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.

The ionisation energy of hydrogen is

Answer»

`1312 KJ "MOLE"^(-1)`
`520 KJ "mol"^(-1)`
`495 KJ "mol"^(-1)`
`1681 KJ "mol"^(-1)`

Answer :A
Discussion
2.

The ionisation energy of He^(+) " is " 8.72 xx 10^(-18)J " atom"^(-1). Calculate the energy of the first stationary state of Li^(2+)

Answer»

Solution :`E_(n) = - (2pi^(2) mZ^(2) e^(4))/(n^(2) h^(2)) = - K (Z^(2))/(n^(2))` (K = constant)
I.E. of `He^(+) = E_(oo) - E_(1) = 0 - (-K (2^(2))/(1^(2)))= 4K`
Hence, `4K = 8.72 xx 10^(-18)J "atom"^(-1)` (GIVEN) or `K = 2.18 xx 10^(-18) J " atom"^(-1)`
For `Li^(2+), Z =3` and for 1st stationary STATE, n = 1
`:. E_(1) = -K (Z^(2))/(n^(2)) = - 2.18 xx 10^(-18) xx (3^(2))/(1^(2)) = - 19.62 xx 10^(-18) J "atom"^(-1)`
Discussion
3.

The ionisation energy of He^(+) is 19.6xx10^(-18)"J atom"^(-1) Calculate the energy of first stationary state of Li^(2+) ion.

Answer»

Solution :Ionisation energy of hydrogen LIKE species = Ionisation energy of `H XXZ^(2)`
`"I.E of "He^(+)="I.E of H"xx4,"I.E of "H=(1.9xx10^(-18))/(4)"______(1)"`
`"I.E of "LI^(2+)="I.E of "Hxx9," I.E of H"=("I.E of "Li^(2+))/(9)"____(2)"`
`"Equating (1) and (2) "("I.E of "Li^(2+))/(9)=(19.6xx10^(-18))/(4)`
`"I.E of "Li^(2+)=(19.6xx10^(-18)xx9)/(4)`
The energy of first stationary state of `Li^(2+)`
`=(-19.6xx10^(-18)xx9)/(4)=-4.41xx10^(-17)"J atom"^(-1)`
Discussion
4.

The ionisation energies of Lithium and Sodium are 520kJ Mol^(-1) and 495kJ Mol^(-1) respectively. The energy required to convert all the atoms present in 7mg of Li vapour and 23 mg of sodium vapour to their respective gaseous cations respectively are

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52 J , 49.5 J
520 j, 495j
49.5 j , 52 j
495 j , 520 j

Answer :B
Discussion
5.

The ionisation energy and electron affinity of an element are 13.0ev and 3.8ev respectively. Its electronegativity is

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2.8
3
3.5
4

Answer :B
Discussion
6.

The ionisation energies from Ga to TI do not decrease due to

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SHIELDING EFFECT
Improper shielding effect
Increase in atomic size
Decrease in NUCLEAR charge

ANSWER :B
Discussion
7.

The ionisation constant of NH_3 is 1.77 xx 10^(-5). What is the 'a' and 'pH' of 0.05M ammonia solution?

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SOLUTION :`alpha=0.018, pH=10.97`
Discussion
8.

The ionisation constant of chloroacetic acid is 1.35xx10^(-3) .What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution ?

Answer»

Solution :PH of 0.1 M chloroacetic ACID `(ClCH_2COOH)` : This is a weak acid and suppose x M of 0.1 M is ionised and equilibrium is established.
So At equilibrium `x=[H_3O^+] = [ClCH_2COO^-]`
`{:(,ClCH_2COOH_((aq)) +H_2O_((l)) hArr , ClCH_2COO_((aq))^(-) + , H_3O_((aq))^(+)),("At equilibrium" , 0.1 M, 0.0 , 0.0),(,(0.1 -x)M,x M, xM),(,approx 0.1, ,):}`
`K_a=([ClCH_2COO^(-)][H_3O^+])/([ClCH_2COOH])`
`therefore 1.35 xx 10^(-3) =((x)(x))/0.1`
`therefore x^2=1.35xx10^(-4)`
`therefore x=1.162xx10^(-2)M=[H_3O^+]`
pH=log `[H^+]`
`=-log (1.162xx10^(-2))`
=-(0.0652-2)
=1.9348 `approx` 1.93
`therefore` pH of this weak base is 1.93
Acid `ClCH_2COOH` and SALT `ClCH_2COONa` present in solution so solution is buffer. The pH of this buffer is calculate by following equation.
`pH=pK_a + "log"([A^-])/([HA])`
`[A^-]=[ClH_2COO^-]=0.1`
[HA]=`[ClCH_2COOH]`=0.1
where `pK_a=-log (K_a)`
=-log `(1.35xx10^(-3))`
=2.8690
pH=2.869 +log `(0.1/0.1)`=2.869
Discussion
9.

The ionisation constant of an acid, K_(a) is the measure of strength of an acid. The K_(a) values of acetic acid, hypochlorous acid and formic acid are 1.74xx10^(-5), 3.0xx10^(-8) and 1.8xx10^(-4) respectively. Which of the following orders of pH of 0.1 mol dm^(-3) solutions of these acids is correct ?

Answer»

acetic acid `gt` hypochlorous acid `gt ` formic acid
hypochlorous acid `gt ` acetic acid `gt ` formic acid
formic acid `gt` hypochlorous acid `gt `acetic acid
fromic acid `gt` acetic acid `gt ` hypochlorous acid

SOLUTION :Greater the IONIZATION constant, greater is the strength of the acid. Hence,
`{:(,HCO OH ,gt,CH_(3)CO OH,gt,HClO),(K_(a),1.8xx10^(_4),,1.74xx10^(-5),,3.0xx10^(-8).):}`
Discussion
10.

The ionisation constant of an acid, K_a is the measure of strength of an acid. The K_a values of acetic acid, hypochlorous acid and formic acid are 1.74 xx 10^(-5), 3.0 xx 10^(-8) and 1.8 xx 10^(-4) respectively. Which of the following orders of pH of 0.1 mol dm^(-3) solutions of these acids is correct ?

Answer»

ACETIC acid `GT` hypochlorous acid `gt` formic acid
hypochlorous acid `gt` acetic acid `gt` formic acid
formic acid `gt` hypochlorous acid `gt` acetic acid
formic acid `gt` acetic acid `gt` hypochlorous acid

Solution : If the acidity or `K_a` value increases, pH decreases, HENCE LIKE this, the ORDER of pH value of the acids is,
`{:("Hypochlorous acid",lt,"Acetic acid",lt , "Formic acid"),((3.8xx10^(-8)),,(1.74xx10^(-5)),,(18xx10^(-4))):}`
Discussion
11.

The ionisation constant of an acid base indicator is 3' 10^(-5). The acid from of the indictor is red and the basic form is blue. By how much must the p^(H) change in order to change the indicator from 75% red to 75% blue?

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`1.95`
`2.95`
`0.95`
`3.95`

SOLUTION :`HIn hArr H^(o+) + In^(o+)`
`K_(In)=K_(n)=([H^(+)][In^(-)])/(HIn),[H^(+)] = Ka(["acid"])/(["BASIC"])`
for `75%` red,
`[H^(+)] = ((3xx10^(-5))(75))/(25) = 9 xx 10^(-5), p^(H) = 4.05`
The change in `p^(H) = 5 - 4.05 = 0.95`
Discussion
12.

The ionic strength of solution containing 0.5 M MgSO_(4)0.1M AlCl_(3) and 0.2(NH_(4))_(2)SO_(4) is

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`0.75`
`1.85`
2
`1.50`

Solution :`MgSO_(4)=0.5xx2=1`
`AlCl_(3)=0.1xx4=0.4`
`(NH_(4))_(2)SO_(4)=0.2xx3=(0.6)/(2)`
Discussion
13.

The ionic radius of H^+ ions and that of other ions is .......... pm and .......... pm respectively.

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`50-200,1.5xx10^(-3)`
`1.5xx10^(-3), 50-100`
`10^(-3),300`
1.5 , 400

ANSWER :B
Discussion
14.

The ionic radius of H^+ ions and that of other ions is ........ pm and ........ pm respectively.

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SOLUTION :`1.5xx10^(-3)` PM , 50-200 pm
Discussion
15.

The ionicradiusofCris minimum inwhichthe followingcompounds ?

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`K_(2) CrO_(4)`
`CrF_(3)`
`CrO_(2)`
`CrC1_(3)`

SOLUTION :AMONGST `K_(2) CrO_(4) , CrF_(3),CrO_(2) ` and `CrC1_(3) ,Cr` has the maximumoxidation numberof + 6IN `K_(2)CrO_(4)` and hencehas theminimum IONICRADIUS.
Discussion
16.

The ionic radius(in Å ) of N^(3-), O^(2-) and F^(-)are respectively .

Answer»

1.71, 1.40 AND1.36
1.71 .1.36 and1.40
1.36, 1.40 and1. 71
1.36 . 1.71and 1.40

Solution :`N^(3-) (7+ 3= 10) ,O^(2-) (8+2 =10)`
and `F^(-) (9+ 1=10) ` areisoelectronicions .
Amongst isoelectronicions, greaterthe negativechargeon theanion , largeris the ionicradius , Thusionicradiidecreae intheorder`N^(3-) (1.71 Å ) lt`
`O^(2-)(1.40 Å) GT F^(-)(1.36 Å) . i.e., `option(a) ISCORRECT .
Discussion
17.

The ionic radii of N^(3-) , O^(2-) , F^(-) " and " Na^(+) follow the order.

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`N^(3-) GT O^(2-) gt F^(-) gt Na^(+)`
`N^(3-) gt Na^(+) gt O^(2-) gt F^(-)`
`Na^(+) gt O^(2-) gt N^(3-) gt F^(-)`
`O^(2-) gt F^(-) gt Na^(+) gt N^(3-)`

ANSWER :a
Discussion
18.

The ionic radii of A^(+) and B^(-) ions are0.98 xx 10^(-10) m and 1.81 xx 10^(-10) m.The coordination number of each ion in AB is

Answer»

2
6
4
8

Solution :Radius RATIO , ` r_(+)/r_(-) = ( 0.98 xx 10^(-10))/( 1.81 xx 10^(-10)) = 0.541`
It liess in the range 0.414 - 0.732. Hence, coordination NUMBER of each ION will be 6, as in the case of NaCl.
Discussion
19.

The ionic radii of A^+ and B^- ions are 0.92xx10^(-10) m and 1.81xx10^(-10) m . The coordination number of each ion in AB is

Answer»

2
6
4
8

Solution :Radius ratio , `r_(+)/r_(-)=(0.98xx10^(-10))/(1.81xx10^(-10))=0.541`
It lies in the RANGE 0.414-0.732 . HENCE, coordination NUMBER of each ION will be 6, as in the CASE of NaCl
Discussion
20.

The ionic product of water is 10^(-14), The H^(+) ion concentration in 0.1 M NaOH soluiion is

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`10^(-11) M`
` 10^(-13) M`
` 10^(-1)M`
` 10 ^(-4) M`

SOLUTION :`[H^(+) ] [OH^(-) ] = 10 ^(-14)`
` [OH^(-) ] =10 ^(-1)=[H^(+) ] =10 ^(-13)M`
Discussion
21.

The ionic product of water at 60^(@)C is 9.55 xx 10^(-14) "mole"^(2) lit^(-2) . The dissociation constant of water at the same temperature is

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`1.09 XX 10^(-15)`
`5.2 xx 10^(-16)`
`1.8 xx 10^(-16)`
`1. 72 xx 10^(-15)`

Answer :D
Discussion
22.

The ionic product of water at 310 K is 2.7xx10^(-14). What is the pH of neutral water at this temperature ?

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SOLUTION :`[H^(+)]=SQRT(K_(w))=sqrt(2.7xx10^(-14))=1.643xx10^(-7)M`
`pH= - LOG [H^(+)]=-log (1.643xx10^(-7))=7-0.2156=6.78`
Discussion
23.

The ionic product of water at 310 K is 2.7 xx10^-14. What is the pH of neutral water at this temperature.

Answer»

SOLUTION :`K_w=[H^+][OH^-]=2.7xx10^-14`
`THEREFORE[H^+]=SQRT(2.7xx10^-14)=1.643xx10^-7`
`therefore pH=-log(1.643xx10^-7)=7-0.2156=6.7844=6.78`
Discussion
24.

The ionic mobility of Li^(+) is less than that of the Na+ ion in solution because

Answer»

Li+ ION has a high MELTING point
Li+ ion has the HIGHEST hydration tendenc
Li+ ion has the highest ionisation enthalpy
Li+ ion has two electrons

Solution :Due to highest hydration enthalpy the mobility of `Li^(+)` is LESS
Discussion
25.

The ionic mobility of alkali metal ions in aqueous solution is maximum for

Answer»

`K^(+)`
`Rb^(+)`
`Li^(+)`
`Na^(+)`.

SOLUTION :Bigger the SIZE, LOWER is the extent of hydration THEREFORE, smaller is the size of the hydrated species and highest is the ionic MOBILITY, i.e., `Rb^(+)` has the maximum mobility .
Discussion
26.

The ionic compounds of alkali metals colourless. Why?

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Solution :ALKALI metals FORM unipositive ions which have STABLE CONFIGURATION of the nearest inert gas. Alkali metal salts are diamagnetic and colourless because they do not have unpaired ELECTRONS.
Discussion
27.

The ionic chloride is

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`PbCl_(2)`
`pbCl_(4)`
`C C l_(4)`
`SiCl_(4)`

Solution :LOWER OXIDATION state.
Discussion
28.

The ion that is pumped out from the cells is

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`NA^(+)`
`K^(+)`
both
none

Answer :A
Discussion
29.

The ion that has pseudo-octet configuration

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`Mn^(2+)`
`Cu^(2+)`
`Zn^(2+)`
`CR^(2+)`

ANSWER :C
Discussion
30.

The ion exchange resin which removes metal ions from hard water consists of giant organic molecule having

Answer»

`-CL` group
`-COOH` group
`-OH` group
`-NH_(2)` group

Answer :B
Discussion
31.

The ion having highest mobility in aqueous solution is

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`Be^(+2)`
`Mg^(+2)`
`CA^(+2)`
`Ba^(+2)`

SOLUTION :DUE to large size `Ba^(+2)` hasless HYDRATED and more ionic mobility
Discussion
32.

The ion acting as an oxidising agent in the reaction, Cr_(2)O_(7)^(2-)+6Fe^(2+)+14H^(+)to2Cr^(3+)+6Fe^(3+)+7H_(2)O is

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`Cr_(2)O_(7)^(2-)`
`FE^(2+)`
`CR^(3+)`
`H^(+)`

Answer :A
Discussion
33.

The iodide content of a solution was determined by titration with sodium thiosulphate crystalline containing 11.2 % impurity. Calculate the normality of iodide ion solution in 250 mL of the iodide solution required 20 mL hypo (42 g hypo is dissolved in 1 litre).

Answer»


SOLUTION :N//A
Discussion
34.

The iodine molecule dissociates into atoms after absorbing light of 4500Å. If one quantum of radiat is absorbed by each molecule, calculate the kinetic energy of iodine atoms (Bond energy I_(2) = 240 kJ mol^(-1))

Answer»

Solution :Bond energy PER molecule of `I_(2) = (240 xx 1000J)/(6.022 xx 10^(23)) = 3.985 xx 10^(-19)J`
Energy absorbed `= (HC)/(lamda) = ((6.626 xx 10^(-34) Js) (3 xx 10^(8) ms^(-1)))/(4500 xx 10^(-10)m) = 4.417 xx 10^(-19)J`
The difference of energy absorbed and energy used for dissociation of bond in `I_(2)` gives the KINETIC energy one `I_(2)` molecule.
`:.` K.E. of one `I_(2)` molecule `= (4.417 xx 10^(-19) - 3.985 xx 10^(-19)) J = 4.32 xx 10^(-20)J`
`:.` K.E. of one iodine atom `= (4.32 xx 10^(-20))/(2) J= 2.16 xx 10^(-20)J`
Discussion
35.

The io that is isoelectronic with CO

Answer»

`O_(2)^(+)`
`CN^(-)`
`O_(2)^(-)`
`N_(2)^(+)`

ANSWER :B
Discussion
36.

The inversion temperature for a gas is given by ________.

Answer»

`(2A)/(RB)`
`(a)/(Rb)`
`(Rb)/(2a)`
`(Rb)/(a)`

ANSWER :A
Discussion
37.

The introduction of neutron in nucleus of an atom would lead to the change i n ........

Answer»

ATOMIC weight
Atomic number
Number of electrons around a nucleus
CHEMICAL natureof the atom

Solution :The mass of a NEUTRON is 1.00867 U.`therefore`INTRODUCTION of neutron in nuclecus will increase mass... i.e. atomic weight
Discussion
38.

The intramolecular hydrogen bonding in molecules lead to ________

Answer»

HIGH MELTING point
low BOILING point
high boiling point
high SOLUBILITY in water

Answer :B
Discussion
39.

The intestinal antiseptic salol is correctly represented as

Answer»




SOLUTION :
Discussion
40.

The internalenergy change ( Delta U )for the reaction CH_(4)(g) + 2O_(2)(g) rarr CO_(2)(g) +2H_(2)O(l)is - 885kJ mol^(-1) at298 K . What is DeltaH at298 K ?

Answer»


ANSWER :`-889.96 KJ MOL^(-1)`
Discussion
41.

The internal resistance to flow in liquid is called

Answer»

Fluidity 
Specific resistance 
VISCOSITY 
SURFACE tension 

SOLUTION :Viscosity is a measure of resistance of FLOW.
Discussion
42.

The internal pressure loss of one mole of vander Waal gas over an ideal gas is not equal to

Answer»

ZERO 
`B^2`
`a/(V^2)`
`b- a/(RT)`

SOLUTION :`RHO = (an^2)/(V^2) rho = (an^2)/(V^2)`
Discussion
43.

The internal energy of a system is ………….. Property and …………….function.

Answer»

INTENSIVE , SATE
Extensive PATH
Intensive , path
Extensive STATE

Answer :D
Discussion
44.

The internal energy change when a system goes from state A to B is 40 KJ/mole. If the system goes from A to B by a reversible path and returns to state A by an irreversible path what would be the net change in internal energy ?

Answer»

`40 KJ`
`gt 40KJ`
`LT 40KJ`
ZERO

ANSWER :D
Discussion
45.

The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is :

Answer»

ion - ion interaction
ion - DIPOLE interaction
London FORCE
Hydrogen BOND

Solution :H - bond is one of the dipole - dipole interaction and dependent on INVERSE CUBE of distance between the molecules.
Discussion
46.

The intermolecular force of attraction present between NH_3 and C_6H_6 are

Answer»

DIPOLE - Dipole 
ION - dipole 
Dipole - INDUCED dipole 
DISPERSION 

SOLUTION :`NH_3` dipole , `C_6H_6` - induced dipole.
Discussion
47.

The intermolecular attraction that is dependent on the inverse cube of the distance between the molecules is

Answer»

Ion-ion interaction
Ion-dipole interaction
London FORCE
Hydrogen bond

SOLUTION :Hydrogen bond is a type of dipole-dipole interaction which is dependent on the inverase cube of the distance between the molecules `(PROP(1)/(R^(3)))` in stationary molecules (e.g. in solides).
Discussion
48.

The intermetallic compound LiAg crystallizes in a cubic lattice in which Li and Ag atoms have C.N. of 8. To which crystal class the unit cell belongs

Answer»

NaCl
CsCl
ZnS
none of these

Solution :InCsCl both CS and CL has C.N. of 8.
Discussion
49.

The intermediate species involved in the acid catalysed dehydration of alcohol is

Answer»

Free radical
Carbocation
Carbanion
Carbene

Solution :Acid catalysed dehydration of ALCOHOL PROCEEDS through a carbocation.
Discussion
50.

The intermediate is produced in which of the following is correct for intermediate?

Answer»

It is a FREE radical
It is carbocation
It is a carbanion
B and C

Answer :B
Discussion