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.

The group of elements with highest second ionisation energy is

Answer»

II A GROUP
ZERO group
VII A group
IA group

Answer :D
Discussion
2.

The group having isoelectronic species i s ....

Answer»

`O^(2-),f^(-),Na^(+).MG^(2+)`
`O^(-),F^(-),Na,Mg^(+)`
`O^(2),F^(-),Na,Mg^(2+)`
`O^(-),F^(-),Na^(+),Mg^(2+)`

Solution :
NOTE : Option A isoelectronic in species .

`therefore` option(B,C,D) are notisoelectronic species
Discussion
3.

The group having isoelectronic species is

Answer»

`O^(2-) , F^(-), Na, Mg^(2+)`
`O^(-), F^(-), Na^(+), Mg^(+)`
`O^(2-), F^(-), Na^(+), Mg^(2+)`
`O^(-), F^(-), Na, Mg`

Solution :`O^(2-) = 8 + 2 = 10 e^(-)1`,
`Na^(+) = 11 - 1 = 10 e^(-), Mg^(2+) = 12 - 2 = 10 e^(-)`
Hence, GROUP (C) contains isoelectronic species.
Discussion
4.

The ground state term symbol for an electronic state is governed by

Answer»

HEISENBERG's PRINCIPLE
Hund's Rule
AUFBAU'sprinciple
pauli's exclusion principle

Solution :It is GOVERNED by Aufbau principle
Discussion
5.

The ground-state electronic configurations listed here are incorrect. Explain what mistake have been made in each and write the correct electronic configurations Al : 1s^(2) 2s^(2) 2p^(4) 3s^(2) 3p^(3) " " B : 1s^(2) 2s^(2) 2p^(5) " " F : 1s^(2) 2s^(2) 2p^(6)

Answer»

Solution :In `._(13)Al, 2p` should be filled first before FILLING of 3S starts
Correct ELECTRONIC CONFIGURATION `= 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(1)`
In `._(5)B`, TOTAL electrons = 5. Hence, E.C. `= 1s^(2) 2s^(2) 2p^(1)`
In `._(9)F`, total electrons = 9. Hence, E.C. `= 1s^(2) 2s^(2) 2p^(5)`
Discussion
6.

The ground state electronic configuration of CO molecule

Answer»

`1sigma^2 2sigma^2 1pi^4 3sigma^2`
`1sigma^2 2sigma^2 3sigma^2 1pi^22pi^2`
`1sigma^2 2sigma^2 1pi^2 3sigma^2 2pi^2`
`1sigma^2 1pi^4 2sigma^2 3sigma^2`

SOLUTION :
E.C for CO : `1sigma^2 2sigma^2 1pi^4 3sigma^2`
Discussion
7.

The ground state electron configurations listed here are incorrect. Identify the mistakes that have been made in each and write the correct electronic configurations. Cl, 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6)

Answer»

Solution :Atomic number of CHLORINE is 17, so its ELECTRONIC configuration WOULD be `1s^2 2^s 2p^6 3s^2 3p^5.` The configuration MENTIONED above is that of Argon.
Discussion
8.

The ground state electron configurations listed here are incorrect. Identify the mistakes that have been made in each and write the correct electronic configurations. Al, 1s^(2) 2s^(2) 2p^(4) 3s^(2) 3p^(3)

Answer»

Solution :3S orbital cannot be filled before complete FILLING of 2P orbital. So, this configuration is INCORRECTLY WRITTEN.
Discussion
9.

The ground state electron configurations listed here are incorrect. Identify the mistakes that have been made in each and write the correct electronic configurations. B: 1s^(2) 2s^(2) 2p^(5)

Answer»

SOLUTION :ATOMIC number of boron is 5, so its electron configuration would be `1s^2 2^s 2p^1,` the configuration written above is that of fluorine.
Discussion
10.

Why is cement clinker obtained from rotary Kiln mixed with gypsum?

Answer»

SOLUTION :CEMENT CLINKER
Discussion
11.

The green house gas is

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`CO_(2)`
`SO_(2)`
`N_(2)`
`H_(2)S`

ANSWER :A
Discussion
12.

The green house effect is because of the

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PRESENCE of GASES, which in general are strong infrared absorbers, in the atmosphere.
presence of `CO_2` only in the atmosphere.
pressure of `O_3` and `CH_4` in the atmosphere.
`N_2 O `and CHLOROFLUOROHYDROCARBONS in the atmosphere.

Solution :Green house gases such as `CO_2`, ozone, methane, the chlorofluorocarbon compounds and water vapour form a thick cover around the earth which prevents the IR rays emitted by the earth to escape. It gradually leads to increase in TEMPERATURE of atmosphere.
Discussion
13.

The green house effect is mainly caused by

Answer»

`NO_2`
`NO`
CO
`CO_2`

SOLUTION :`CO_2`
Discussion
14.

The green coloured borax bead obtained from copper salts is

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CUPRIC METABORATE
COPPER orthoborate
Copper boride
Cuprous metaborate

SOLUTION :
Discussion
15.

The 'green' chemical used in household cleaners to remove stains and also a favourite dressing on salads is

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hydrochloric ACID
citric acid
vineger
water

SOLUTION :Vinegar (acetic acid ) is used as HOUSEHOLD CLEA
Discussion
16.

The greatest percentage of CO isin

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coal GAS
producer gas
water gas
OIL gas

Solution :Water gas `= 40%` CO
Coal gas = 3% CO
produce gas = 33% CO
Discussion
17.

The greatest percentage of CO is in

Answer»

coal GAS
producer gas
water gas
oil gas

Solution :Water gas = 40% CO
Coal gas = 3% CO
PRODUCE gas = 33% CO
Discussion
18.

The greatest affinity for haemoglobin is shown by

Answer»

NO
CO
`O_2`
`CO_2`

ANSWER :B
Discussion
19.

The gravitational work done by an object is …………

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Qv
fx
PV
mgh

ANSWER :D
Discussion
20.

The graph relates in K_(eq) Vs 1/T for a reaction. The reaction must be :

Answer»

exothermic
endothermic
`DeltaH` is NEGLIGIBLE
spontaneous

Solution :`Delta G^(@)=RT -RTl nk, l nk =(-DeltaH^(@))/(RT)+(deltaS^(@))/(R)`
`:.` slope= `(-DeltaH^(@))/(R)` = .+. Ve `IMPLIES DeltaH^(@) LT 0`
`:.` exothermic
Discussion
21.

The graph that does represent the behaviour of an ideal gas is :

Answer»




ANSWER :B::C
Discussion
22.

The graph of viscosity coefficient (T|) and absolute temperature (T) is ________

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STRAIGHT line passing through origin 
Straight line PARALLEL to TEMPERATURE axis 
Straight line with (+) ve slope 
EXPONENTIAL graph 

Solution :`eta ` is an exponential graph.
Discussion
23.

The graph of atomic number (Z) Vs atomic radius (r) is shown in figure 1, 2, 3 and 4 Which graph correctly represents the element of 2^(nd) period and 1^(st) group respectively. choose your options from below.

Answer»

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

SOLUTION : Atomic radius goes on DECREASING with increasing of atomic NUMBER

Atomic radius increases with atomic number in a GROUP
Discussion
24.

Thegraph of P vs V is given at temperature and number of moles : The correct relationship is/are

Answer»

`T_(1) GT T_(2) gt T_(3)`
`T_(1) lt T_(2) lt T_(3)`
`n_(1) gt n_(2) gt n_(3)`
`n_(3) gt n_(2) gt n_(1)`

Solution :Both relationships are correct.
Discussion
25.

The graph ………… is known as isothermal.

Answer»

`p to V`
`PV to V`
`V to (1)/(p)`
`p to (1)/(V)`

ANSWER :C
Discussion
26.

The graph between momentum p and de-Broglie wavelength lambda of photon is

Answer»




SOLUTION :`LAMBDA =(H)/(SQRT(2mKE))`
Discussion
27.

The grade of 2.95 M H_2O_2 in volume ofpercentage is ____ and ____ respectively.

Answer»

33,10
35,12
28,10
36,12

SOLUTION :V=M X 11.2
=2.95 x 11.2 = 33.04 (VOLUME)
%w/v = `(Mxx34)/10 = (2.95xx34)/10`=10.03 %w/v
Discussion
28.

The gold number of some colloids are given below: {:("Colloid","Gold number"),(A,0.01),(B,2.5),(C,20):} The protective nature of these colloids follows the following order

Answer»

`C gt B gt A`
`A gt B gt C`
`A=B=C`
`B gt A gt C `

Solution :LESSER the GOLD number of a colloid, greater is the protective NATURE.
Discussion
29.

The given following elements belongs to which block? Which of this is having the maximum metallic property ? K, Na , Li , Rb

Answer»

Solution : Rb :(They are ALKALI METAL and top to BOTTOM METALLIC property increases.)
Discussion
30.

The given figure shows a chnge of state A to state C by rtwo paths ABC and AC for an ideal gas. Calculate. a. The path along which work done is least. b. The internal enegry at C if the internal enegry of gas at A is 10J and amount of heat supplied to change its state to C through the path AC is 200J. c. The amount of heat supplied to the gas to go from A to B, if inetrnal enegry of gas at state B is 10J.

Answer»

Solution :We know, Path `CA:` Isothermal compression
Path `AB`: Isobaric expansion
Path `BC`: Isochroic change
Let `V_(i)` and `V_(f)` be initial volume and final volume at repective points.
For temperature `T_(1) ("for" C): PV - nRT_(1)`
`2 xx 10 = 1 xx 0.0821 xx T_(1)`
`T_(1) = 243.60K`
For temperature `T_(2)` (For `C` and `B): (P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))`
`(2xx10)/(T_(1)) = (20 xx 10)/(T_(2))`
`(T_(2))/(T_(1)) = 10`
`T_(2) = 243.60 xx 10 = 2436.0K`
Path `CA: w = + 2.303 nRT_(1) "log"(V_(i))/(V_(f))`
`= 2.303 xx 1xx2xx 243.6"log" (10)/(1)`
`=+1122.02 CAL`
`DeltaE = -0` for isotermal compression, also `Q = W`
Path `AB, w =-P (V_(f) - V_(1))`
`=- 20 xx (10 - 1) =- 180 L-atm`
`= (-180 xx 2)/(0.0821) =- 4384.9 cal`
Pth `BC: w =- P(V_(f) - V_(1)) = 0`
(`:' V_(f) - V_(1) = 0)` since volume is constant
For monoatomic gas, heat chnage at constant volume `=qv = DeltaU`
Thus, for path `BC, q_(v) = C_(v) xx n xx DeltaT = DeltaU`
`:.q_(v) = (3)/(2)R xx1 (2436 - 243.6)`
`= (3)/(2) xx2xx1xx (2192.4) = 65772.2 cal`
Since process involves cooling `:.q_(v) = DeltaU =- 6577.2 cal`
Also in path `AB`, the initial enegry in state `A` and state `C` is same. Thus,, during path `AB`, an INCREASE in internal enegry equivalent of change in internal enegry during path.
`BC` should take place. Thus, `DeltaU` for path `AB = + 6577.2 cal`
Now `q` for path `AB = DeltaU - w_(AB) = 6577.2 + 6577.2 cal`
`= 10962.1 cal`
Cycle: `DeltaU = 0`,
`q =- w =- [w_(path CA) +w_(pathAB) +w_(pathBC)]`
`=- [+1122.02 +(-4384.9+0]`
`:. q =- w = + 3262.88 cal`
Discussion
31.

The given figure shows the Maxwell distribution of molecular speeds of a gas at three different temperatures T_1, T_2 and T_3. The correct order of temperatures is:

Answer»

`T_1 GT T_2 gt T_3`
`T_1 gt T_3 gt T_2`
`T_3 gt T_2 gt T_1`
`T_2 gt T_3 gt T_1`

ANSWER :D
Discussion
32.

The given electronic configuration of C in excited state This configuration can form

Answer»

TWO `PI` bond only
One `sigma`-bond and three `pi`-bonds
Two `sigma` bond and two `pi`-bond
Two `sigma` bond only

Solution :`H overset(sigma)(-) underset(underset(sp)(uarr))(C )underset(uarr)(-=) underset(underset(sp)(uarr))(C)-H`
Two `sigma` and two `pi` bonds
Discussion
33.

The Gibb's energy for the decomposition of Al_(2)O_(3) at 500^(@)C is as follows (2)/(3)Al_(2)O_(3) rarr (4)/(3)Al+O_(2) Delta_(f) = 960 kJ mol^(-1) The potential difference needed for the electrolytic reduction of aluminium oxide (Al_(2)O_(3)) at 500^(@)C us at least

Answer»

`4.5 V`
`3.0 V`
`2.5 V`
`5.0V`

Solution :The half CELL reactions are :
`{:("At anode: " 2O^(2-)+4e^(-)rarrO_(3)"]" xx 3),("At anode: " Al^(3+) rarr Al + 3e^(-)"]" xx 4),(ul(BAR("Net REACTION:" 4AL^(3+)+6O_(2-) rarr 3O_(2)+4Al))),(or 4//3 Al^(3+) + 2O^(2-) rarr O_(2)+4//3Al):}`
`:. n = (12)/(3) = 4`
`Delta G^(@) = -NFE^(@)`
`960 xx 1000 = -4 xx 96500 xx E^(@)`
[Here `Delta G^(@) = +960 kJ mol^(-1) = 960 xx 1000 J mol^(-1), n = 4, F = 96500 "coulomb mol"^(-1)]`
or `E^(@) = -2.48 V`
Potential difference needed `= 2.5 V`
Discussion
34.

The geometry of H_(3)O^(+) ion is

Answer»

Planar
Triangular
PYRAMIDAL
TETRAHEDRAL

ANSWER :C
Discussion
35.

The geometry of H_(2)S and its dipole moment are

Answer»

ANGULAR and non-ZERO
Angular and zero
LINEAR and non-zero
Linear and zero

Answer :A
Discussion
36.

The geometry of 'C' in diamond is

Answer»

Planar
Linear
Tetrahedral
Octahedral

Answer :B
Discussion
37.

The geometry of a molecule depends upon the repulsions between electron pairs present in the valance shell of the central atom. LP - LP repulsions are greater than LP-BP which are greater than BP - BP. A pi bondcan not decide the geometry of a molecule because pi bond has no independent exsistance. Deviations in bond angle are due to the presence of lone pairs. In which of the following orientation of electron pairs around the central atom is tetrahedral ?

Answer»

`NH_3`
`CH_(3)^(-)`
`CH_(3)^(+)`
Both`NH_3` and ` CH_(3)^(-)`

ANSWER :D
Discussion
38.

The geometry of a molecule depends upon the repulsions between electron pairs present in the valance shell of the central atom. LP - LP repulsions are greater than LP-BP which are greater than BP - BP. A pi bondcan not decide the geometry of a molecule because pi bond has no independent exsistance. Deviations in bond angle are due to the presence of lone pairs. Square planar species among the following is

Answer»

`XeF_4`
`NiCl_(4)^(-)`
Both `XeF_4` and `NiCl_(4)^(2-)`
`CH_4`

ANSWER :A
Discussion
39.

The geometry of a molecule depends upon the repulsions between electron pairs present in the valance shell of the central atom. LP - LP repulsions are greater than LP-BP which are greater than BP - BP. A pi bondcan not decide the geometry of a molecule because pi bond has no independent exsistance. Deviations in bond angle are due to the presence of lone pairs. Which of the following molecule is linear in shape ?

Answer»

`I_(3)^(-)`
`CO_(2)`
Both ` I_(3)^(-) ` and ` CO_(2)`
`SO_2`

ANSWER :B
Discussion
40.

The geometry of a molecule depends upon the repulsions between electron pairs present in the valance shell of the central atom. LP - LP repulsions are greater than LP-BP which are greater than BP - BP. A pi bondcan not decide the geometry of a molecule because pi bond has no independent exsistance. Deviations in bond angle are due to the presence of lone pairs. Which of the following differs from the rest in geometry ?

Answer»

`BF_3`
`SO_3`
`SO_(3)^(2-)`
`CO_(3)^(2-)`

ANSWER :C
Discussion
41.

The geometry of a complexspecies can be understood from the knowledge of type ofhybridisation of orbitalsof central atoms. The hybridisation of orbitals of central atom. The hybridisationof orbitalsof centralatom in [Be(OH)_(4)]^(-) and the geometry of the complex are respectively

Answer»

`SP^(3)`, TETRAHEDRAL
`sp^(3)`square planar
`sp^(3)d^(2)`, OCTAHEDRAL
`DSP^(2)`, square planar

Answer :A
Discussion
42.

The geometry of a complex species can be understood from the knowledge of type of hybridization of orbitals of central atom. The hybridization of orbitals of central atom in [Be(OH)_4]^(-)and the geometry of the complex are respectively ..........

Answer»

`sp^3` , tetrahedral
`sp^3`, square PLANAR
`sp^3d^2` , octahedral
`dsp^2` , square planar

SOLUTION :In `[B(OH)_4]^-` ,Boron is surrounded by 4 bonds pair and has no lone pair. Geometry and HYBRIDIZATION of central atom is based on BOND pairs and lone pairs around the central atom. `[B(OH)_4]^(-)`is tetrahedral and Boron is `sp^3` hybridized.
Discussion
43.

The geometrical shape of carbocation is

Answer»

LINEAR
TETRAHEDRAL
PLANAR
PYRAMIDAL

SOLUTION :N/A
Discussion
44.

The geometrical shape of carbocation is .............

Answer»

Linear
tetrahedral
Planar
Pyramidal

Answer :C
Discussion
45.

The general trend in the properties of elements of carbon family shows that, with the rise in atomic number.

Answer»

The tendency TOWARDS catenation INCREASES 
The tendency to show `+2` oxidation STATE increases 
The metallic character decreases 
The tendency to form complexes with covalency higher than four decreases. 

ANSWER :B
Discussion
46.

The generalouter electronicconfigurationoftransitionmetals is

Answer»

`NS^(2)nd^(1-10)`
`ns^(2)np^(1)(n-1)d^(1-10)`
`ns^(2)np^(6)(n-1)d^(1-1)`
`ns^(0-2) (n-1)d^(1-10)`

Solution :NA
Discussion
47.

The generaloutershellelectronicconfiguration of d- block element is ……..

Answer»

SOLUTION :`(n-1) d^(1-10) NS^(0-2) ` where `n=4 -7`
Discussion
48.

The general molecular formula which represents the homologousseries of alkanols is

Answer»

`C_(N)H_(2N+2)O`
`C_(n)H_(2n)O_(2)`
`C_(n)H_(2n)O`
`C_(n)H_(2n+1)O`

Answer :A
Discussion
49.

The general formula of primary alcohol is:

Answer»

`GT CHOH`
`-underset(|)overset(|)(C)-OH`
`-CH_(2)OH`
`gt C(OH)_(2)`

Solution :`overset(1^(@))(C)H_(2)-OH`
Discussion
50.

The general formula of arenes is:

Answer»

`C_(N)H_(2N)`
`C_(n)H_(2n-4)`
`C_(n)H_(2n+2)`
`C_(n)H_(2n-6)`

ANSWER :d
Discussion