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.

Number of isotopes of oxygen are:

Answer»

One
Three
Two
Zero

Answer :B
2.

Number of isomers of molecular formula C_(2)H_(2)Br. Is :

Answer»

1
2
3
4

Answer :C
3.

Number of isomers possible for [Co(en)_(2)Cl(NO_(2))]Cl are

Answer»

SOLUTION :Number of isomers for `[M(A-A)_(2)bc]^(npm)` of isomers POSSIBLE.
4.

Number of isomers formed by the coordination compound with the formula MABCD is

Answer»


SOLUTION :3 ISOMERS.
5.

Number of ions which are identified by dil. HCl from the following. (i) SO_(4)^(2-) "" (ii) CO_(3)^(2-) "" (iii) SO_(3)^(2-) "" (iv) HCO_(3)^(-) (v) SO_(3)^(2-) "" (vi) NO_(3)^(-) "" (vii) CH_(3)CO O^(-) "" (viii) PO_(4)^(3-)

Answer»


Answer :5 (II, iii, IV, V, vii)
6.

Number of isomeric ethers with molecular formula C_5H_12O are

Answer»

4
6
8
10

Answer :B
7.

Number of intramolecular aldol condensation product is:

Answer»

1
2
3
4

Answer :C
8.

Number of incorrect statement are- (A) The pi bond between metal and carbonyl carbon reduces the bond order of C-O in carbon monoxide. (B) dz^(2) orbital of central metal atom/ion is used in dsp^(2) hybridisation. (C) CN^(-) is a pi_(-) acid Ligand. (D) All negative ligands are stronger than neutral ligends.

Answer»


ANSWER :2
9.

Number of hydroxyl groups present in pyrosulphuric acid is:

Answer»

3
4
2
1

Answer :D
10.

number of hyperconjugable hydrogen atoms present in major product.

Answer»


SOLUTION :
11.

Number of hydrogen ions present in 10 millionth part of 1.33 cm^(3) of pure water at 25^(@)C is

Answer»

6.023 million
60 million
8.01 million
80.23 million

Solution :Now `[H^(+)]=10^(-7)` mole/litre
Now 1000 ml CONTAINS `10^(-7)` mole `H^(+)`
1 ml contains `10^(-7)/1000` mole `H^(+)`
`1.33xx10^(-7)` ml - contains `1.33xx10^(-17)` 10 million `=10^(-7)` so, `10 "million"^(TH)` PART of `1.33 cm^(3)=1.33xx10^(-7)` ml
so, no. of `H^(+)` ions `=1.33xx10^(-17)xxN_(A)`.
12.

Number of HlO_4molecular required to complete oxidation one mole of glucose is

Answer»

4
5
6
none

Answer :C
13.

Number of hydrogen ions present in 10 millionth part of 1.33 cm of pure water at 25°C is

Answer»

6.023 MILLION
60 million
8.01 million
80.23 million.

Solution :Now, `[H^(+)] = 10^(-7) mol//L`
Now, 1000 ML contains `10^(-7)` mole `H^(+)` ions
1 mL contains `10^(-7)/1000` mole `H^(+)`
`1.33 xx 10^(-7)` mL contains `1.33 xx 10^(-17)` mole `H^(+)` 10 million `=10^(-7)`
so, 10 million th part of 1.33 `cm^(3) = 1.33 xx 10^(-7)` mL
So, no. of `H^(+)` ions `=1.33 xx 10^(-17) xx N_(A)`
14.

Number of halide ions among (F^(-),Cl^(-),Br^(-),I^(-)) which change their oxidation number on heating with MnO_(2)+conc.H_(2)CO_(4)

Answer»


ANSWER :`0003`
15.

Number of haloform form reaction is given by following compounds are x.

Answer»

Solution :1.5
`H_(3)C-OVERSET(OH)overset(|)CH-CH_(3)""H_(3)C-overset(O)overset(||)C-CH_(3)""H_(3)C-overset(O)overset(||)C-Cl`
`H_(3)C-overset(O)overset(||)C-CH_(2)-overset(O)overset(||)C-CH_(3)""H_(3)C-overset(O)overset(||)C-OH""H_(3)C-overset(O)overset(||)C-NH_(2)`
Then VALUE of `(x)/(2)` is 1.5
16.

Number of H^(+) ions present in 250 ml of lemon juice of pH = 3 is

Answer»

`1.506 xx 10^(22)`
`1.506 xx 10^(23)`
`1.506 xx 10^(20)`
`3.012 xx 10^(21)`

ANSWER :C
17.

Number of H bonds present between adenine and thymine in formation of Nucleic acid

Answer»


SOLUTION :In between Adenine and THYMIN TWO hydrogen bonds FORMED
18.

Number of geometrical isomer(s) of square planar complex [RhCl(PPh_(3))(H_(2)O)(CO)] is :

Answer»

0
2
3
4

Answer :C
19.

Number of fused six membered rings in sterol ?

Answer»


SOLUTION :STEROL is 3-6 MEMBERED & 1-5 membered SUBSTANCE
20.

number of fractions obtained after fractional distillation of product mixture.

Answer»

SOLUTION :N//A
21.

Number of functional groups in the given compound is

Answer»


4
5
6

Solution :NA
22.

Number of following substituents those are deactivating but ortho and para directing.

Answer»


SOLUTION :NA
23.

Number of Faraday's required to generate one gram atom Cacl_2 is

Answer»

1
2
3
4

Answer :B
24.

Number of Faradays required to liberate 2 g of hydrogen is __________.

Answer»

2
4
6
1

Answer :A
25.

Number of faradays of electricity required to liberate 12 g of hydrogen is

Answer»

1
8
12
16

Answer :C
26.

Number of faradays of electricity required to liberate 12 g of hydrogen is :

Answer»

1
8
12
16

Solution :A Faraday is defined as the AMOUNT of electric charge in ONE mole of electrons. Since 1F of electricity is REQUIRED to liberate 1g of hydrogen thus, 12 Faraday of electricity is required to liberate 12 G of hydrogen.
27.

Number of faraday required to reduce a mole of Fe^(3+) to Fe^(2+)are:

Answer»

1
2
3
4

Answer :A
28.

Number of Faraday required to liberate 8g of H_2 is :

Answer»

8
16
4
2

Answer :A
29.

Number of Faraday needed to deposit 0.1 mole of copper from Cu(II) sulphate solution are :

Answer»

`0.1`
`0.2`
`0.05`
`0.5`

ANSWER :B
30.

Number of electrons transferred in each case when KMnO_(4) acts as an oxidising agent and converts into MnO_(2), Mn^(2+), Mn(OH)_(3) and MnO_(2)^(-) are respectively

Answer»

3, 5, 4 and 1
4, 3, 1 and 5
1, 3, 4 and 5
5, 4, 3 and 1

Answer :A
31.

Number of electrons transferred in each case when KMnO_4 acts as an oxidising agent to give MnO_2, Mn^(2+), Mn_2(O)_3 and MnO_4^(2-) respectively are :

Answer»

3, 5, 4 and 1
4, 3, 1 and 5
1, 3 , 4 and 5
5, 4, 3 and 1

Answer :A
32.

Number of electrons liberated on cathode by passing 1 ampere current for 60 second during electrolysis is . . . .(Charge on electrone =1.60xx10^(-19)C)

Answer»

`3.75xx10^(20)`
`7.48xx10^(23)`
`6XX10^(23)`
`6xx10^(20)`

SOLUTION :`Q=n*E^(-)`
`Ixxt=n*e^(-)`
`n=(1xx60)/(1.6xx10^(-19))=3.75xx10^(20)e^(-)`
33.

number of electrons present in M shell of an element with atomic number 26 in its M^(3+) state will be

Answer»

zero
8
13
14

Answer :C
34.

Number of electrons lost during oxidation of 0.355 g of Cl^(-) are :

Answer»

`0.01`
`0.01N_(A)`
`0.02N_(A)`
`0.01/(2N_(A))`

ANSWER :B
35.

Number of electrons present in the outermost orbit of Fe atom is:

Answer»

3
1
2
4

Answer :C
36.

Number of electronsinvolved in the electrodepositedof 63.5 g of Cu from a solution of CuSO_(4) is

Answer»

`6.022 XX 10^(23)`
`3.011 xx 10^(23)`
`12.044 xx 10^(23)`
`6.022 xx 10^(22)`

ANSWER :C
37.

Number of electrons in the nucleus of an element of atomic number 14 is

Answer»

14
7
7
0

Answer :D
38.

Number of electrons in outer most shell of Ce (Z = 58)

Answer»


Solution :`CE-[Xe]4F^(1)5D^(1)6s^(2)`
39.

Number of electrons in 3.6mg of NH_(4)^(+) are:

Answer»

`1.2XX10^(21)`
`1.2xx10^(20)`
`1.2xx10^(32)`
`2xx10^(-3)`

Solution :Number of electrons in one ION of `NH_(4)^(+)`=10
`=(3.6xx10^(-3))/(18)xx6.023xx10^(23)=1.2xx10^(20)`
`therefore` Number of electrons in 3.6mg `NH_(4)^(+)=1.2xx10^(20)xx10`
`1.2xx10^(21)`
40.

Number of electron in an atom having n=4 , m= 1 and m_(s)=-1//2 are

Answer»

16
8
32
6

Solution :NUMBER of electron in `n=4.2xx4^(2)=32`
HALF of these i.e 16 have `m_(s)=-1//2`
41.

Number of electrons gained by Pd in [PdCl_4]^(-2):

Answer»

4
8
10
Zero

Answer :B
42.

Number of electrons involved in the reduction of Cr_2O_7^(2-) ion in acidic solution to Cr^(3+) is:

Answer»

3
4
2
6

Answer :D
43.

Number of dipeptide which can be formed by : Glycine, Alanine, Leucine, Phenylalanine.

Answer»


SOLUTION :N//A
44.

Number of dichloro derivatives of tetramethylbutane is

Answer»


SOLUTION :
45.

Number of diastereomers possible for

Answer»

1
2
4
0

Solution :N//A
46.

Numberofdi - substitiutedisomers of theproduct of thereaction ,CH_(3) CH_(2) CH_(3) +Br_(2)wouldbe :

Answer»

2
1
3
4

Answer :D
47.

Number of diasterwomer of given compound :

Answer»

2
3
4
6

Solution : 3 DIASTEREOMERS.
48.

Number of d'electrons present in M shell of Ag+ ion?

Answer»

10
20
18
16

Answer :A
49.

Number of dative bonds (including sigma & pi) in P_4O_10 molecule is

Answer»

<P>

Solution :
each P = O consists of twodative BONDS
`4 - sigma ` DATIVE bonds
and `4-pi` dative bonds
so total 8 - dative bonds
50.

Number of dative bonds in the complex CoCl_(3).5NH_(3) is

Answer»

5
6
3
4

Answer :B