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.

Give ozonized structure for following : (i) Propene (ii) Ethene (iii) 1-butene (iv) 2-butene (v) Benzene (vi) Buta-1,3-diene

Answer»

Solution :(i) Ozonized structure of propene `(CH_(3)CHC=CH_(2))`

(ii) Ozonized structure of ethene `CH_(2) = CH_(2)`

(iii) Ozonized structure of 1-butene `(CH_(2)=CHCH_(2)CH_(3))`

(iv) Ozonized structure of 2-butene `(CH_(3)CH=CHCH_(3))`

(v) Ozonized structure of BENZENE. It has THREE C=C and so its ozonized sructure is `C_(6)H_(6)(O_(3))_(2)`

(vi) Ozonized structure of Buta-1,3-diene `(CH_(2)=CH-CH=CH_(2))` is dioznonized `C_(4)H_(6)(O_(3))_(2)`
Discussion
2.

Give oxidation state of gallium.

Answer»

SOLUTION :GALLIUM have +1 and +3 oxidation STATE.
Discussion
3.

Give origin of group 13 elements

Answer»

Solution :Boron is a typical non-metal
Aluminium is a metal but shows MANY chemical similarities to boron.While gallium , indium ,thallium and nihonium are almost exclusivelymetallic in character.
Boron is a fairly rare element , it mainly occurs as orthoboric acid `(H_3BO_3)`, borax `(Na_2B_4O_7. 10H_2O)` and kernite `(Na_2B_4O_7 . 4H_2O)`.
In India borax occurs in Puga valley (Ladakh) andSambhar Lake (Rajasthan) .The abundance of boron in earth crust is less than0.0001% by mass.
There are two ISOTOPIC forms of boron `.^10B` (19%) and `.^11B` (81%)
Aluminium is the most abundant metal and the THIRD most abundant element in the earth.s crust (8.3% by mass) after oxygen (45,5%) and Si (27.7%)
Bauxite `(Al_2O_3 . 2H_2O)` and cryolite `(Na_3AlF_6)` are the important minerals of aluminium.In India it is found as MICA in Madhya Pradesh, Karnataka , Orissa and Jammu.
Gallium , indium and thallium are less abundant elements in nature .
Discussion
4.

Give order of repulsion of electron with central atom in covalent molecule.

Answer»

Solution :`({:("REPULSION of "),("NON boding "),("electronpair"),("and non bonding"),(e^(-)"pair"):}) gt ({:("Repulsion of "),("non boding "),("pair"),("bonding "e^(-)),("pair"):}) lt ({:("Repulsion of "),("non bonding "),(e^(-) "pair" ),("bonding " e^(-) ),("pair "):})`
i.e., `({:(LP - lp),("repulsion"):}) gt ({:(lp - bp),("repulsion"):}) gt ({:(bp - bp),("repulsion"):})`
Here, lp = nonbonding `e^(-)` pair
bp = bonding `e^(-)` pair
Discussion
5.

Give order of stability for different alkenes.

Answer»

Solution :`R_(2)C=CR gt R_(2)C = CHR gt RCH = CHR ("trans") gt`
`R_(2)C = CH_(2) gt RCH = CHR ("cis") gt RCH = CH_(2) gt CH_(2)=CH_(2)`.
Discussion
6.

Give order of energy of solid, liquid and gas.

Answer»

Solution :ENERGY ORDER : Solid `LT` Liquid `lt` GAS
Discussion
7.

Give order of kinetic energy of solid, liquid, gas.

Answer»

Solution :ORDER of kinetic ENERGY : SOLID `lt` LIQUID `lt` Gas
Discussion
8.

Giveorbitalpictureof followingelementsNa MgAl ,SiP , S , CI , Ar , K , Ca

Answer»

Solution :
NOTE: where`NE^(10)`MEANS`1s^(2)2s^(2) 2p^(6) ` and`(Ar)^(18)` means`1s^(2)2s^(2) 2p^(6) 3S^(2) 3p^(6) `
Discussion
9.

Give only reaction for detection of nitrogen in organic compound

Answer»

SOLUTION :In fusion with Na: `N + Na + C rarr NaCN`
`6CN^(-) Fe^(2+) rarr [Fe(CN)_(6)]^(4-)`
`Fe^(2+) rarr Fe^(3+)` and then
`4Fe^(3+) + 3[Fe(CN)_(6)]^(4-) overset(xH_(2)O)rarr Fe_(4) [Fe(CN)_(6)]_(3). x H_(2)O`
Discussion
10.

Give one molecule that does not obey octet rule.

Answer»

SOLUTION :HYDROGEN MOLECULE.
Discussion
11.

Give one method for industrial preparation and one forlaboratory preparationof CO and CO_(2) endcarbonmonoxide.

Answer»

Solution :`{:("CARBON monoxide",, "Carbon dioxide"),("Industrial" : 2C(s) + O_(2)(g) overset("Limited")underset("air")rarr2CO(g),,,C(s)+O_(2)(g)overset("Excess")underset("air")rarrCO_(2)(g)),("Laboratory": underset("Formic acid")(HCOOH)overset(H_(2)SO_(4))rarrCO+H_(2)O,,,CaCO_(3)(s)+HCL(AQ) rarr CaCl_(2)(aq) + CO_(2)(g) + H_(2)O(L)) :}`
Discussion
12.

Give one method for industrial preparation and one for laboratory preparation of CO and CO_2 each.

Answer»

Solution :Carbon monoxide (CO):
CO is commercially prepared by passing steam over hot coke. The reaction involved is as follows:
`2C_((s)) + O_2 underset"COMBUSTION"overset"Incomplete"to 2CO_((g))`
In the laboratory, CO is prepared by the dehydration of formic ACID with conc. `H_2 SO_4` at 373 K. The reaction involved is as follows:
`underset"Formic acid"(HCOOH) overset(H_2SO_4) toCO+ H_2O`
Carbon dioxide `(CO_2)`:
In the laboratory, `CO_2` can be prepared by the action of dilute hydrochloric acid on calcium carbonate. The reaction involved is as follows:
`CaCO_3 + 2HCl to CaCl_2 + CO_2 + H_2O`
`CO_2` is commercially prepared by heating limestone. The reaction involved is as follows :
`C_((s)) + O_(2(g)) underset"combustion"overset"COMPLETE"to CO_(2(g))`
Discussion
13.

Give one important uses of following compounds. (i) NaHCO_(3) (ii) Slaked lime (iii) NaOH

Answer»

Solution :(i) Sodium bicarbonate is USED as ANTACID. (ii) slaked lime is used for WHITE washing (iii) NAOH is used in manufacture of soap.
Discussion
14.

Give one Important pratical application of photoelectric effect

Answer»

Solution :In automatic 'electric EYE' door openers, when a person come near to the door, the interuption LIGHT BEAM is CAUSED by the person. As a result, the door automatically OPENS
Discussion
15.

Give one important ore of each of sodium and potassium .

Answer»

SOLUTION :NA : NACL and K : KCL .
Discussion
16.

Give one example with structure and name for each of the following compounds. (a) Primary haloalkane (b) Secondary haloalkane (c) Tertiary haloalkane

Answer»

SOLUTION :(a) Primary HALOALKANE`UNDERSET("CHLOROETHANE")(CH_3-underset(H)underset(|)overset(H)overset(|)C- Cl)"("1^@" Haloalkane)"`
(b) Secondary haloalkane`underset("2-chloropropane")(CH_3-underset(H)underset(|)overset(H)overset(|)C -CH_3)(2^@"Haloalkane")`
(c ) Tertiary haloalkane
`underset("2-Iodo-2-methylpropane")(CH_3-underset(I)underset(|)overset(CH_3)overset(|)C -CH_3)(3^@"Haloalkane")`
Discussion
17.

Give one example of everyday life in which there is gas solution equilibrium .

Answer»

SOLUTION :SODA- WATER BOTTLE.
Discussion
18.

Give one example of each of the following reactions : (i) Wurtz Reaction(ii) Wurtz - Fitting reaction

Answer»

Solution :(i) Wurtz Reaction : It involves conversion of alkyl halides into ALKANE . Example `CH_(3) Cl + 2 NA + CH_(3) Cl overset("DRY ETHER ") (to) 2 NaCl + C_(2) H_(6)`
(ii) Wurtz - FITTING reaction : It involves the reaction of an aryl halide and alkyl halide to form the corresponding hydrocarbon.
Discussion
19.

Give one example of each of the following: (i) Isotope of ._(17)^(35)Cl (ii) Isobar of ._(18)^(40)Ar (iii) Isotone of ._(6)^(14)C

Answer»

SOLUTION :(i) `._(17)^(37)Cl` (ii) `._(19)^(40)K`(iii) `._(8)^(16)O`
Discussion
20.

Give one example of a reversible reaction taking place in aqueous solution.

Answer»

Solution :`CH_(3)COOH (l) + C_(2) H_(5) OH (l) HARR CH_(3) COOC_(2)H_(5) (l) + H_(2)O (l)`
Discussion
21.

Give one example of a compound containing ionic, covalent and coordinate bond.

Answer»

SOLUTION :`[H_(3)NtoH^(*)]+CL^(-)=NH_(4)Cl` AMMONIUM chloride.
Discussion
22.

Give one example for the relation between K_(p)andK_(c) for a reaction, if Deltan= 0.

Answer»

SOLUTION :`N_(2(G))+O_(2(g))hArr2NO_((g))`
Discussion
23.

Give two examples for each of the following type of organic compounds. (i) non-benzonoid aromatic, (ii) aromatic heterocyclic, (iii) alicyclic and (iv) aliphatic open chain.

Answer»

SOLUTION :(i)
Discussion
24.

Give one example for heterocyclic aromatic compound.

Answer»


ANSWER :PYRIDINE.
Discussion
25.

Give one example each for homogeneous and heterogeneous equilibrium.

Answer»

SOLUTION :HOMOGENOUS equilibrium: `N_(2(g))+O_(2(g))hArr2NO_((g))`
HETEROGENOUS equilibrium: `CaCO_(3(S))hArrCaD_((s))+CO_(2(g))`
Discussion
26.

Give one chemical reaction to show that : (i) Tin (II) is a reeducing agent whereas Pb(II) is not. (ii) Tin (II) chloride is a reducing agent.

Answer»

Solution :(i) Both tin and lead show two oxidation state of `+2` and `+4` due to inrt pair effect . But the inert pair effect. But the inert pair effectis more prominaent in case of Pb than in Sn. In other words +2, oxidation state of Sn is less stablethan its +4oxidation state. Therefore, `Sn(II)`acts asa reducing agent and GETS cor. erted into more stable `Sn (IV)` by losing two electrons. For example it reduces `Fe^(3+)` to `Fe^(2+)` ions.
`2Fe^(+) + Sn^(2+) RARR 2Fe^(2+) + Sn^(4+)`
In contrast`+2` oxidation state of Pb is more stable than its +4 oxidation state. In other words, `Pb(II)` does not lose electrons easily and HENCE does not acts as a redusingagent.
(ii) Being areducingagent,tin (II) CHLORIDE reeducesmercuric chloride `(HgCl_(2))` to mercurous chloride `(Hg_(2)Cl_(2))` and ferric salt to ferrous salts.
`SnCl_(2) + 2HgCl_(2) rarr SnCl_(4) + Hg_(2)Cl_(2)`
`SnCl_(2) + 2FeCl_(3) rarr SnCl_(4) + 2 FeCl_(2)`
Discussion
27.

Givenumberofunpairedelectronin Ni^(2+)

Answer»

0
1
3
2

Answer :D
Discussion
28.

Give number of sigma and pi bonds in above compounds of questions (i).

Answer»

Solution :8 C-H `SIGMA` bond + 5 C-C `sigma` bond = TOTAL 13 `sigma` bond and `3pi` BONDS are present.
Discussion
29.

Give number of pi bond and sigma hond of the following. (i) Methane (ii) Ethane (iii) Ethene (iv) Ethyne (v) dioxygen (vi) dinitrogen (vii) Benzene (viii) Ozone

Answer»

SOLUTION :
Discussion
30.

Give number of H atoms in 0.046 g ethanol.

Answer»

`6xx10^(20)`
`1.2xx10^(21)`
`3XX10^(21)`
`3.6xx10^(21)`

ANSWER :A::B::C::D
Discussion
31.

Give number of H atom attach with terminate carbon in

Answer»

Solution :At the end of the C-H is PRESENT MEANS, at LEFT handside, carbon has triple BOND. So, it has one hydrogen other right end indicates the `CH_(3)`
Discussion
32.

Give number of carbon and its name in alkane

Answer»

Solution :(A) Number of carbon - Prefixies: The name of such compounds are based on their carbon carbon chai and this carbon chain is decided by number of carbon

(B) 1 to 10 carbon containing alkane and its name and formula: The name of such compounds are based on their chain structure adn with suffix .-ane. and CARRY a prefix indicating the number of carbon atoms present in teh chain. e.g. If 4 carbon present then .But. + suffix .ane. i.e., the name is Butane. Some formula and names of alkane series are as under. The alkanes are differ from each other by merely the number of `-CH_(2)` group in the chain.
The general formula of alkane: `C_(n) H_(2N + 2)`
Prefix: Alk: Alk.Suffix: ane Name : Alkane
Discussion
33.

Give num ber of protons, electrons and neutrons respectively in ._(16)^(32)S^(2-)

Answer»

16,16,16
16,14,16
16,16,14
16,16,18

Answer :D
Discussion
34.

Give nitration products of toluene and benzoic acid ?

Answer»

SOLUTION :
Discussion
35.

Giveno. of energystate,orbitalandelectronsin N orbital .

Answer»

4,12,32
4,16 ,30
4,16 ,32
4,32,64

Solution :E=hv `XX 6.022 xx 10^(23) J mol^(-1)`
Discussion
36.

Give name of two drugs used in treatment of cancer.

Answer»



SOLUTION :TAXOL and CISPLATIN.
Discussion
37.

Give name of group-2 elements.

Answer»

Solution :The group-2 ELEMENTS comprise BERYLLIUM, MAGNESIUM, calcium, STRONTIUM, barium and radium.
Discussion
38.

Give name of group 13 elements.

Answer»

SOLUTION :Boron (B), Aluminum (Al), GALLIUM (Ga), INDIUM (In) and Thallium (Th) are group 13 ELEMENTS. They have `ns^2 np^1` general electronic configuration.
Discussion
39.

Give name of Cao.

Answer»

CALCIUM Oxide
Quick lime
Both (A) and (B)
NONE

ANSWER :C
Discussion
40.

Give name of following

Answer»

Solution :(a) (i) Methyl CYCLOPENTANE (ii) 1-cyclopentyl HEXANE
(B) (i) Methyl cyclopropane (ii) 1-cyclopropyl butane
(c ) 2-(Butyl) CYCLOHEXANE
(d) 3, 5-cyclohex -1-ene
Discussion
41.

Give name of element, symbol & electronic configuration of 1^(st)group.

Answer»

Solution :Elements in the same VERTICAL column or group have similar valence shell electronic configurations, the same number of electrons in the outer orbitals and similar properties.
e.g. The group 1 elements (alkali METALS `ns^(1)` have his valence shell electronic CONFIGURATION as shown below :

THSU it can be see that the properties of an element have PERIODIC dependence upon its atomic number and not on relative atomic mass .
Discussion
42.

Give name of 4 compounds of polymerized substance. And give their polymer name.

Answer»

Solution :(i) `UNDERSET("BENZENE")underset(darr)("Acetelene")` (ii) `underset("polyethelene")underset(darr)("Ethelene")` (III) `underset(PVC)underset(darr)("Vinyl chloride")`
(iv) `underset("Polypropene")underset(darr)("PROPENE")` (v) `underset("POLYBUTADIENE")underset(darr)("Butadiene")`
Discussion
43.

Give name and structure of cancer causing hydrocarbon.

Answer»

SOLUTION :
Discussion
44.

Give name and structure of aromatic hydrocarbon with molecular formula C_(12)H_(10) which contain two benzene ring.

Answer»

Solution :`C_(12)H_(10)` POSSESS two benzene RING. So two `C_(6)H_(5)` structure is possible, such two `C_(6)H_(5)` compound on FUSION will gives biphenyl.
Discussion
45.

Give name and structure of aromatic compound with C_(10)H_(8) molecular formula. And give the number of sigma and pi bonds present in the structure.

Answer»

Solution :`C_(10)H_(8)` is aromatic compound. So it has 14 LESS Hydrogen that `C_(10)H_(22)`, it possess two RINGS and 5 DOUBLE BOND, the compound is nepthelene.
There are 11 C-C sigma bond and 8 C-H sigma bonds so total 19 sigma bonds are present in it.
It has 5 double bond. So it possess five C=C `pi` bond.
Discussion
46.

Give moles of 50 gAl_(2) (SO_(4))_(3) .........

Answer»

0.083
0.952
0.481
0.146

Answer :A::D
Discussion
47.

Give molecular formula for washing soda.

Answer»

`Na_(2)CO_(3)*10H_(2)O`
`Na_(2)CO_(3)*H_(2)O`
`Na_(2)CO_(3)`
`NaHCO_(3)`

ANSWER :A
Discussion
48.

Give molecular structure of SO_(2) and calculate formal charge.

Answer»

SOLUTION :
BONDING electron : 6
Nonbonding electron : 12
Total electron : 18
`N_(A)` = Non.bonding electron
`N_(B)` = Bonding electron
Total valence electron : 6 + 12 = 18
Formal charge of O (1) = `({:("velence"),("electron"):}) - (N_(A) + (1)/(2) N_(b))`
= 6 - `[ 4 + (1)/(2) (4)] = 6 -6 = 0 `

Formal charge of O (2) = `6 - [ 6 + (1)/(2) (2) ] , `
Formal charge of S = ` 6 - [ 2 + (1)/(2) (6) ] = 6 - 5 = +1 `
Discussion
49.

Give molarity and strength % w/v accordingly in 30 volume of H_2O_2 solution.

Answer»

2.678 M, 18.22%
2.678 M, 9.11%
5.356 M, 9.11%
5.356 M, 18.22%

ANSWER :A
Discussion
50.

Give MO diagram and explain Ne_(2) molecule is not possible.

Answer»

Solution :`Ne_(2) (z = 10) 1S^(2) 2s^(2) 2p^(6)`. In `Ne_(2)` valence orbital 8 electrons and affected in bond structure are 16 electrons.
ELECTRON configuration in MO for `Ne_(2) ` :
KK `(sigma 2s)^(2) (sigma^(**) 2s) (pi 2p_(z))^(2)(pi 2p_(x))^(2) = (pi 2p_(y))^(2) (pi^(**)2p_(x))^(2)= (pi^(**) 2p_(y))^(2) (sigma^(**) 2p_(z))^(2) `
Bond order `= (1)/(2) (N_(b) - N_(a)) = (1)/(2) (10 - 10)` = 0
BO is zero, So it is unstable means does not exist. energy DIAGRAM for `Ne_(2)` .
Discussion