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 brown ring formed in the test of nitrates with FeSO_(4) solution and concentrated H_(2)SO_(4) has the formula ….. .

Answer»

SOLUTION :`[FE(H_(2)O)_(5)NO]^(2+)`
Discussion
2.

The brown ring less test is performed for the qualitative detection of

Answer»

Bromides
Iodides
Nitrates
Nitrite

Answer :C,d
Discussion
3.

The brown ring complex is formulated as [Fe(H_2O)_(5)NO^(+)]SO_4The Oxidation state of Fe is X what is the value of X?

Answer»


Solution :`[FE(H_2O)_5no]SO_4 RARR x = x+ 0 +1 - 2=0 rArr x =1`
Discussion
4.

The brown ring complex compound is formulated as [Fe(H_2O)_5(NO)]SO_4. The oxidation state of iron is:

Answer»

`+1`
`+2`
`+3`
`+4`

ANSWER :A
Discussion
5.

The brown ring complex compound is formulated as [Fe(H_(2)O)_(5)(NO)]SO_(4). The oxidation state of iron is :

Answer»

`+1`
`+2`
`+3`
0

Answer :B
Discussion
6.

The brown ring complex compound is formulated as [Fe(H_2O)_5NO]SO-4 The oxidation states of iron is

Answer»

`+1`
`+2`
`+3`
`+4`

SOLUTION :`[Fe(H_2O)_5NO]SO_4 hArr X + 5(0) + 1-2 = 0 rArr x =1`
Discussion
7.

The brown ring complex compound is formu = lated as [Fe(H_(2)O)_(5)NO]SO_(4). Theoxidation state of iron is :

Answer»

`+1`
`+2`
`+3`
0

Answer :B
Discussion
8.

The brown righ test for NO_(2)^(-) and NO_(3)^(-) is due to the formation of complex ion with a formula

Answer»

`[FE(H_(2)O)_(6)]^(2+)`
`[Fe(NO)(CN)_(5)]^(2+)`
`[Fe(H_(2)O)_(5)NO]^(2+)`
`[Fe(H_(2)O)(NO)_(5)]^(2+)`

Solution :`[Fe(H_(2)O)_(5)NO]^(2+)` ION is formed
Discussion
9.

The brown compound formed in the ring test for nitrates contains the ion

Answer»

`[Fe(H_(2)O)_(5)NO]^(3+)`
`[Fe(H_(2)O)_(5)NO]^(2+)`
`[Fe(H_(2)O)_(5)NO]^(4+)`
`[Fe(H_(2)O)_(5)NO]`

ANSWER :B
Discussion
10.

The brown precipitate formed in step C consists of

Answer»

`HG(NH_(2)I`
`Hg(NH_(2))NO_(3)`
`HgO, Hg(NH_(2II))I`
`Ag_(2)O`

Solution :C) brown precipitate DUE to `HgO.Hg(NH_(2))I` WSIT me due to the test.
Discussion
11.

The brown gas formed when HNO_(3) is reduced by metals is

Answer»

`N_(2)O`
`N_(2)O_(3)`
`NO_(2)`
NO

ANSWER :C
Discussion
12.

The brown complex obtained in the detection of nitrate radical is formulated as [Fe(H_2 O)_5 NO]SO_4. What is the oxidation number of Fe in this complex?

Answer»

`+1`
`+2`
`+3`
0

Solution :Thebrowncomplexobtained in detectionofnitrateradical is
`[Fe(H_2 O)_5 NO [ SO_4]or [Fe(H_2O)_5NO]^(2+) SO_(4)^(2-)`
thecolourof thecomplexis dueto changetransfer. Thiscomplexformallycontains`Fe(+1)`and `NO^(+)`sinceFe (II)gainsone electron fromNOwhichchangesto `NO^(+)`
Discussion
13.

The brown colour of ring in NO_(3)^(-) test is of

Answer»

`FeSO_(4)`
`[FE(H_(2)O)_(5)(NO)]SO_(4)`
`FeSO_(4).NO_(2)`
none

Answer :B
Discussion
14.

The Bronsted acids in the reversible reaction are HCO_(3)^(-)(aq.) + OH^(-) (aq.) CO_(3)^(2-)(aq.) + H_(2)O

Answer»

`OH^(-)` and `CO_(3)^(2-)`
`OH^(-)` and `H_(2)O`
`HCO_(3)^(-)` and `H_(2)O`
`HCO_(3)^(-)` and `CO_(3)^(2-)`

Answer :C
Discussion
15.

The bronstad acid which gives the weakest conjugate base is :

Answer»

HF
`H_2S`
`H_2O`
HCl

Answer :D
Discussion
16.

The bromination of anisole is carried out in the presence of…….

Answer»

`CH_(3)COCH_(3)`
`CH_(3)COOH`
`CH_(3)CH_(2)OH`
`CH_(3)CN`

Solution :`CH_(3)COOH`
Discussion
17.

The bromination of aniline produces

Answer»

2- bromoaniline
4-bromoaniline
2, 4, 6-tribromoaniline
2, 6-dibromoaniline

Solution :
Discussion
18.

The bromination of acetone which occurs in acid solution is represented by the equation: CH_(3)COCH_(3)(aq) + Br_(2)(aq) to CH_(3)COCH_(2)Br(aq) + H^(+)(aq) + Br^(-1)(aq) The following kinetic data was obtained for the given reaction concentration: Initial rates of disappearance of of Br_(2)(Ms^(-1) 5.7 xx 10^(-5), 5.7 xx 10^(-5) 1.2 xx 10^(-4), 3.1 xx 10^(-4) Based on these data, the rate equation is:

Answer»

RATE = `k[CH_(3)COCH_(3)][Br_(2)][H^(+)]^(2)`
Rate = `k[CH_(3)COCH_(3)][Br_(2)][H^(+)]`
Rate = `k[CH_(3)COH_(3)][H^(+)]`
Rate = `k[CH_(3)COCH_(3)][Br_(2)]`

Solution :c) From the first two experiments, it is clear that when the concentration of `Br_(2)` is doubled, the initial rate of disappearance of `Br_(2)` remains unchanged. This means that order of reaction with RESPECT to `Br_(2)` is zero. The PROBABLY rate LAW for the reaction will be:
`k[CH_(3)COCH_(3)][H^(+)]`
Discussion
19.

The bromination of acetone that occurs in acid solution is represented by this equation. CH_(3)COCH_(3)Br(aq)+Br_(2)(aq) to CH_(3)COCH_(2)Br(aq)+H^(+)(aq)+Br^(-)(aq) These kinetic data were obtained for given reaction concentrations. Initial concentrations, M {:([CH_(3)COCH_(3)],[Br_(2)],[H^(+)],"Initial rate, disap"),(,,,"pearance of " Br_(2)),(0.30,0.05,0.05,5.7xx10^(-5)),(0.30,0.10,0.05,5.7xx10^(-5)),(0.30,0.10,0.10,1.2xx10^(-4)),(0.40,0.05,0.20,3.1xx10^(-4)):} Based on these data, the rate equation is

Answer»

`"RATE"=k[CH_(3)COCH_(3)][Br_(2)]`
`"rate"=k[CH_(3)COCH_(3)][Br_(2)][H^(-)]^(2)`
`"rate"=k[CH_(3)COCH_(3)][Br_(2)][H^(-)]`
`"rate"=k[CH_(3)COCH_(3)][H^(+)]`

Answer :D
Discussion
20.

The bromination of acetone that occurs in acid solution is represented by this equation CH_(3)COCH_(3) (aq) + Br_(2) (aq) to CH_(3) COCH_(2) Br(aq) + H^(+) (aq) + Br^(-) (aq) These kinetic data were obtained for given reaction concentrations .Initial concentrations , M {:([CH_(3)COCH_(3)] ,, [Br_(2)] ,, [H^(+)]) , (0.30 ,, 0.05 ,, 0.05) , (0.30 ,, 0.10 ,, 0.05) , (0.30 ,, 0.10 ,, 0.10) , (0.40 ,, 0.05 ,, 0.20):} Initial rate , disappearance of Br_(2), Ms^(-1)

Answer»

`5.7 xx 10^(-5)`
`5.7 xx 10^(-5)`
`1.2 xx 10^(-4)`
`3.1 xx 10^(-4)`

SOLUTION :In experiment(c) , rate doesn't depend on the concentration of `Br_(2)` . So , rate EXPRESSION will not include `[Br_(2)]`.
Discussion
21.

The bromination of acetone that occurs in acid solution is represented by CH_(3)COCH_(3)(aq.)+Br_(2)(aq.)to CH_(3)COCH_(2)Br(aq.)+H^(+)(aq.)+Br^(-)(aq.) These kinetic data were obtained for given reaction concentration Initial concentration {:("Initial concentration",,,"Initial rate of"),(,,,"disappearance"),([CH_(3)COCH_(3)],[Br_(2)],[H^(+)],"of "Br_(2)),(0.30,0.05,0.05,5.0xx10^(-5)),(0.30,0.10,0.5,5.0xx10^(-5)),(0.30,0.10,0.10,1.0xx10^(-4)),(0.90,0.05,0.10,3.0xx10^(-4)):} Based on these data, order of reaction will be -

Answer»


Solution :From (1) and (2),
`[CH_(3)COCH_(3)]` and `[H^(+)]` = constant
`[Br_(2)]` = doubled
RATE = UNCHANGED
`therefore` Order w.r.t `Br_(2) = 0`
POSSIBLE rate LAW (By option)
`r = k [CH_(3)COCH_(3)][H^(+)]`
Discussion
22.

The broad spectrum antibiotic among the following is :

Answer»

Chloramphenicol
PARACETAMOL
Bithional
Morphine.

Solution : Chloramphenicol is broad SPECTRUM ANTIBIOTIC.
Discussion
23.

The B.P. of alcohols are much higher than the ethers of comparable molecular masses due to

Answer»

interamolecular H - BONDING
intermolecular H - bonding
dipole - dipole ATTRACTION
Hitler - London FORCES

ANSWER :B
Discussion
24.

The branch of chemistry which with the reaction rates and reaction mechanism is called

Answer»

THERMOCHEMISTRY
PHOTOCHEMISTRY
analytical CHEMISTRY
chemical kinetics

Solution :It is the DEFINITION of chemical kinetics.
Discussion
25.

The branch of chemistry which deals with the reaction rates and. reaction mechanism is called :

Answer»

THERMOCHEMISTRY
PHOTOCHEMISTRY
both a and b
Chemical kinetics

Answer :D
Discussion
26.

The b.p. of a solution of 5g of sulphur in 100g of carbon disulphide is 0.476^(@) above that of pure solvent. Determine the molecular formula of sulphur in this solvent. The b.p. of pure carbon disulphide is 46.30^(@)C and its heat of vaporisation is 84.1 calories per gram.

Answer»

Solution :We have, `K_(b)=(RT^(2))/(1000l_(v))=(2xx(46.30+272)^(2))/(1000xx84.1)=2.425`
`:.m=(DeltaT_(b))/(K_(b))=(0.476)/(2.425)`
And also `m=("moles of solute")/("wt.of solvent in grams")xx1000`
`=(5)/(M)xx(1000)/(100)=(50)/(M)`
where M is the mol. wt. of SULPHUR
Thus, `(50)/(M)=(0.476)/(2.425)`, `M=255`
Since the atomic wt. of sulphur is 32 and its mol.wt. is `255`, therefore, number of sulphur atoms associated to form a SINGLE molecule is `(255)/(32)=7.96` , i.e., 8. Hence the formula is `S_(8)`.
Discussion
27.

The Bouveault-Blanc reduction involves

Answer»

`C_(2)H_(5)OH//Na`
`LiAIH_(4)`
`C_(2)H_(5)MgX^(-)`
`Zn//HCl`

SOLUTION :`underset("Ethyl butyrate")(C_(3)H_(7)COOC_(2)H_(5))overset(Na//C_(2)H_(5)OH)to underset("Butyl alcohol")(C_(3)H_(7)CH_(2)OH)`
Discussion
28.

The Born -Haber cycle for formation of rubidium chloride (RbCl) is given below (the enthalpies are in K Cal mol"^(-1) ) Find the value of X.

Answer»


SOLUTION :`-105=20.5+96+28.75-159.5+x`
`x=-90.75" KCAL MOL"^(-1)`
Discussion
29.

The borax bead is chemically :

Answer»

`B_2O_3 +NaBO_2`
`Na_2B_4O_7`
`Na_3BO_3`
`B_2O_3`

ANSWER :A
Discussion
30.

The boron, carbon, nitrogen and oxygen have their increasing ionisation enthalpy in the order

Answer»

N,O,B,C
B,C,O,N
B,N,C,O
O,B,C,N

Answer :B
Discussion
31.

Thebonds presentin thestatemetn is/aretrue?

Answer»

FOUR equivalentCr-O BONDS only.
six equivalent CR-O bonds andone O- O bond
Eight equivalentCr-O bonds and ONE Cr-Cr bond.
six equivalentCr-O bonds andone Cr-O- Cr bonds

Solution :
Six equivlent Cr-O bonds and one Cr-O-Cr bond.
Discussion
32.

The bonds present inN _2O_5 are :

Answer»

Only ionic
Only COVALENT
Covalent and COORDINATE
Covalent and ionic

Answer :C
Discussion
33.

The bonds present in the structure of dichromate ion are

Answer»

four equivalent CR - O bonds only
Six equivalent Cr- O bonds and one O-O BOND
six equivalent Cr-O bonds and one Cr -Cr bond
six equivalent Cr -O bonds and one Cr - O - Cr bond

Solution :IN `Cr_(2)O_(7)^(2-)`

There ar six equivalent Cr-O bonds and one Cr-O-Cr bond.
Discussion
34.

The bonds present in N_(2)O_(4) are

Answer»

only IONIC
only COVALENT
covalent and coordinate
ionic and coordinate

Answer :C
Discussion
35.

The bonds present in N_(2)O_(5) are :

Answer»

Only IONIC
Only COVALENT
Covalent and ionic
Covalent and COORDINATE

ANSWER :D
Discussion
36.

The bonds is K_(4)[Fe(CN)_(6)] are

Answer»

all IONIC
all covalent
ionic, covalent and coordinate
ionic and covalent.

Solution :Ionic bond between `4K^(+)` and `[Fe(CN)_(6)]^(4-)` coordinate bond between `Fe^(2+)` and `6CN^(-)` covalent bond between C and N to form `C-=N^(-)`
Discussion
37.

The bonds formed by transition metals in lower oxidation states are .......... while those formed in higher oxidation states are ............

Answer»

SOLUTION :IONIC, covauent
Discussion
38.

The bond that determines the secondary structure of proteins is ……. .

Answer»

COORDINATE bond
covalent bond
HYDROGEN bond
peptide bond

ANSWER :C
Discussion
39.

The bond that determines the secondary structure of a protein is

Answer»

co-ordinate BOND
covalent bond
HYDROGEN bond
ionic bond.

Solution : Hydrogen bond.
Discussion
40.

The bond stability of O_(2)^(+),O_(2),O_(2)^(-),O_(2)^(2-) varies in the order

Answer»

`O_(2)^(2-)gtO_(2)^(-)gtO_(2)gtO_(2)^(+)`
`O_(2)^(-)gtO_(2)gtO_(2)^(2-)gtO_(2)^(+)`
`O_(2)^(+)gtO_(2)gtO_(2)^(-)gtO_(2)^(2-)`
`O_(2)^(+)gtO_(2)^(-)gtO_(2)^(2-)gtO_(2)`

SOLUTION :Bond order `O_(2)^(+)=2.5` bond order `O_(2)=2` Bond order `O_(2)^(-)=1.5` bond order `O_(2)^(2-)=1`
Bond order `prop` STABILITY the bond stability of `O_(2)^(+), O_(2),O_(2)^(-), O_(2)^(2-)` varies in the following order `O_(2)^(+)gtO_(2)gtO_(2)^(-)gtO_(2)^(2-)`
Discussion
41.

The bond that determines the secondary structure of a protein is :

Answer»

COORDINATE bond
IONIC bond
Hydrogen bond
Covalent bond.

ANSWER :C
Discussion
42.

The bond (s) persent inquaternary ammonium salt is /are.

Answer»

covalent
ionic
covalent , CO-ordinateand ionic
covalent and co - ORDINATE

ANSWER :C
Discussion
43.

The bonds present in pernitric acid are:

Answer»

IONIC BONDS
COVALENT bonds
Semipolar bonds or DATIVE bonds
Co-ordinate and covalent bonds

Answer :D
Discussion
44.

The bond order of the underlined species: NOHSO_4 is.

Answer»


SOLUTION :`NOHSO_4` EXISTS as `NO^+` and `HSO_4^(-),NO^+` is derivative of oxygen and isoelectronic with `O_2^(2+)` Bond ORDER is 1/2(10-4)=3
Discussion
45.

The bond order order is maximum in:

Answer»

`H_2 `
`H_2^+`
`He_2`
`He_2^+`

ANSWER :A
Discussion
46.

The bond order of NO molecule is

Answer»


ANSWER :2.5
Discussion
47.

The bond order of NO is 2.5 while that of NO^(+) is 3. Which of the following statements is true for these two species?

Answer»

BOND LENGTH in `NO^(+)` is GREATER than in NO
Bond length in NO is greater than in `NO^(+)`
Bond length in `NO^(+)` is EQUAL to that in NO
Bond length in unpredictable.

Answer :B
Discussion
48.

The bond order of individual carbon-carbon bonds in benzene is

Answer»

ONE
TWO
Between one and two
One and two , ALTERNATELY

ANSWER :C
Discussion
49.

The bond order of CO molecule on the basis of molecular orbital theory is :

Answer»

Zero
2
3
1

Answer :C
Discussion
50.

What is the bond order of carbon molecule.

Answer»

1
2
0
3

Solution :`KK (sigma_(2s))^(2)(sigma_(2s)^(**))^(2)(pi2p_(x))^(2)=(p2 p_(y))^(2)`
BOND order `=(6-2)/(2)=2`.
Discussion