 K and having concentration 1*5.9" M "PCl_(3), 1*59" M "Cl_(2) and 1*41" M "PCl_(5) hArr PCl_(3) + Cl_(2).

Answer»


ANSWER :`1*79`
Discussion
13.

PCl_5, PCl_3 and Cl_2 are at equilibrium at 500 K and having concentration 1.59 M PCl_3, 1.59 M Cl_2 and 1.41 M PCl_5. Calculate K_c for the reaction , PCl_5 hArr PCl_3 + Cl_2 .

Answer»

SOLUTION :`{:("Equilibrium reaction :",PCl_5 hArr, PCl_3 + ,Cl_2),("Equilibrium CONCENTRATION:",1.41 M, 1.59 M, 1.59 M):}`
Equilibrium concentration is mol `L^(-1)` means molarity (M).
`therefore K_c=([PCl_3][Cl_2])/([PCl_5])`
`=((1.59 M)(1.59 M))/(1.41 M)` = 1.7930 M
Additional calculation -1 : Equilibrium constant of reverse reaction.
`K._c=1/K_c=1/1.793 = 0.5577 M^(-1)`
Additional calculation -2: Calculation `K_p` .
R=0.0831 bar L `mol^(-1) K^(-1), K_p = K_c(RT)^(Deltan)`
Where, `K_c`= 1.7930 mol `L^(-1)`
R=0.0831 bar L `mol^(-1) K^(-1)`
T=500 K
`Deltan`=(1+1) -1 =+1
`therefore K_p`=(1.7930 mol `L^(-1)` ) [(0.0831 bar L `mol^(-1) K^(-1)` ) (500 K) ]
= 74.499 bar `approx` 74.5 bar
Thus, `K_p gt K_c` because `Deltan gt 0`
Discussion
14.

The degree of dissociation is 0.4 at 400 K and a I atm for the gaseous reaction: PCI_(5(g)) hArr PCl_(3(g)) +Cl_(2(g)) : assuming ideal behaviour of gases, calculate the density of cquilibrium mixture at 400K and latm

Answer»

SOLUTION :`4.53 G L^-1`
Discussion
15.

PCl_(5) is kept in a closed container at a temperature of 250 K the equilibrium concentrations . PCl_(5) , PCl_(3) and Cl_(2) are 0.045 moles L^(-1) , 0.096 moles L^(-1) , 0.096 moles L^(-1) respectively . The value of equilibrium constant for the reaction PCl_(5) (g) hArr PCl_(3) (g) + Cl_(2) (g) will be .............

Answer»

Solution :0.205
`K_(C) = ([PCl_(3)] [ Cl_(2)])/([PCl_(5)]) = (0.096 XX 0.096)/(0.045) = 0.2048 = 0.205`
Discussion
16.

PCl_(5) is ionic in nature in the solid state. Give reasons. Or Solid phosphorus pentachloride behaves as an inoic compound. Explain.

Answer»

SOLUTION :`PCl_(5)` CONDUCTS electricity in the molten state. This means that in the solid state, it exists as `[PCl_(6)]^(+)[PCl_(6)]^(-)` in which the CATION is tetrahedral and the anion is octahedral
`2PCl_(5)to[PCl_(4)]^(+)[PCl_(6)]^(-)` On melting these ions become free to move and hence `PCl_(5)` conducts electricity in the molten state.
Discussion
17.

PCl_(5) is known but NCl_(5) is not known. Or Nitrogen does not form pentahalide.

Answer»

Solution :Electronic configuration of P is `1s^(2)2S^(2)s^(6)3s^(2)" "3p_(x)^(1)3p_(y)^(1)3p_(z)^(1)3D^(0)`. Thus, P has empty 3d orbitals to

which the 3s electron can be excited to have five half-filled orbitals needed for formation of `PCl_(5)`. Thus `PCl_(5)` is known.
In contrast, electronic configuration of N is `1s^(2)2s^(2)" "2p_(x)^(1)2p_(y)^(1)2p_(z)^(1)`. Since the valence shell of nitrogen has n=2, therefore, it cannot have d-orbitals. However, if one of the 2s ELECTRONS is excited to 3s orbital, five

half-filled orbitals needed to FORM `NCl_(5)` can still be obained. But such an excitation is thermodynamically not favoureble since the energy needed for excitation is more than the energy expected to be generated during the formation of two additionl P-Cl bonds. Therefore, nitrogen does not form `NCl_(5)`. In other words, `NCl_(5)` is unknown.
Discussion
18.

PCl_(5) dissociates as follows in a closed reaction vessel PCI_(5(g)) harr PCl_(3(g)) + Cl_((g)). If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl_(5) is x, the partial pressure of PCI_(3) will be

Answer»

<P>`[(x)/(x-1)]P`
`[(x)/(1-x)]P`
`[(x)/(x+1)]P`
`[(2x)/(1-x)]P`

SOLUTION :
`P_(Cl_(3))=(x)/(1-x)P`
Discussion
19.

PCl_(3),Cl_(2),PCl_(5) are in equilibrium in a closed vessel at 500K. The equilibrium concentration are 1.6 mol L^(-1), 1.6"mol" L^(-1)and1.4"mol"L^(-1) respectively. Calculate K_(c)andK_(p)" for "PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g).

Answer»

Solution :`K_(C)=([PCl_(3)][Cl_(2)])/([PCl_(5)])=(1.6xx1.6)/1.4=1.828`
`DELTAN=2-1=1`
`thereforeK_(p)=K_(c)(RT)^(Deltan)" BECOMES "=1.828(0.0831xx500)^(1)=75.95`.
Discussion
20.

PCI_(5) (g)hArrPCI_(3)(g) + CI_(2)(g) DeltaG_(f)^(circ)[PCI_(5)(g)]=-74kcal//mol DeltaG_(f)^(circ)[PCI_(5)(g)]=-60kcal//mol The, calculate value of equilibrium constant for dissociation of PCI_(5)(g) at 727^circC temperature. (In2=0.7)

Answer»

`2^10`
`2^-10`
`2^-20`
`2^(+20)`

ANSWER :B
Discussion
21.

PCI_5dissociates to 25% in a three litre flask. To what volume the container must be changed such that PCI_5 dissociates to 50% extent

Answer»

18
12
6
24

Solution :
`K_(C)=((calpha)(calpha))/(C(1-alpha))=(Calpha^(2))/(1-alpha)=(nalpha^(2))/(V(1-alpha))`
`K_(C)` is constant
`:. (alpha^(2))/(V_(1)(1-alpha))=(f^(2))/(V_(2)(1-beta))`
`((0.25)^(2))/(3(1-0.25))=((0.5)^(2))/(V_(2)(1-0.5)), V_(2)=18`
Discussion
22.

PbS+4H_(2)O_(2)rarrPbSO_(4)+4H_(2)O. In this reaction PbS undergoes

Answer»

oxidation
reduction
both
none of these

Answer :A
Discussion
23.

PbS+4H_(2)O_(2)toPbSO_(4)+4H_(2)O. The reaction of H_(2)O_(2) with is an example of. .. .reaction.

Answer»

addition
oxidation
reduction
acidic

ANSWER :B
Discussion
24.

PbOandPbO_(2) react with HCl according to following chemical equations (i) 2PbO+4HCl to2PbCl_(2)+2H_(2)O (ii) PbO_(2)+4HCl toPbCl_(2)+Cl_(2)+2H_(2)O Why do these compounds differ in their reactivity ?

Answer»

SOLUTION :(i) `overset(+2)(2Pb)overset(-2)(O)+overset(+1)(4H)overset(-1)(CL) tooverset(+2)(2Pb)overset(-1)(Cl_(2))+overset(+1)(2H_(2))overset(-2)(O)`
OXIDATION number of element does not change so this REACTION is not redox reaction. It is acid- base reaction.
(ii)
In given reaction `Pb^(+4)` is convert into `Pb^(+2)`, so `PbO_(2)` is oxidising agent and `Cl^(-)` is oxidise in `Cl_(2)`.
Discussion
25.

PbO andPBO_(2) react with hci according tofollowing chemical equation 2 Pbo +HCI rarr 2 PbCI_(2)+2H_(2)O PBO_(2)+4 HCI rarr PbCI_(2)+CI_(2)+2H_(2)O why do these compounds differ in their reactivity ?

Answer»

Solution :Writing the O.N of each element above its symbol in the following reaction s we have
in raction (i) O.N of none of the atoms undergo change THEREFORE it is not a ardox reeaction it is fact an acid base reaction since `PbO_(2)` being a basic with HCIto form `PbCI_(2)` and `H_(2)O`
`PbO+2HCIrarrPbCI_(2)+H_(2)O`
in reaction (ii) Pb in `PbO_(2)` is present in +4 oxidation state since DUE to inert pari EFFECT +2 oxide state of Pb is moire stable therefore `PbO_(2)` ACTS as an oxidisning agent as RESUTL it oxidises `CDI^(-) to CI_(2)` itself gets reduced to `Pb^(2+)` in tother words reaction of `PbO_(2)` with HCI is a redox reaction `PbO_(2) +4 HCI rarrPbCI_(2) +CI_(2)+2H_(2)O`
Thus the two compuntds differ in their reactivity beacuse they undergo different types of reactions
Discussion
26.

PbO_(2) is a stronger oxidising agent than SnO_(2).

Answer»

Solution :In `PbO_(2)` and `SnO_(2)`, BOTHLEAD and tin are present in `+4`-oxidationstate. But due to STRONGER inert pair EFFECT,`Pb^(2+)`ions is more stable than `Sn^(2+)`ions . In other words, `Pb^(4+)` ions i.e. `PbO_(2)`,is more easily reduced to `Pb^(2+)`ions than `Sn^(4+)`ions are reduced to`Sn^(2+)` ions. Thus, `PbO_(2)` acts a strongeroxidising agent than `SnO_(2)`.
Discussion
27.

PbO_(2) is

Answer»

ACIDIC
Basic
Neutral
Amophoteric

Solution :`PbO_(2)` amphoreric
Discussion
28.

PbO_(2) can act as an oxidising agent.

Answer»

Solution :In `PbO_(2)` and `SnO_(2)`, bothlead and tin are present in `+4`-OXIDATIONSTATE. But due to stronger inert PAIR effect,`Pb^(2+)`ions is more stable than `Sn^(2+)`ions . In other words, `Pb^(4+)` ions i.e. `PbO_(2)`,is more EASILY REDUCED to `Pb^(2+)`ions than `Sn^(4+)`ions are reduced to`Sn^(2+)` ions. Thus, `PbO_(2)` acts a strongeroxidising agent than `SnO_(2)`.
Discussion
29.

Pb(NO_(3))_(2) on heating gives a brown gas which undergoes dimerization on heating. Identify the gas.

Answer»

Solution :`Pb(NO_(3))_(2)` on heating produces nitrogen dioxide `(NO_(2))`. It is a brown gas. Being an odd electron species on COOLING it UNDERGOES dimerization to form dinitrogen tetroxide `(N_(2)O_(4))` which is COLOURLESS.
`underset("Lead nitrate")(2PB(NO_(3))_(2))overset(Delta)to2PbO+underset(("Brown gas"))underset("Nitrogen dioxide")(4NO_(4))+O_(2),underset(("Brown"))(2NO_(2))overset("Cooling")tounderset(("Colourless"))(N_(2)O_(4))`
Discussion
30.

PbCl_(4)exists, but PbBr_(4)and PbI_(4)do not. This is because of

Answer»

INABILITY of bromine and IODINE to oxidise `Pb^(2+) " to" Pb^(4+)`
Bromide and iodide are bigger in size
More electropisitive character of `Br_(2)and I_(2)`
Chlorine is a GAS, but bromine is a LIQUID and iodine is a SOLID .

Answer :A
Discussion
31.

PbCl_(2) has a solubility product of 1.7xx10^(-8). Will a precipitate of PbCi_(2) form when 0.010 mole of lead nitrate and 0.010 mole of potassium chloride are mixed and water added upto 1 litre ?

Answer»


Solution :As final volume of the solution = 1 L, therefore `[Pn^(2+)] = 0.01 M = 10^(-2) M, [ Cl^(_)] = 0.01 M = 10^(-2) M`
Ionic product of `PbCl_(2) = [Pb^(2+)][Cl^(-)]^(2) (10^(-2))^(2) = 10^(_6)`
which is greater than `K_(sp)`. Hence , ppt. of `PbCl_(2) ` will be formed.
Discussion
32.

Pb_(3)O_(4) is regarded as a compound oxide of PbO and PbO_(2). How many parts of PbO_(2), are present in it?

Answer»


Solution :`Pb_(3)O_(4)` is a MIXTURE of 2 parts of PbO and 1 PART of `PbO_(2)`
Discussion
33.

Pay load is defined as the difference between the mass of the displaced air and the mass of the balloon. Calculate the say load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27^(@)C (Density of air =1.2" kg m"^(-3) and R=0.0833 bar dm^(3) K^(-1)mol"^(-1) )

Answer»

Solution :Radius of the balloon =10 m
`:.` Volume of the balloon`=(4)/(3)pir^(3)=(4)/(3)XX(22)/(7)xx(10 m)^(3)=4190.5" m"^(3)`
Volume of He filled at 1.66 bar and `27^(@)C=4190.5" m"^(3)`
CALCULATION of mass of He :
PV=nRT`=(w)/(M)RT`
or `"" w=("MPV")/("RT")=((4xx10^(-3)" kg mol"^(-1))(1.66"bar")(4190.5xx10^(3)"DM"^(3)))/((0.083" bar dm"^(3)" K "^(-1)"mol"^(-1))(300K))=1117.5" kg"`
Total mass of the balloon along with He `=100+1117.5=1217.5" kg"`
Maximum mass of the air that can be displaced by balloon to GO up =Volume xxDensity
`=1490.5" m"^(3)xx1.2" kg "m^(-3)=5028.6" kg"`
`:. " Pay load" =5028.6-1217.5" kg" =3811.1" kg"`
Discussion
34.

Pb and Sn are extracted from their chief ore by

Answer»

CARBON REDUCTION and SELF reduction
self reduction and carbon reduction
electrolysis and self reduction
self reduction and electrolysis

Answer :B
Discussion
35.

Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27^(@)C. (Density of air = 1.2 kg m^(-3) and R = 0.083 bar dm^(3)K^(-1)mol^(-1)).

Answer»

Solution :Calculation of balloon.s VOLUME :
Radius of balloon r = 10 m
Volume of balloon `= (4)/(3)pi r^(3)=(4)/(3)((22)/(7))(10 m)^(3)`
`= 4190.48 m^(3)`
Calculation of weight of displaced AIR :
`THEREFORE` Volume of He gas in balloon = Volume of balloon
`= 4190.48 m^(3)`
`~~4190.5 m^(3) = V`
`therefore (("The weight of air"),("displaced by balloon"))=(("Volume of"),("balloon (V)"))xx(("Air of "),("density"))`
`=(4190.5 m^(3))(1.2 kg m^(-3))`
`= 5028.6` kg air
Calculation of weight (w) of He gas in balloon :
where, Volume of Balloon (V) `= 4190.5 m^(3)=4190.5xx10^(3)`
Pressure of `H_(2)` gas in `dm^(3)` in balloon (p) = 1.66 bar
Temperature (T) of He in Balloon,
`(T)=(27+273)=300K`
Moles of He in Balloon `= ("Weight (w)")/("Molecular Mass (M)")`
Gas Constant (R ) = 0.083 bar `dm^(3)K^(-1)mol^(-1)`
Molecular mass (M) of He gas = 4.0 gm `mol^(-1)`
`= 4.0xx10^(-3)kg mol^(-1)`
`pV = nRT=(wRT)/(M)`
`therefore w=(pVM)/(RT)`
`therefore w=(("1.66 bar")(4190.5xx10^(3)dm^(3))(4.0xx10^(-3)kg mol^(-1)))/(("0.083 bar dm"^(3)K^(-1)mol^(-1))(300K))`
`= 1117.47 kg ~~ 1117.5 kg`
Total calculation of balloon and He :
Total weight = (weight of balloon) + (weight of He)
`= 100 kg + 1117.5 kg`
`= 1217.5 kg`
Calculation of pay LOAD :
Pay load `= (("Mass of air"),("displaced"))-(("Total weight of"),("He nad balloon"))`
`= 5028.6 kg - 1217.5 kg`
= 3811.1 kg
Discussion
36.

Pay load is defined as the difference between the mass of displaced air and the mass of the balloon Calculate the pay-load when a balloon of radius 10m mass 100 kg is filled with helium at 1.66 bar at 27^(@)C (Density of air= 1.2 kg m^(-3) and R = 0.083 bar dm 3 K^(-1) mol^(-1)) .

Answer»

SOLUTION :Volume of BALLOON
`= 4/3 PI r^3 =4/3 xx 22/7 xx (10)^3 = 4190.5 m^3`
`= 4190.5 xx 10^3 dm^3`
Mass of helium FILLED in the balloon
`PV = m/M xxRxxT`
or`1.66 xx 4190.5 xx 10^3=m/4 xx 0.083 xx 300`
or `m= (1.66 xx 4190.5 xx 10^3 xx 4)/(0.083 xx 300)`
= `1117466.7 g = 1117.47 kg`
`:. " Total mass of the balloon " = 100 +1117.47`
= `1217.47 kg`
Volume of air displaced by balloon `= 4190.5 m^3`
`:." Mass of displaced air = volume " xx "density"`
`= 4190.5 xx 1.2`
`= 5028.6 kg`
`:." Pay load of the balloon " = 5028.6 - 1217.47`
`= 3811.1 kg`
Discussion
37.

Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27^@C. (Density of air= 1.2 kg m^-3 and R=0.083 bar dm^3K^-1mol^-1)

Answer»

Solution :Volume of balloon (V) `4/3 pi r^3 = 4/3 xx 22/7 xx 10^(3)`
`= 4190.50 m^3`
`= 4190.5 xx 10^(3) DM^(3)`
MASS of displaced air = volume of air `xx` density of air
`= 4190.5 xx 1.2 = 5028.6 kg`
Mass of He in balloon = `(PVM)/(RT) = (1.66 xx 4190.5 xx 10^3 xx 4)/(0.083 xx 300)`
`= 1117 xx 10^(3) g = 1117 kg`
Mass of filled balloon =`100 + 1117 = 1217 kg`
Pay load = mass of displaced air - mass of balloon
`= 5028.6 - 1217 = 3811.6 kg`
Discussion
38.

Explain Pauli's Exclusion principle with an example.

Answer»

Nucleus of an atom contains no negative charge
Electrons MOVE in circular orbits around the nucleus
Electrons OCCUPY orbitals of lowes energy
All the four quantum numbers of TWO electrons in an atom cannot be EQUAL

Solution :Pauli.s PRINCIPLE
Discussion
39.

Pauling's electronegativity values for elements are useful in predicting

Answer»

Polarity of the MOLECULES
POSITON in the E.M.F. series
Coordination numbers
DIPOLE moments

ANSWER :A
Discussion
40.

Pauling give method to calculate univalent ion radii by assuming that (i) In ionic crystal (let M^(+)X^(-)) cations and anions are is contact of eac other and sum of their radii is equal to interionic distance, i.e., d_((M^(+)-X^(-))) = r_(M^(+) +r_(X^(-)) (ii) The radius of an ion having noble gas configuration is inversely proportional to the effective nuclear charge felt at the periphery of the ion, i.e., r_((M^(+))) = (C)/(Z_(eff(M^(+)))) and r_((X^(-))) = (C)/(Z_(eff(X^(-)))) Here C is constant of proportionality whose value depends on electronic configuration of ion. Thus, d_((M^(+)-X^(-))) =(C)/(Z_(eff(M^(+)))) +(C)/(Z_(eff(X^(-)))) pm Z_(eff) is the effective nuclear charge whose value can be calculated by the formula: Z_(eff) = Z - sigma. Here sigma is shielding constant and for neon, the value of sigma when calculate by Slater's rule is found to be 4.5 The value of constant C for NaF crystals is [given that interionic distance of NaF = 231 pm]:

Answer»

231
115.5
614.5
307.25

Answer :C
Discussion
41.

Pauling's electronegativity is based on

Answer»

Electron AFFINITY
Ionisatin POTENTIAL
Both IP and EA
BOND ENERGIES

Answer :D
Discussion
42.

Pauli exclusion principle helps to calculate the maximum number of electrons that can be accommodated in any

Answer»

orbital
subshell
shell
All of these

ANSWER :A
Discussion
43.

Paste is

Answer»

SUSPENSION of solid in a LIQUID
Mechanical DISPERSION of a solid in liquid
Colloidal SOLUTION of a solid in solid
None of these

Solution :Suspension of solid in a liquid.
Discussion
44.

Passing H_(2)S gas into a mixture of Mn^(2+), Ni^(2+), Cu^(2+) and Hg^(2+) ions in an acidified aqueous solution precipitates

Answer»

CuS and HgS
MnS and CuS
MnS and NiS
NiS and HgS

Solution :`H_(2)S` in acidic MEDIUM PRECIPITATES RADICALS of Group 2 of quantitative analysis, i.e. `Cu^(2+), HG^(2+)`, (and `Pb^(2+), Bi^(3+), Cd^(2+), As^(3+), Sb^(3+), Sn^(2+)`).
Discussion
45.

Passivity is exhibited by i) Be ii)Mgiii)Aliv)B

Answer»

I,ii
ii,iii
I,iii
ii,iv

Answer :3
Discussion
46.

Passing H_(2)S gas into a mixture of Mn^(2+), Ni^(2+), Cu^(2+) and Hg^(2+) ions in an acidified aqueous solution, precipitates :

Answer»

MNS and CUS
CuS and HgS
MnS and NiS
NiS and HgS

Solution :`Cu^(2+)` and `Hg^(2+)` ions BELONG to group II of basic radicals. In the acidic solution, both are precipitated as CuS and HgS having black colour. `Mn^(2+)` and `Ni^(2+)` ions belong to group IV of basic radicals and remain in solution
Discussion
47.

Passenger aeroplane cabins is artificially pressurised since

Answer»

PRESSURE DECRASES with the increase in altiude
pressure increases with the increases in altitude
temperature increases with the increase in altitude
none of the above

ANSWER :A
Discussion
48.

Passage: Whenall the coefficients ina balancedchemical equationare multipliedbya constant factor X the equilibrium constant (originallyK) becomes K^J.Similarly,whenbalancedequations are addedtogether, the equilibrium constant for the combined process is equal to the product of the equilibrium constants for eachstep.Equilibrium constant of the reversedreactionis numericallyequal tothe reciprocal of the equilibrium constant of the original equation. Unit of K_p = ("atm")^(Deltan) , Unit of K_c =("mol" L^(-1))^(Deltan) The equilibrium constants for the following reactions at 1400 K are given: 2H_2O_((g)) harr 2H_(2(g))+O_(2(g)) , K_1=2.1 xx 10^(-13) 2CO_(2(g)) harr 2CO_((g))+O_(2(g)) , K_2=1.4 xx 10^(-12) Then the equilibrium constant K for the reaction, H_(2(g))+CO_(2(g)) harr CO_((g)) + H_2O_((g)) is

Answer»

2.04
20.5
2.6
8.4

Solution :EQUATION `(b-a)/(2)=C :. K_(3)=SQRT((K_(2))/(K_(1)))=2.6`
Discussion
49.

Passage: Whenall the coefficients ina balancedchemical equationare multipliedbya constant factor X the equilibrium constant (originallyK) becomes K^J.Similarly,whenbalancedequations are addedtogether, the equilibrium constant for the combined process is equal to the product of the equilibrium constants for eachstep.Equilibrium constant of the reversedreactionis numericallyequal tothe reciprocal of the equilibrium constant of the original equation. Unit of K_p = ("atm")^(Deltan) , Unit of K_c =("mol" L^(-1))^(Deltan) Consider the two reactions: XeF_(6(g))+H_(2)O_((g)) harr XeOF_(4(g))+2HF_((g)), K_(1)""XeO_(4(g))+XeF_(6(g)) harr XeOF_(4(g))+XeO_(3)F_(2(g)), K_(2) Then the equilibrium constant for the following reaction will be XeO_(4(g))+2HF_((g)) harr XeO_(3)F_(2(g))+H_(2)O_((g))

Answer»

`K_(1) // K_(2)^(2)`
`K_(1) // K_(2)`
`K_(1)^(2) // K_(2)`
`K_(2) // K_(1)`

SOLUTION :equation b-a = C `:. K_(3)=(K_(2))/(K_(1))`
Discussion
50.

Passage: Whenall the coefficients ina balancedchemical equationare multipliedbya constant factor X the equilibrium constant (originallyK) becomes K^J.Similarly,whenbalancedequations are addedtogether, the equilibrium constant for the combined process is equal to the product of the equilibrium constants for eachstep.Equilibrium constant of the reversedreactionis numericallyequal tothe reciprocal of the equilibrium constant of the original equation. Unit of K_p = ("atm")^(Deltan) , Unit of K_c =("mol" L^(-1))^(Deltan) Consider the reactions: (i) CO_((g))+H_2O_((g)) harr CO_(2(g))+H_(2(g)) , K_1 (ii) CH_(4(g))+H_2O_((g)) harr CO_((g))+3H_(2(g)) , K_2 (iii) CH_(4(g))+2H_2O_((g)) harr CO_(2(g))+4H_(2(g)) , K_3 Which of the following is correct ?

Answer»

`K_3=K_1//K_2`
`K_3=K_1^2//K_2^3`
`K_3=K_1K_2`
`K_3=K_1sqrt(K_2)`

SOLUTION :EQUATION a+b=c `:. K_(3)=K_(1).K_(2)`
Discussion