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.

Na_(2)CO_(3) cannot be used in place of (NH_(4))_(2)CO_(3) for the precipitation of V group, because

Answer»

`NA^(+)` INTERFERES in the detection of V group
conc. of `CO_(3)^(2-)` is very low
Na will REACT with acid radicals
Mg will be precipitated.

Answer :D
Discussion
2.

Na_(2)CO_(3) cannot be used in place of (NH_(4))_(2)CO_(3) for the percipitation of V group because

Answer»

Na will react with acid radicals
Conc. of `CO_(3)^(2-)` is very low
Mg will be percipitated
`Na^(+)` INTERFERES in the detection of V group

Answer :C
Discussion
3.

Na_2CO_3 cannot be used in place of (NH_4)_2,CO_3, for precipitation of group Vcations because

Answer»

Na INTERFERES in the detection of group V
Concentration of `CO_3^(2-)` is very LOW
`Na^+` will react with ACID RADICALS
`Mg^(2+)` will precipitate as `MgCO_3`, if present in the mixture

Answer :D
Discussion
4.

Na_(2)CO_(3) cannot be used in place of (NH_(4))_(2)CO_(3) for precipitation of group V cations because

Answer»

`NA^(+)` INTERFERES in the detection of group V
concentration of `CO_(3)^(2-)` is very low
`Na^(+)` will REACT with acid radicals
`Mg_(2+)` will precipitate as `MgCO_(3)` il present in the mixture

Solution :`Na_(2)CO_(3)` is highly ionizable electrolyte which PRODUCES very high concentration of `CO_(3)^(2-)` ions. As a result ionic PRODUCT of `MgCO_(3)` may increase its K, and it may get precipitated along with the radicals of group V.
Hence, (D) is the correct answer.
Discussion
5.

Na_2CO_(3) can be manufactured by Solvay's process but K_(2)CO_(3) cannot be prepared because

Answer»

`K_(2)CO_(3)` is more soluble
`K_(2)CO_(3)` is LESS soluble
`KHCO_(3)` is more soluble than `NaHCO_(3)`
`KHCO_(3)` is less soluble than `NaHCO_(3)`

ANSWER :C
Discussion
6.

Na_2CO_3can be manufactured by Solvay's process butK_2CO_3 can not be prepared because

Answer»

`K_2CO_3` is more SOLUBLE
`K_2CO_3` is LESS soluble
`KHCO_3` is more soluble than `NaHCO_3`
`KHCO_3` is less soluble than `NaHCO_3`

ANSWER :C
Discussion
7.

Na_(2)B_(4)O_(7)overset(Delta)toNaBO_(2)+B_(2)O_(3) NaBO_(2)+H_(2)O_(2)+H_(2)O to compound A(aq)The compound A the sum of oxidation states of all the oxygen atoms is X. Then the value of (X) is

Answer»


SOLUTION :`2NaBO_(2)+2H_(2)O_(2)+6H_(2)OtoNa_(2)[(OH)_(2)B(O-O)_(2)B(OH)_(2)]`sodium peroxoborate

`X=[(-2)xx4+(-1)xx4]=-12` , HENCE `|X|=12`
Discussion
8.

Na_(2)[Be(OH)_(4)] is formed when

Answer»

BEO reacts with NAOH solution
Be reacts with NaOH solution
both are correct
NONE is correct

Answer :C
Discussion
9.

Na^(23) is more stable isotope of Na. Find out the process by which ._(11)Na^(24) can undergo radioactive decay.

Answer»

`beta^(c-)-` emission
`alph-`emission
`beta^(o+)-`emission
`K` electron capture.

Solution :Isotopic `_(11)Na^(24)` is less stable than `._(11)Na^(23)` because it shows radioactive decay Less stability of `Na^(24) w.r.t Na^(23)` also based upon `13//11(n/p)` ratio. Higher the value higher will be unstability. So it is disintegrated to ATTAIN stability).
`underset(Less stabl e)(._(11)Na^(24)) rarr underset(Stabl e)(._(11)Na^(23))+ underset("Neutron")(._(0)n^(1)`
This neutron on DECOMPOSITION gives proton and `beta-` particle`(._(-1)e^(0))`
`._(0)n^(1) rarr underset(Prot on)(._(1)H^(1)) ` or`._(1)P^(1)+underset(beta-partic l e)(._(-1)e^(0)`
Hence, isotopicsodium is CHANGED into sodium by means of emission of `beta-` particle and a proton `i.e., ` by `beta-` emission.
Discussion
10.

Na^(22) has half life of 2.68 years . It decays both by positron emission and electron capture with a ratio of 86% of the former to 14% of the later . The half life for positron emission in years (nearly) is ……….

Answer»


SOLUTION :
`t_(1//2)= (0.693)/(lambda_(1)) = (0.693)/(0.2224) = 3.116 -= 3, Y_(A) = (lambda_(1))/(LAMBDA) xx 100 = 86 l lambda = (0.693)/(2.68) ………. (2)`
`Y_(B) = (lambda_(2))/(lambda) xx 100 = 14 , lambda = (lambda_(1) + lambda_(2)) ……... (3) implies (lambda_(1))/(lambda_(2)) = (86)/(14) ...... (1)` From (1) , (2) and (3)
`lambda = (lambda_(1)+ lambda_(2)) = (0.693)/(2.68) , (lambda_(1) + (14 lambda_(1))/(80) ) = 0.2586 , lambda_(1) (1 + (14)/(80)) = 0.2586`
`lambda_(1) = (0.2586 xx 86)/(100) = 0.2244 , t_(1//2) = (0.693)/(lambda_(1)) = (0.693)/(0.2224) = 3.116 -=3`
Discussion
11.

Na""^(22) has half life of 2.68 years. It decays both by positron emission and electron capture with a ratio of 86% of the former to 14% of the later. The half life for positron emission in years (nearly) is

Answer»


SOLUTION :`LAMDA""_(1)/lamda""_(2)=14/86,(lamda=2.303/tlog"N""_(0)/N),`
`lamda=lamda""_(1)+lamda""_(2)`
`0.693/2.68=(lamda""_(2)xx14)/86+lamda""_(2)02`
`0.258=lamda""_(2)(14/86+1)`
`0.258=lamda""_(2)(100/86)`
lamda""_(2)=(86xx0.258)/(100),t""_(1//2)=0.693/lamda""_(2)`
Discussion
12.

Na_2 O and P_4 O_(10) on heating at 1000^@C yields

Answer»

<P>SODIUM tripolyphosphate : `Na_(5)P_(3)O_(10)`
Sodium tripolyphosphate : `Na_(5)P_(3)O_(10).6H_(2)O`
sodium dihydrogenphosphate : `Na_(2)HPO_(4)`
Sodium hydrogenphosphate: `Na_(2)HPO_(4)`

Solution :`Na_(5)P_(3)O_(10)`
Discussion
13.

N_A spheres of radius 'R' are melted in x xxN_A smaller spheres, so that when fcc lattice is generated from smaller spheres the edge length is formed to be 2 5/6 R.Find x.

Answer»


SOLUTION :`N_A xx 4/3 piR^3=x xxN_Axx4/3 pi R^3 implies r=R/ROOT(3)(x)`
In fcc
`sqrt2a=4r. " " implies " " sqrt2xx2^(5//6)R=(4R)/root(3)(x)`
`2^(5/6+1/2)=2^2/x^(1//3)implies x^(1//3)=2^(2 4/3)implies x^(1//3)=2^(2//3) implies x=4`
Discussion
14.

Na^(+), Mg^(2+), Al^(3+) and Si^(4+) are isoelectronic. The order of their ionic size is

Answer»

`Na^(+) GT MG^(2+) LT Al^(3+) lt SI^(4+)`
`Na^(+) lt Mg^(2+) gt Al^(3+) gt Si^(4+)`
`Na^(+) gt Mg^(2+) gt Al^(3+) gt Si^(4+)`
`Na^(+) lt Mg^(2+) gt Al^(3+) lt Si^(4+)`

Solution :Amongst isoelectronic ions, the size of the cation decreases as the magnitude of the charge increases.
Discussion
15.

Na reacts rapidly with :

Answer»

`1^@`ALCOHOL
`2^@` alcohol
`3^@`alcohol
None

Answer :A
Discussion
16.

Na is used in reduction of Zn salt because

Answer»

`E^(@)Zn_("OX") GT E^(@)Na_("ox")`
`E^(@)Zn_("ox") lt E^(@)Na_("ox")`
`E^(@)Zn_("ox") = E^(@)Na_("ox")`
`E^(@)Zn_("red") lt E^(@)Na_("red")`

Solution :Element with SMALLEST SOP displaces element with high SOP from its salt.
Discussion
17.

Na is used in the reduction of Znsalt because

Answer»

`E_(Zn("oxi"))^(@) gt E_(Na("oxi"))^(@)`
`E_(Zn("RED"))^(@) lt E_(Na("red"))^(@)`
`E_(Zn("oxi"))^(@) lt E_(Na("oxi"))^(@)`
Both a and b

Solution :Metal having higher OXIDATION potential (or lesser `E_("red")^(@)`) can REPLACE metals having lesser oxidation potential
`because` oxidation potential of Nais more than that of Zn
`therefore` Na is USED to reduce Zn salts
Discussion
18.

Na and Mg crystallize in bee and fcc type crystals respectively, then the nwnber of atoms ofNa and Mg present in the unit cell of their respective crystal is

Answer»

4 and 2
9 and 14
14 and 9
2 and 4

Solution :In BCC- POINTS are at corners and one in the centre of the unit cell.
Number of atoms per unit cell`=8xx(1)/(8)+1=2`. ,brgt In fcc-points are at the corners and ALSO centre of the six faces of each cell.
Number of atoms per unit cell`=8xx(1)/(8)+6xx(1)/(2)=4`.
Discussion
19.

Na and Mg crystallize in BCC and FCC type of crystals respectively, then the number of atoms of Na and Mg present in the unit cell of their respective crystals is

Answer»

4,2
9,14
14,9
2,4

Answer :D
Discussion
20.

When Na and Li placed in dry air we get:-

Answer»

`NAOH,Na_2O,Li_2O`
`Na_2O,Li_2O`
`Na_2O,Li_2O,Li_3,NH_3`
`Na_2O,Li_3ON,Li_2O`

ANSWER :D
Discussion
21.

Na // alcoholis a good

Answer»

Dehydratingagent
Oxidisingagent
Reducingagent
Catalyst

Solution :`C_(2) H_(5) OH + N/A toC_(2) H_(5) N/A + [H]`
Discussion
22.

Na^(+) and Ag^(+) differ in

Answer»

`Na_(2)CO_(3)` is thermally stable while `Ag_(2)CO_(3)` decompose into Ag, `CO_(2) and O_(2)`
Na forms complexes, `Ag^(+)` does not
AgCl is water soluble, NACL is INSOLUBLE
NaCl and AgCl do not give colour in FLAME when ignited

ANSWER :A
Discussion
23.

N_2O_4 molecule is completely changed into 2NO_2 molecules at:

Answer»

`-10^@` C
`140^@ - 150^@ C`
`420^@` C
`-40^@` C

Answer :B
Discussion
24.

N_2O_4 is 25% dissociated at 37^@C and one atmosphric pressure. Calculate (i) K_p and (ii) per cent dissociation at 0.1 atmospheric pressure and 37^@C.

Answer»


ANSWER :(i)0.267 ATM (II) 63.25%
Discussion
25.

N_(2)O_(4) is

Answer»

ACIDIC and paramagnetic
Acidic and dimagnetic
Basic and paramagnetic
Basic and dimagnetic

Answer :B
Discussion
26.

N_2O_2 hArr NO,K_1,1/2 N_2 +1/2O_2hArr NO,K_22NO hArr N_2+O_2k_3 , NO hArr 1/2 N_2 +1/2O_2 K_4

Answer»

`K_1 xx K_3 =1`
`sqrt(K_1)xxK_4=1`
`sqrt(K_3)xxK_2=1`
none of the above

Answer :A::B::C
Discussion
27.

N_(2)O_(2)_(g) rarr 2NO is a first -order reaction in terms of the concentration of N_(2)O_(2)(g) . Which of the following is valid. [N_P(2) O_(2) ] being constant ?

Answer»

`[NO] = [ N_(2) O_(2) ]_(0) e^(-KT)`
`[NO]=[N_(2)O_(2) ]_(0)(1-e^(-kt))`
`[NO]=[N_(2)O_(2)]_(0)(e^(-kt)-1)`
`[NO]=2[N_(2)O_(2)]_(0)( 1- e^(-kt))`

Answer :D
Discussion
28.

N_(2)O_(3) exists pure only in the solid state at low temperature. Above its melting point, it dissociates to :

Answer»

`NO and NO_(2)`
`N_(2)O_(5) andO_(2)`
`N_(2)O and NO_(2)`
`N_(2)O and O_(2)`.

SOLUTION :`N_(2)O_(3)HARR NO+NO_(2)`
Discussion
29.

N_(2)O is formed when:

Answer»

moist FE reacts with NO
`SN^(+2)`reacts with conc. `HNO_(3)` in presence of HCI
 cold DIL. `HNO_(3)` reacts with Cu and Zn
by all the reactions

Answer :D
Discussion
30.

N_(2)O is formed when ........

Answer»

Moist Fe REACTS with NO
`Sn^(2+)` reacts with conc. `HNO_3` in presece of HCl
COLD DIL `HNO_3` reacts with Zn
All the above

Solution :a) `2NO+H_(2)O+FerarrN_(2)+Fe(OH)_2`
b) `4SnCl_(2)+2HNO_(3)+3HClrarr 4SnCl_(4)+N_(2)O+5H_(2)O`
c) `4Zn+10HNO_3 rarr 4Zn(NO)_3(2)+N_(2)O+5H_(2)O`
Discussion
31.

N_(2)O is obtained by the reaction of

Answer»

Cu with dil `HNO_(3)`
Cu with CONC. `HNO_(3)`
Zn with dil `HNO_(3)`
Zn with conc.`HNO_(3)`

ANSWER :3
Discussion
32.

N_2O is formed on reaction with dil, HNO_3 with:

Answer»

Cu
Hg
Ag
Fe

Answer :D
Discussion
33.

N_(2(g))hArr+NH_(3(g)) In the reaction given above, the addition of small amount of on inert gas at constnat pressure will shift the equlibrium towards which side of

Answer»

LHS (Left Hand Side)
RHS (RIGHT Hand Side)
Neither side
Either side

Solution :ADDITION of inert gas at constant PRESSURE shift EQULIBRIUM where no. of moles are more.
Discussion
34.

N_(2)H_(4) and H_(2)O_(2) show similarity in ..........

Answer»

Hybridisation of central ATOMS
Oxidising nature
Reducing nature
MOLAR mass

SOLUTION :
Discussion
35.

N_2H_2 reduces IO_3^(-) //H^+ ......

Answer»

to `I^(+)`
with `I_2` as an intermediate by violet colour in `"CCL"_4` LAYER
to `I^(-)`
indicated by blue colour with starch

Solution :`3N_(2)H_(4)+2IO_(3)^(-) rarr 3N_(2)+2I^(THETA)+6H_(2)O`
Discussion
36.

N_(2(g))+3H_(2(g))hArr2NH_(3)(g)+22k.cal. The activation energy for the forward reaction is 50 k. cal What is the activation energy for the backward reaction ?

Answer»

`-72` K.CAL
`-28` k. cal
`+28` k.cal
`+72`k.cal

Answer :D
Discussion
37.

N_2(g) + 3H_2(g) hArr 2NH_3(g), /_\H =-93.5 kj what will happen when helium gas is added to the vessel at constant valume:

Answer»

more `NH_3` is FORMED
less `NH_3` is formed
no effect
NONE of these

Answer :C
Discussion
38.

N_(2)+O_(2)hArr2NO-Q cal In the above reaction which is the essential condition for the higher production of NO

Answer»

HIGH temperaturee
High pressure
LOW temperaturee
Low pressure

SOLUTION :`N_(2)+O_(2)hArr2NO,Q` cal
The above reaction is endothermic so for higher PRODUCTION of NO, and the temperature should br high.
Discussion
39.

N_2 O is used as an anesthetic in dental surgery.

Answer»


ANSWER :T
Discussion
40.

N_(2) is found in a litre flask under 100 kPa pressure and O_(2) is found in another 3 litre flask under 320 kPa pressure. If the two flasks are connected, the resultant pressure is

Answer»

310 kPa
210 kPa
420 kPa
265 kPa

Solution :`P_(1)V_(1)+P_(2)V_(2)=P_(3)(V_(1)+V_(2))`
`100xx1+320xx3=P_(3)(1+3)` or`P_(3)=265` kPa
Discussion
41.

N_(2) gas will not be evolved upon reaction of HNO_(2) with which of the following amines

Answer»

`1^(@)`
`2^(@)`
`3^(@)`
Both B and c

Solution :Primary amines form alcohols with nitrous ACID `(NaNO_(2)+HCl)`. Nitrogen gas (with brisk effervescence) is eleminated and clear solution is obtained which is not so in the CASE of SECONDARY and tertiary amines.
`underset(1^(@)"amine")(RNH_(2))+HONOrarrunderset("Alcohol")(ROH)+N_(2)uparrow+H_(2)O`
`underset(2^(@)"amine")(R_(2)NH)+HONOrarrunderset("N-Nitrosodialkylamine")(R_(2)N NO+H_(2)O)`
Discussion
42.

N_(2) exerts a pressure of 0.987 bar. The mole fraction of N_(2) ia (K_(H)=74.48 k bar)

Answer»

`7.648 xx 10^(-3)`
`9.87 xx 10^(-5)`
`1.3 xx 10^(-5)`
`2.6 xx 10^(-5)`

Solution :`P_(N2)=K_(H) xx X_(N2)`
`:. X_(N2)=P_(N2)/K_(H)=(0.987"barr")/(76480"bar")=1.29 xx 10^(-5)`
Discussion
43.

N_2 combines with metal to form:

Answer»

Nitritie
Nitrate
Nitride
Nitrosyl chloride

Answer :C
Discussion
44.

N_(2) combines with metal to form

Answer»

Nitride
Nitrate
Nitrite
Nitrosyle chloride

SOLUTION :Nitrogen react with metal to form a nitride.
`6LI +N_(2) overset(450^(@)C) to 2Li_(3)N` (LITHIUM nitride)
Discussion
45.

N_(2) DOES NOT show property of catenation because :

Answer»

it has no 'd' orbitals in the valence shell
inter electronic REPULSION between NON bonding electrons is GREATER
`N-=N`has HIGH BOND enthalpy
Nitrogen has very high ionization enthalpy.

Answer :B
Discussion
46.

N_(2) and O_(2) are converted into monocations N_(2)^(+) and O_(2)^(+) respectively. Which of the following statements is wrong?

Answer»

In `N_(2)^(+)`, the N-N bond weakens
In `O_(2)^(+),` the O-O bond order increases
In `O_(2)^(+)`, paramagnetism decreases
`N_(2)^(+)` becomes diamagnetic

Solution :`N_(2)^(+)` contains `[sigma(2p_(z))]^(1)` and HENCE is PARAMAGNETIC
Discussion
47.

N_(2) and O_(2) are converted to monocations N_(2)^(+) and O_(2)^(+) respectively, which is wrong statement:

Answer»

In `N_(2)^(+)`, the N-N BOND weakens
In `O_(2)^(+)`, the O-O bond order increase
In `O_(2)^(+)`, the paramagnetism DECREASE
`N_(2)^(+)` becomes diamagnetic

Answer :D
Discussion
48.

N_(2) and O_(2) are converted to mono cations N_(2)^(+) and O_(2)^(+) respectively. Which is wrong ?

Answer»

In `N_(2)^(+), N-N` bond WEAKENS
In `O_(2)^(+)`, the `O-O` bond ORDER increases
In `O_(2)^(+)`, the paramagnetism DECREASES
`N_(2)^(+)` becomes diamagnetic.

Answer :D
Discussion
49.

N_(2)+3H_(2)hArr2NH_(3). Which is correct statement, if N_(2) is added at equlibrium condition

Answer»

The equlibrium will shift to forward direction because according to `II^(ND)` law of thermodynamics the entropy must increases in the direction of spontaneous reaction
The condition for equlibrium is `G_(N_(2))+3G_(H_(2))=2G_(NH3)` where G is Gibber free energy per mole of the gaseous species masured at that PARTIAL PRESSUER. The condition of speical measured at that partial pressuer.The condition of equlibrium is unaffected by the use of catalyst, which increase the rate of both the forward and backward REACTIONS to the same EXTENT.
The catalyst will increase the rate of forward reaction by `alpha` and that of backward reaction by `beta.`
Catalyst will not alter the rate of either of the reaction

Answer :B
Discussion
50.

N_(2)+3H_(2)hArr2NH_(3)+ heat. What is the effect of the increase of temperature on the equilibrium of the reaction

Answer»

Equlibrium is unaltered
Reactin rate does not change
Equlibrium is shifted to the left
Equlibrium is shifted tothe right

Solution :`N_(2)+3H_(2)hArr2NH_(3)+` HEAT
Formation of ammonia is exothermic REACTION so increase in TEMPERATURE favours BACKWARD reaction and equlibrium is shifted to the left.
Discussion