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 dative bonds around Mg^(2+) ion in chlorophyll is

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2
3
4
6

Answer :A
2.

Number of crystal systems having only 2 types of bravais lattices = x, number of crystal system having at least two interfacial angles equal = y and number of crystal systems having all the three edge lengths equal = z. Then find the value of x xx y xx z.

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ANSWER :24
3.

Number of dative bonds around Ag^+ ion in [Ag(NH_3)_2]^+

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2
3
4
6

Answer :A
4.

Number of Cr-O sigma bonds in dichromate ion Cr_(2)O_(7)^(2-) is

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6
7
8
4

Answer :C
5.

Number of Cr-O sigma bonds in dichromate in Cr_(2)O_(7)^(2-) is

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SOLUTION :`""^(Ө)O- OVERSET(O)overset(||)UNDERSET(O)underset(||)Cr-O-overset(O)overset(||)underset(O)underset(||)Cr-O^(Ө)`
6.

Numbers of coulombs required to deposit 90 gm of aluminium, when the electrode fraction is,Al^3+3e^(-)toAl.....9.65times10^4,8.68times10^5,9.65times10^5,6.95.

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`9.65 XX 10^4`
`8.68 xx 10^5`
`9.65 xx 10^5`
`6..95`

ANSWER :C
7.

Number of corner/oxygen atoms shared per tetrahedron in 3-D silicate and pyroxene silicate are ........and .......respectively-

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2,5,3
4,2
4,3
2,2

Answer :A
8.

Number of complexes that are paramagnetic in nature with number of unpairedare : 1.[MnCl_4]^(2-)2.[Mn_2(CO)_10] 3.[V(CO)_6]^(-)4. 5.[Pt(NH_3)_4Cl_2]6.[Co(NH_3)_2(H_2O)_4]Cl_2 7.[Ni(CN)_4]^(2-)8.[Cu(H_2O)_4]^(2+) 9. K_3[Cr[CN)_5]

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SOLUTION :`[MnCl_4]^(2-),[Ni(H_2O)_6]Cl_2,[Co(NH_3)_2(H_2O)_4]Cl_2,K_3[CR(CN)_6]`
9.

Number of CO groups that can be attached to chromium is

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Solution :APPLYING EFFECTIVE Atomic Number RULES, number of groups that can be attached to Cr=6.
10.

Number of following compounds in which central atom has "+6" oxidation state XeF_2, XeF_6, XeO_3, XeO_4, XeF_4,XeOF_4, XeO_2F_2

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SOLUTION :
11.

Number of chlorine atoms present gammaxine

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SOLUTION :
12.

Number of chlorine atoms in sucrolose

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SOLUTION :SUCRALOSE in trichlolo SUCROSE
13.

Number of chlorine atoms in D.D.T

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SOLUTION :DDThas 5CL ATOMS
14.

Number of chlorides satisfying secondary valency in CoCl_3 . 4NH_3

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1
2
3
6

Answer :B
15.

Number of chiral isomers are:

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SOLUTION :
16.

Number of chiral carbon atoms in the compound W, X, Y and Z respectively would be :

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Answer :W has 3, X has zero, Y has 2 and Z has only ONE CHIRAL centres.
17.

Number of carbonyl compounds which form stable geminal diol in below is x then value of x/3 is

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SOLUTION :1.33
18.

Number of chelate rings in [Fe(edta)]^(-)=a Numberof chelate rings in [Co (en)_2(NH_(3))Cl]^(2+)=b Number of chelate rings in [Ni(dmg)_2]=c Number of chelate rings in brown ring complex [Fe(H_2O)_5NO]SO_4=d Calculate the value of (a+b+c+d) is ....

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SOLUTION :a=5, b=2 , c=2,d=0`IMPLIES` (a+b+c+d)=9
19.

Number of chiral C-atom in glucose are …......

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4
5
3
6

Answer :A
20.

Number of carbon atoms in parent carbon chain of iso-octane is :

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6
4
5
8

Answer :B
21.

Number of carbon atom present linearly in C_(3)O_(2) :-

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2
3
6
1

Solution :`N-=C-C-=C-C-=N`
[FOUR CARBON ATOMS PRESENT LINEARLY]
22.

Number of C = O groups in thymine and in X are equal. Now, X is

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CYTOSINE 
URACIL 
ADENINE 
guanine 

Answer :B
23.

Number of bond pairs and lone pairs respectively that the central atom in xenon difluoride has

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2,6
2,3
2,4
2,2

Answer :B
24.

Number of beta particles emitted during the decay of ""_(92)^(235) U to ""_(82)^(207)Pb

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Solution :`""_(92) U^(235) to ""_(82) Pb^(207) + a(""_(2) alpha^(4)) + b(""_(-1) beta^(0)) , a= ((235-206)/(4)) = 7 alpha`
`92 = 82 + (2 xx 7) - b , b = 14 + 82 - 92 = 14- 10 = 4 beta`
25.

Number of B-O-B bond in borax

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Solution :Borax CONTAINS the tetranuclear UNITS, i.e.,
`[B_(4)O_(5)(OH)_(4)]^(2-)`

Structure of `[B_(4)O_(5)(OH)_(4)]^(2-)` ion
26.

Number of atoms of oxygen present in 10.6 g Na_(2)CO_(3)will be

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`6.022 xx 10^(22)`
`12.04 xx 10^(22)`
`1.806 xx 10^(23)`
`31.80 xx 10^(28)`

Solution :Molar mass of `Na_(2)CO_(3) = 2 xx 23 + 12 + 3 xx 16 = 106 g mol^(-1)`
`THEREFORE 10.6 g Na_(2)CO_(3) =0.1 mol`
`=0.1 xx 6.02 xx 10^(23)` molecules
`=3 xx 0.1 xx 6.02 xx 10^(23)` atoms of `O = 1.806 xx 10^(23)`
27.

Number of atoms of oxygen present in 10.6 g of Na_(2)CO_(3) will be

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`6.02xx10^(22)`
`12.04xx10^(22)`
`1.806xx10^(23)`
`31.80xx10^(28)`

Solution :Molar MASS of `Na_(2)CO_(3) = 2xx23+1216xx3`
`=46+12+48`
`=106gmol^(-1)`
`:.` 10.6 g of `Na_(2)CO_(3)-=0.1` MOL of `Na_(2)CO_(3)`
`=0.3` mol of oxygen atoms
`=0.3xx6.02xx10^(-23)` oxygen atom
`=1.806xx10^(23)` oxygen atom
28.

Number of atoms in one molecule of sulphur

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`1.8`
`2.4`
`3.3`
`4.1`

ANSWER :A
29.

Number of atoms in 558.5 g Fe (molar mass Fe= 55.85 g mol^(-1))

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twice than in 60 g carbon
`6.023 XX 10^(22)`
HALF that of 8g He
`558.6 xx 6.023 xx 10^(23)`

SOLUTION :558.5 g Fe = 10 moles
`=10 xx 6.023 xx 10^(23)`atoms.
60 g C = 5 moles `=5 xx 6.023 xx 10^(23)` atoms.
30.

Number of atoms in 588.6 g Fe (atomic mass of Fe = 55.86 g "mol"^(-1) ) is

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twice that in 60 g carbon
`6.023 XX 10^22`
HALF that of 8g He
`558.6 xx 6.023 xx 10^23`

Solution :No. of atoms ` = (558.6)/(55.86) xx 6.02 xx 10^23`
` = 10 xx 6.02 xx 10^23`
No . OFATOMS in 558.6 g = twice the no. of atoms in 60 g C .
31.

Number of atoms in 4.25 g of NH_(3) is:

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`6.023xx10^(23)`
`4xx60.23xx10^(23)`
`1.7xx10^(24)`
`4.5xx6.023xx10^(23)`

ANSWER :A
32.

Number of Aromatic compounds –

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1
4
3
2

Solution :AROMATIC COMPOUND MAKE loop-current.
33.

Number of atoms 558.5g Fe (At. Wt. of Fe=55.85g "mol"^(-1)) is:

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twice that in 60 G CARBON
`6.023xx10^(23)`
HALF that of 8 g He
`55.5xx6.023xx10^(23)`

ANSWER :A
34.

Number of angular nodes for 4 d orbital is

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4
3
2
1

Answer :C
35.

Number of Amino acids present in Aspartame ?

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SOLUTION :PHENYLALANINE + ASPARTIC ACID
36.

Number of amines among the following which give positive carbylamine test are C_(2)H_(5)NH_(2), (C_(2)H_(5))_(2)NH

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ANSWER :3
37.

Number ofamine isomer of the molecular formula C_(4)H_(11)N is/are structural isomers

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Solution :`C-C-C-C-NH_(2) C-C-UNDERSET(NH_(2))underset(|)(C)-C C-OVERSET(C)overset(|)(C)-C-NH_(2)`
`C-underset(NH_(2))overset(C)overset(|)(C)-C C-underset(C)underset(|)(C)-overset(H) overset(|)(N)-C C-C-underset(H)underset(|)(N)-C-C C-underset(C)underset(|)(N)-C-C underset(C)underset(|)(C)-C-overset(H)overset(|)(N)-C`
38.

Number of alpha- particles emitted per second by a radioactive element falls to 1//32 of its original value in 50 days. The half-life-period of this element is

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5 days
15 days
10 days
20 days

Solution :`T = 50` days, `t_(1//2) = ?, N_(0) = 1, N = (1)/(32)`
`N = N_(0) xx ((1)/(2))^(n) or (1)/(32) = 1 xx ((1)/(2))^(n)`,
or `((1)/(2))^(5) = ((1)/(2))^(n) or n = 5`
`T = t_(1//2) xx 2, or t_(1//2) = (50)/(5) = 10` days.
39.

Number of alpha - particles produced during the decay process of U - 235 to Pb - 207

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Solution :`""_(92) U^(235) to ""_(82) PB^(207) + a ( ""_(2) He^(4)) + b (""_(-1) beta^(0)) , a = (235 - 207)/(4) = (28)/(4) = 7 alpha`
40.

Number of aldehydes isomers of butanone is

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SOLUTION :`CH_3 - OVERSET(O) overset(||)C-CH_2 -CH_3 ` BUTANONE`CH_3 -CH_2-CH_2 - overset(O ) overset(||)C-H `,
41.

Number of acylon product(s) formed during this reaction. Me-overset(O)overset(||)C-OEt+Et-overset(O)overset(||)C-OMeunderset(liq.NH_(3))overset(Na)to

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ANSWER :8
42.

Number of acyclic structural isomers represented by molecular formula C_(4)H_(10)O is:

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7
6
8
5

Solution :
43.

Number of acidic hydrogen in Aspirin

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SOLUTION :1 ACIDIC HYDROGEN
44.

Number of acidic hydrogen atom in butyne-1 is:

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ANSWER :C
45.

Number of acidic hydrogen in 1-butyneis

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1
2
3
4

Answer :A
46.

Number of 3 central 2 electron bonds present in B_(2)H_(6) is

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ANSWER :2
47.

Number of 2^(@)C nitrogens present in given compound are

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2
3
4
5

Answer :A
48.

Number average molecular mass overset-(M_n) of synthetic and weigth average mol. Mass overset-(M_w) of synthetic polymer are related as :

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`OVERSET-(M_n)ltoverset-(M_w)`
`overset-(M_n)gtoverset-(M_w)`
`overset-(M_n)=overset-(M_w)`
`M_ngt M_w`

ANSWER :A
49.

Number average molecular mass, bar(Mn) and weight average molecular mass (barMw) of synthetic polymers are related as

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`bar(M_n) = (barM_(W))^(1//2)`
`bar(M_n) = barM_w`
`barM_w GT barM_n`
`barM_w lt barM_n`

SOLUTION :`PDI =(barM_w)/(bar(M_n))`
For synthertic polymer, PDI > 1
`:. barM_w gt barM_n`
50.

Nueleicacids are bio polymers of ………….

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SOLUTION :NUCLEOTIDES