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 number of electrons involved in the reduction of nitrate ion to hydrazine is ........

Answer»


Solution :`NO_(3)^(-) rarr N_2H_4`
To BALANCE N atoms, multiple `NO_3^(-)` by 2,we have
`2overset(+5)(NO_3^(-))rarroverset(-2)(N_2H_4)`
`{:("Total O.N "= 2x×15 ,"Total O.N" = 2 xx -2),(=+10 ,=-4):}`
To balance O.N. ADD `14e^(-) " to " ` L.H.S
we have `2NO_(3)^(-)+ 14e^(-) rarr N_2H_4`
2.

The numberof electrons involved in the half reaction Cr_(2)O_(7)^(2-)rarr2Cr^(3+) is

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3
6
5
10

Answer :B
3.

The number of electrons in the atom which has 20 protons in the nucleus

Answer»

20
10
30
40

Solution :NUCLEUS of 20 PROTONS ATOM have 20 ELECTRONS
4.

The number of electrons in [K^(40)]^(-1) are

Answer»

18
19
20
40

Solution :K (Z = 19) has 19 electrons. Number of electrons in `K^(-1)` is 19 + 1 = 20.
5.

The number of electrons in one molecule of CO_2 are

Answer»

22
44
66
88

Solution :ONE MOLECULE of `CO_2` have 22 ELECTRONS.
6.

The number of electrons in an atom with atomic number 105 having (n +l) = 8 is

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15
17
19
21

Solution :The electronic configuration of element with atomic number 105 is
`1s^(2), 2s^(2) 2p^(6), 3s^(2) 3p^(6) 3d^(10), 4s^(2) 4p^(6) 4d^(10) 4F^(14), 5s^(2) 5p^(6) 5d^(10) 5f^(14), 6s^(2) 6p^(2) 6d^(3) 7s^(2)`
For `5f, (n + L) = 5 + 3 = 8`, ELECTRONS present = 14
For `6d, (n + l) = 6 + 2 = 8`, Electrons present = 3
7.

The number of different spatial arrangements for the orbital with 1 = 2 is

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1
3
5
7

Solution :`m_l =-L ` to `l=-2, -1 , 0 , + 1, +2`
8.

The number of dichloro isomeric products possible of dichorination of neo-pentane is

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Solution :
2,3,4,6, have chiral carbon this is have OPTICAL isomers. So FOUR optical isomers so TOTAL 5 + 4 = 9.
9.

Total number of stereoisomers of the following compound is CHCl=CH-CH(CH_(3))C_(2)H_(5)

Answer»

2,2
1,1
1,2
2,1

Solution :DIASTEREOMERIC pair (cis and TRANS) = 1
Enantiomeric pair for ONE chiral centre (R and S) = 1
10.

The number of d-electron retained in Fe^(2)("At no. of" Fe =26) ion is.

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3
4
5
6

Answer :D
11.

The number of d-electrons retained in Fe^(3+) (At. No. of Fe = 26) ion is

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3
4
5
6

Answer :C
12.

The number of d-electrons in Fe^(2=) (Z = 26)is notequal to the number p of electrons in which one of the following?

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p-electronsinCl (Z= 17)
d - electronsin FE (Z =26)
p- electronsin Ne (Z =10)
s- electronsin MTG (Z = 12)

Solution :p-electronsinCl (Z= 17)

`Fe^(2+) ` has 6 d-electrons.
(Option - A) has 11 p - el electrons , which are not equal to six d-electrons in `Fe^(2-)`
13.

Thenumber of d-electrons in Fe^(2+)(Z= 26)is notequal to thenumber of theelectronsin whichone ofthe following?

Answer»

d- ELECTRONSIN Fe (Z= 26)
p- electronsin Ne(Z = 10 )
s-electrons in Mg (Z= 12 )
p -electrons inC1 (Z =17)

Solution :(a) E.C of Fe (Z=26)
`= 1S^(2) 2S^(2)2p^(6)3s^(2)3p^(6)3D^(6)4s^(2)`
`:. `E.C. of `Fe^(2+)= 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(6)`
Ithas sixp- electrons.
( c) E.C. of Mg (Z=12)= `1s^(2)2s^(2)2p^(6)3s^(2)`
It haseleven( 6+5)p-electrons .
NowC1 haseleven p-elements but `Fe^(2+)` has six d- electrons. Thereforeoption (d)iscorrect.
14.

The number of d-electrons in Fe^(2+) (Z = 26) is not equal to the number of electrons in which one of the following ?

Answer»

d-electrons in Fe (Z = 26)
p-electrons in Ne (Z = 10)
s-electrons in Mg (Z = 12)
p-electrons in CL (Z = 17)

Solution :`._(26)Fe = 1s^(2) 2s^(2) 2P^(6) 3s^(2) 3P^(6) 3d^(6) 4s^(2)` (d - electrons = 6)
`Fe^(2+) = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(6)` (d-electrons = 6)
`._(10)Ne = 1s^(2) 2s^(2) 2p^(6)` (p-electrons = 6)
`._(12)Mg = 1s^(2) 2s^(2) 2p^(6) 3s^(2)` (s-electron `= 2 + 2 + 2 = 6`)
`._(17)Cl = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(5)` (p-electrons `= 6 + 5= 11`)
15.

The number of correct option is (a) P_(2)O_(5) gtZnO gtMgO gt Na_(2)O_(2) (acidic strenght) (b) TI_(2)O_(3) gt TI_(2)O gtGa_(2)O_(3) gt AI_(2)O_(3)(basic strenght) ( c) MnO gt P_(2)O_(5) gt CrO_(3) gt Mn_(2)O_(7) (ionic character) (d) H_(2)OgtHF gtNH_(3) (melting point) (e) H_(2)O gtHF gtNH_(3) (boiling point) .

Answer»


Solution :(B) Wrong Correct ORDER `TI_(2)O gtTI_(2)O_(3) gtGa_(2)O_(3)`
` gtAI_(2)O_(3)`
Wrong Correct order `H_(2)Ogt NH_(3) gtHF` .
16.

The number of conformational isomers possible of n-butane about C_(2)-C_(3) bond

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ONE
FOUR
Three
Infinite

Solution :Infinite DIHEDRAL ANGLES are possible.
17.

The number of compounds showing enantiomers among the following compounds is Butan-2-ol, 2-Hydroxypropanoic acid, 2-Methylhexane, 2-Chlorobutane, 2-Bromo-2-chlorobutane, 2-Methylbutanoic acid, isopropyl chloride

Answer»


SOLUTION :The COMPOUNDS having chiral CARBON and therefore, SHOWING enantiomers are: Butan-2-ol, 2-Hydroxypropanoic ACID, 2-Chlorobutane, 2-Bromo-2-chlorobutane.
18.

The number of compounds which show tautomerism among the following is

Answer»


ANSWER :6
19.

The number of compounds which are more reactive towards electrophilic substitution than p-xylene? Benzene, Toluene, Mesitylene, o-xylene, m-xylene.

Answer»

SOLUTION :ELECTROPHILIC SUBSTITUTION
20.

The number of collisions depends on A) mean free pathB) pressure C) temperature

Answer»

<P>A, C 
A,B
B,C
A, B, C 

SOLUTION :Frequency of COLLISIONS DEPENDS on `lambda, P , T`.
21.

The number of cis-trans isomers with molecular formula C_2BrClFI is

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SOLUTION :SIX
22.

The number of chiral compounds produced upon monochlorination of 2-methylbutane is :

Answer»

2
4
6
8

Answer :B
23.

The number of chain isomers for C_(5)H_(12) is

Answer»

1
2
3
4

Answer :C
24.

The number of chain isomers possible for hydrocarbon C_(5)H_(12) is

Answer»

3
5
4
6

Solution :`C_(5)H_(12): UNDERSET("n-Pentane")(CH_(3)CH_(2)CH_(2)CH_(2)CH_(3))`
`underset("2-Methylbutane (iso-Pentane)")(CH_(3)-underset(CH_(3))underset(|)(C)H-CH_(2)-CH_(3))""underset("2,2-Dimethylpropane (neo-Pentane)")(CH_(3)-underset(CH_(3))underset(|)OVERSET(CH_(3))overset(|)(C)-CH_(3))`
25.

The number of carbon monoxide molecular present in 1 dm^3 at N.T.P. is _________ .

Answer»

`6.02xx10^(23)`
`6.02xx10^(22)`
`0.269xx10^(22)`
`2.69xx10^(22)`

SOLUTION :22.4 `dm^3` of 'CO' CONTAINS `6.02xx10^(23) ` at N.T.P.
Hencenumber of CO MOLECULES PRESENT in
1 `dm^3` at N.T.P `=(6.02xx10^(23))/(22.4) = 2.69 xx10^(22)`
26.

The number of C-C sigma bonds present in 1-butyne is

Answer»

2
3
4
5

Answer :B
27.

The number of bridge bonds, the maximum number of planar atoms and the number of electrons involved in the formation of bridge bonds in diborane are x, y and z respectively, then (x+y–z) = ?

Answer»


Solution :x = 2 (B-H-B bonds)
y = 6 (`2B, 4H_(t)`atoms)
`z=4 , x+y-z=2+6-4=4`
28.

The number of bridge hydrogen atoms in diborane is

Answer»

1
2
3
4

Answer :B
29.

Thenumber of atoms presents in 20 g of calciumwill equal to the number of atoms presents in (20 g Ca = (1)/(2) Ca) (Ca = (6.023 xx 10^(23))/(2) = 3.012 xx 10^(23))

Answer»

`12 G C`
`12.15 g Mg`
`24.0 g C`
`24.3 gMg`

Solution :`24.3 g Mg = 1 mol, so 12.15 g = (1)/(2) mol`
30.

The number of atoms presentin 0.1 mole of P_(4) (at mass = 31) are

Answer»

`2.4 XX 10^(23)` atoms
same as in 0.05 mole of `S_(8)`
`6.02 xx 10^(22)` atoms
same as in 3.1 g of phosphorus

Solution :a) `therefore` 1 mole of `P_(4)` contains phosphorus atoms
`=4 xx 6.02 xx 10^(23)`
`therefore` 0.1 mole of `P_(4)` contains phosphorous atoms
`=4 xx 6.02 xx 10^(23) xx 0.1`
`=2.4 xx 10^(23)`atoms
31.

The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following element contains the greatest number of atoms ?

Answer»

4g He
49g Na
0.40g Ca
12g He

Solution :To compare the number of atoms, we need to CALCULATE the moles as all are monatomic and HENCE, moles `XX N_(A) =` number of atoms.
Moles of `4 G He = 4/4 = 1 ` mol
`46 g Na = (46)/(23) = 2 `mol
`0.40 g Ca =(0.40)/(40) = 0.1 mol`
`12 g He = (12)/(4) = 3 `mol
Here, 12 g He contains greatest number of atoms as it possess maximum number of moles
32.

The number of atoms present in 16 g of oxygen is:

Answer»

`6.02 XX 10^(11.5)`
`3.01 xx 10^(23)`
`3.01 xx 10^(11.5)`
`6.02 xx 10^(23)`

ANSWER :D
33.

The number of atoms present in 142g of Chlorine is

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`6xx110^(23)`
`1.2xx10^(24)`
`2.4xx10^(24)`
`3.6xx10^(24)`

ANSWER :C
34.

The number of atoms present in 1 g of hydrogen is the same as that in 1 g of oxygen.

Answer»


ANSWER :F
35.

The number of atoms present in 1 g of hydrogen gas is the same as are present in

Answer»

4 g of Helium
32 g of Oxygen
7 g of Nitrogen
24 g of Carbon

Solution :`2 g " ofHydrogen "= 6xx10^(23)" molecules of "H_2`
`-=2xx6xx10^(23)" ATOMS of "H_2`
`1 g " of HYDROGEN "-= 6xx10^(23)" atoms of Hydrogen "`
(a) 4 g He `-= 6xx10^(23)" atoms of He"`
(b) `32 g" of "O_2 -=2xx6xx10^(23)" atoms of oxygen"`
`= 12xx10^(23)" atoms of oxygen"`
( C) `7 g " of " N_2 -= (2xx6xx10^(23)xx7)/(28)" atoms of " N_2`
`=3xx10^(23)" atoms of " N_2`
(d) `24 g " of " C -=(6xx10^(23)xx24)/(12)`
`= 12xx10^(23)" atoms of carbon"`
Correct answer is (a).
36.

The number of atoms of hydrogen present in 1.5 mole of H_(2)O is

Answer»

1N
2N
3N
0.5N

Answer :C
37.

The number of atoms of 0.03 g of aluminium is nearly (Al = 27)

Answer»

`6.68 xx10^(20)`
`6.68xx10^(21)`
`6.68 xx10^(22)`
`6.68xx10^(23)`

SOLUTION :The number of ATOMS in 0.3 g of AL
`=(6.02xx10^(23))/(27) xx0.03`
`= 0.668xx10^(21) = 6.68xx10^(20)`
38.

The number of atoms is 100 g of a fcc crystal with density ="10.0 g/cm"^3 and cell edge equal to 200 pm is equal to

Answer»

`5xx10^24`
`5xx10^25`
`6xx10^23`
`2xx10^25`

Solution :`RHO=(ZxxM)/(a^3XX10^(-30)xxN_0)` or `M=(10xx(200)^3xx10^(-30)xx6xx10^23)/4=12`
Thus, 12 g CONTAIN=`N_0=6xx10^23` ATOMS
`therefore` 100 g will contain `=(6xx10^23)/12xx100=5xx10^24`
39.

The number of atoms in a face centred cubic unit cell is:

Answer»

4
3
2
1

Answer :A
40.

The number of atoms in 4.25g of NH_(3) is approximately :

Answer»

`1 XX 10^(23)`
`2XX 10^(23)`
`4 xx 10^(23)`
`6 xx 10^(23)`

SOLUTION :4.25 g of `NH_(3)` (= 0.25 mol)
1 mole of `NH_(3)` have atoms
`= 4 xx 6.022 xx 10^(23)`
0.25 mole of `NH_(3)` have atoms
`= 4 xx 0.25 xx 6.022 xx 10^(23)`
`=6.022 xx 10^(23)`.
41.

The number of atoms in 100g of a fcc crystal with density=10gcm^(-3) and cell edge as 200gm, is equal to

Answer»

`3xx10^(25)`
`5XX10^(24)`
`1XX10^(25)`
`2xx10^(25)`

Solution :(b) We KNOW that, d `(Z.M)/(nxxa^(3))`
`N=(4xx100)/(10xx(200xx10^(-10))^(3))`
=`5xx10^(24)`
42.

The number of atoms in 0.1 mole of triatomic gas is :

Answer»

`6.026 xx 10^(22)`
`1.806 xx 10^(23)`
`3.6 xx 10^(23)`
`1.8 xx 10^(22)`

Solution :No. of atoms of `= N_(A) xx` No. of moles `xx 3`
`= 6.022 xx 10^(23) xx 0.1 xx 3`
`=1.806 xx 10^(23)`.
43.

The number of atoms in 0.1 mole of a triatomic gas is (N_(A) = 6.02 xx 10^(23) "mol"^(-1))

Answer»

`6.026 XX 10^(22)`
`1.806 xx 10^(23)`
`3.600 xx 10^(23)`
`1.800 xx 10^(22)`

Answer :B
44.

The number of atoms in 0.004 g of magnesium is close to

Answer»

24
`2 xx 10^(20)`
`10^(20)`
`6.02 xx 10^(23)`

Solution :`24.31 g` (gram atomic mass) of MG CONTAIN `6.022 xx 10^(23)` atoms.
45.

The number of atoms collinear in H_(3)C-C-=C-C-=C-CH_(3) is

Answer»

Four
Six
TEN
Eight

Solution :Six CARBON ATOMS present in the same PLACE.
46.

The number of benzene isomers for C_(8)H_(10) ?

Answer»

1
2
3
4

Answer :D
47.

The number of angular nodal planes are same in the orbitals:

Answer»

3P and 4p
3s and 4d
4s and 3p
4s and 3D

SOLUTION :Angular NODES = `l`
48.

The number of alkene (structural) isomers possible, each of which on hydrogenation gives 2-methyl pentane

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SOLUTION :4-Structural ALKENES POSSIBLE
49.

The number of alkali metals which is (are) radioactive.

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SOLUTION :FR is RADIO ACTIVE.
50.

How many structural isomers exist with the formula C_(4)H_(10)O?

Answer»

7
6
5
3

Answer :D