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.

When Benzene is treated with chlorine under the influence of u.v. light, a solid material of molecular wt. 291 is formed. Quantitative analysis gives an empirical formula of CHCI. What statement(s) is/are correct regarding above information

Answer»

Product is aromatic
Product can be separated into six isomeric COMPOUNDS ONE of which has been used as insecticide
Product formed by addition reaction
Product CONTAIN nearly 73% chlorine

Solution :
Discussion
2.

Which of the following gases is used for welding

Answer»

Methane
Ethane
Acetylene
Ethene

Solution :`2CH-=CH + 5O_(2) rarr 4CO_(2) + 2H_(2)O Delta H = -1300 KJ`
The combustion of acetylene is HIGHLY exothermic and the HEAT PRODUCED during the combustion can be used for welding PURPOSES in the form of oxy acetylene flame.
Discussion
3.

The spin of the electron

Answer»

increases the angular momentum
decreases the angular momentum
can be forward (clockwise) relative to the direction of the path of the electron
can be BACKWARD (anti-clockwise) relative to the direction of the path of the electron

SOLUTION :SPIN MOTION is relative
Discussion
4.

What type of semiconductor is obtained when silicon is doped with arsenic ?

Answer»

SOLUTION :n-type SEMICONDUCTOR
Discussion
5.

What is the meaning, "Dipole moment is a vector quantity." How it is express ?

Answer»

Solution :Dipole moment is a VECTOR quantity, I.e., it has MAGNITUDE as well as direction.
It is representedby an arrow with its TAIL at the positive CENTRE and BEAD pointing towards the negative end. `rarr`
Discussion
6.

Which of the following metal carbonates is decomposed on heating?

Answer»

`MgCO_3`
`Na_2CO_3`
`K_2CO_3`
`Pb_2CO_3`

SOLUTION :`MgCO_3`
Discussion
7.

what is molal depression constant ? Does it depend on natureof the solute?

Answer»

SOLUTION :` K_f ` =molal freezing POINT depression constant or cryoscopic constant.
` DELTA T_f =K_f m,` Where ` Delta T_f ` =depression in freezing point m= molality of the solution
` K_f ` = cryoscopic constant
` 1_fm,=1 Delta T_f =K_f`
i.e. cryoscopic constant depends on the MOLAR concentration of the solute particles ,` K_f` is directly PROPORTIONAL to the molar concentration of the solute particles
` Delta T_f= (K_f xx W_B xx 1000 )/( M_B xx W_A ) `
` W_B= ` mass of the solute `W_A `=mass of solvent `M_B ` =molecular mass of the solute.
Discussion
8.

What happens when HI reacts with tert. Butyl alcohol?

Answer»

SOLUTION :`UNDERSET("Tert butyl alcohol")(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-OH+HI) underset(H_(3)PO_(4))overset(KI)to underset("Tert butyl IODIDE")(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-I+H_(2)O)`.
Discussion
9.

Which type of elements form interstitial hydrides?

Answer»

s-BLOCK and p-block
p-block only
d and F block
s-block only

SOLUTION :d and f block
Discussion
10.

Which of the following is not correct for the velocity of electron?

Answer»




ANSWER :A
Discussion
11.

What are the atomic numbers of elements whose outermost electrons are represented by (i) 3S^(1) (ii) 3P^(5)?

Answer»


ANSWER :(i) SODIUM (II) CHLORINE
Discussion
12.

Which bond angle theta would result in the maximum dipolemoment for the traitomic molecule XY_(2)

Answer»

`THETA =90^@`
`theta =120^@`
`theta=150^@`
`theta =180^@`

ANSWER :A
Discussion
13.

Which of the following is/are corret statements ?

Answer»

`Mg` burns in air releasing dazzling light rich in UV RAYS
`CaCl_(2) * 6H_(2)O` when mixed with ice gives freezing mixture
Mg' cann't form complexes
"Be" can form complexes due to is very SMALL size

Solution :Mg. can form complexes
Eg : CHOROPHYLL pigment
Discussion
14.

Which of the following statement (s)are wrong.

Answer»

one MOL of photons having ebnergy 400 KJ/mol will always breaks 4 male bonds of a molecule `A^(2)`
where A-A bond dissocitaion energy id is 100 KJ/mol.
Two bubls areemitting light having wavelenght ` 2000Å` ans `3000Å` respectively . If the bubls A and
B ar 40 watt and 300 watt respectively then the ratio ofnumber of photons EMITTED by A and B per dsy
is `1//2`.
When an electronmake transitionalfrom higher to lower orbit, photons(s) is /are emitted.
Ratio of time periodof electron in H-atom is `1^(st)` and `3^(rd)` level is `(9)/(1).

Solution :(A) One mole photons will breakone bond
(b)`(n_(A))/(n_(B))=(40xx2000)/(30xx3000)=(8)/(9)`
(D) Time PERIOD`n^(3)`
`(T_(1))/(T_(3))=(1)/(9)`
Discussion
15.

Which of the following species has lowest first ionization potential ?

Answer»

O
`O_(2)`
`O_(2)^(+)`
`O_(2)^(-)`

Solution :In O-atom , the ELECTRON is to be removed from 2p
while in `O_(2) , O_(2)^(+) and O_(2)^(-)` , electron is to be
removed from`pi^(**) (2p_(x)) or pi^(**) (2p_(y))` which has
higher ENERGY and hence less stable . Further , `O_(2)^(-)`
is least stable DUE to GREATER number of ELECTRONS
in the antibonding molecularorbitals .
Discussion
16.

We have 1.6 g CH_(4),1.7gNH_(3)and1.8gH_(2)O select correct the alternate(s).

Answer»

There are equal number of moles of each reactant
Total number of ATOMS in `CH_(4)gtNH_(3)gtH_(2)O`
Total number of H-atoms are in the ratio of `4:3:2`
Total number of C-atoms in `CH_(4)lt` that of N-atoms in `NH_(3)lt` that of O-atoms in `H_(2)O`

Solution :moles = `(1.6)/(10),(1.7)/(17),(1.8)/(18)`
atoms = `0.1N_(0)xx5,0.1N_(0)xx4,0.1N_(0)xx3`
H-atoms = `0.1N_(0)xx4,0.1N_(0)xx3,0.1N_(0)xx2`
no. of `C=N=O`
Discussion
17.

Which of the following is orrect potential energy diagram for the given chain propagating step. underset(B.E.=435kJ//mol)(CH_(3)-H+F)tounderset(B.E.569kJ//mol)(CH_(3)+H-F) DeltaH^(@)=-32kcal//mol

Answer»




ANSWER :D
Discussion
18.

Which one of the following is the correct quadraticform of Ostwald's dilution law equation ?

Answer»

`alpha^(2)C+alphaK-K=0`
`alpha^(2)C-alphaK-K=0`
`alpha^(2)C-alphaK+K=0`
`alpha^(2)C+alphaK+K=0`

SOLUTION :ACCORDING to Ostwald'sdilution law,
`K=(C alpha^(2))/(1-alpha)`
or `K-K alpha = C alpha^(2)`
or, `C alpha ^(2) + K alpha-K=0`
or, `alpha^(2)C + alpha K - K=0`
Discussion
19.

Write thermo chemical equation for vaporization of ethanol

Answer»

Solution :`C_(2)H_(5)OH(1)toC_(2)H_(5)OH(G),DeltaH=+QkJ`
Discussion
20.

Which of the following compound(s) gives brownish-black precipitate, on boiling with Na_(2)CO_(3) to their respective aqueous salt solution ?

Answer»

`PB(OH)_(2).2PbCO_(3)`
`Hg_(2)CO_(3)`
`Ag_(2)CO_(3)`
`(NH_(4))_(2)CO_(3)`

Answer :B::C
Discussion
21.

Which of the following is shown the Lewis formation of N_(2) ?

Answer»




Solution : In `N_(2)` molecule TRIPLE bond present between two nitrogen and one-one non BONDING electron pair present on both N.:N `EQUIV N`:
Therefore in lewis structure there are THREE bonding electron pair of triple bond present between two N and one-one electron pair present on both N.
Discussion
22.

When copper oxide is strongly heated with hydrogen it reduced to give metal. If the weight loss suffered by the oxide is 14.9g and weight of water formed is 16.78g, calculate the equivalent weight of oxygen.

Answer»


ANSWER :8
Discussion
23.

Valency of an element indicates

Answer»

COMBINING POWER with hydrogen
acidity
electrons in the OUTERMOST orbit
None of these

Answer :A
Discussion
24.

Which of the following is most reactive towards electrophilic substitution reaction ?

Answer»

ANILINE
nitrobenzene
Benzoic ACID
ACETANILIDE

ANSWER :A
Discussion
25.

Which is the correct statement for the given acids ?

Answer»

Phosphinic acid is a MONOPROTIC acid while phosphonic acid is a diprotic acid
Phosphinic acid isa diprotic acid while phosphonic acid is a monoprotic acid
Both are diprotic acids
Both are triprotic acids

Solution :Phosphinic aicd, `H_(3)PO_(2)` (monoprotic), phosphonic acid, `H_(3)PO_(3)` (diprotic),
Discussion
26.

What is sigma bond ? How many sigma bonds are present in C_2H_2 ?

Answer»
Discussion
27.

underset((A))(CH_(3)-underset(Br)underset(|)overset(CH_(3))overset(|)(C)-CH_(3))+underset((B)(CH_(3)-overset(CH_(2)-Br)overset(|)(C)H-CH_(3)) CH_(3)-overset(CH_(3))overset(|)(C)H-CH_(3)overset(Br_(2))to the percentage yield of 'A' is (The relative reactivities of 1^(@),2^(@),3^(@)- hydrogens 1:3.8:5)

Answer»

0.36
0.64
0.72
0.28

Answer :A
Discussion
28.

Which out of Cu^(2+), Fe^(2+) and Cr^(3+) has highest paramagnetism and why ?

Answer»

Solution :`Cu^(2+)` has one, `FE^(2+)` has four and `Cr^(3+)` has three unpaired electrons. Hence, `Fe^(2+)` has HIGHEST PARAMAGNETISM
Discussion
29.

Which one of the following is the likely bond angles of sulphur tetrafluoride molecule ?

Answer»

`120^(@) , 80^(@)`
`109^(@).28`
`90^(@)`
`89^(@), 117^(@)`

Answer :d
Discussion
30.

Which acts as a reducing agent ?

Answer»

`HNO_3`
`KMnO_4`
`H_2SO_4`
` (COOH)_2`

SOLUTION :`Hoverset(N)O_3,overset(+7)(KMnO_4),overset(+6)(H_2SO_4)`
ELEMENT it its HIGHEST O.S acts as O.A
`{:(COOH),(|),(COOH):},2x-8+2=0,x=+3`
Discussion
31.

Which metals are used as an anode and a cathode in Denial cell ?

Answer»

SOLUTION :Cu ACTS as cathode and ZN acts as an anode.
Discussion
32.

Which of the following gas has maximum ?

Answer»

Methane
ETHANE
Ethelene
Acetetene

Solution :Number of C and H atoms are MAXIMUM in ethane.
Discussion
33.

What happens when a ferromagnetic substance is heated to high temperature ?

Answer»

SOLUTION :On heating to high TEMPERATURE, FERROMAGNETIC substances changes to paramagnetic. This is DUE to randomisation of domains (SPINS ) on heating
Discussion
34.

What is extensive property ?

Answer»

Solution :A property WHOSE value depends on the QUANTITY or SIZE or value of MATTER is called extensive property.
Discussion
35.

Write the expression which gives the energy of electron in the nth shell of hydrogen like particles

Answer»

SOLUTION :`E_(n) = - (2pi^(2) mZ^(2) e^(4))/(n^(2)h^(2))` where Z is the atomic number of the particles and m and e are the mass and CHARGE of the ELECTRON.
Discussion
36.

X' is a colourless salt giving following reactions : 'X' can be

Answer»

`AlCl_(3)`
`ZnCl_(2)`
`ZN (CH_(3)COO)_(2)`
`Zn Br_(2)`

Solution :`3H_(2)S + underset((dil))(2HNO_(3)) rarr 2NO + 4H_(2)O + underset(("YELLOW turbidity"))(3S)`
Discussion
37.

Which gas has greater contribution in global warming ?

Answer»

`CH_4`
CFC
`NO_2`
`CO_2`

SOLUTION :`CO_2`
Discussion
38.

The volume of a concentrated H_(2)SO_(4), mixed with 0.5 N KOH to prepare 150 cm^(3) of 0.2 N KOH. Solution is :

Answer»

`50 cm^(3)`
`60 cm^(3)`
`70 cm^(3)`
`80 cm^(3)`

SOLUTION :N//A
Discussion
39.

Which one of the following when dissolved in water, gives a solution with pH less than seven at 298 K ?

Answer»

`CH_(3)COCH_(3)`
`C_(6)H_(5)OH`
`C_(6)H_(5)NH_(2)`
`C_(2)H_(5)OH`

Solution :`C_(6)H_(5)OH`
Discussion
40.

Which method is used to estimate sulphur?

Answer»

LASSAIGNE's test
Oxide test
Carius method
Kjedahl's method

Answer :C
Discussion
41.

Which is not related to lanthanide contraction

Answer»

Atomic radii of 4d and 5d series elements are SIMILAR
4d and 5d series elements have DIFFERENT properties
I.P. VALUES of 4d series elements are less than that of 5d series
Lanthanides exhibit +3 oxidation state

Answer :B
Discussion
42.

Which of the following outer electronic configurations corresponds to the d-block of the periodic table ?

Answer»

`3s^2 3p^3`
`3d^5 4s^1`
`3s^2 3p^6 4s^2`
`5d^1 6s^2`

ANSWER :B::D
Discussion
43.

What is the actual configuration of copper(Z=29)?Explain about its stability.

Answer»

SOLUTION :Copper (Z=29)
Expected configuration :`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(9)4S^(2)`
Actual configuration:`1s^(2)2s^(2)2p^(6)3s^(2)3d^(10)4s^(1)`
The reason is that electronic configuration[Ar]`3d^(10)4s^(1)` and not[Ar]`3d^(9)4s^(2)` due the symmetrical distribution andexchange ENERGIES of d electrons.
Symmetry leads to stability.The FULL filled configuration have symmet distribution of electrons and hence they are more stable than unsymmetrical configuration.
Discussion
44.

What happens when acethylene undergoes ozonolysis ?

Answer»

SOLUTION :When ACETYLENE unndergoes OZONOLYSIS,
Discussion
45.

What is the mass in grams of : (i) 6.022 xx 10^(23) atoms of oxygen (ii) 1.0 xx 10^(23) molecules of H_(2)S (iii) 6.022 xx 10^(23) molecules of oxygen ?

Answer»


SOLUTION :(i) `6.022 xx 10^(23)` atoms of OXYGEN have mass =16.0 G
`6.022xx 10^(23)` MOLECULES of `H_(2)S` have mass = 34.0 g
`1.0 xx 10^(23)` molecules of `H_(2)S` have mass `= (34.0xx10^(23))/(6.022xx10^(23))=5.645g`
(iii) Mass of `6.022 xx 10^(23)` molecules of oxygen `(O_(2))=32g`
Discussion
46.

Which of the following is/are chiral?

Answer»

Celing fan
Hockey stick
Screw
Kid's tricyle

Solution :N//A
Discussion
47.

Write the postulate of Kossol for formation of Ionic bond.

Answer»

Solution :In the periodic Table, the highly electronegative halogens and the highly electropositive alkali metals are separated by the noble gases.
The formation of a negative ion from a halogen atom and a positive ion from an alkali metal atom is associated with the gain and loss of an electron by the respective atoms.
The negative and positive ions thus FORMED attain stable noble gas electronic configurations. The noble gases (with the EXCEPTION of helium which has a duplet of electrons) have a particularly stable outer shell configuration of EIGHT (octet) electrons `ns^(2) np^(6)`.
The negative and positive ion are stabilized by electrostatic attraction.
Kossel.s postulation provide the basis for modem concepts REGARDING ion-formation by electron transfer and the formation of ionic crystalline COMPOUNDS.
Discussion
48.

Which of the following pair of reactants are involved the acid-catalyzed aldol reaction?

Answer»




ANSWER :3
Discussion
49.

Why BBr_(3) is a strongerLewis acid as compared to BF_(3) throughflorine is more electronegative than bromine ?

Answer»

Solution :The B atom in `BF_(3)`or `BBr_(3)`has only six electron inits valenceshell andhence can accepta pair ofelectronsto completeits octet.Therefore,both `BF_(3)` and `BBr_(3)` at as Lewisacids. But in `BF_(3)`,the sizes of empty `2p-`orbitalof B ANDTHE 2p-orbital ofF containing thelone pairof electronsare almostidenticaland henceeffective `ppi - ppi`boundingoccurs. As aresult,the lone pair of F is donate to Batom and hence the electrons deficiencyof borondecreases. Incontrast, in `BBr_(3)`.the size of 4p-orbitalov Br containingthe lone pair of electronsis much bigger than the empty 2p-orbitalof Band hencedonationof lone pairof electronsof Br toB does not occurto an significationextent. As a result, the electron deficiencyof B is much higherto `BBr_(3)`than in `BF_(3)`, and hence`BBr_(3)` i a stronger Lewisand than `BF_(3)`.
Discussion
50.

Which of the two, NO_(2)CH_(2)CH_(2)O^(oplus) or CH_(3)CH_(2)O^(oplus) is expected to be more stable and why?

Answer»

SOLUTION :
(I) is more stable than (II) because`NO_(2)` group has `-I` effect and HENCE it tends to disperse the `-ve` charge on the O-atom. In CONTRAST, `CH_(3)CH_(2)` has `+I` effect.It THEREFORE, tends to intensify the `-ve` charge and hence destablizes it.
Discussion