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.

In order that a compound may show geometrical isomerism A Compound must have a chiral carbon (B) Each of the two doubly bonded carbon atoms must have different atoms or groups (C ) Compound must have a plane of symmetry

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only A
only B
only C
None

ANSWER :B
2.

In order of decreasing contribution to the green house effect, the three main gases are _____, __________ and ______

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ANSWER :`CO_2 (50%) , CH_4 (19%) , CFC(17%)`
3.

In open vessel sugar solution hArr "sugar"_((s)) which information is obtained in this reaction ?

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SOLUTION :Here saturated solution of SUGAR is given. So at given equilibrium is ESTABLISHED between SOLID sugar and dissolve sugar.
4.

In one organic compound the weight proportion of C, H and N is 9 : 1 :3.5respectivily. Its molecular mass is 108 mg/mole. Find the molecular formula.

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Answer :`C_(6)H_(8)N_(2)`
5.

In one organic compound the proportion of C, H & Cl is 10%, 0.84% and 89.2 % respectively.Its empirical formula is.....

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`C Cl_(4)`
`CHCl_(3)`
`CH_(3)CL`
`CH_(2)Cl_(2)`

Solution :`C:H:Cl=(10)/(12):(0.84)/(1) : (89.2)/(35.5)`
`=0.83:0.84:2.51`
`= (0.83)/(0.83) : (0.84)/(0.83) : (2.51)/(0.83)`
`= 1:1:3`
`:.CHCl_(3)`
6.

In one organic compound percentage of C& H are 54.55 and 9.06. What will be its epirical formula.

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ANSWER :`C_(2)H_(4)O`
7.

In one of the following compounds, the oxidation number of sulphur is not a whole numbers

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`Na_(2)S_(4)O_(6)`
`H_(2)SO_(5)`
`H_(2)SO_(4)`
`Na_(2)S_(2)O_(3)`

SOLUTION :In `Na_(2)OVERSET(+2.5)(S_(4)O_(6)), 'S' O.S = +2.5`
8.

In one experiment, certain amount of NH_(4)I(s) was heated rapidly in a closed container at 375^(@)C. The following equilibrium was established: NH_(4)I(s)hArrNH_(3)(g)+HI(g) Excess of NH_(4)I(s) remained unreacted in the flask and equilibrium pressure was 304mm of Hg. After some time, the pressure started increasing further owing to the dissociation of HI. 2HI(g)hArrH_(2)(g)+l_(2)(g) K_(C)=0.010 calculate final pressure.

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`331mm "of" HG`
`335mm "of" Hg`
`662mm "of" Hg`
`151 MM "of" Hg`

SOLUTION :N//A
9.

In one discharge tube, H_(2) gas is taken and in the other O_(2) gas is taken. Will the electrons an positive ions in the cathode rays and in the anode rays be same or different ?

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SOLUTION :ELECTRONS will be same but POSITIVE IONS will not be same
10.

In one compound 54.55 % C, 9.09 % H and 36.36 % O are present. Its empirical formula is.....

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`C_(3)H_(5)O`
`C_(4)H_(8)O_(2)`
`C_(2)H_(4)O_(2)`
`C_(2)H_(4)O`

Answer :B::C::D
11.

In oil fire extinguisher, the compound used pyrene is chemically ………………. .

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SOLUTION :`C Cl_4`
12.

In OF_(2) ,number of bond pairs and lone of electrons are respectively

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`2,0`
`2,8`
`2,10`
` 2, 9 `

SOLUTION :
13.

In O_(2) molecule, the correct order of molecular orbitals is

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`PI2 pygt pi2 pz`
`pi 2 PY ^(**) = pi2pz`
`sigma 2s LT sigma 2s^(**)`
`sigma 2s ^(**)gt sigma2px ^(**)`

Answer :C
14.

In O_(2), H_(2)O_(2) and O_(3) the correct order of oxygen -oxygen bond length is

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`O_(2)gtO_(3)gtH_(2)O_(2)`
`O_(3)gtH_(2)O_(2)gtO_(2)`
`H_(2)O_(2)gtO_(3)gtO_(2)`
`O_(2)gtH_(2)O_(2)gtO_(3)`

Answer :C
15.

In O_(2)^(-) and O_(2)^(2-) which has more bond order ?

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SOLUTION :`O_(2)^(-) (1.5) and O_(2)^(2-) ` (1.0) THEREFORE BOND ORDER in `O_(2)^(-)`
16.

In nuclear reactors, heavy water is used as a

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Fuel
Projectile
Moderator
Coolent

Answer :C
17.

In nuclear transmutation tritium is obtained from

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`""_7N^14`
`""_7C^14`
`""_LI^6`
`""_8O^16`

ANSWER :C
18.

In nomenclature of cyclic compound…… prefix is written before the name of simple chain containing alkane.

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ANSWER :CYCLO
19.

In NO_(3)^(-) ion, the number of bond pairs and lonepairs of electrons on nitrogen atom are

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2 ,2
3, 1
1, 3
4, 0

Solution :`NO_(3)^(-) = [underset(. .) overset(. . ) O= underset(. .) underset(":O:")underset(|)N- underset(. .)overset(. . )O: ] ^(-)`, Bond PAIRS on N = 4, LONE pairs = 0
20.

In NO_(3)^(-)ion, the number of bond pairs and lone pa1rs of electrons on nitrogen atom are

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2,2
3,1
1,3
4,0

Solution :4,0
In N-atom, are = 5 valence electrons = 5
As the presence of one negative charge, NUMBER of valence electrons = 5 + 1 = 6 one O-atom forms TWO bond (= bond) and two O-atom shared with two electrons of N-atom
So, 3 O-atoms shared with 8 electrons of N-atom.
`therefore `Number of bond PAIRS (or shared pairs) = 4 Number of lone pairs = 0
21.

In {:(-NO_(2),-NH_(2),-SO_(3)H),(I,II,III):}, the decreasing order of -I effect is

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`I GT II gt III`
`I gt III gt II `
`III gt II gt I`
`III gt I gt II `

SOLUTION :N/A
22.

In nitric oxide (NO), the oxidation state of nitrogen is

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`-2`
`+1`
`-1`
`+2`

SOLUTION :`NO,x-2=0impliesx=+2`
23.

In nitration of benzene ……is electrophile.

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NITRATE ions
Nitrite ions
Nitronium ions
Nitro

Answer :C
24.

In neutralisation of KI by AgNO_(3) positive charge is due to adsorption of

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`AG^(+)` ions
Ag
I ions
Both (B) and (C)

ANSWER :A
25.

In nature, non metals are found in the

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native STATE
reduced form
oxidised form
form of alloys with metals.

Solution :In nature , non-metals are found in native state `(N_2, O_2, `C ETC .), reduced form (`CH_4, H_2S, NH_3 etc .), oxidised form (`CO_2, H_(2)O` etc.)
26.

In natue a decay chain sereis starts with ._(90)h^(232) and finally terminates at ._(82)Pb^(208). A throium ore sample was found to contain 8xx10^(-5) mL of He At STP and 5xx10^(7)g of Th^(232). Find the age of ore sample assuming that sourcesof He to be only due to decay of Th^(232) Also assumecomplete retention of He within the ore, t_(1//2) for Th^(232) = 1.39xx10^(10) year.

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Solution :We KNOW,
`._(90)Th^(232) rarr ``._(82)Pb^(208) + 6 ``._(2)He^(4) + 4 ``._(-1)e^(0)`
`:' 6xx22400 mL` He is formed by `232g Th` decay
`:. 8xx10^(-5) mL` He is formed by
`= (232xx8xx10^(-5))/(6xx22400)g Th` decay
`= 1.38xx10^(-7)g Th` decay
At `t = t`, sample has `Th = 5xx10^(-7) g prop N`
At `t = 0`, sample has
`Th = 5xx10^(-7) + 1.385 xx 10^(-7) prop N_(0) = 6.38xx10^(-7)g`
For `Th` decay `:' t = (2.303)/(LAMBDA) log_(10) (N_(0))/(N)`
`= (2.303xx1.39xx10^(10))/(0.693) log_(10) (6.38xx10^(-7))/(5x10^(-7))`
`=4.89xx10^(9)` year
27.

In naturally available hydrogen, the percentage of tritium is

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`99.986%`
`0.014%`
`7X10^(-16)%`
`0.02%`

ANSWER :C
28.

In NaCl is doped with10^(-4) mol % ofScCl_(2) , the concentration of cation vacancies will be(N_(A) = 6.02 xx 10^(23) mol^(-1))

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` 6.02 xx 10^(14) mol^(-1)`
`6.02 xx 10^(15) mol^(-1)`
` 6.02 xx 10^(16) mol^(-1)`
` 6.02 xx 10^(17)mol^(-1) `

Solution :For each ` Sr^(2+)`ion INTRODUCED, one catio vacancy is created because` 2 Na^(+)`ions are REMOVED and one vacant site is occupied by ` Sr^(2+)` . Doping with ` 10^(-4) "mol"%" of " SrCl_(2)`means 100 moles of NaCl are doped with ` 10^(-4)` mole of ` SrCl_(2)`
`SrCl_(2)` doped per mole of NaCl = `10^(-4) //100`
` 10^(-6)` mole
=`10^(-6) xx ( 6.02 xx 10^(23)) Sr^(2+)` ions
` = 6.02 xx 10^(17) Sr^(2+)` ions
Hence, concetration of cation vacancies
` 6.02 xx 10^(17) "mol"^(-1)`
29.

In NaCl crystals , Cl^- ions are in FCC arrangement. Calculate the number of Cl^- ions in its unit cell

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SOLUTION :`Cl^-` ion per UNIT CELL =`8xx1/8` (from corners ) + `6xx1/2` (from face CENTRES)=4
30.

In NaCl crystal, Ci^(-1) ions are I FCc arrangement, Calculate the number ofCl^(-)ions in its unit cell.

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Solution :`CL^(-)` ION per unit CELL = ` 8 xx 1/8 "( from corners)" + 6 xx 1/2`( form face CENTRES ) = 4.
31.

In NaCl, Cl^- ions are present in the _________ structure whereas Na^+ ions are present in the _______ voids.

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ANSWER :FCC, OCTAHEDRAL
32.

In Na and K metals, which salts are high proportion in the ash of shrubs ?

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`PO_(4)^(3-)`
`CO_(3)^(2-)`
`NO_(3)^(1-)`
`SO_(4)^(2-)`

Answer :B
33.

In mutarotation of alpha-D Glucose hArr beta-D Glucose equilibrium constant is 1.8. At equilibrium what will be the percentage of alpha-D Glucose ?

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35.8
64.3
55.6
44.4

Solution :`{:(,alpha-"D-GLUCOSE" hArr, BETA-"D-Glucose"),("INITIAL",1,0),("At EQUILIBRIUM" , 1-alpha,alpha):}`
`K=alpha/(1-alpha)`=1.8
`alpha=1.8-1.8alpha`
`therefore 2.8 alpha-1.8`
`therefore alpha`=0.642
`therefore 1-alpha`=0.358
So, at the equilibrium of `alpha`-D Glucose =35.8%
34.

In multi-electron atom, 4s - orbital is lower in energy than

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3D -orbital
3p - orbital
2s - orbital
2p - orbital

Answer :A
35.

In molecule, the formal charges of oxygen atoms 1,2,3 are respectively

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`-1,0,+1`
`0,-1,+1`
`0,+1 ,-1`
`+1,0,-1`

ANSWER :D
36.

In modern periodic table, if an element has electronic configuration 1s^(2), 2s^(2), 2p^(6), 3s^(2), 3p^(4), then what will be the atomic number of element, which is just below the element in its group ?

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18
34
45
24

Solution :34
Electronic configuration of the element
`1S^(2) 2s^(2) 2p^(6) 3s^(2) 3P^(4)`
` :. ` TOTAL electrons in it = 16 and Z = 16
Which INDICATES element is S(Z = 16) of`III^(rd)` Period, and`16^(th)` /VIB GROUP. Element just below S in same group is Se having atomic number 34.
37.

In Millikan's oil drop experiment, we make use of:

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OHM's law
Ampere's law
Stoke's law
Faraday's law

Answer :C
38.

In Millikan's experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge in the oil drop is -1.282 xx 10^(-18)C, calculate the number of electrons present in it.

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Solution :ELECTRONS present = `(-1.282 xx 10^(-18)C)/(-1.6022 xx 10^(-19)C) = 8`
39.

in milikanexperimentstaticelectricchargeon theoildrops has beenobtainedby shiningX-raysif thestaticelectric chargeon theoildropis-1.282xx 10^(18) Ccalculatethe numberof electronpresencton it.

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Solution :Staticelectricchargeon theoil drop(q ) =1.282 `XX 10^(18) C`
Chargeon oneelectron(e ) = 1.6022 `xx 10^(19) C`
NUMBEROF electron= `(q ) /( e )`
`(1.282 xx 10^(18) C)/( 1.6022xx 10^(19) C)`
8.00
40.

In [Mg(H_(2)O)_(6)]^(2+) Mg hybridiasationis

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`d^(2)SP^(3)`
`sp^(3)d^(2)`
`dsp^(2)`
`sp^(3)`

Answer :1
41.

In MgH_2oxidation number of hydrogen is:

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`-1`
`+1`
`-2`
`+2`

Solution :`Mgoverset(-1)(H_2)` , In METALLIC HYBRIDES , oxidation state of HYDROGEN is -1.
42.

In meta-aluminates, co-ordination number of aluminium is

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3
6
1
4

Solution :In meta-aluminates, co-ordinates number of aluminium is 6.
`UNDERSET("Meta-aluminate ion")([AL(OH)_(4)(H_(2)O)_(2)]^(-))`
43.

In melting of ice, which one of the conditions will be more favorable ?

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HIGH TEMPERATURE and high PRESSURE
LOW temperature and low pressure
low temperature and high pressure
high temperature and low pressure

ANSWER :A
44.

In measurement of BOD_x, is generally taken as

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SOLUTION :We generally MEASURE `BOD_5` where 5 represents number of DAYS for which the sample is kept, i.e.amount of `O_2` consumed in 5 days
45.

In long form of periodic table, an element has electronicconfiguration 1s^(2)2s^(2)2p^(6)3s^(2)3p^(3). Thus what will be the atomic number of element which is just below this element in its group ?

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<P>23
16
33
47

Solution :Electronic CONFIGURATION is given as : `1S^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(3)`
` :. ` TOTAL electrons = 15
` :. ` Element is P of group 15 or VB The element just below to P is As As has ATOMIC number = 33 and As is of group 15 or VB
46.

In long form of periodic table, an elem ent has electronic configurationIs^(2) 2s^(22)p^(6)3s^(2)3p^(3) Thus what will be the atomic no. of elem ent ? Which is just below this elem ent in its group ?

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23
16
47
33

Solution :The GIVEN CONFIGURATION indicates Z = 15, and outerm ost configuration `3s^(2) 3p^(3)` ..... indicates group `15^(th) , row 3^(rd)therefore`Next row `4^(th)` and group `15^(th)` element will behaving Z = (15 + 18) = 33
47.

In lithium aluminium hydride molecule,how many hydrogen atoms are present ?

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SOLUTION :`LiAlH_4` It has FOUR HYDROGEN ATOMS
48.

In Leibig's method 0.24 g of organic compound on combustion with dry oxygen produced of 0.62 g of CO_(2) and 0.11 g of H_(2)O. Determine the percentage composition of the compound.

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SOLUTION :Mass of organic COMPOUND `=m=0.24g`
Mass of carbon dioxide formed `=0.62 g`
Mass of water formed `=0.11 g`
Percentage of carbon `=12/4xx0.62/0.24xx100=70.45`
Percentage of hydrogen `=2/4xx0.11/0.24xx100=5.09`
Percentage of oxygen `=[100-(70.5+5.0)]=24.46`
49.

In Lassaigne's test, when both N and S are present blood red colour is obtained. This is due to the formation of

Answer»

FERRIC ferrocyanide
ferric cyanide
ferric thiocyanate
ferric hydroxide.

Solution :`Na + C + N+S to Na CNS`
`Fe^(3+) SCN^(-) to [Fe (SCN)_3]`
Ferric thiocyanate
50.

In Lassaigne's test, thiourea is converted into

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NACN
`Na_(2)S`
`Na_(2)SO`
NaCNS

Answer :D