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.

Silica is added to roasted copper pyrites during smelting in order to remove

Answer»

CUPROUS oxide
cuprous sulphide
ferous oxide
ferous sulphide

Answer :C
Discussion
2.

Silica is a covalent solid, but dry is a molecular solid. Why ?

Answer»

Solution :In silica, a chain of `SiO_(4)` tetrahedra is PRESENT in a NETWORK 3D-structure.
In dry ice, DISCRETE molecular units of `CO_(2)` are present. Between molecules, only weak dispersion forces exist.
Discussion
3.

Silanes (Si_(H)H_(2n+2)) compounds are less stable (highly reactivcs) than the corresponding hydrocarbons (C_(n) H_(2n+2)). Why?

Answer»

The Si-Si bond is slightly WEAKER than the C - C bond
The Si-H bond are weaker than C-H bond
Silicon atoms have LARGER and therefore provide greater surface area for attack by NUCLEOPHILES
Silicon atoms have low-lying d-orbitals that can ACT as acceptors of electron pairs from nucleophiles

Answer :A::B::C::D
Discussion
4.

Signa of Delta G for the melting of ice is negative at

Answer»

265K
270K
277K
274K

Solution :`DELTA G lt 0` for melting above M.P `= 0^(@)C`
Discussion
5.

sigma bond is stronger compare to p bond.

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Solution :In a `SIGMA` BOND the overlapping is along the axis and is maximum EXTENT. Where as `pi` bond overlapping is sidewise and overlappingis MINIMUM.
Discussion
6.

Sigma bonds are :

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CYLINDRICALLY symmetrical about the BOND axis
Perpendicular to the bond axis
asymmetrical about the bond axis
NONE of the above

ANSWER :A
Discussion
7.

Sigma bond is not f orme in which overlapping of the following ? (i) 2p_(y) - 2p_(y) (ii) 2p_(z)- 2p_(z)(iii) 2s - 2p_(x)(iv) 2p_(x) - 2p_(x)

Answer»

SOLUTION :(i) and (IV)
Discussion
8.

[SiF_6]^(2-) is known whereas [SiCl_6]^(2-) not. Give possible reasons.

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Solution :The MAIN reasons are : (i) Six large chloride ions cannot be ACCOMMODATED around `Si^(4+)` due to LIMITATION of its size. (II) Interaction between LONE pair of chloride ion and `Si^(4+)` is not very strong.
Discussion
9.

SiF_6^(2-)is known while SiCl_6^(-2)is not - explain.

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SOLUTION :Six LARGE CHLORIDE IONS cannot be accommodated AROUND `Si^(4+)` due to limitation of its size. Interaction between lone pair of chloride ion and `Si^(4+)` is not very strong.
Discussion
10.

SiF_(4) on hydropysis gives X and Y. SiCl_(4) on hydrolysis gives X and Z. Covalency of central atom in Y is C_(1) and that of halogenin Z is C_(2). Then C_(1) + C_(2) is:

Answer»


ANSWER :7
Discussion
11.

[SiF_(6)]^(2-)is knownwheares [SiCl_(6)]^(2-)not. Givepossiblereasons.

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Solution :The mainreasons are :
(i) Due to smallersize ofF as comparedto Cl, six SMALL Fatoms can be easily accomodatedaround Si atom but six largeCl atoms cannot.
(II) The lone PAIR of electrons on F are present in a smaller 2p-orbitalbut in Cl they are present on a LARGER 3p-orbital.Therefore, interaction of F lone pair electrons with d-orbitalsof siliconare stronger than that of Cl lone pairs.
Discussion
12.

SiF_4 has octet, but it has residual bonding ability. Comment.

Answer»

Solution :SILICON has four valance electons and forms four bonds with fluorine atoms. `SiF_(4)` has octet configuration. It is an ELECTRON PAIR acceptor, due to the AVAILABILITY of empty VALENCE d-orbitals.
Discussion
13.

SiCl_(4) overset(H_(2)O)(to) X overset("Heat")(to) Y overset(NaOH)(to) Z find X,Y,Z.

Answer»

`{:(SiO_(2), Si,NASI):}`
`{:(Si(OH)_(4),SiO_(2),Na_(2)SiO_(3)):}`
`{:(Si(OH)_(4),Si,SiO_(2)):}`
`{:(SiO_(2),SiCI_(4),Na_(2)SiO_(3)):}`

Solution :`SiCI_(4) + H_(2) O to UNDERSET((X))(si(OH)_(4))`
Discussion
14.

S.I. unit for dipole moment is

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e.s.u -cm
Coulomb -cm
coulomb -metre
e.s.u-metre

Answer :C
Discussion
15.

Silicon has diagonal relationship with

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Sulphur
Boron
Phosphorous
Carbon

Solution :
Discussion
16.

Si has digonal relationship with

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Sulphur
Boron
Phosphorus
Carbon

Answer :B
Discussion
17.

Shows which type of isomerism

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FUNCTIONAL GROUP of isomerism
geometrical isomerism
metamerism
Position isomers

Answer :A
Discussion
18.

Shows which type of isomerism

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Functional GROUP isomerism
Geometrical isomerism
METAMERISM
Position isomerism

Solution :They are not geometrical ISOMERS because structured formula is DIFFERENT. They are metamers because nature of alkyl group is different.
Discussion
19.

Shows which type of isomerism

Answer»

functional group of isomerism
geometrical isomerism
Metamerism
Position isomers

Solution :Alkyl group is different
`-overset(O)overset(||)(C)-O-` is bivalent group which has different alkyl groups in both compounds. As we know that ACCORDING to definitions of matamerism "when alkyl group is different around to bivalent to TRIVALENT atom 1 group compounds termed as metamers.

Although in second compound TRANS from is present but strucutre is ALSO present so it will be metamers not geometrical isomer.
Discussion
20.

Show the polarisation of carbon-magnesium bond in the following structure. CH_(3)-CH_(2)-CH_(2)-CH_(2)-Mg-X

Answer»

Solution :SINCE electronegativity of Mg (1.2) is MUCH LOWER than that of C (2.5), therefore , Mg carries a PARTIAL positive charge while the carbon attached to it carries a partial negative charge, i.e.,
Discussion
21.

Show the polarisation of carbon -magnesium bond in the following structure. CH_(3)-CH_(2)-CH_(2)-CH_(2)-Mg-X

Answer»

Solution :Since electronegativity of MG (1, 2) is much lower than that of C(2.5), therefore, Mg carries a partical POSITIVE charge while the carbon attached to it carries a partical negative CHORE.
Discussion
22.

Show the heterolysis of covalent bond by using curved arrow notation and ocmplete the following equation . Identify the nucleophile is each case . CH_(3) - OCH_(3) + HI to

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SOLUTION :
1 from HI ACTS as a NUCLEOPHILE
Discussion
23.

Show the heterolysis of covalent bond by using curved arrow notation and ocmplete the following equation . Identify the nucleophile is each case .CH_(3) - Br + KOH to

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SOLUTION :
`OH^(-)` from KOH is the nuclephilc
Discussion
24.

Show the oxidising property of H_(2)O_(2) with PbS.

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Solution :Hydrogen peroxide oxidises LEAD sulphide to lead SULPHATE in a NEUTRAL medium.
`PbS+3H_(2)O_(2)toPbSO_(4)4H_(2)O`
Discussion
25.

Show the heterolysis of covalent bond by using curved arrow notation and complete thefollowing equations.Identify the nucleophile in each case. (i) CH_(3) -Br + KOH to (ii) CH_(3) - OCH_(3) + HI to

Answer»

SOLUTION :
Discussion
26.

Show the heterolysis of covalent bond by using curved arrow notation and complete the following equations.Identify the nucleophile is each case. CH_(3)-Br+KOHrarr

Answer»

SOLUTION :
`OH^(-)` from KOH is the NUCLEOPHILE.
Discussion
27.

Show the heterolysis of covalent bond by using curved arrow notation and complete the following equations.Identify the nucleophile is each case. CH_(3)-OCH_(3)+HIrarr

Answer»

SOLUTION :
I from HI ACTS as a NUCLEOPHILE.
Discussion
28.

Show the electron distribution in the molecular orbitals of the cycloheptatrienyl cation. Strategy: Use the polygon-andj-crcle method for deriving the relative energies of the Mos.

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Solution :The polygon-and-circle medthod shows that the three lowest ENERGY `MOs` are below the DIAMETER and are bonding The FOUR HIGHEST energy ones are above the diameter and are antibonding `MO s`
Discussion
29.

Show the distribution of electrons in oxygen atom (atomic number 8) using orbital diagram

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SOLUTION :
Discussion
30.

Show the distribution of electrons in oxygen atom (atom ic num ber 8) using orbital diagram .

Answer»

SOLUTION :
Discussion
31.

Show the direction of inductive effect in following compounds

Answer»


ANSWER :`(##RES_CHM_GOC_I_E01_001_A01##)`
Discussion
32.

Show that the sum of mole fraction of a solution is equal to one.

Answer»

Solution :Consider a solution CONTAINING two COMPONENTS A and B whose mole fractions are `x _(A) and x _(B)` respectively. Let the number of moles of two components A and B are `n_(A) and n_(B)` respectively.
`(n _(A))/(n _(A) + n _(B)), x _(B) = (n_(B))/(n _(A) + n _(B))`
` x _(A) + x _(B) = ( n _(A))/(n _(A) + n _(B)) + (n_(B))/( n _(A) + n _(B)) =1`
Discussion
33.

Show that the reaction : CO(g) + 1/2O_(2)(g) to CO_(2)(g) at 300 K is spontaneous and exothermic, when standard entropy change is -0.094 kJ"mol"^(-1).The standard Gibbs energies of formation of CO_(2) and CO are -394.4 and -137.2 kJ"mol"^(-1) respectively.

Answer»

Solution :`Delta_(R)G^(@) = Delta_(f)G^(@)("products") - Delta_(f)G^(@)("REACTANTS")`
` = [Delta_(f)G^(@(CO_(2))] - [Delta_(f)G^(@)(CO) + 1/2Delta_(f)G^(@)(O_(2))]`
`= -394.4 - [-137.2 + 0]`
` = -257.2 kJ`
Since `Delta_(r)G^(@)` is negative, the REACTION is spontaneous. Now
`DELTAG^(@) = DeltaH^(@) -TDeltaS`
` -257.2 = DeltaH^(@) -[300 xx(-0.094)]`
`therefore DeltaH^(@) = -257.2 -28.2 = -285.4 kJ`
Since `DeltaH^(@)` is negative, the reaction is exothermic.
Discussion
34.

Show that the role of cone, sulphuric acid is not a dehydrating agent, but used for production of nitronium cation.

Answer»

Solution :Nitration mixture is a mixture of conc. nitric ACID and conc. sulphuric acid. Sulphuric acid acts as an acid and donates a proton to nitric acid. The protonated nitric acid, gives NITRONIUM cation by the loss of water molecule.
`HHSO_(4)+HNO_(3) to HSO_(4)^(-)+H_(2)O^(+) -NO_(2) , H_(2) OVERSET(+)(O) NO_(2) to NO_(2)^(+)+ H_(2)O`
Benzene gives nitorbenzene by electrophilic substitution.
`C_(6)H_(6)+NO_(2)^(+) to C_(6)H_(5)NO_(2)+H^(+) , H^(+)+HSO_(4)^(-) to H_(2)SO_(4)`
Discussion
35.

Show that the reaction CO+1//2O_2 to CO_2 at 300 K is spontaneous. The standard Gibbs free energies of formation of CO_2 and COare -394.4 and -137.2 "kJ mole"^(-1)respectively.

Answer»

Solution :`CO+1//2O_2 to CO_2`
`DeltaG_"(REACTION)"^0=sum G_(f"(products)")^0-sum G_(f"(reactants)")^0`
`DeltaG_"(reaction)"^0=[G_(CO_2)^0]-[G_(CO)^0+1/2G_(O_2)^0]`
`DeltaG_"(reaction)"^0`=-394.4+[137.2+0]
`DeltaG_"(reaction)"^0=-257.2 "kJ MOL"^(-1)`
`DeltaG_"(reaction)"` of a reaction at a given temperature is NEGATIVE hence the reaction is spontaneous
Discussion
36.

Show that the formation of sodium choride from gaseous sodium and gaseous chlorine is a redox reaction

Answer»

SOLUTION :`Na(G) RARR Na^(+)(g)+e^(-)]xx2`
`CI_(2)(g)+2E^(-)rarr2cI^(-)(g)`
`2Na(g)+cI_(2)rarr2Na^(+)CI^(-)(s)`
Discussion
37.

Show that the cirumference of the Bohr orbit for hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron moving around the orbit.

Answer»

SOLUTION :According to Bohr POSTULATE of angular momentum, `mvr= n h/(2PI) or 2 pi r=n h/(mv)` ...(i)
According to de Broglie equation, `lambda=h/(mv)` ...(ii)
Substituting this VALUE in eqn. (i) we get `2 pi r= n lambda`
Thus, the cirumference `(2pi r)` of the bohr orbit for hydrogen atom is an integral mutiple of de Broglie wavelength.
Discussion
38.

Show that the data given below is in agreement with the law of constant composition {:("Exp. No.","Mass of iron","Mass of iron oxide"),("(i)","2.7812 g","3.9756 g"),("(ii)","3.0499 g","4.3596 g"),("(iii)","4.4913 g","6.4202 g"):}

Answer»

Solution :Experiement No. (i)
MASS of iron = 2.7812 g , Mass of iron oxide = 3.9756 g
PERCENTAGE of iron `= ((2.7812g))/((3.9756g))xx100=70%`
Percentage of oxygen `= 100 - 70 = 30 %`
Experiment No. (ii)
Mass of iron = 3.0499 , Mass of iron oxide = 4.3596 g
Percentage of iron `= ((3.0499))/((4.3596g))xx100=70%`
Percentage of oxygen `= 100-70=30%`
Experiment No. (iii)
Mass of iron = 4.4913 g , Mass iron oxide = 6.4202 g
Percentage of iron `= ((4.4913g))/((6.42402g))xx100=70%`
Percentage of oxygen `= 100 - 70 = 30 %`
Since the RATIO by weights of iron and oxygen in the THREE oxides is the same, the DATA is according to Law of constant Composition.
Discussion
39.

Show that the circumference of the Bohr orbital for the hydrogen atom is an integral multiple of the de Broglie wave lenght associated with the electron revolving arround the nucleus.

Answer»

Solution :In order for the electron wave to EXIST in phase, the circumference of the ORBIT should be an integral MULTIPLE of the wavelength of the electron wave. Their electron wave is out phase.
`mvr = nh//2pi`
Where `mvr` = ANGULAR momentum
Where `2 pi r` = circumference of the orbit
Circumference of the orbit = `n LAMBDA`

`2 pir = n lambda`
`n = 3, n = 4`.
Discussion
40.

Show that the circumference of the Bohr orbit for the hydrogen atom is an intergral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

Answer»

Solution :According to Bohr.s THEORY,
`mvr=(nh)/(2pi)`(n=1,2,3,…….so on)
` or 2pir=(nh)/(mv) or mv=(nh)/(2pir)`……..(i)
According to de Broglie equation,
`GAMMA=h/(mv) or mv=(h)/(gamma)`…………..(ii)
COMPARING (i)and (ii),
`(nh)/(2pir)=h/(pi)or 2pir=ngamma`
THUS,the circumference`(2pir)`of the Bohr orbit for hydrogen ATOM is an into the de Broglie wavelength.
Discussion
41.

Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wave length associated with the electron revolving around the nucleus.

Answer»

SOLUTION :In ORDER for the electron WAVE to exist in phase,the circumference of the orbital should be an intergral multiple of the wavelength of the electron wave.otherwise ,the electron wave is out of phase.
mvr=ng/2pi
`2pir = N lambda`
where mvr=angular momentum
where `2pir`=circumference of the orbit.
n= 3, n=4
Discussion
42.

Showthat thecircumferencefo the Bohrorbitfor the hydrogenatom isan integralmultipleof thede-Brogliewavelengthassociatedwiththe electronrevolvingaroundthe orbit.

Answer»

Solution :ForBohr.smodelangularmomentumofelementis
`m_(e)vr = (nh)/(2pi)`
circumference `=2pi R = (nh)/(m_(e ) v)`
and `m_( e)v= (nh)/(2pi r)`
Accordingot de - BROGLIE
`lambda= (h)/(mv) ` andmv `=(h)/(lambda)`
as (i) and(II)
mv = `(nh)/( 2pi r)= (h)/( lambda ) :., (N)/(2pi r)= (1)/(lambda):., 2pi r= n lambda` Bohr.sorbitcircumference `(2pi r ) = n ` de- Brogliewavelength
In this wayde broylicwave lengthis associatedwthBohr.sorbitcircumference.
Discussion
43.

Show that solubility of a sparingly soluble salt M^(2+)A^(2-) in which A^(2-) ion undergoes hydrolysis is given by: S=sqrt(K_(SP)(1+([H^(+)])/(K_(2))+([H^(+)]^(2))/(K_(1)K_(2)))) where K_(1) and K_(2) are the dissociation constant of acid HA_(2).K_(SP) is solubility product of MA.

Answer»
Discussion
44.

Show that six hydrogen atoms in benzene are identical.

Answer»

Solution :When benzene ADDS THREE molecules of chlorine in the presense of SUNLIGHT gives benzen hexachloride.
`C_(6)underset("Benzene")(H_(6)) + 3Cl_(2) overset("sunlight")(rarr) underset("Benzene HEXS Chlonde(BHC)")(C_(6)H_(6)Cl_(6))`
Discussion
45.

Show that presence of three double bonds in benzene.

Answer»

Solution :When benzene adds THREE molecules of hydrogen in presence of nickel catalyst of formcyclohexane
`C_(6)H_(6) + underset("Benzene")(3H_(2)) underset(180^(@)C)overset(" "Ni" ")(rarr) underset("CYCLOHEXANE")(C_(6)H_(12)`
Discussion
46.

Show that K_p = K_c with two examples

Answer»

Solution :`K_p +K_c (RT)^(Delta)g`. If `Delta n_g = 0. K_p = K_c` EXAMPLES
`(i)H_(g) + l_(2)(g) hArr 2HI(g)`
`Delta n_g =2-2=0`
`(ii) N_(2)(g) + O_(2)(g) hArr 2NO(g)`
`Delta n_(g) = 2-2 =0`
Discussion
47.

Show that in the isothermal expansion of an ideal gas , DeltaU = 0 and DeltaH = 0

Answer»

Solution :(a) For one moole of an ideal GAS, `C_(V) = ( dU)/( DT) ` or ` dU =C_(v) dT`
For a finite change, `DELTAU = C_(v) DeltaT`
For ISOTHERMAL process, `T=` constant so that`DeltaT =0`. Hence, `DeltaU = 0`
(b)`DeltaH = DletaU + Delta(PV) =DeltaU+ Delta(RT) =DeltaU+ R DeltaT`
But `DeltaU =0` ( proved above) and `DeltaT = 0` ( for isothermal process) , `:. DeltaH =0`
Discussion
48.

Show that if the measurement of the uncertainty in the location of the particle is equal to its de Broglie wavelength,the minimum uncertinty in its velocity (/_\v) is equal to 1/4pi of its velocity(V)

Answer»

Solution :If, uncertainty in position`=/_\x=gamma`,the vale of uncertainty in velocity=`v/(4PI)`
Heisenberg.s PRINCIPLE states that
`/_\x./_\v.m=h/(4pi)`………….(i)
de Broglie equation states that
`gamma=h/(mv)`………..(2)
`:.h=gamma.m.v`……..(3)
`/_\x=h/(/_\v.4pi)`.............(4)
`/_\X=(gammaxxm.v)/(/_\v.4pi.m)`
`if/_\x=gamma`
`/_\v=(CANCEL(x.m).v)/(cancel(x).4pi.cancel(m))=v/(4pi)`.
Discussion
49.

Show that ground state energy of electron in hydrogen atom is equal to the first excited state energy of electron in He^(+) ion, assuming that their Rydberg's constants are equal

Answer»

SOLUTION :`E_(n) = - R_(H) (Z^(2))/(n^(2)) = - 13.60 (Z^(2))/(n^(2)) eV`
For H-atom in the ground state, `Z = 1, n =1, :. E_(1) = -13.60eV`
For `He^(+)` ion in the first EXCITED state, `Z = 2, n = 2, E_(2) = - 13.60 xx (2^(2))/(2^(2)) = - 13.6 eV`
HENCE, the desired result is proved.
Discussion
50.

Show that free energy (Delta G) as a criterion for spontaneous and non-spontaneous processes.

Answer»

Solution :Effect of temperature on the spontaneity of chemical reaction using Gibb's-Helmholtz equation : `Delta G = Delta H - T.Delta S`
Case I : For an exothermic process, `Delta H` is negative.
(a) If `Delta S` is positive, `Delta G` is always negative. The process is always spontaneous.
(b) If `Delta S` is negative,
(i) If the temperature is LOW such that `T Delta S lt Delta H`, then `Delta G` is negative. The process is spontaneous.
(ii) If the temperature is high such that `T Delta S gt Delta H`, then `Delta G` is positive. The process is not spontaneous
Thus, an exothermic process which is not spontaneous at a GIVEN temperature may become spontaneous at low TEMPERATURES.
Case II : For an endothermic process, `Delta H` is positive.
(a) If `Delta S` is negative, `Delta G` is always positive. `:.` The process is not spontaneous
(b) If `Delta S` is positive,
(i) If the temperature is low such that `TDelta S lt Delta H`, then `Delta G` is positive. The process is not spontaneous.
(ii) If the temperature is high such that `T Delta S gt Delta H`, the `Delta G` is negative. The process is spontaenous.
Thus, an endothermic process which is not spontaneous at a given temperature may become spontaneous at high temperatures.
Case III :When any process attains equilibrium, `T Delta S gt Delta H, Delta G = 0`
i.e., `Delta G = Delta H - T Delta S = 0`or `Delta H = T Delta S` at equilibrium
For a chemical reaction, `A + B rarr C + D`, Gibb's ENERGY change is given by, `Delta G = SigmaG_(2) - Sigma G_(1)`. It is the difference between the total free energies of all the PRODUCTS and the total free energies of all the reactants.
Discussion