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.

Find the incorrect pair?

Answer»

`HNO_3,H_2SO_4`
`Al(OH)_3,MG(OH)_2`
`CH_3COOH,HCOOH`
`H_2O,OH^-`

Answer :B
Discussion
2.

find the incorrect order of acidic strength of following.

Answer»




SOLUTION :
Discussion
3.

Find the identify of compound [N] in the following reaction sequences: H=C-Na+CH_(3)CH_(2)CH_(2)CH_(2)Br to [A], underset(1"mole")[A]underset(1"mole")((Cl_(2)) to [B]

Answer»

cis-1,2-dichlorobutane
trans-2,3-dichloro-2-hexene
cis-1,2-dichloro-1-hexene
trans-1,2-dichloro-1-hexene

Solution :N//A
Discussion
4.

Find the incorrect match.

Answer»

FINE dust of ALUMINIUM - Paints and Lacquers
Reducing agent in the manufacture of dye-stuffs,paints etc.-Copper dust
For cutting tools and crushing MACHINES -Chorme steel
For making cables, AUTOMOBILES and aeroplane parts -Nickel steel

Solution :Zinc dust is used as a reducing agent in the manufacture of dye-stuffs, paints, etc.
Discussion
5.

Find the incorrect match

Answer»

`Al_(2)Cl_(6)` : 3c-4e BOND is PRESENT
`Al_(2)(CH_(3))_(6)` : All carbon atoms are `sp^(3)` hybridised
`I_(2)Cl_(6)` : Nonplanar
`B_(2)H_(6)` : Nonpolar

Answer :C
Discussion
6.

Find the incorrect match :-

Answer»

`Al_(2)Cl_(6)` : 3C-4e BOND is PRESENT
`Al_(2)(CH_(3))_(6)` : All carbon atoms are `sp^(3)` hybridised
`I_(2)Cl_(6)` : NONPLANAR
`Al_(2)Br_(6)` : Nonplanar

Answer :C
Discussion
7.

Find the heat change in the reaction : NH_(3)(g)+HCl(g)rarr NH_(4)Cl(s) from the following data {:(NH_(3)(g)+aq rarr NH_(3)(aq)",",Delta H=-8.4"K. Cal."),(HCl(g)+aq rarr HCl (aq)",",Delta H=-17.3"K. Cal."),(NH_(3)(aq)+HCl(aq)rarrNH_(4)Cl(aq)",",Delta H=-12.5 "K. Cals."),(NH_(4)Cl(s)+aq rarr NH_(4)Cl(aq)",",Delta H=+3.9 "K.Cal."):}

Answer»

`-42.1`
`-34.3`
`+34.3`
`+42.1`

Solution :`NH_(3(G))+HCl_((g))rarr NH_(4)Cl_((s)), Delta H = ?`
DESIRED EQUATION= EQ(i) + eq(ii) + eq(iii) - eq(iv)
`Delta H=(-8.4)+(-17.3)+(-12.5)-(3.9)`
`Delta H =-42.1 KCal`
Discussion
8.

Find the final product 'E' CH_(2)OH-CH=CH_(2)overset((i)OsO_(4))underset((ii)NaSO_(3)H)rarr(A)overset(KHSO_(4))rarr (B) overset(Zn-Hg//"conc. HCl")rarr (C ) overset(NBS//hv)underset(C Cl_(4))rarr (D ) overset((CH_(3))_(2)"CuLi")rarr (E )

Answer»

But - 2 - ene
But -1 - ene
Isopentene
Pen - 2 - ene

Answer :B
Discussion
9.

The formal charges on the three oxygen atoms in O_(3) molecule are

Answer»


Solution :
CENTRAL ATOM MARKED as `1=6-2-3=+1 THEREFORE` FORMAL charge = 1
Discussion
10.

Find the equilibrium constant for the reaction, In^(2+)+Cu^(2+) rarr In^(3+)+Cu^(+), at 298 K Given: E_(Cu^(2+)//Cu^(+))^(@) = 0.15 V , E_(In^(3+)//In^(+))^(@) = -0.42 V and E_(In^(2+)//In^(+))^(@) = -0.40V

Answer»

Solution :`CU^(2+) + e RARR Cu^(+) E_(1)^(@) = 0.15 V`...(1)
`In^(3+) + 2e rarr In^(+) E_(2)^(@) = -0.42 V` ...(2)
`In^(2+) + 2e rarr In^(+) E_(3)^(@) = -0.40 V` ...(3)
By eqs. `(2)` and (3), a third half cell reaction can be GIVEN as
`In^(3) ++ e rarr In^(2+)`
`1 xx F xx E_(4)^(@) = E_(2)^(@) xx 2 F - e_(3)^(@) xx 1 xx F`
or `E_(4)^(@) = 2 xx (-0.42) - (-0.40) xx 1 = -0.44 V`
For two half reactions, the NET redox change as asked is
`Cu^(2+) + In^(+) rarr Cu^(+) + In^(3+)`
`E_(cell)^(@) = E_(1_(RP_(Cu^(2+)//Cu^(+)))^(@)) + E_(4OP_(IN^(2+)//IN^(3+)))^(@)`
`[E_(OP_(In^(2+)//In^(3+))) = -E_(RP)^(@) = + 0.44]`
`= 0.15 + 0.44 = 0.59V`
Also, `E_(cell)^(@) = (0.059)/(1)logK_(c)`
`0.59 = (0.059)/(1)logK_(c)`
`:. K_(c) = 10^(10)`
Discussion
11.

find the entropy change when 2 moles of ideal gas at 27 ^(@)C temperature is expanded reversiby from 2 Lto 20 L .

Answer»

92 . 1
0
4
9.2

Solution :`DeltaS=2.303nR LOG((V_(2))/V_(1))= 2.303xx2xx2xxlog(20/2)=9.2`
Discussion
12.

Find the enthalpy of S-S bond from the following data. {:((i),C_(2)H_(5)-S-C_(2)H_(5) (g),Delta_(f)H^(@)=-147.2" kJ/mol"),((ii),C_(2)H_(5)-S-S-C_(2)H_(5)(g),Delta_(f)H^(@)=-201.9" kJ/mol"),((iii),S(g),Delta_(f)H^(@)=222.8" kJ/mol"):}

Answer»


Solution :`C_(2)H_(5) - S - C_(2) H_(5) + S RARR C_(2)H_(5) - S - S - C_(2) H_(5)`
`Delta H = B.E.` of `S - S` BOND
`= - 201.9 - [222.8 - 147.2] = - 277.5`
Discussion
13.

Find the enthalpy of S – S bond from thedata. S (g) = Delta H_(f)^(@) = 222.8 kJ /mol

Answer»

SOLUTION :227.5 kJ/mol
Discussion
14.

Find the enthalpy of S – S bond from thedata. C_(2) H_(5)- S - S - C_(2) H_(5) (g) Delta H_(f)^(@) = -201.9 kJ/mol

Answer»

SOLUTION :227.5 kJ/mol
Discussion
15.

Find the enthalpy of dissociation of H_2C_2O_4 from the given data: Enthalpy of neutralisation of strong acid and strong base is -13.7 kcal mol^(-1) and that of oxalic acid by a strong base is -25 kcal mol^(-1)

Answer»

`-11.3 KCAL MOL^(-1)`
`2.4 kcal mol^(-1)`
`1.2 kcal mol^(-1)`
`11.7 kcal mol^(-1)`

ANSWER :B
Discussion
16.

Find the EAN value of cetral atom of ferrocene [Fe(pi-C_(5)H_(5))_(2)]

Answer»


SOLUTION :26-2+12=36
Discussion
17.

Find the e.m.f. of the following cell at 18^(@)C taking the degree of dissocution of 0.2 N AgNO_(3) and 0.05 AgNO_(3) solution as 0.75 and 0.95 respectively. Ag|0.2N AgNO_(3)||0.05 AgNO_(3)|Ag

Answer»


ANSWER :`-0.029 V ;`
Discussion
18.

Find the end product in the given reaction. CH_3CONH_3overset(Br_2KOH)rarrXunderset(Delta)(CHCl_3//KOH) " Y"overset(LiAlH_4)rarrZ

Answer»

`CH_3CH_2NH_2`
`CH_3NH_2`
`C_2H_5overset(+)N-=overlineC`
`(CH_3)_2NH`

ANSWER :D
Discussion
19.

Find the empirical formula of chromium oxide containing 68.4% of chromium. (Cr=52, O=16)

Answer»

Solution :Let the weight of chromium oxide be 100g.
`THEREFORE` Weight of chromium= 68.4gand weight of oxygen =31.6g
`{:(therefore "moles of chromium"=(68.4)/(52)=1.32),("and moles of oxygen"=(31.6)/(16)=1.98):}}`
Relative NUMBER of moles of Cr and O atoms
`= ("moles of Cr")/("moles of O")= (1.32)/(1.98)= (132)/(198) = (2 xx 66)/(3 xx 66)= (2)/(3)` (by inspection only).
The empirical formula is `Cr_(2)O_(3)`.
Discussion
20.

Find the distance between two Po atoms that lie along a body diagonal, given that a = 0.336 nm.

Answer»

Solution :The coordinates of the two atoms are (0, 0, 0) and (1, 1, 1) respectively. THUS,
`L = 0.336 [(1-0)^(2) + (1-0)^(2) + (1-0)^(2)]^(1/2)`……….(Eqn.2)
=0.582 NM
Discussion
21.

Find the distance between the body-centred atom and one corner atom in sodium, a = 0.424 nm.

Answer»

Solution :The atom at the centre of the cube is SUPPOSED to touch the two nearest corner atoms. The longest diagonal at which all these three atoms LIE will be of LENGTH 4r.`therefore` distance between the centre atom and the corner atom=2r
`=2 xx (sqrt(3)a)/4 = (2 xx sqrt(3) xx 0.424)/4 NM = 0.367 nm`
Discussion
22.

Find the difference in the EAN value, if the NO ligand is considered as netural and if NO ligand is considered as positive ligand in [Fe(H_(2)O)_(5)NO]SO_(4) complex

Answer»


Solution :In complex `[Fe(H_(2)O)_(5)NO]SO_(4)` If NO is neutral then oxidation state of Fe will be +2
If NO is POSITIVE then oxidation state of Fe will be +1 So EAN VALUE will be same.
Discussion
23.

Find the difference in EAN of the complexes : [Ti(sigma-C_(5)H_(5))_(2)(pi-C_(5)H_(5))_(2)]" and "[Cr(C_(6)H_(6))_(2)]

Answer»

2
4
zero
3

Solution :ENA `implies [TI^(+4)(sigma-C_(5)H_(2))_2)(pi-C_(5)H_(5))_(2)]`
`=22-4+2xx2+6xx2=34]`
ENA`implies [Cr(C_(6)H_(6))_(2)]=24+6xx2=36`
Discussion
24.

Find the difference in the EAN of the complexes [Fe(CN)_(6)]^(3-)" and "[CuCl_(2)]^(-)

Answer»


SOLUTION :`|(26-3+12)-(29-1+4)]`
Discussion
25.

Find the degree of dissociation of HF in 1 M aqueous solution. The value of K for the ionic equilibrium HF=H^(+)+F^(-) is 7.2xx10^(-4).

Answer»

Solution :HF dissociates in water to form `H^(+)` and `F^(-)` ions. On reaching the equilibrium we have
`HF HARR H^(+)+F^(-)`
Thus one mole of HF taken initially dissociates to yield 1 mole of `H^(+)` and 1 mole of `F^(-)`.
If x be the degree of DISSOCIATION, the concentration TERMS at equilibrium are :
`[HF]=(1-x)"mol/L"`
`[F^(-)]=xmol//L`
`[H^(+)]=x mol//L`
Substituting these values in the equilibrium expression, we have
`K=7.2xx10^(-4)=([H^(+)][F^(-)])/([HF])=((x)/(x))/((1-x))`
If x is very small COMPARED to 1, we can write :
`7.2xx10^(-4)=(x^(2))/(1.00)`
`thereforex=(7.2xx10^(-4))^(1//2)`
Thus the degree of dissociation of HF in 1 M solution is `2.7xx10^(-2)`
Discussion
26.

Find the critical constant ( P_(c ) , V_(c )and T_(c )) in terms of A and B, also find compressibility factor ( z) for the following equation of state. PV= RT-( A)/( V ) + ( 2B) /( V^(2)) where && and B are constant, P =pressure and V=molar volume.

Answer»


ANSWER :`V_(C ) - (6B)/( A), T_(C ) = (A^(2))/( 6RB) , P_(C ) = ( A^(3))/( 108 B ^(2)), ` compressibility FACTOR `= (P_(C ) V_(C ))/( RT_(C ) ) =(1)/(3)`
Discussion
27.

Find the de Broglie wavelength of the electron in the first orbit.

Answer»

SOLUTION :`3.3 XX 10^(-10) m`
Discussion
28.

Find the correct option in which both ligands form same number of rings with central metal ion:

Answer»

`Ox^(2),Cl^(-)`
`en, gly^(-)`
`PY, Br^(-)`
none of these

Solution :In (B) option both form ONE ring with central METAL ION.
Discussion
29.

Find the composition of an acidicbuffer mixture made up of HAAND naA of total molarity 0.29 having ph 4.4 and K_a = 1.8xx 10^(-5) in terms of concentration of salt and acid respectively:

Answer»

0.09 M and 0.20 M
0.20 M and 0.09 M
0.1 M and 0.19 M
0.19 M and 0.10 M

Answer :A
Discussion
30.

Find the correct match out of following :

Answer»

`[Co(NH_(3))_(4)(H_(2)O)Cl]Cl_(2)=" Tetraammineaquachloridocobalt(III) chloride "`
`K_(2)[Zn(OH)_(4)]=" POTASSIUM tetrahydroxidozincate(II) "`
`K_(3)[AL(C_(2)O_(4))_(3)]=" Potassium trioxalatotaluminate(III) "`
`[CoCl_(2)(en)_(2)]Cl=" Dichloridobis(ethane-1, 2-diamine)cobalt(III) chloride "`

Answer :A::B::C::D
Discussion
31.

Find the charge of 48 g of Mg^(2+) ions in coulombs

Answer»

`2.4 XX 10^(23)C`
`6.82 xx 10^5 C`
`3.86 xx 10^5 C`
`1.93 xx 10^5 C`

ANSWER :C
Discussion
32.

Find the change of EAN values when the complex (A) is changing into (B) {:(Na_(2)[Fe(CN)_(5)NO]+Na_(2)S to Na_(4)[Fe(CN)_(5)NOS]),(""(A)"" (B)):}

Answer»


SOLUTION :EAN does not CHANGE.
Discussion
33.

Find the charge of 1 g ion of N^(3-) in coulomb.

Answer»


ANSWER :`2.894xx10^(5)` COULOMB.
Discussion
34.

Find the charge in coulombs on 1 g-ion of N^(3-)

Answer»

SOLUTION :CHARGE on one ion of `N^(3-)=3xx1.6xx10^(-19)`coulombs
One g-ion`=6.02xx10^(23)`ions
`therefore`charge on 1 g-ion of `N^(3-)=3xx1.6xx10^(-19)xx6.02xx10^(23)`
`=2.89xx10^(5)`coulombs
Discussion
35.

Find the charge in coulombs required to convert 0.2 mole VO_(3)^(2-) into VO_(4)^(3-)

Answer»

`1.93 xx 10^(4)`
`9.65 xx 10^(4)`
`1.93 xx 10^(5)`
`9.65 xx 10^(5)`

SOLUTION :CHARGE = `0.2 xx` 1 Faraday = `0.2 xx 96500` coulombs
`= 19300 = 1.93 xx 10^(4)` coulombs .
Discussion
36.

Find the bond enthalpy (in kJ/mol) of one "three centre two electron bond" in B_(2)H_(6) {B-H-B rarr 2B(g)+H(g)} from the given data. {:(DeltaH_(f)^(@) [BH_(3)(g)]=100" kJ/mole",,DeltaH_(f)^(@) [B_(2)H_(6)(g)]=36" kJ/mole"),(DeltaH_("atm")[B(s)]=565" kJ/mole",,DeltaH_("atm")=[H_(2)(g)]=218" kJ/mole"):}

Answer»

SOLUTION :
Discussion
37.

Find the change in magnetic moment when [Ni(H_(2)O)_(6)]^(2+) is converted into [Ni(NH_(3))_(6)]^(2+)

Answer»


Solution :Both of these are `sp^(3)d^(2)` hybridised and outer orbital COMPLEX having configuration of `t_(2G)^(6)e_(g)^(2)`. HENCE, change in magnetic moment is zero.
Discussion
38.

Find the change in oxidation numbers in the following conversion : Cr in K_(2)CrO_(4) rarr K_(2)Cr_(2)O_(7)

Answer»


ANSWER :ZERO
Discussion
39.

Find the bond energy of S-S bond from the following data: {:(C_(2)H_(5)-S-C_(2)H_(5)(g),,,,Delta_(f)H^(@)=-147KJmol^(-1)),(C_(2)H_(5)-S-S-C_(2)H_(5)(g),,,,Delta_(f)H^(@)=-201KJmol^(-1)),(S(g),,,,Delta_(f)H^(@)=222 KJ mol^(-1)):}

Answer»

Solution :Given that
(i) `4C(s)+5H_(2)(g)+S(s)rarrC_(2)H_(5)-S-C_(2)H_(5)(g)""Delta_(F)H^(@)= -147KJ mol^(-1)`
(ii) `4C(s)+5H_(2)(g)+2S(s)rarrC_(2)H_(5)-S-S-C_(2)H_(5)(g)""Delta_(r )H^(@)-201 KJ mol^(-1)`
Substracting Eq.(i) from Eq.(ii), we get
`C_(2)H_(5)-S-C_(2)H_(5)(g)+S(s)rarrC_(2)H_(5)-S-S-C_(2)H_(5)(g)"" Delta_(r )H^(@)=-54KJmol^(-1)`
ADDING to this, the following equation
`S(g)rarrS(s)""Delta_(r )H^(@)=-222 KJ mol^(-1)`
we get
`C_(2)H_(5)-S-C_(2)H_(5)(g)+S(g)rarrC_(2)H_(5)-S-S-C_(2)H_(5)(g)""Delta_(r )H^(@)= -276 KJ mol^(-1)`
In the least equation `276 KJ` of heat evolved because of the `S-S` bond formation.
Hence, the bond enthalpy of `S-S` is `276 KJ mol^(-1)`
Method-2 Diagrammatically, we may represent the above calculation as follows:
According to Hess's LAW
`Delta_(r )H^(@)=` "Enthalpy INVOLVED in bond breaking `Delta_(vap)H^(@)(s)`-Enthalpy involved in bond making
`Delta_(r )H^(@)=(2DeltaH_(C-c)+10DeltaH_(C-H)+2DeltaH_(c-s)+Delta_(vap)H(S)]+(-2DeltaH_(c-c)-10DeltaH_(C-H)-2DeltaH_(c-s)-DeltaH_(s-s)]`
`Delta_(vap)H^(@)(S)-DeltaH_(s-s)`
or `DeltaH_(s-s)=Delta_(vap)H^(@)(S)-DeltaH^(@)`
`=Delta_(vap)H^(@)(S)-[Delta_(r )H^(@)(C_(2)H_(5)-S-S-C_(2)H_(5)]=Delta_(f)H^(@)(C_(2)H_(5)-S-C_(2)H_(5))]`
`=[222-{-201-(-174)}]KJmol^(-1)=276KJ mol^(-1)`
Discussion
40.

Find the anion (s)

Answer»

`SO_(3)^(2-)`
`S_(3)^(2-),S^(2-)`
`SO_(3)^(2-),CO_(3)^(2-)`
`S_(2)O_(3)^(2-)`

SOLUTION :
Discussion
41.

Find the anion (s) :

Answer»


`S_(2)O_(3)^(2-)`

SOLUTION :
Discussion
42.

Find the amount of time expressed in units of t_(1//2)at which A//A_0= 0.125.

Answer»

SOLUTION :`N = 2^(-n) N_0 , n`= no. of `t_(1//2)`
3
Discussion
43.

Find the age of an ancient Egyptian wooden article (in years) from the given information. (i) Activity of 1g of carbon obtained from ancient wooden article =7 counts/min/g (ii) Activity of 1 g carbon obtained from fresh wooden sample =15.4 counts per min/g (iii) Precentage increases in level of C^(14) due to nuclear explosions in past 100 years is 10% (iv) t_(1//2) of ._(6)^(14)C=5770 years

Answer»

`5.770xx10^(3)`
`16.87xx10^(3)`
2488
none of these

ANSWER :A
Discussion
44.

Find sum of stereoisomer of the following compound.

Answer»


Solution :(a) `2^(5) rArr32`
(B) `2^(5) rArr32`
(c ) `a=2^(n-1)=2^(8-1)=2^(7)=128 , m=2^((n//2-1))=2^((8//2-1))=2^((4-1))=2^(3)=8`
(d) `2^(9)=512`
Summing up all stereoisomer: `underset((a))32+underset((b))32+(128+8)+underset((d))512=712`
Discussion
45.

Find structure of compound B:

Answer»




ANSWER :A
Discussion
46.

Find plane of symmerty, centre of symmerty and axis of symmetry (if possible) in the following molecules.

Answer»


ANSWER :(i) POS present and COS, AOS ABSENT.(ii) POS present and COS, AOS absent.
(III) POS, AOS present and COS absent.(iv) POS, COS present and AOS present.
(v) POS present and COS, AOS absent.(vi) POS, COS and AOS absent.
Compound `(##RES_CHM_SIM_E01_014_A01##)` has a single chiral CENTRE which has no plane of symmetry.
Discussion
47.

Find R/S configuration of following compounds.

Answer»


ANSWER :(I) R(II) S(III) R(IV) (R, R)
Discussion
48.

Find pK if K is the overall complex dissociation equilibrium constant for [Cu(NH_(3))_(4)]^(2+) ion, given that beta_(4) for this complex is 2.0xx10^(13) :

Answer»

13.3
12.7
13.7
12.3

Solution :`K=(1)/(beta_(4))=(1)/(2)xx10^(-13)`
`pk=-log K=13+log 2=13.3`
Discussion
49.

Find pH of solution prepared by mixing 100ml, 0.1M Na_(3),PO_(4), and 200m/, 0.1M HCI. P^(k_(a1)),P^(k_(a2))& P^(k_(a3))"of " H_(3)PO_(4) are 3,7 & 10 respectively )

Answer»


ANSWER :5
Discussion
50.

Find out X in the following reaction X + CH_(3) OH overset(conc. H_(2)SO_(4)) to C_(2)H_(5)COOCH_(3) + H_(2)O

Answer»

`C_(2)H_(5) -X`
`(CH_(3)CO)_(2)O`
`CH_(3)COOH`
`C_(2)H_(5)COOH`

Solution :`C_(2)H_(5)COOH+CH_(3)OH overset("pyridine") to C_(2)H_(5)COOCH_(3)+H_(2)O`
Discussion