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.

Iron and oxygen combine to form three oxides, FeO, Fe_(2)O_(3) and Fe_(3)O_(4). Which law does it prove?

Answer»

SOLUTION :LOW of MULTIPLE PROPORTIONS.
2.

IUPAC name of K_3[Al(C_2O_4)_3]

Answer»

POTASSIUM alumino oxalte
potassium trioxalato ALUMINATE (III)
potassium ALUMINIUM OXALATE (III)
potassium trioxalato aluminate (VI)

Answer :C
3.

Iron and sulphuric acid

Answer»

ELECTRONS are gained by `Fe^(2+)`
Electrons are LOST by `Fe^(2+)`
There is no loss or GAIN of electrons
Iron hydroxide precipitates

Answer :B
4.

Iron and nickel are used to make an electrochemical cell by using slat bridge to join a half-cell containing 1.0 M solution of Fe^(2+)(aq) in which a strip of iron has been immersed to a second half-cell which contains 1.0 M Ni^(2+)(aq) solution in which a strip of nickel has been immersed. A voltmeter is connected between the two metal strips. (i) In which cell does reduction occurs? (ii) Write the half-cell reactions involved. (iii) Which metal is the anode? (iv) In which direction are the electrons passing through the voltmeter? (v) What would be effect ont he voltmeter reading if Fe^(2+) concentration were increased? (vi) What will be the voltmeter readind when the cell reaches equilbirium? Given that the standard electrode potential of Fe^(2+)//Fe and Ni^(2+)//Ni electrodes are -0.44 and -0.25 volt respectively.

Answer»


Answer :(i) Nickel half cell
(ii) `NI^(2+)+2e^(-)toNi`
(iii) Fe is anode (iv) IRON to nickel
(v) Voltmeter reading decreases (VI) V=0
5.

(i)(r)+(p)to NaClO_2 + O_2 (ii)IO_3^(-) + I^(-) overset(H^+)to (q) (iii)Acidic solution of ClO_2^(-) to (s) + (r) (disproportionation reaction ) (p)used as air purifier and CO_2 absorber (q) estimated by hypo and it forms violet colour vapour (it self) (r)is yellow colour gas (s) gives chromylchloride test. which option is correct .

Answer»

(p)=`Na_2O_2 , (Q) =l_2`
`(q)=I_2 O_5 , (R)=Cl^-`
`(p)=Na_2O, (r) =Cl_2O`
`(r)=Cl_2 , (s)=Cl_2 O_3`

Solution :`ClO_2+Na_2O_2to NaClO_2+O_2`
`IO_3^(-) + I^(-) to l_2 + H_2O`
`underset((+3))(5ClO_2^(-))OVERSET(H^+)to underset((+4))(4ClO_4) + underset((-1))(Cl^-) + H_2O`
`A=Na_2O_2 , B=I_2, C=ClO_2 , Cl^-`
6.

IUPAC nameof isoproyl amineis

Answer»

2-aminopropane
2-propanamine
2,2-dimethylethanamine
dimethanamine

ANSWER :B
7.

Iron -49 decay by positron emission with a half-life of 0.08 seconds . Nuclide produced in this decay has atomic number x^2. Find x .

Answer»


Solution :`""_(26) Fe^(49) to ""_(+1) E^(0) + ""_(25) X^(49)`
8.

Iproniazid is used as

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antiallergic
antimalarial
sedative
antiseptic

Answer :C
9.

IUPACname of isobutyric acid is

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3- methylpropionic ACID
2- methylpropanoic acid
3- methylbutanoic acid
isobutanoic acid

ANSWER :B
10.

IP_(1), IP_(2), IP_(3), IP_(4) and IP_(5) for an element is 7.1, 14.3, 34.5, 46.8 and 162.2 eV respectively then element is :-

Answer»

Na
SI
F
CA

SOLUTION :
VALENCE electron for this element =4
11.

(i)P+C("carbon")+Cl_(2)toQ+CO uarr , (ii)Q+H_(2)OtoR+HCl (iii)BN+H_(2)O to R+NH_(3) , (iv)Q+LiAIH_(4) to S+LiCl+ AlCl_(3) (v)S+H_(2)toR+H_(2) uarr , (vi)S+NaH to T (P,Q,R,S and T do not represent their chemical symbols) Compound T is used as a/an:

Answer»

<P>OXIDISING agent
complexing agent
bleaching agent
reducing agent

Solution :(i)`B_(2)O_(3)(P)+3C+3Cl_(2) to 2BCl_(3) (Q) +3CO uarr`
(ii)`BCl_(3)(Q)+3H_(2)O to H_(3)BO_(3)(R)+3HCl`
(iii)`BN+3H_(2)O to H_(3)BO_(3) ( R) + NH_(3)`
(iv)`4BCl_(3)(Q)+3LiAlH_(4) to 2B_(2)H_(6)(S)+3LiCl+3AlCl_(3)`
(v)`B_(2)H_(6)+6H_(2)O to 2H_(3)BO_(3) (R)+6H_(2) uarr`
(vi)`B_(2)H_(6)(S)+2NaH to 2NaBH_(4)(T)`
12.

(i)P+C("carbon")+Cl_(2)toQ+CO uarr , (ii)Q+H_(2)OtoR+HCl (iii)BN+H_(2)O to R+NH_(3) , (iv)Q+LiAIH_(4) to S+LiCl+ AlCl_(3) (v)S+H_(2)toR+H_(2) uarr , (vi)S+NaH to T (P,Q,R,S and T do not represent their chemical symbols) Compound S is: (I) an odd compound (II)(III) an electron deficient compound (IV) ahybridized compound Choose the correct code:

Answer»

III
I,III
II,III,IV
I,II,IV

Solution :(i)`B_(2)O_(3)(P)+3C+3Cl_(2) to 2BCl_(3) (Q) +3CO uarr`
(ii)`BCl_(3)(Q)+3H_(2)O to H_(3)BO_(3)(R)+3HCl`
(iii)`BN+3H_(2)O to H_(3)BO_(3)(R) + NH_(3)`
(iv)`4BCl_(3)(Q)+3LiAlH_(4) to 2B_(2)H_(6)(S)+3LiCl+3AlCl_(3)`
(v)`B_(2)H_(6)+6H_(2)O to 2H_(3)BO_(3) (R)+6H_(2) uarr`
(VI)`B_(2)H_(6)(S)+2NaH to 2NaBH_(4)(T)`
13.

(i)P+C("carbon")+Cl_(2)toQ+CO uarr , (ii)Q+H_(2)OtoR+HCl (iii)BN+H_(2)O to R+NH_(3) , (iv)Q+LiAIH_(4) to S+LiCl+ AlCl_(3) (v)S+H_(2)toR+H_(2) uarr , (vi)S+NaH to T (P,Q,R,S and T do not represent their chemical symbols) Compound Q has: (I) zero dipole moment (II) a planer trigonal structure (III) an electron deficient compound (IV) a Lewis base Choose the correct code.

Answer»

I,IV
I,II,IV
I,II,III
I,II,III,IV

Solution :(i)`B_(2)O_(3)(P)+3C+3Cl_(2) to 2BCl_(3) (Q) +3CO uarr`
(ii)`BCl_(3)(Q)+3H_(2)O to H_(3)BO_(3)(R)+3HCl`
(iii)`BN+3H_(2)O to H_(3)BO_(3)(R) + NH_(3)`
(iv)`4BCl_(3)(Q)+3LiAlH_(4) to 2B_(2)H_(6)(S)+3LiCl+3AlCl_(3)`
(v)`B_(2)H_(6)+6H_(2)O to 2H_(3)BO_(3) (R)+6H_(2) uarr`
(vi)`B_(2)H_(6)(S)+2NaH to 2NaBH_(4)(T)`
14.

I.P is maximum for :-

Answer»

LI
NE
Be
B

ANSWER :A
15.

(i)Ortho silicate""(a)"Thortveitite" (ii)Pyro silicate""(b)Beryl (iii)Cyclic silicate""(c)Quartz (iv)"Tecto silicate"""(x)Phenacite

Answer»


ANSWER :A::B::C::D
16.

IOPAC name for given compound is

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3-Hydroxy butane-1,3,5-trioic ACID
3-Hydroxy -4-carboxy hexan-1,6-dioic acid
2-Hydroxy butane-1,3,4-tricarboxylic acid
3-Hydroxy butane-1,2,4-tricarboxylic acid

Answer :D
17.

Ionosphere is

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above ozone layer
below ozone layer
below 15 KM in earth crust
none of these.

Solution :Ionosphere is PART of atmosphere that is ionized by solar radiation. Ozone layer is in STRATOSPHERE below the ionosphere.
18.

IUPAC name of Isobutyraldehyde is

Answer»

Butanal
Methyl PROPANAL
Ethyl ethanal
Methyl butanal

ANSWER :B
19.

Ionization potential values of nobles gases decrease down the group with increase in atomic size. Xenon forms binary fluorides by the direct rection of elements.Identify the correct statement(s) from below.

Answer»

Only the heavier noble GASES form such compounds
It happens because the noble gases have higher ionizaiton energies.
It happens because the compound are formed with electronegative ligands.
Octet of electrons provide the stable arrangments.

SOLUTION :Having high ionization energy, INERT gases do not form compounds. Inert gas compounds are hypervalent in nature. So no question of octet fill up through compound FORMATION.
20.

Ionization potential of Na would be numericlally the same as

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ELECTRON affinity of `NA ^(+)`
ELECTRONEGATIVITY of `Na^(+)`
Electron affinity of He
IONIZATION potential of Mg

Answer :A
21.

IUPAC name of isobutyl halide is

Answer»

2-halopropane
2-halo-2-methylpropane
1-halo-2-methylpropane
2-halobutane

Answer :C
22.

Ionization energy of nitrogen is more than oxygen because

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Nucleus has more attraction for electrons
HALF -filled p-orbitals are more stable
Alkaline EARTH metals
Alkali metals

ANSWER :B
23.

IUPAC name of isobutyl amine is

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2-methylpropan - 2- AMINE
2- methylprop -2-amine
2-methylbutan -2- amine
PROPAN - 2 - amine

ANSWER :C
24.

Ionization energies of five elements in kcal/mol are given below : Which of the following pair represents elements of same group ?

Answer»

Q, R
P, Q
P, S
Q, S

Solution :
25.

Ionization energy of an element is:

Answer»

EQUAL in magnitude but opposite in sig to the electron gain ENTHALPY of the cation of the element
Same as electron AFFINITY of the element
Enegy required to remove one VALENCE electron from an ISOLATED gseous atom in its ground state
Equal in magnitude but opposite in sign to t no gain enthalpy of the anion of the element

Answer :A::C
26.

IUPAC name of hexamethylene diamine is

Answer»

HEXAMETHYLENE 1,6-diamine
1,6 -diaminohexamethylene
1,6- diamno hexane
1,6 -hexane-diamine

ANSWER :D
27.

Ionization energy is influenced by

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size of ATOM
charge of NUCLEUS
ELECTRONS present in INNER shells
NONE of the above

Answer :B::C
28.

Ionization energies of five elements in kcal/mol are given below : Which is non-metal (excluding noble gas) ?

Answer»

P
Q
S
T

Solution :
29.

Ionization energies of five elements in kcal/mol are given below : The element having most stable oxidation state +2 is ?

Answer»

<P>P
Q
R
S

SOLUTION :
30.

IUPAC name of H_(2)[PtCl_(6)] is

Answer»

Hydrogen hexachloroplatinate (IV)
dihydrogen hexachloroplatinate (IV)
hydrogen hexachloroplatinic (IV) ACID
hexachloroplatinic (IV) acid

Solution :`H_(2)[PtCl_(6)]` is an acid, not satl HENCE its name, Hexachloroplatinic (IV) acid.
31.

Ionization energies of five elements in kcal/mol are given below : Which element is a noble gas ?

Answer»

P
T
R
S

SOLUTION :
32.

Ionization energies of five elements in kcal/mol are given below : The element form stable unipositive ion?

Answer»

P
Q
T
S

SOLUTION :
33.

Ionization energies of five elements in kcal/mol are given below: Q. Which of the following pair represents elements of same group?

Answer»

Q,R
P,Q
P,S
Q,S

Solution :Both Q and R SHOWS group in `I.E_(2)` values ANND BELONG to alkali metals.
34.

Ionization energies of five elements in kcal/mol are given below: Q. Which is a non-metal (excluding noble gas)?

Answer»

P
Q
R
S

Solution :HIGH I.E. VALUES SHOW its non-metallic NATURE.
35.

Ionization energies of five elements in kcal/mol are given below: Q. Which element is a noble gas?

Answer»

P
T
R
S

Solution :T has ABNORMALLY HIGHER `I.E_(1)` VALUE.
36.

Ionization energies of five elements in kcal/mol are given below: Q. Which element form stable unipositve ion?

Answer»

P
Q
R
T

Solution :There is suddenn JUMP in `I.E_(2)` of `Q`, i.e., unipositive has noble GAS configuration.
37.

Ionization energies of five elements in kcal/mol are given below: Q. The element having most stable oxidation state +2 is?

Answer»

Q
R
S
T

Solution :`I.E_(3)` of S is ABNORMALLY HIGHER.
38.

Ionization energies of five elements in kcal/mol are given below: Q. If Q reacts with fluorine and oxyge, the molecular formula of fluoride and oxide will be respectively:

Answer»

`QF_(3), Q_(2)Q_(3)`
`QF,Q_(2)O`
`QF_(2),QO`
none of these

Solution :Q is alkali METAL as it show GROUP in `I.E_(2)` value.
39.

Ionization energies of five elements in kcal/mol are given below : If Q reacts with fluorine and oxygen, the molecular formula of fluoride and oxide will be respectively :-

Answer»

`QF_(3), Q_(2)Q_(3)`
`QF, Q_(2)O`
`QF_(2), QO`
NONE of these

Solution :
40.

Ionisation potential of noble gases are____.

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SOLUTION :very HIGH
41.

IUPAC name of H_3C - HC(Br)_2 is

Answer»

Ethylidene bromide
Gem - dibromide
Any of the above
1,1-Dibromo ETHANE

ANSWER :D
42.

Ionisation potential of Hydrogen atom is 13.6 eV. Hydrogen atom in the ground state are excited by monochromatic light of energy 12.1 eV. The spectral lines emitted by hydrogen according to Bohr's theory will be:

Answer»

One
Two
Three
Four

Answer :C
43.

IUPAC name of Glycine is

Answer»

2-amino PROPANOIC acid
2-amino BUTANOIC acid
Amino ETHANOIC acid
2-amino pent-I, 5-dioic acid

Answer :C
44.

Ionisation potential of fluorine is abnormally high. It is due to

Answer»

Its HIGH EN value
Its high EA value
Its small size
Its BIG size

Answer :C
45.

Ionisation potentail values of d-block elements as compared to ionization potential potential value of f-block elements are

Answer»

Higher
Equal
lower
All of these

Solution :REMOVAL of electron is EASIER in f-block ELEMENTS due to more SHIELDING.
46.

IUPAC name of following compound is (CH_3)_3C-O-CH_2CH(CH_3)_2

Answer»

2 -(2-methyl -1-propoxy) -2-methylpropane
1 -(2-methyl -2-propoxy)- 2-methylpropane
1 -(2-propoxy) -2-methylpropane
2 -(2-propoxy) -2-methylpropane

Answer :B
47.

Ionisation of K[Ag(CN)_2] will give :

Answer»

`K^+`and`[Ag(CN)_2]^-` ion
KCN and AgCN
`K^+,Ag^+,CN^-`
None

Answer :A
48.

IUPAC name of formic acid is:

Answer»

Methanoic acid
Ethanoic acid
Ethanedioic acid
Methanedioic acid

Answer :A
49.

Ionisation enthalpy of group 15 elements I. decreases down the group due to gradual increase in atomic size. II. Is higher than that of group 14 elements in the corresponding periods. III. The order of successive ionisation enthalpies is triangle_(i)H_(1)lttriangle_(i)H_(2)lttriangle_(i)H_(3) Which of the above statements is/are ture?

Answer»

I and II
II and III
I and III
I, II and III

Solution :N//A
50.

Ionisation isomerism is possible with

Answer»

`[Co(NH_3)_6]Cl_3`<BR>`[Co(NH_3)_3 Cl_3]`
`[Co(NH_3)_5Cl]Br`
`[Co(H_2O)_6]Cl_3`

Answer :C