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.

Which of the following can form more than two mononitration products?

Answer»

BENZENE
o-xylene
m-xylene
p-xylene

SOLUTION :m-xylene is form more than two MONONITRATION PRODUCTS
Discussion
2.

Which is the correct order number for bond length of C -O in CO, CO_(3)^(2-) and CO_(2) ?

Answer»

`CO_(2) LT CO_(3)^(2-) lt CO`
`CO lt CO_(3)^(2-) lt CO_(2)`
`CO lt CO_(2) lt CO_(3)^(2-)`
`CO_(3)^(2-) lt CO_(2) lt CO`

SOLUTION :`CO lt CO_(2) lt CO_(3)^(2-)`
Discussion
3.

Which of the alkali metal is having least melting point ?

Answer»

Na
K
Rb
Cs

Solution :MELTING point decreases with INCREASE in the size of atom of the METAL because as the size increases the STRENGTH of metallic bonding decreases.
Discussion
4.

Which of the following will have lowest value of K_(sp) at room temperature?

Answer»

`Be(OH)_(2)`
`MG(OH)_(2)`
`Ca(OH)_(2)`
`Ba(OH)_(2)`

Solution :`Be(OH)_(2)` is least soluble in water hence, it will have lowest value of `K_(SP)`.
`Be(OH)_(2)HARR Be^(2+)+2OH^(-)`
`K_(sp)=[Be^(2+)][OH^(-)]^(2)`
Discussion
5.

What is the compound formed when CO combines with blood ?

Answer»

SOLUTION :CARBOXYHEMOGLOBIN (HBCO)
Discussion
6.

Which substance cannot be reduced by H_(2)O_(2)

Answer»

`KMnO_(4)//H_(2)SO_(4)`
`K_(2)Cr_(2)O_(7)//H_(2)SO_(4)`
`Ag_(2)O`
`Fe^(3+)`

Answer :D
Discussion
7.

Write the Nernst equation at 298 K for the electrode reaction 2H^(+)(0.1M)+2e^(-)toH_(2)(g)

Answer»

SOLUTION :`E=E^(@)+(0.059)/(2)log_(10)(0.1)^(2)`
Discussion
8.

What is sp hybridization ? Give characteristics of it.

Answer»

Solution :Deinition: Atomic orbitals having less difference in energy of same atom ns & np overlap & form two equal orbitals is called SP-hybridization.
Characteristics :
In each sp-orbital 50% s character and 50%p-character is observed.
Both equivalent sp orbitals ARRANGE on Z axis at `180^(@)` in opposite direction to each other.
The two sp hybrids point in the opposite direction along the Z-axis with PROJECTING POSITIVE lobes and very small negative lobes. Which provides more effective overlapping RESULTING in the formation of stronger bonds.
Such a molecule in which the central atom is sp hybridised and linked directly to two other central atoms possesses lineargeogmetry.
Discussion
9.

Which of the following is not true about covalent compounds ?

Answer»

They UNDERGO molecular REACTIONS
They possess low melting and BONDING points
They undergo ionic reactions
They exhibit ISOMERISM

Answer :C
Discussion
10.

Which of the following is the best description of the behaviour of bromine in the reaction given below?

Answer»

PROTON ACCEPTOR only
Both oxidised and reduced
Oxidised only
Reduced only

Solution :It is a DISPROPORTIONATION REACTION.
Discussion
11.

Write the electronic configuartion of Cu^(+) (At. No. of Cu= 29)

Answer»

SOLUTION :`1 s^(2) 2s^(2) 2P^(6) 3s^(2) 3P^(6) 3d^(10)`
Discussion
12.

What happens when CO_(2)is passed through lime water (i) for a short duration (ii) for a long duration ?

Answer»

SOLUTION :`(i) Ca(OH)_(2)+CO_(2) RARR underset(("white PPT")) (CaCO_(3))+H_(2)O`
(ii) `CaCO_(3)+CO_(2)+H_(2)O rarr underset(("loluble"))(Ca(HCO_(3))_(2)) `
Discussion
13.

Which one of the following is vinyl carbocation

Answer»

`C_6H_5-OVERSET(o+)(CH_2)`
`CH_(2)=CH-overset(o+)(CH_2)`
`CH_(2)=overset(o+)(CH)`
All of these

Answer :C
Discussion
14.

What is volume of n mole gas ? Derive the formula of combined gas low.

Answer»

Solution :Volume of n mole gas : ..At constnat temperature and constant pressure any volume of n mole gas remains equal... because of -
According to pV = nRT
`therefore V = (nRT)/(p)=` constant….(Eq. - i)
[`because` Taking close constant n, R, T, p]
Combined gas equation :
Ideal gas equation : pV = nRT
`therefore (pV)/(T)=nR=` constant .....(Eq. -II)
[`because n =` fixed mole and R = gas constant]
This is the RELATIONSHIP for the simultaneous VARIATION of the variables. If temperature (T), volume (V) and pressure (p) of a fixed amount of gas vary from `T_(1), V_(1)` and `p_(1)` to `T_(2), V_(2)` and `p_(2)` then we can write -
`(p_(1)V_(1))/(T_(1))=(p_(2)V_(2))/(T_(2))""`.....(Eq. -iii)
This (Eq. -iii) is known as combined gas law.
Uses : This equation have SIX variable. If out of six, values of five variables are known, the value of unknown variable can be calculated from the equation of combined gas.
Discussion
15.

Which of the following is used in baking powder ?

Answer»

CITRIC ACID
lactic acid
TARTARIC acid
malonic acid

Solution :SEE tartaric acid
Discussion
16.

Which of the following is an electophile?

Answer»

`SO_(3)`
`CN^(-)`
`H_(2)O`
`NH_(3)`

ANSWER :A
Discussion
17.

Which is half reduction reaction in the following reaction (Basic medium). P_(4)+OH^(-)toPH_(3)+H_(2)PO_(2)^(-)

Answer»

`P_(4)toH_(2)PO_(2)^(-)`
`P_(4)toPH_(3)`
`OH^(-)toPH_(3)`
`PH_(3)toP_(4)`

Answer :B
Discussion
18.

What is the weight of oxygen that is required for the complete combustion of 2.8 kg of ethylene?

Answer»

2.8kg
6.4kg
96.0kg
9.6kg

Answer :D
Discussion
19.

Why are beryllium halides polymeric in nature ?

Answer»

SOLUTION :Beryllium chloride is an ELECTRON deficient MOLECULE since Be has only two covalent bonds and hence only four electrons in the valence shell. Therefore , to complete its octet , it can accept two electron pairs from Cl atoms of neighbouring MOLECULES forming polymeric structure . THUS , beryllium chloride has polymeric structure due to its electron deficient nature .
Discussion
20.

Write structures of various carbocations that can be obtained from 2-methylbutane. Arrange these carbocations in order or increasing stability.

Answer»

Solution :Four possible carbocations can ARISE from 2-methylbutane. These are :
`{:(CH_(3)-CH-CH_(2)-CH_(3) OVERSET(-H^(+))(rarr)CH_(3)-CH-CH_(2)-overset(o+)(C)H_(2)""CH_(3)-CH-overset(o+)(C)H-CH_(3)""CH_(3)-overset(o+)(C)-CH_(2)-CH_(3)""overset(o+)(C)H_(2)-CH-CH_(2)-CH_(3)),("|""|""|""|""|"),(""CH_(3)""CH_(3)""CH_(3)""CH_(3)""CH_(3)),(""Ioverset(@)((1))""IIoverset(@)((2))""IIIoverset(@)((3))""IVoverset(@)((1))):}`
Order of increasing stabilities of the carbocations is : `(I) lt (IV) lt (II) lt (III)`
EXPLANATION. We know that the order of relative stabilities of carbocations is : tertiary `gt` secondary `gt` primary. In this CASE, carbocations (III) and (II) are tertiary and secondary respectively. carbocations (I) and (IV) are both primary carbocations. Out of the two, structure (IV) is more stable because the electron releasing - `CH_(3)` group (-I effect) is present at `alpha`-position w.r. to the carbon atom carrying positive charge. In structure (I), it is present at the `beta`-position. the increasing order of stabilities of carbocations is : `(I) lt (IV) lt (II) lt (III)`.
Discussion
21.

Write the structure of all the isomers of dichloroethene. Which one of them will have zero dipole-moment?

Answer»

SOLUTION :There are THREE ISOMERS of DICHLOROETHENE.
`UNDERSET("1,1-dichlorethene")(underset(H-C-H)underset(||)(Cl-C-Cl))""underset("cis-1,2-dichloroethene")(underset(H-C-Cl)underset(||)(H-C-Cl))""underset("trans-1,2-dichloroethene")underset(Cl-C-H)underset(||)(H-C-Cl)`
Discussion
22.

Which one of the following orders is not in accordance with the property stated against it?

Answer»

`F_2 gt Cl_2 gt Br_2 gt I_2` (Oxidising POWER) 
`HI gt HBr gt HCI gt HF `(Acidic property in water) 
`F_2 gt Cl_2 gt Br_2 gt I_2` (Electron negativity) 
`F_2 gt Cl_2 gt Br_2 gt I_2` (Bond DISSOCIATION energy) 

Answer :D
Discussion
23.

Whichf the following species does not acts as a nucleophile ?

Answer»

`ROH `
ROR
`PCl_(3)`
`BF_(3)`

Solution :N/A
Discussion
24.

Which one of the following arrangements not truly represent the property indicated against it?

Answer»

`Br_(2) lt Cl_(2) lt F_(2)` , electronegativity
`Br_(2) lt F_(2) lt Cl_(2)`, electron affinity
`Br_(2) lt Cl_(2) lt F_(2)`, bond ENERGY
`Br_(2) lt Cl_(2) lt F_(2)`, OXIDISING power

Answer :C
Discussion
25.

Which of the following adopts normal spinal structure ?

Answer»

CsCI
`MgAl_(2)O_(4)`
FeO
`CaF_(2)`

Solution :NORMAL spinal is `MgAl_(2)O_(4)`.
Discussion
26.

Which ofthe following alkynes on treatment with H_2 92 mole)/pt gives an optically inactive compound ?

Answer»

3-methyl-1-pentyne
4-methyl-1-hexyne
3-methyl-1-heptyne
NONE of these

SOLUTION :
Discussion
27.

What are the conditions for getting maximum yiels of NH_(3) by Haber's process ?

Answer»

Solution :HIGH concentrations of `N_(2) and H_(2)`, LOW temperature , high pressure.
Discussion
28.

Which of the following is not an example of tautomeric equilibrium?

Answer»

`CH_(2)=CH-underset(H)underset(|)C=OiffCH_(2)-CH=underset(H) underset(|)C-O^(-)`
`-N=N-underset(|)NHiff-NH-N=N-`

`-NH-underset(|)C=O iff-N=underset(|)C-OH`

Solution : No acidd `ALPHA`-HYDROGEN
Discussion
29.

What I a semiondutor ? Mention the two main types of semiconductor.

Answer»

SOLUTION :Solids which have conductivity inbetween those of conductors and insulators, i.e, RANGING from `10^(-6) " to " 10^(4)ohm^(-1) m^(-1)` are called SEMICONDUCTORS. The Two main types are n-type and p-type.
Discussion
30.

Which of the following reaction(s) give Hofmann product as the major product?

Answer»

treatment of 2-fluorobutane with MeONa
treatment of 2-chloro BUTANE with t-BuO-`K^(+)`
treatment of butan-2-ol with con. `H_2SO_4 // DELTA `
treatment of 3-amino pentane with excess of `CH_3` – I FOLLOWED by HEATING with moist `Ag_2O`

Solution :
Discussion
31.

Which complex is formed by carbon monoxide gas with hemoglobin ?

Answer»

SOLUTION :Carbonxy HEMOGLOBIN COMPLEX
Discussion
32.

Which of the following water smple has maxiumum degree of hardness?

Answer»

`9.5g "of" MgCl_(2)"in"10^(5)KG" of water "`
`11.1 g "of"CaCl_(2)"in"10^(5)kg"os water "`
`6.8 "of"CaSO_(4)"in"10^(4)kg "of water"`
`1.2 g "of" MgSO_(4)"in"10^(4)"of water"`

Answer :C
Discussion
33.

Which of the following groups will increase the acidity of phenol?

Answer»

`-NO_(2)`
`-CN`
`-X` (HALOGENS)
All

Solution :All ELECTRON withdrawing groups will increases ACIDIC nature of phenol.
Discussion
34.

Which two of the following reactions are possible? (i) FeCl_(3)+SnCl_(2)to2FeCl_(2)+SnCl_(4) (ii) 2FeCl_(2)+SnCl_(4)to2FeCl_(3)+SnCl_(2) (iii) PbCl_(2)+SnCl_(4)toSnCl_(2)+PbCl_(4) ltBrgt (iv). SnCl_(2)+PbCl_(4)toPbCl_(2)+SnCl_(4)

Answer»

`i` and II
`i` and III
`i` and iv
ii and iv

Answer :D
Discussion
35.

Which one of the following gases has the highest critical temperature ?

Answer»

NITROGEN
Ammonia
Water vapour
Carbon dioxide

Solution :The most EASILY LIQUEFIABLE gas/vapour has the HIGHEST CRITICAL temperature.
Discussion
36.

Which is the correct order of hydration enthalpy of alkali metals?

Answer»

`Li^(+) gt Na^(+) gt K^(+) gt Rb^(+) gt Cs^(+)`
`Cs^(+) gt Rb^(+) gt K^(+) gt Na^(+) gt Li^(+)`
`Cs^(+) gt Rb^(+) gt K^(+) gt Li^(+) gt Na^(+)`
`Na^(+) gt Li^(+) gt K^(+) gt Rb^(+)gt Cs^(+) `

Solution :
The hydration enthalpy of alkali metal ions decreases with increase in their ionic sizes
`Li^(+) gt Na^(+) gt K^(+) gt Rb^(+) gt Cs^(+)`
The degree of hydration of `Li^(+)` is the HIGHEST, DUE to which LITHIUM salts are MAINLY hydrated, e.g. LiCl, `H_(2)O`
Discussion
37.

Which of the following electrodes will act as anodes, when connected to Standard Hydrogen Electrode ?

Answer»

`Al//Al^(3+) ""E^@=-1.66V`
`Fe//Fe^(2+) ""E^@=-0.44V`
`Cu//Cu^(2+) ""E^@=-0.34V`
`F_2(G)2F^(-)(aq) " "E^@=+2.87V`

SOLUTION :The electrodes having negative ELECTRODE potentials are stronger reducing agents than `H_2` gas and therefore , will ACTS as anodes.
Discussion
38.

The solution containing Mn^(2+), Fe^(2+), Zn^(2+) and Hg^(2+) concentration is 10^(-3) M. If K_(sp) of Mns, FeS, ZnS an HgS are respectively 10^(-15), 10^(-2), 10^(-4), 10^(-54) than which one is the first precipitate by H_2S ?

Answer»

FeS
HgS
MnS
ZnS

Solution :`K_(SP)` is LEAST , PRECIPITATION will be FAST.
Discussion
39.

Which of the following statements is correct for benzene ?

Answer»

All the SIX carbon atoms and all the six hydrogen atoms lie in different planes.
The ring SYSTEM consists of three localised `C-C pi`-bonds.
The ring system consists of a delocalised `pi`-molecular ORBITAL containing six electrons.
Each C atom is in a state of `sp^3`hybridisation.

Answer :C
Discussion
40.

What are pH and pOH values of a neutral solution at a temperature at which K_(w)=10^(-13)?

Answer»

Solution :`pK_(W)=pH + POH `. But `pK_(w)=13`. Also, for neutral solution, pH = pOH . Hence, pH = pOH = 6.5.
Discussion
41.

Write the preparation of ethane.

Answer»

Solution :(i) `underset("Ethene")(CH_(2)=CH_(2)) underset("Pt/Pd/Ni")overset(+H_(2), "Dihydrogenation")rarr CH_(3)-CH_(3)`
(ii) `underset("ETHYL chloride")(CH_(3)CH_(2)Cl) underset(-HCL)overset(+H_(2), Zn, H^(+) " Reduction")rarr CH_(3)-CH_(3)`
(iii) `underset("Bromo")(2CH_(3)Br) underset(-2NaBr)overset("2Na, DRY ether Wurtz reaction")rarr CH_(3)-CH_(3)`
(iv)
(v)
(vi) `2CH_(4) underset(-HCl, +Cl)overset(hv)rarr 2CH_(3) underset("combustion")overset("FREE RADICAL")rarr CH_(3)-CH_(3)-`
(vii)
Discussion
42.

The uncertainty in the position of an electron ("mass "9.1 xx 10^(-28)g) moving with a velocity of 3.0 xx 10^(4)cm^(-1) accurate up to 0.011%, will be

Answer»

0.192 cm
7.68 cm
0.175 cn
3.84 cm

Solution :`DELTA X . M Delta V ge h/(4pi) , Delta V = (0.001)/(100)= 11 XX 10^(-5) ` cm/sec
Discussion
43.

What is bond enthalpy? How they relate with bond strength?

Answer»

SOLUTION :The bond ENTHALPY is defined as the minimum amount of energy required to break one mole of a PARTICULAR bond in molecules in their GASEOUS state. Larger the bond enthalpy STRONGER will be the bond
Discussion
44.

Which of the following electronic configuration represent the element in ground state?

Answer»

`1s^2 2s^1 2p^1`
`1s^2 2s^2 2p^1`
`1s^2 2s^1 2p_x^1 2p_y^1 2p_z^1`
`1s^2 2s^2 2p^6 3s^2 3p_x^1 3p_z^1 3d^1`

ANSWER :B
Discussion
45.

Which of the following Benzene rings contain o-p directing groups ?

Answer»


ANSWER :A,B,C,F,H
Discussion
46.

Which of the following group is not an electron withdrawing group ?

Answer»

`-NO_(2)`
`-C_(2)H_(5)`
`-COOH`
`-C Cl_(3)`

Answer :B
Discussion
47.

Which of the following is responsible to rule out the existence of definite paths or trajectories of electrons

Answer»

Pauli’s exclusion principle.
Heisenberg’s UNCERTAINTY principle.
Hund’s rule of MAXIMUM multiplicity.
Aufbau principle

Solution :Heisenberg’s uncertainty principle. Werner Heisenberg, a German physicist in 1927, given uncertainty principle which states that it is not POSSIBLE to determine simultaneously, the exact position and exact momentum of an electron. The important implications of the Heisenberg uncertainty principle is that it rules out existence of definite PATHS or trajectories of electrons and other similar particles
Discussion
48.

Which of the following has the least mass?

Answer»

one gram atom of magnesium
0.9 MOLE nitric oxide
22.4 LIT of NITROGEN at STP
`6.02 xx 10^(24)` molecules of oxygen

Answer :A
Discussion
49.

Why are group 1 elements called alkali metals ?

Answer»

SOLUTION :Because their HYDROXIDES are STRONGLY ALKALINE.
Discussion
50.

What is meant by molar heat of vaporization?

Answer»

Solution :The molar heat of vapourisation is DEFINED as the change in enthalpy when one mole of liquid is converted into VAPOUR or GASEOUS STATE at its boiling point.
Discussion