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.

Consider the following compounds (1)CH_3CH_2CH_2CH_2OH (2)CH_3CH_2CHOHCH_3 (3)(CH_3)_3COH These compounds are dehydrated by treatment with H_2SO_4 .The correct sequence of increasing order of the reactivity of three compounds towards dehydration is

Answer»

3 lt 1 lt 2
1 lt 2 lt 3
2 lt 1 lt 3
1 lt 3 lt 2

Solution :The increasing order of stabilities of CARBONIUM IONS is `1^@ lt 2^@ lt 3^@` . SINCE carbonium ion is formed as an intermediate in the dehydration of ALCOHOL, so EASE of dehydration in the increasing order is `1^@ lt 2^@ lt 3^@` alcohols.
Discussion
2.

Consider the following compounds and Calculate the value of ((P^(2)-Q^(2))/(R+S)) (i)BeF(ii)ICl(iii)BeF_(3)(iv)BrF_(5)(v)ICl_(3)(vi)IF_(3)(vii)IF_(5)(viii)IF_(7), Where P: Total number of polar compounds. Q: Total number of planar compounds. R: Total number of non-polar compounds. S: Total number of non-planar compounds.

Answer»


ANSWER :`6`
Discussion
3.

Consider the following compounds 1. CH_(3)CH_(2)CH_(2)Br "" 2 CH_(2)CHBrCH_(3) 3. (CH_(3))_(3)CBr These compounds are dehydrohalgenated by treatement with strong base under indentical condition. The correct sequence of reactivity of these compounds in the given reaction is

Answer»

`1 gt 2 gt 3`
`2 gt 1 gt 3`
`3 gt 1 gt 2`
`3 gt 2 gt 1`

Solution :As we proceed along a series of alkyl halide from `1^(@) " to" 2^(@) "to" 3^(@)`, th STRUCTURE becomes more branched at the carbon carrying the halogen atom. This increases branching has two result.
i. It providees a greater number of `beta`- hydrogenes for ATTACK by base, and hence, a more favorable probability factor towards elimination.
i It leads to more HIGHLY branched, more stable alkene. Thereofore, in dehydro halogenation, the increasing order of reactivity of alkyl halides is
`3^(@) gt 2^(@) gt 1^(@) gt OR3 gt 2 gt 1 `
Discussion
4.

Consider the following compound (X): Q. When one equivalent of both phenylmagnesium halide and C_(2)H_(6)OH are reacted separately with compound (X), then the correct major products are:

Answer»




Solution :SINCE one EQUIVALENT is taken, G.R. will give acid base reaction with `-COOH` GROUP and acid halide being more reactive than `-COOH` group reacts with alcohol by `S_(N)2Th` mechanism.
Discussion
5.

Consider the following compound : Which of the above compounds reacts with NaHCO_(3)" giving " CO_(2)

Answer»



`CH_(3)OVERSET(O)overset(||)C COOH`

SOLUTION :N/A
Discussion
6.

Consider the following compound A (below) Select the correct statement (s)

Answer»

It is more acidic that `CH_(3)OH`
it is more acidic than `CH_(3)overset(O)overset(||)(C)OH`
it REACTS very fast with LUCAS reagent
It is a diacidic BASE

Solution :`-`Compound which donate `H^(o+)` are ACID.
`-` Acidity of acids increases as stability of conjugate base increases.
`-` DUE to resonance stabilization of conjugate base of compound, it is more acidic than `CH_(3)-OH` & `CH_(3)-COOH`.
Discussion
7.

Consider the following compound 1) tyrocine 2) terephthalic acid 3) adipic acid 4) glucanic acid which can form zwitter ion?

Answer»

only 2
1, 2, 3
only 1
1, 2, 3, 4

Answer :C
Discussion
8.

Consider the following complexies [V(CO)_(6)]^(-),[Cr(CO)_(6)] and [Mn(CO)_(6)]^(+). Then incorrect statement (s) about metal carbonyls is / are

Answer»

C-O' bond is strongest in the cation and weakest in the ANION
C-O' bond order is less in the cation than in anion
C-O' bond LONGER in the cation than in anion or neutral CARBONYL.
M-C' bond order is higher in the cation than in anionic or neutral carbonyl.

Answer :B::C::D
Discussion
9.

Consider the following complexes : (P) K_(2)PtCl(6)""(Q) PtCl_(4).2NH_(3) (R) PtCl_(4).3NH_(3)""(S) PtCl_(4).5NH_(3)

Answer»

256, 0, 97, 404
404, 0, 97, 256
256, 97, 0, 404
404, 97, 256, 0

Solution :`(P)toK_(2)[PtCl_(6)],`
`(Q)to[Pt(NH_(3))_(2)Cl_(4)],`
`(R )to[Pt(NH_(3))_(3)Cl_(3)]Cl,`
`(S)to[Pt(NH_(3))_(5)Cl]Cl_(3)`
Discussion
10.

Consider the following complexes (P) [Cr(CO)_(x)]""(Q) [Cr(CO)_(x-1)PF_(3)] If PF_(3) is better pi acceptor than CO, what will be the order of bond length of CO in complexes (P) and (Q) :

Answer»

`PgtQ`
`QgtP`
`P=Q`
can not be compared

Solution :As `PF_(3)`, is a better `pi`-acceptor than CO, so back BONDING in`pi^(*)` of CO`DARR` THUS, CO B.L. SHORTENS.
Discussion
11.

Consider the following complexes. (P) [Cr(CO)_(5)(PF_(3))]""(Q) [Cr(CO)_(5)(PCl_(3))] Select the correct options.

Answer»

Both P and Q have longer CO BOND LENGTH than free CO molecule.
P has shorter CO bond length than Q.
P has longer Cr-C bond length than Q.
`PF_(3)` is a better `pi-acceptor` than `PCl_(3)`.

Solution :Due to back donation C-O bond length in given compounds is longer than free COmolecule. `PE_(3)` is better `pi`-acceptor than CO, hence back donation is more favoured towards `PF_(3)`rather than CO and `PF_(3)` is is better `pi`-acceptor than `PCl_(3)` hence `pi`-donation is more favourable in (P) than(Q). Hence,CO length in (P) is shorter than (Q). Due to LESS `pi`-donations towards CO or C bond is longer in (P) than (Q).
Discussion
12.

Consider the following complexes: (i) [FeIF(CN)(H_(2)O)(en)] (ii) [MoCl_(2)F_(2)(gly)]^(2-) Then, calculate value of |x-y| (where x and y are total number of possible optically active isomers in (i) and (ii) complex respectively).

Answer»


Solution :(i) `[FeIR(CN)(H_(2)O)(en)]`
x=12 (active isomer)
(ii) `[MoCl_(2)F_(2)(GLY)]^(-)`
y=4 (active isomer)
x-y=12-4
Discussion
13.

Consider the following complexes : [NiCl_(4)]^(2-), [Ni(CN)_(4)]^(2-), [Ni(CO)_(4)]^(0) What is the similarity in all three complexes ?

Answer»

Oxidation NUMBER and EAN of METAL
Geometry and hybridisation
Magnetic nature and number of `SIGMA` bonds
COORDINATION number and TYPE of metal

Answer :D
Discussion
14.

Consider the following complexes compounds (i) [Cu(NH_(3))_(4)][Cu(NO_(2))_(4)] (ii) [Cr(py)^(6)][Cr(SCN)_(6)] (iii) [Co(NH_(3))_(5)(NO_(2))][Pt(SCN)_(4)] If x_(1),x_(2),x_(3) are more possible coordination isomers of given complex compounds respectively then calculate value of x_(1)^(2)+x_(2)^(2)+x_(3)^(2).

Answer»


ANSWER :41
Discussion
15.

Consider the following complexes 1) K_2PtCl_62) PtCl_4 . 2NH_3 3) PtCl_4 . 3NH_3 4) PtCl_4. 5NH_3 Their electrical conducatance in aqueous solutions are

Answer»

256, 0, 97 and 404 RESPECTIVELY
404, 0, 97 and 256 respectively
256, 97, 0 and 404 respectively
404, 97, 256 and 0 respectively

Answer :A
Discussion
16.

Consider the following complexes : 1. Fe(CO)_(5) 2. Ni(CO)_(4) 3. K_(4)[Fe(CN)_(6)] 4. Na_(3)[Cu(SCN)_(4)] Which of the above complexes obey EAN rule ?

Answer»

1, and 2 only
2, 3 and 4 only
1, 2, 3 and 4
none of the above

Solution :All the GIVEN COMPLEXES follow EAN rule.
Discussion
17.

Consider the following complex ions, P, Q and R P=[FeF_(6)]^(3-), Q=[V(H_(2)O)_(6)]^(2+) and R=[Fe(H_(2)O)_(6)]^(2+) The correct order of the complex ions, according to their spin only magnetic moment (in B.M.) is

Answer»

`RltQltP`
`QltRltP`
`RltPltQ`
`QltPltR`

SOLUTION :`P=[FeF_(6)]^(3-)` has `Fe^(3+)`, i.e., `3d^(5)` (5 unpaired electron)
`Q=[V(H_(2)O)_(6)]^(2+)` has `V^(2+)`, i.e., `3d^(3)` (3 unpaired electrons)
`R=Fe(H_(2)O)_(6)]^(2+)` has `Fe^(2+)`, i.e., `3d^(6)` (4 unpaired electrons)
Hence, the order of spin only magnetic MOMENT is `QltRltP`.
Discussion
18.

Consider the following complex ions. [CrCl_(6)]^(3-)""[Cr(H_(2)O)_(6)]^(3+) [Cr(NH_(3))_(6)]^(3+)""[Cr(CN)_(6)]^(3-) The correct order of Delta_(0) is :

Answer»

<P>`P GT Q gt Rgt S`
`P LT R lt Q lt S`
`P lt Q lt R lt S`
`P lt Q lt S lt R`

Solution :`"Strength of ligand" prop Delta_(0),CN^(-)gtNH_(3)gtH_(2)OgtCl^(-)`
Discussion
19.

Consider the following complex ions, P, Q and R. P=[FeF_(6)[^(3-),Q=[V(H_(2)O)_(6)]^(2+) and R=[Fe(H_(2)O)_(6)]^(2+) The correct order of the complex ions, according to theirspin-only magnetic moment values (in B.M.) is

Answer»

`R lt Q lt P`
`Q lt R lt P`
`R lt P lt Q`
`Q lt P lt R`

Solution :The electronic configuration of central ion in complex IONS P, Q and R are

Higherthe no. of unpaired ELECTRON(s), higher will be magnetic moment. THUS the correct order of SPIN only magnetic moment is QltRltP.
Discussion
20.

Consider the following complex entities. [FeF_(6)]^(3-), [Fe(H_(2)O)_(6)]^(2+), [Mn(CN)_(6)]^(4-), [Co(gly)_(3)], [Co(EDTA)]^(-), [Pt(NH_(3))_(2)Cl_(4)], [PtCl_(2)(en)_(2)]^(2+) If W, X, Y and Z are the total number of diamagnetic, paramagnetic, capable of exhibiting stereoisomerism and coloured complex entities respectively, find WXYZ. If W,X,Y and Z are 2, 4, 3, 6 respectively, write your answer as 2436.

Answer»


Solution :`W to [CO(gly)_(3),[Co(EDTA)]^(-),[Pt(NH_(3))_(2)Cl_(4)],[PtCl_(2)(EN)_(2)]^(2+)`
`X to FeFe_(6)^(3-),Fe(H_(2)O)_(6)^(2+),Mn(CN)_(6)^(4-)`
`Y to [Co(gly)_(3)],[Co(EDTA)]^(-)[Pt(NH_(3))_(2)Cl_(4)],[PtCl_(2)(en)_(2)]^(2+)`
`Z to "Except" FeF_(6)^(3-)`, all are coloured.
Discussion
21.

Consider the following complex [Co(NH_3)_5CO_3]ClO_4 The coordination number, oxidation number, number of d-electrons and number of unpaired d-electrons, respectively, on the metal are

Answer»

6,3,6,0
7,2,7,1
7,1,6,4
6,2,7,3

Answer :A
Discussion
22.

Consider the following complex: [Co(NH_(3))_(5)CO_(3)]ClO_(4) The coordination number, oxidation number, no. of d-electrons an dnumber of unpaired d-electrons on the metal are respectively:

Answer»

6,2,7,3
7,2,7,1
5,3,6,4
6,3,6,0

Answer :D
Discussion
23.

Consider the following complex [Co(NH_(3))_(5),CO_(3)]CiO_(4). The coordination number, oxidation number, number of d-electrons and number of unpaired d-electrons on the metal are respectively.

Answer»

6,3,6,0
7,2,7,1
7,1,6,4
6,2,7,3

Solution :`[CO(NH_(3))_(5)CO_(3)]ClO_(4)`. SINX monodentate ligands are attached to Co HENCE C.N. of Co=6.
O.N.`=x+5xx(0)+1xx(-2)+1xx(-1)=0therefore x=+3`, electronic configuration of `Co^(3+)[AR]3d^(6)4s^(0)` hence number of d electrons is 6. all d-electrons are paired due to strong ligand hence unpaired electron is zero.
Discussion
24.

Consider the following comparisons of basic nature of different compound Identify correct comparision

Answer»

I
II
III
IV

Solution :RESONANCE `UARR` basic NATURE `darr`
Discussion
25.

Consider the following complex compounds: (i) [Pt(NH_(3))_(2)(SCN)_(2)] (ii) [Co(NH_(3))_(3)(NO_(2))_(3)] (iii) [Pt(en)Cl_(2)] (iv) [Cr(en)_(2)Br_(2)]^(+) (v) [Rn(en)_(3)]^(3+) (vi) [CoCl_(2)Br_(2)] Then calculate sum of total number of geometrical isomers in all above complex compounds.

Answer»


ANSWER :6
Discussion
26.

Consider the following: [Co(CO_(3))(NH_(3))_(5)]ClO_(4) mark the correct option:

Answer»

`{:("COORDINATION no.","OXIDATION no.","No. of d-electrons","UNPAIRED d-electron"),(6,3,6,0):}`
`{:("Coordination no.","Oxidation no.","No. of d-electrons","Unpaired d-electron"),(7,2,7,1):}`
`{:("Coordination no.","Oxidation no.","No. of d-electrons","Unpaired d-electron"),(7,1,6,4):}`
`{:("Coordination no.","Oxidation no.","No. of d-electrons","Unpaired d-electron"),(6,2,7,3):}`

ANSWER :A
Discussion
27.

Consider the following changes : M_((s))toM_((g)) "" ……..(1) M_((s)) +M_((g))^(2+) +2e^("ϴ")""……..(2) M_((g))to M_((g))^(+) +e^("ϴ") ""………(3) M_((g))^(+) to M_((g))^(+2) +e^("ϴ") "".........(4) M_((g)) to M_((g))^(2+) +2e^("ϴ")"".........(5) The second ionization energy of M_((g)) could be calculated from the energy values associated with :

Answer»

`1+3+4`
`2-1+3`
`1+5-3`
`5-3`

Solution :`M_((G)) UNDERSET(IE_(1)) overset(E^(-)) to M_((g))^(+) underset(IE_(II)) overset(e^(-))to M_((g))^(+2)`
`M_((g)) toM_((g))^(+2) +2e^(-)`
`M_((g)) to M_((g))^(+) +e^(-)`
--------------------------------------
`""to M_((g))^(+2) - M_((g))^(+) +e^(-)`
or `M^(+) to M^(+2) +e^(-)`
Discussion
28.

Consider the following chemical reaction HCOOH(aq.)overset(X, Delta)rarr CO(g)+H_(2)O(g) Formic acid In the above reaction (X) is :

Answer»

Conc. `H_(2)SO_(4)`
Conc. HCl
Conc. `HNO_(3)`
Dil. `HNO_(3)`

Solution :Bond order `ALPHA` STABILITY
`{:(""N_(2)gt N_(2)^(+)gt N_(2)^(-)gt N_(2)^(-2)),("B.O. 32.52.52"):}`
`N_(2)^(-)` is less stable than `N_(2)^(+)` due to presence of one `E^(-)` in ABMO (more energy orbital)
Discussion
29.

Consider the following cell reaction: Fe_((s))+O_(2(g))+4H_((aq))^(+)to2Fe_((aq))^(2+)+2H_(2)O_((l)) E^(@)=1.67V At[Fe^(2+)]=10^(-3)M,P_((O_(2)))^(0)=0.1atm and pH=3, the cell potential at 25^(@)C is

Answer»

<P>1.47V
1.77V
1.87V
1.57V

Solution :`E=E^(o)-(0.059)/(4)"log"([Fe^(2+)]^(2))/([H^(+)]^(4)P_(O_(2)))`
`=1.67-(0.06)/(4)"log"((10^(-3))^(2))/((10^(-3))^(4)xx0.1)=1.67-(0.03)/(2)"log"10^(7)`
`=1.67-(0.03)/(2)xx7=1.67-0.105=1.565=1.57V`
Discussion
30.

Consider the following cell reaction : 2Fe_((s)) + O_(2(g)) + 4H_((aq))^(+) to 2Fe_((aq))^(2+) + 2H_2O_((l)), E^(0) = 1.67 V At[Fe^(2+)] = 10^(-3) M, P_((O_2)) = 0.1 atmand pH=3 , the cell potential at25^(@) Cis

Answer»

`1.47 V `
`1.77 V `
`1.87 V `
`1.57 V `

ANSWER :D
Discussion
31.

Consider the following cell reaction : 2Fe_(s) + O_(2(g)) + 4H_((aq.))^(+) rarr 2Fe_((aq.))^(2+) + 2H_(2)O_((l)), E^(@) = 1.67 V At[Fe^(2+)] = 10^(-3)M, P(O_(2)) = 0.1 atm and pH = 3, the cell potential at 25^(@)C is:

Answer»

`1.47 V`
`1.77 V`
`1.87 V`
`1.57 V`

SOLUTION :`2Fe_(s) + O_(2(g)) + 4H_((AQ.))^(+) rarr 2Fe_((aq.))^(2+) + 2H_(2)O_((L))`
`n = 4` (no. of moles of electron involved)
From Nernst's equation,
`E_(cell) = E_(cell)^(@) - (0.059)/(n)log Q`
`= 1.67 - (0.0591)/(4)log'((10^(-3))^(2))/(0.1 XX (10^(-3))^(4))`
`= 1.67 xx 0.106`
`= 1.57 V`
Discussion
32.

Consider the following cell reaction. 2Fe(s)+O_(2)(g)+4H^(+)(aq)rarr2Fe^(2+)(aq)+2H_(2)O(l), E^(@)=1.67V At [Fe^(2+)]=10^(-3)M,P(O_(2))=0.1 atm and pH=3, the cell potential at 25^(@)C is

Answer»

`1. 77 V`
`1.57 V`
`1. 87 V`
`1.47 V`

Answer :B
Discussion
33.

Consider the following cell reaction: 2Fe(2)+O_(2)+4H^(+)(aq)to2Fe^(2+)(aq)+2H_(2)O(l),E^(@)=1.67V at [Fe^(2+)]=10^(-3)M,P(O_(2))=0.1 atm and pH=3, the cell potential at 25^(@)C is,

Answer»

1.47
1.77
1.87
1.57

Solution :Applying nernst equation to the given reaction,
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([FE^(2+)]^(2))/(P_(O_(2))XX[H^(+)]^(4))`
Fot the given reaction, n=4 and pH=3 MEANS `[H^(+)]=10^(-3)M`
`thereforeE_(cell)=1.67-(0.0591)/(2)"log"((10^(-3))^(2))/(0.1xx10^(-3))^(4)`
`=1.67-(0.0591)/(4)log10^(7)=1.67-0.10=1.57V`
Discussion
34.

Consider the following cell reaction: 2Fc(s) +O_(2)(g) +4H^(+) (aq)to 2Fe^(2+) (aq)+ 2H_(2) O (l)E ^(@)=1.67V At [Fe^(2+)]=10^(-3)M, p(O_(2))=0.1 atm and pH =3, the cell poteintial at 25^(@)C is

Answer»

`1.47 V`
`1.77 V`
`1.87 V`
`1.57 V`

Solution :Here `n=4, and [H^(4)]=10^(-3)(aspH=3)`
Applying Nernst equation
`E=E^(@)-(0.059)/(n)log""([FE^(2+)]^(2))/([H^(+)]^(4) (po_(2)))`
`=1.67 -(0.059)/(4) log ""((10^(-3))^(2))/((10^(-3))^(4) xx0.1)`
`1.67 (0.059)/(4) log 10^(7) =1.67 =0.103=1.567V`
Discussion
35.

Consider the following carbonyl complex compounds. (i) Mo(CO)_(x) (ii) H_(y)[Cr(CO)_(5)] and (iii) Ru_(3)(CO)_(s) Then calculate value of |x+y-x|.

Answer»


ANSWER :8
Discussion
36.

Consider the following bromides The correct order of S_N1 reactivity is

Answer»

` A GT B gt C`
` B gt C gt A `
` B gt A gt C `
` C gt B gt A `

ANSWER :B
Discussion
37.

Consider the following bromides: The correct order towards S_(N^(1)) reactivityis

Answer»

BgtCgtA
BgtAgtC
CgtBgtA
AgtBgtC

Solution :the order of reactivity is based on the stability of the carbocation which are formed in all the three cases in `S_(N^(1))` MECHANISM. These are:
.
Both B and C are secondary carbocations but (B) is more STABLE tha (C) since it is allylic in nature.
.
Discussion
38.

Consider the following Born-Haber's cycle: (Where DeltaH_(1),DeltaH_(2),DeltaH_(3),DeltaH_(4),DeltaH_(5) and DeltaH_(6) are in kJ/mol) Then according to given information the correct statement is/are:

Answer»

`DeltaH_(2) and DeltaH_(3)` are always POSITIVE
`DeltaH_(1)=2DeltaH_(2)+(3)/(2)DeltaH_(3)+2DeltaH_(4)+3DeltaH_(5)+DeltaH_(6)`
Second ELECTRON gain enthalpy of X is NEGATIVE
`DeltaH_(1)` must be negative for formationn of `B_(2)X_(5)`(s)

Solution :`impliesDeltaH_(1)=2DeltaH_(2)+(3)/(2) DeltaH_(3)+2DeltaH_(4)-3DeltaH_(5) implies`second electron gain enthalpy of X will be positive.
Discussion
39.

Consider the following Borh-Habber's cycle for formation of MX_(3)(s). Then calculate value (q_(1))/(50), here q_(1) is electron affinity of X(g) in kJ/mol.

Answer»


SOLUTION :`-750=150+(3)/(2)xx200+350+q-1000`
`q=3xxDeltaH_(BG)=-550,`
`q_(1)=(550)/(3)=183,(q_(1))/(50)=3.66~~4.0`
Discussion
40.

Consider the following bond angles alpha= O-O-O inozone, beta=P-P-P in P_(4) (white), gamma=N-N-N in azide anion (N_(3)^(-)), delta=C-C-C diamond, then

Answer»

`alpha+beta=gamma`
`beta+delta gt gamma `
`(delta)/(beta) gt (gamma)/(delta)`
`gamma-alpha=alpha-beta`

Solution :
`=(109^(@)28.)/(60)=(180)/(109^(0)28.)`
Discussion
41.

Consider the following balanced chemical equation: 2A+4B_(2)+6C_(3)rarrP+2Q If initially 6.023xx 10^(24) atoms orf A, 448L of B_(2) gas at NTP and 960 gm of C_(3) gas are taken, which of the following is incorrect? [Given : Atomic mass of C=8]

Answer»

total number of species in final mixture is 4
total number of species in final mixture is 3
A and `B_(2)` both are present in limited amount amount.
`C_(3)` is in excess.

Solution :`n_(A)=10`
`n_(B)=20`
`n_(C)=(960)/(24)=40`
`2A+4B_(2)+6C_(3)rarrP+2Q`
Limiting reagent `rarrA & B_(2)`
so only 3 species will be left in final mixture.
Discussion
42.

Consider the following Baeyer-Villiger oxidation.

Answer»

a.`EWG` in PERACID facilitates the reaction.
b.Strong `bar(e)`-dobating group migrates.
c.The MIGRATING group order is `3^(@) "alkyl" gt "Phenyl" gt H gt 2^(@) "alkyl" gt 1^(@) "alkyl" gt Me`.
d.The migrating group order of SUBSTITUTED phenyl group is

Solution :All the statements are explanations in themselves.
Discussion
43.

Consider the following atomic orbitals : Which of the following statement(s) is /are correct regarding given orbital ?

Answer»

It is a GERADE atomic orbital
It has zero nodal plane
CIRCULAR electron density is PRESENT in XY plane
Opposite lobes of orbital have same SIGN of wave function `(psi)`

Solution :It has two angular nodes modified in shape of nodal cones.
ONE lobe is dumb-bell shape while the other is in the shape of circular ring (in XY plane)
Discussion
44.

Consider the following attraction (a) HCl------HCl (b) CH_4------CH_4 (c) NH_4^(+) ------- H_2O (d) He------He The correct order of strength of given attraction is

Answer»

C GT B gt a gt d
c gt a gt b gt d
c gt d gt a gt b
c gt b gt d gt a

ANSWER :B
Discussion
45.

Consider the following arrangement volume of liquid in container I and II at equlibrium are:

Answer»

I=100 ML, II=100mL
I=200mL,II=0mL
I=0 mL, II=200mL
I=110 mL, II=90mL

Answer :B
Discussion
46.

Consider the following aqueous solution

Answer»

`0.1M` GLUCOSE solution
`0.1M Na_(2)SO_(4(AQ))`
`0.1MNaCl_((aq))`
`0.1MNa_(3)PO_(4(aq))`

Solution :`DeltaT_(B)prop` No. of moles of solute PARTICLES in aqueous solution.
Discussion
47.

Consider the following antibiotics (i) Erythormycin(ii) Ofloxacin (iii) Chloroamphenicol(iv) Penicillin The pair of bactericidal antibiotics is

Answer»

i - iii
ii - iv
iii- iv
i - iv

Answer :B
Discussion
48.

Consider the following alkyl halides 1. (CH_3)_3 "CCH"_2Br 2. ClCH_2CH=CH_2 3. ClCH_2 CH_2 CH_3 4. BrCH_2CH_2CH_3 Arrange these alkyl halides in decreasing order of reactivity in Williamson reaction.

Answer»

`2 GT 1 gt 3 gt 4`
`2 gt 1 gt 4 gt 3`
`1 gt 2 gt 3 gt 4`
`2 gt 4 gt 3 gt 1`

ANSWER :D
Discussion
49.

Considerthe followingamines(1) n-butyl amine (2) ethyl dimethylamine (3) diethyl dimethylamine the correctsequenceof boilingpoint is

Answer»

`1 GT3 gt 2`
`1 gt 2 gt 3`
`2 gt 3 gt 1`
` 2 gt1 gt 3`

Answer :A
Discussion
50.

Consider the following acids (1) CH_(3) COOH ""(2) Cl_(2) CHCOOH (3) ClCH_(2)COOH""(4) Cl_(3)C COOH The correct sequence of acidity is

Answer»

`1 gt 4 gt 3 gt 2`
` 1 gt 3 gt 2 gt 4`
` 1 gt 2 gt 3 gt 4`
` 4 gt 2 gt 3 gt 1`

Solution :Inductive effect diminishes with increases in the number of `sigma`- bonds through which the effect operates. This MAKES `ClCH_(2)COOH` more ACIDIC than `ClCH_(2)CH_(2)COOH`. An electronegative atom Cl is an electron withdrawing group and acid strengthener.
THEREFORE, `ClCH_(2)COOH` is a stronger acid than `CH_(3)COOH` . Two chlorine atoms are more electron withdrawing than single Cl atom. So, `Cl_(2)CHCOOH` stronger acid than `ClCH_(2)COOH`.
Discussion