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 auto-reduction process is not used in the metallurgy of…

Answer»

HG 
CU 
PB 
Fe

Answer :D
Discussion
2.

Theauto- reductionprocess is notusedin themetallurgyof

Answer»

Hg
Cu
Pb
FE

Solution :Feisobtainedbyreductionof` Fe _2O_3`with carbonwhileall othersare obtainedbyauto-reduction.
Discussion
3.

The electrophile involved in the sulphonation of benzene is :

Answer»

`SO_(3)`
`SO_(3)^(+)`
`HSO_(4)^(-)`
`SO_(3)H^(+)`

ANSWER :A
Discussion
4.

The attacking species in the aromatic sulphonation is:

Answer»

`SO_3H^+`
`H_3SO_4^+`
`HSO_4^-`
`SO_3`

ANSWER :D
Discussion
5.

The attacking reagent in electrophilic sulphonation of benzene is

Answer»

`SO_2`
`SO_3`
`SO_4^(2-)`
`HSO_3^(-)`

ANSWER :B
Discussion
6.

The attacking reagent in electrophilic sulphonation of benzene is :

Answer»

`SO_(4)^(2-)`
`SO_(3)^(2-)`
`SO_(2)`
`SO_(3)`

ANSWER :D
Discussion
7.

The atoms the face centre is being shared by ……………………………… unit cells.

Answer»

4
8
2
6

Answer :C
Discussion
8.

The atomicity of yellow phosphorous is ____

Answer»


Solution :`P_4` - YELLOW (or) WHITE PHOSPHOROUS - ATOMICITY is 4
Discussion
9.

The atomicity of white Phosphorous is ‘x' and the P-hat(P)-P bond angle in the molecule is 'y'. What are 'x' and 'y' ?

Answer»

`X = 4 , y = 90^@`
`x = 4 , y = 60^@`
`x = 4 , y = 120^@`
`x = 4 , y = 180^@`

ANSWER :B
Discussion
10.

The atomicity of sulphur in orthorhombic sulphur is-

Answer»

8
6
4
2

Answer :A
Discussion
11.

The atomic weights of two elements A and B are 20 and 40 respectively. Which of the following statements are correct for these two elements?

Answer»

x g of A contains y ATOMS which is EQUAL to atoms present in x g of B.
x g of A contains y atoms which is equal to atoms present in 2x g of B.
At STP, x L of monoatomic gas A is equal to xL of monoatomic gas B.
At STP, x L of monoatomic gas A weighs y g and y g monoatomic gas B has volume xy.

Solution :Atomic WEIGHT
`underset(1 mol)(20)A underset(1 mol)underset(40)B`
Suppose 10 g of A = 1/2 mol of A = 1/2 mol of B = 20 g of B = 2x g of B
Hence, XG of A and 2x g of B will contain same number of atoms.
x L of A = yg
Suppose, 11.2 mL of A = 1/2 mol of A = 10 g of A
`=x/2` L of B at STP
Discussion
12.

The atomicitiesof oxygen and sulphur are different. What is the ratio of the atomicities of sulphur to oxygen ?

Answer»


Solution :Atomocity of `O_(2)=2` , Atomocity of SULPHUR `=8:.` RATIO `=(8)/(2)=4`
Discussion
13.

The atomic weights are expressed in terms of atomic mass unit. Which one of the following is used as a standard ?

Answer»

`.^(1)H_(1)`
`.^(12)C_(6)`
`.^(16)O_(8)`
`.^(35)Cl_(17)`

Solution :`""^(12)C_(6)` used as a STANDARD in the expression of atomic weights in term of amu.
Discussion
14.

The atomic weights of carbon, nitrogen and oxygen are 12,14 and 16 respectively. Among the following pairs of gases, the pair that will diffuse at the same rate is :

Answer»

1. CARBON DIOXIDE and NITROUS oxide
2. Carbon dioxide and nitrogen peroxide
3. Carbon dioxide and carbon monoxide
4. Carbon dioxide and NITRIC oxide

Answer :A
Discussion
15.

The atomicweight of Fe is 56 The weightof Fe deposited fromFeCI_(3)solutionby passing0.6 faraday of electricity is

Answer»

5.6 g
11.2 g
22.4 g
33.6 g

Solution :`FE^(3+) +3e^(-) rarr Fe`
`therefore` 3 MOLESOF electrondeposite = 1 moleof Fe
`therefore`0.6 molesof electron deposite `=1/3xx0.6` moleof Fe
`because` 1 moleof Fe =56 g (atomicweightof fe =56)
`therefore`0.2 moleof Fe `=(56)/(1) xx0.2 =11.2 g`
Discussion
16.

The atomic weight of noble gases is obtained by using the relationship

Answer»

ATOMIC WEIGHT=equivalent weightXvalency
Atomic weight=equivalent weight/valency
Atomic weight=valency/equivalent weight
`2xxVD`= MOLECULAR weight =atomic weight

Answer :D
Discussion
17.

The atomic weight of Al is 27. When a current of 5 Faradays is passed through a solution of Al^(++) ions, the weight of al deposited is

Answer»

27gm
36gm
45gm
39gm

Solution :`Al^(3+)+3E^(-)toAl`
`E_(Al)=(27)/(3)=9`
`W_(Al)=E_(Al)XX"No. of FARADAYS"=9xx5=45gm`.
Discussion
18.

The atomic weight of Al is 27. When a current of 5 Faraday is passed through a solution of Al^(3+) ions, the wt. of Al depostited is :

Answer»

27g
36 g
45 g
9 g

Answer :C
Discussion
19.

The atomic weight of a metal (M) is 27 and its equivalent weight is 9, the formula of its chloride will be

Answer»

`MCl_3`
`MCl_9`
`M_3 Cl_4`
`MCL`

ANSWER :A
Discussion
20.

The atomic sizesof Fe,Co and Ni are nearlysame, Explain with reason. Or Atomic size of 3dseries elements from chromium to copper is almost the same. Give reason.

Answer»

Solution :As we MOVE from LEFT to right alonga TRANSITIONSERIES, the nuclear change increases which tends to decrease the size but the addition of ELECTRONS in thed-subshellincreases the screening effect which counter balances the effect of increased nuclear CHARGE.
Discussion
21.

The atomic sizes are not significantly different for the series of elements

Answer»

BI, Na, K, Pb
Na, Mg,Al,Si
O,S,SC,Te
Cr,Mn,Fe,CO

Solution :The decrease in size of inner d-subshell due to added electrons and their shielding EFFECT on the outer most electrons from the nuclear charge almost compensate for `Cr, Mn, Fe` and Co.
Discussion
22.

The atomic reactor when used to generate electricity is termed breeder reactor.

Answer»


ANSWER :F
Discussion
23.

The atomic radius of strontium ( Sr ) is 215 pm and it crystallizes in FCC . Edge length of the cube is

Answer»

`430` PM
`608.2` pm
`496.53` pm
`304.1` pm

Answer :B
Discussion
24.

The atomic radius of palladium is 1.375 A. The unit cell of palladium is a face-centred cube. Calculate the density of palladium.

Answer»


ANSWER :12.01 g/cc
Discussion
25.

The atomic radius of Nb is closest to :

Answer»

TA
Zr
NI
Ag

Solution :Due to lanthanoid CONTRACTION.
Discussion
26.

The atomic radius of cd is closest to :

Answer»

AG
Hg
Cu
NI

ANSWER :B
Discussion
27.

The atomic radius of an ion which crystallizes in fcc structure is 9/7 Å. Calculate the lattice constant.

Answer»


SOLUTION :Atomic radius, `r = 9/7 Å =9/7 xx 10^(-10)` m.
For FCC STRUCTURE, `r = a//(2sqrt2)` or lattice constant, = `9/7 xx 10^(-10) m xx 2 xx 1.4142 = 3.6 xx 10^(-10)` m.
Discussion
28.

The atomic radius of Ag is closest to:

Answer»

Ni
Cu
AU
Hg

Solution :Because of Lanthanide contraction, an increase in Zeff is observed and so, the size of Au INSTEAD of being greater, as is expected, turns out to be SIMILAR to that of Ag.
Discussion
29.

The atomic radius of a face centred cubic cell is:

Answer»

`a/2`
`(sqrt2.a)/4`
`(sqrt3.a)/4`
`a/4`

ANSWER :B
Discussion
30.

The atomic radius of a body centred cubic cell is:

Answer»

`a/2`
`(sqrt2a)/4`
`(SQRT3A)/4`
`a/4`

ANSWER :B
Discussion
31.

The atomic radius of 5d elements and 4d elements are nearly same due to…..

Answer»

SOLUTION : LANTHANOID CONTRACTION
Discussion
32.

The atomic radius in a face-centred cubic cell is:

Answer»

`a/2`
`(SQRT(2)a)/4`
`(sqrt(3)a)/4`
`a/4`

Answer :B
Discussion
33.

The atomic radii of Nb and Ta are comparable due to __________.

Answer»

SOLUTION :LANTHANOID CONTRACTION
Discussion
34.

The atomic radii of Fe, Co and Ni are __________ .

Answer»

SOLUTION :ALMOST EQUAL
Discussion
35.

The atomic radii of __________ and __________ series of transition elements are comparable.

Answer»

SOLUTION :SECOND, THIRD
Discussion
36.

The atomic radii in periodic table among elements from right to left

Answer»

DECREASES
Increases
Remain constant
First decreases and then increases

Answer :B
Discussion
37.

The atomic radii from Cr to Cu is almost identical because of

Answer»

Increasing NUCLEAR CHARGE from Cr to Cu
Repulsion among increased electrons
Increased SCREENING effect to nullify increased nuclear charge
All the above

Solution :Increased screening effect to nullify increased nuclear charge.
Discussion
38.

The atomic orbitals are progressively filled in order of increasing energy. This principle is called:

Answer»

HUND's rule
Aufbau principle
Exclusion principle
de-Broglie rule

Answer :B
Discussion
39.

The atomic orbital not allowed in quantum theory is

Answer»

2p
2s
3f
1s

Answer :C
Discussion
40.

The atomic numbers of the metallic and non-metallic elements which are liquid at room temperature respectively are:

Answer»

55,87
33,87
35,80
80,35

Answer :D
Discussion
41.

The atomic numbers of other elements which lie in same group as the tenth element in the periodic table are

Answer»

18, 32, 54, 86
8, 18, 36, 84
2, 18, 30, 36
2, 18, 36, 54

Solution :Tenth ELEMENT is a member of noble gases (Group 18 or zero). The atomic numbers of other elements of this group are `2 (He), 18(Ar), 36(Kr), 54(Xe), 86(RN) and Og(118)`.
Discussion
42.

The atomic numbers of chromium and iron are 24 and 26 respectively. Which one of the following complexes exhibits paramagnetic character due to electronic spin?

Answer»

`[Fe(CO)_5]`
`[CR(NH_3)_6]^(3+)`
`[Fe(CN)_6]^(4-)`
`[Cr(CO)_6]`

ANSWER :B
Discussion
43.

The atomic number of Sn is50. The shape of gaseous SnCl_(2) molecule is

Answer»

`Cl-Sn-Cl`


Solution :In `SnCl_(2),` Sn is `SP^(2)` hybridized. As such it has two bond PAIRS and one lone pair of ELECTRONS. Therefore, TITS structure is as :
Discussion
44.

The atomic number of Ni and Cu are 28 and 29 respectively. The electronic configuration 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10) represents

Answer»

`Cu^(+)`
`Cu^(2+)`
`Ni^(2+)`
Ni

Solution :ELECTRONIC CONFIGURATIONS of :
`Ni-[AR]3d^(8)4s^(2), "" Ni^(2+)-[Ar]3d^(8)`
`Cu-[Ar]3d^(10)4s^(1), ""Cu^(+)-[Ar]3d^(10)`
`Cu^(2+)-[Ar]3d^(9)`
Discussion
45.

The atomic number of elements lie in the range of

Answer»

88 to 101
89 to 102
90 to 103
91 to 104

Answer :C
Discussion
46.

The atomic number of element having pseudo inert gas configuration in it's atomic state is

Answer»

46
45
47
48

Answer :A
Discussion
47.

Which element has pseudo inert gas electronic configuration ?

Answer»

46
45
47
48

Answer :A
Discussion
48.

The atomicnumber of cerium (Ce) is 58. The correct electronic configuration of Ce^(3+) ion is

Answer»

`[Xe]4F^(1)`
`[Kr]4f^(1)`
`[Xe]4f^(13)`
`[Kr]4d^(1)`

Solution :`Z = 54` is `[Xe]=Z= 56` is `BA = [ Xe] 6s^(2)`
`Z =57` ( La) is `[Xe]5d^(1)`( electron 5d in place of expected 4f)
But in `Z= 58` ( Ce ) , electron enters 4f and `5d^(1)` also shifts to 4f. Hence, electronic configuration of `Ce( 58) ` is `[Xe] 4f^(2) 5d^(0) 6s^(2)`.
`:. Ce^(3+) = [Xe]4f^(1)`
Discussion
49.

The atomic number of cerium (Ce) is 58. The correct electronicconfiguration of Ce^(3+) ion is

Answer»

`[Xe]4f^(1)`
`[Kr]4f^(-1)`
`[Xe]4f^(-13)`
`[Kr]4d^(1)`

Solution :
For Z = 56, it will be `[Xe]6s^(2)`
In LANTHANUM (Z = 57), electrons should enter in 4f but to maintain stable xenon core it will go into 5d(DIFFERENCE is little in between 4f and 5d.)
But in cerium (Ce, Z = 58), ELECTRON enters into 4f.
`therefore""Ce^(3+)=[Xe]4f^(1)`
Discussion
50.

The atomic number of cobalt is 27 . The EAN of cobalt in Na_(3)[Co(NO_(2))_(4) Cl_(2)] is

Answer»

35
24
36
34

Solution :EAN of Co in `[Co (NO_(2))_(4) Cl_2]^(3-)` is EAN = `(27 + 4 XX 1 + 2 xx 1) + 3 = 36`
Discussion