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 litres of CO_(2) represented by 4.4 g of CO_(2) at S.T.P are :

Answer»

2.4 litres
2.24 litres
44 litres
22.4 litres

Solution :44 G (MOLAR mass) of `CO_(2)` REPRESENT
= 22.4 L at S.T.P
4.4 of `CO_(2)` represent
`= ((22.4L))/((44g))xx(4.4g)=2.24L`.
Discussion
2.

The liquefaction behaviour of temporary gases like CO_(2) approaches that of N_(2),O_(2) (permanent gases ) as we go,

Answer»

below CRITICAL TEMPERATURE
above critical temperature
below ABSOLUTE zero
above absolute zero

Answer :B
Discussion
3.

The liquefaction behaviour of temporary gases approacches that of perrmanent gases as we go

Answer»

below CRITICAL temperature
above critical temperature
above ABSOLUTE zero
below absolute zero

Answer :B
Discussion
4.

The linear structure is assumed by

Answer»

`SnCl_(2)`
`NCO^(-)`
`NO_(2)^(+)`
`CS_(2)`

Solution :`CNO^(-),NO_(2)^(+) and CS_(2)` all have LINEAR structure
while `SnCl_(2)` is ANGULAR .
Discussion
5.

The linear structure is assumed by :

Answer»

`SnCI_(2)`
`CO_(2)`
`NO_(2)^(+)`
`CS_(2)`

ANSWER :B::C::D
Discussion
6.

The linearshape of CO_(2) is due to "……………."

Answer»

`SP^(3)` hybridisation of carbon
`sp` hybridisation of carbon
`ppi-ppi`BONDING between carbon and oxygen
`sp^(2)`hybridisation of carbon

Solution :The linear SHAPEOF `CO_(2)`is due to sp-hybridization of carbon and `ppi-ppi`bonding between carbon and oxygen. Thus, option (b) and (c ) ar` CORRECT.
Discussion
7.

The linear shape of CO_2 is due to ……

Answer»

`sp^3` hybridization of carbon.
sp hybridization of carbon.
`ppi-ppi` bonding between carbon and oxygen.
`sp^3` hybridization of carbon.

Solution :Hybridization of C in `CO_2` is sp and the STRUCTURE is linear. Carbon is linked to two oxygen atoms through DOUBLE bonds, one is sigma and other one is `ppi`-bond (p-p bonding).
Structure of `CO_2` is of following TYPE:
`O{:(ul(p-p(pi))),(bar(sp-p(6))):}C{:(ul(p-p(pi))),(bar(sp-p(6))):}O`
Here , `2 sigma`-bonds and `2 pi` - bonds
HYBRIDISATION of `CO_2` is sp so its shape is linear.
Discussion
8.

The lightest metal among the following is

Answer»

Be
Li
H
Na 

ANSWER :B
Discussion
9.

The lightest element in the periodic table is

Answer»

LITHIUM
Fluorine
Hydrogen
HELIUM

ANSWER :B
Discussion
10.

The light oil fraction of coal tar mainly contains..........,..............., and ..........

Answer»

SOLUTION :BENZENE, TOLUENE, XYLENE
Discussion
11.

The Li^(2+) ion is a hydrogen like ion that can BE DESCRIBED BY THE Bohr model.Calculate the Bohr radius of the third orbit and calculate the energy of an electron in 4^(th) orbit.

Answer»

Solution :`Li^(2+)`=HYDROGEN LIKE ion.
Bohr radius of the third orbit =`r_3`=?
`r_3=((0.529)xx3^(2)N^(2))/(Z)`A where n=shell number,Z= atomic number.
`r_3=(0.529xx3^(2))/(2)`[because for lithliumZ=3,n=3]
`=(0529xx9)/(3)``r_3+1.587A`
`E_n=((-13.6)Z^(2))/(n^(2))eV at OM^(-1)`
`E_4`=Energy of the fourth orbit=?
`E_4=((-13.6)Z^(2))/(4^(2))`
`=(-13.6xx9)/(16)=-7.65eVatom^(-1)`
`E_4=-7.65eVatom^(-1)`
Discussion
12.

The Li^(2+) ion is a hydrogen like ion that can be described by the Bohr model. Calculate the Bohr radius of the thired orbital and calculate the energy of an electron in 4^(th) orbital.

Answer»

SOLUTION :`r_(n) = ((0.529)n^2)/(z) Å ""E_(n) = (-13.6 (z^2))/(n^2) ev "atom"^(-1)`
for `Li^(2+) z = 3`
Bohr RADIUS for the THIRD orbit `(r_3)`
`=((0.529)(3)^2)/(3)`
`= 0.529 xx 3`
`= 1.587 Å`
Energy of an electron in the fourth orbit
`(E_4) = (-13.6(3)^2)/((4)^2)`
`= -7.65 eV atom^(-1)`.
Discussion
13.

The Lewis dot structure of carbonate ion is …………..

Answer»

SOLUTION :
Discussion
14.

The Lewis dot structure can be written by following the steps given below. Which one is incorrect?

Answer»

DRAW the skeletal structure of the molecule.
CALCULATE the specific NUMBER of valence electrons of all the atoms in the molecule.
Draw a single bond between the atoms in the skeletal structure of the molecule.
Distribute the remaining valence electrons as pairs .

Answer :b
Discussion
15.

Why does the Lewis acid strength of boron halides follow the order B Br_3gt BCl_3 gtBF_3 ?

Answer»

`BF_3 GT BCl_3 gt B Br_3`
`B Br_3 gt BCl_3 gt BF_3`
`BCl_3 gt BF_3 gt B Br_3`
`BCl_3 gt B Br_3 gt BF_3`

SOLUTION :`B Br_3 gt BCl_3 gt BF_3`
Discussion
16.

The Lewis acid strength of BF_3 is less than that of BCl_3 Why?

Answer»


ANSWER :1
Discussion
17.

The level of thermal energy in a substance is known as ...... .

Answer»

Entropy
Temperature
Heat energy
Quantity of Heat

Answer :B
Discussion
18.

The letters n, l and m proposed by Bohr, Sommerfeld and Zeeman respectivelt for quantization of angular momentum in classical physics were later on obtained as the results of solution of Schrodinger wave equation based on quantum mechanics. The term n, l, m are named as principal quantum number, azimuthal quantum number and magnetic number respectively. The fourth quantum number i.e., spin quantum number (s) was given by Uhlenback on the basis of two spins of electrons. The first two quantum number also decides the nodes of an orbital. Which statement about energy level in H-atom is correct ?

Answer»

Only n and L DECIDE ENERGY level
Only .l. DECIDES energy level
Only n decides energy level
n, l and m decide energy level

Answer :C
Discussion
19.

The level of CO gas in air that causes immediate death is

Answer»

I0 PPM
100ppm
500ppm
1000 ppm

Solution :1000 ppm
Discussion
20.

The letters n, l and m proposed by Bohr, Sommerfeld and Zeeman respectivelt for quantization of angular momentum in classical physics were later on obtained as the results of solution of Schrodinger wave equation based on quantum mechanics. The term n, l, m are named as principal quantum number, azimuthal quantum number and magnetic number respectively. The fourth quantum number i.e., spin quantum number (s) was given by Uhlenback on the basis of two spins of electrons. The first two quantum number also decides the nodes of an orbital. The numerical value Psi_(4,3,0) denotes

Answer»

3d-ORBITAL
4F-orbital
2s-orbital
4d-orbital

Solution :`Psi` represent an orbital and `Psi_(4,3,0) ` has N= 4, l= 3 , i.e. , 4f - orbital
Discussion
21.

The lengthof a unit cell in the NI Crystal is 0.352 nm. The diffraction of X-rays of 0.154 nm wavelength (lambda) from a Ni crystal occurs at 22.2^(@), 55.9^(@) and 38.2^(@) . By using Bragg's law, (nlambda=2d "sin"theta) , and assuming that the diffractions are first order (n=1), the distance are calculated to be 0.204 nm , 0.176 nm and 0.124 nm The various structures for the Ni crystal are represented as: Which of the following statements is correct?

Answer»

I represents sc- type structure
II represents bcc-type structure
III represents both sc-and bcc-type structure.
All FIGURES represent FCC-type structure.

Solution :All there DATA re CONSISTENT with an fcc CRYSTAL.
Discussion
22.

The lengthof a unit cell in the NI Crystal is 0.352 nm. The diffraction of X-rays of 0.154 nm wavelength (lambda) from a Ni crystal occurs at 22.2^(@), 55.9^(@) and 38.2^(@) . By using Bragg's law, (nlambda=2d "sin"theta) , and assuming that the diffractions are first order (n=1), the distance are calculated to be 0.204 nm , 0.176 nm and 0.124 nm The various structures for the Ni crystal are represented as: The distance (d) of 0.176 nm represent which structure?

Answer»

I
II
III
All

Solution :The DISTANCE 0.176nm is half the unit cell length (a/2) which CORRESPONDING to the perpendicular distance between the layers of atoms in ONE face of the unit cell and the CENTER (fig I).
Discussion
23.

The lengthof a unit cell in the NI Crystal is 0.352 nm. The diffraction of X-rays of 0.154 nm wavelength (lambda) from a Ni crystal occurs at 22.2^(@), 55.9^(@) and 38.2^(@) . By using Bragg's law, (nlambda=2d "sin"theta) , and assuming that the diffractions are first order (n=1), the distance are calculated to be 0.204 nm , 0.176 nm and 0.124 nm The various structures for the Ni crystal are represented as: This distance (d) of 0.124 nm represent which structure?

Answer»

I
II
III
All

Solution :The DISTANCE 0.124 nm CORRESPONDS of times the unit cell dimension, i.e., one -fourth of the distance from one edge of the cell to the opposite edge (figII. The plane intersects the face-centered atoms of the two SIDES. The plane including the edge atoms is equivalent OT the plane across the center of the unit cell.
Discussion
24.

The lengthof a unit cell in the NI Crystal is 0.352 nm. The diffraction of X-rays of 0.154 nm wavelength (lambda) from a Ni crystal occurs at 22.2^(@), 55.9^(@) and 38.2^(@) . By using Bragg's law, (nlambda=2d "sin"theta) , and assuming that the diffractions are first order (n=1), the distance are calculated to be 0.204 nm , 0.176 nm and 0.124 nm The various structures for the Ni crystal are represented as: The distance (d) of 0.204 nm represent which structure?

Answer»

I
II
III
All

Solution :The distance 0.204 pm comes out to be" `a//sqrt(3)"` which CORRESPONDS to the perpendiucular distance between a corner of UNIT cell and the plane of the THREE adjacent CORNERS. (FIG III) .
Discussion
25.

The length of a rectangle is 15 inches and the breadth 12 inches. Calculate its area in SI units.

Answer»


ANSWER :`0.116 m^(2)`
Discussion
26.

The length of a rectangle is 10.4 cm and the breadth is 5.661 cm. Find its area.

Answer»


ANSWER :`58.9 CM^(2)`
Discussion
27.

Length cannot be measured by ……………

Answer»

SPECTROSCOPIC method
x-ray DIFFRACTION method
electron-diffraction method
all the above

Solution :all the above
Discussion
28.

The least stable hydride of 15th group is

Answer»

`NH_(3)`
`PH_(3)`
`AsH_(3)`
`BiH_(3)`

ANSWER :D
Discussion
29.

The least stable conformer of ethane is …………….form

Answer»

SOLUTION :clipsed
Discussion
30.

The least stable carbonium ion is

Answer»




Solution :`-OH` group when present at meta POSITION instead of ACTIVATING is DEACTIVATING SINCE at meta position there is no resonance. So it is LEAST stable.
Discussion
31.

The least stable carbonate of alkali metal is

Answer»

`Cs_(2)CO_(3)`
`Na_(2)CO_(3)`
`K_(2)CO_(3)`
`Rb_(2)CO_(3)`

SOLUTION :As the SIZEOF the cation increases , M-O bond weakens and C-O bond STRENGTHENS . Since `Li^(+)` has the smallest size AMONG alkali metals , therefore , Li-O bond is the strongest and C-O bondis the weakest . As a result , `Li_(2)CO_(3)` DECOMPOSERS most readily to give `Li_(2)O` and `CO_(2)`.
Discussion
32.

The least soluble in water among the following is:-

Answer»

`MgCrO_(4)`
`SrCrO_(4)`
`CaCrO_(4)`
`BaCrO_(4)`

ANSWER :4
Discussion
33.

The least soluble hydroxide in H_(2)O is

Answer»

`LIOH`
`NAOH`
`RBOH`
`CSOH`

ANSWER :A
Discussion
34.

Theleast soluble compound (salt)of the following is

Answer»

`CSCL(K_(sp)=10^(-12))`
` HgS(K_(sp)=1xx 10^(-52))`
` PbCl_2 (K_(sp)=1.7xx 10^(-5))`
` ZNS (K_(sp)=1.2 xx 10^(-23))`

Solution :Salt with LOW S is LEAST soluble
Discussion
35.

The most soluble hydroxide among the following is

Answer»

`LIOH`
`NAOH`
`RBOH`
`CSOH`

ANSWER :A
Discussion
36.

The least soluble carbonate among the following is

Answer»

`Na_(2)CO_(3)`
`K_(2)CO_(3)`
`CaCO_(3)`
`Rb_(2)CO_(3)`

ANSWER :C
Discussion
37.

The least soluble alkali metal carbonate is

Answer»

`K_(2)CO_(3)`
`Na_(2)CO_(3)`
`Li_(2)CO_(3)`
`Cs_(2)CO_(3)`

Answer :C
Discussion
38.

The least reactive alkene towards dil. HCl is

Answer»


`CH_(3)-CH = CH_(2)`

`CH_(2) = CH_(2)`

Solution :
COMPOUND I FORMS most stable CARBONIUM ion
Discussion
39.

The least random state of H_(2)O system is

Answer»

Ice
Liquid water
Steam
Randominess is same in all

Solution :ENTROPY of `X LT L lt G`
Discussion
40.

The least priority functional group among the following is

Answer»

`-OH`
`-C -= C-`

`-NH_(2)`

ANSWER :B
Discussion
41.

The least number of carbon atoms in alkane forming chain isomers is

Answer»

3
1
2
4

Answer :D
Discussion
42.

The least ionic chloride is formed by

Answer»

MG
CA
Be
Sr

Solution :Being small in size and more ELECTRONEGATIVE , Be forms least ionic or more covalent CHLORIDE .
Discussion
43.

The lead of leadpencils melts at

Answer»

`2000^(@)C`
`350^(@)C`
`3170^(@)C`
`75^(@)C`

Solution :The LEAD ofpencilsis ACTUALLY graphite which has a high m.p.of `3170^(@)C`.
Discussion
44.

The laws of chemical combination are the basis of the atomic theory. State and explain the law of conservation of mass

Answer»

SOLUTION :Matter can neither be created nor be DESTROYED `CaCo_3overset"heat"toCaO+CO_2` 100 g of `CaCO_3` on the heating given 56 GRAM of CAO and 44 gram of `CO_2`
Discussion
45.

The laws of chemical combination are the basis of the atomic theory. Name the law of chemical combination illustrated by the pair of compounds, CO and CO_2

Answer»

SOLUTION :LAW of MULTIPLE PROPORTION
Discussion
46.

The laws of chemical combination are the basis of the atomic theory Calculate the molarity of a solution containing 8g of NaOH in 500 mL of water

Answer»

SOLUTION :MOLARITY `(M)=(W_Bxx1000)/(M_BxxV)=(8xx1000)/(40xx500)=8000/20000=0.4M`
Discussion
47.

The law that relates the pressure and volume of gases is

Answer»

BOYLE's
CHARLES law
Dalton
none of the above

ANSWER :A
Discussion
48.

The law of triads is obeyed by

Answer»

FE, CO, Ni
C, N,O
He, Ne, Ar
Al, Si, P

Answer :A
Discussion
49.

The law of triads is not obeyed by

Answer»

Ca, Sr, Ba
CI, Br, I
Li, Na, K
 FE,Co, Ni

Answer :D
Discussion
50.

The law of multiple proportions was observed for the pair

Answer»

`CO_(2),NO_(2)`
`N_(2)O,NO_(2)`
`H_(2)O,H_(2)S`
`H_(2)S,SO_(2)`

Solution :`N_(2)O` and `NO_(2)` follows LAW of multiple PROPORTION
Discussion