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.

Pair of smallest and largest species of radii among Mg, Al, Mg^(+2), Al^(+3) is :-

Answer»

`MG^(+2), Mg`
`Al^(+3), Mg`
`Mg^(+2), Al`
`Al^(+3), Al`

ANSWER :B
Discussion
2.

Paris green is:

Answer»

`CU(CH_3COO)_2`
`Cu_3(AsO_3)_(2).2H_2O`
`Cu(CH_3COO)_(2).3Cu(AsO_2)_2`
`CO(AlO_2)_2`

ANSWER :C
Discussion
3.

Pair of isobar is :

Answer»

`._(6)^(77)C,_(7)^(13)N`
`._(6)^(13)C,_(7)^(14)N`
`._(6)^(14)C,_(8)^(15)N`
NONE of these

Answer :A
Discussion
4.

Pair of compound which is planar :-

Answer»

<P>`[NI(CN)_(4)]^(-4), [PtCl_(4)]^(-2)`
`[NiCl_(4)]^(-4), [Ni(CN)_(4)]^(-2)`
`[PtCl_(4)]^(-2), [Ni(CO)_(4)]`
`[Ni(CN)_(4)]^(-2), [Rh(CO)_(2) (P Ph_(3))_(2)]^(+)`

Answer :D
Discussion
5.

paints and varnish contain

Answer»

COCONUT OIL
MUSTARD oil
Linseed oil
Kerosene

ANSWER :C
Discussion
6.

Packing efficiency is the percentage of total space filled by the particles.Which of the following lattices has the highest packing efficiency? Simple cubic lattice, body centered cubic lattice, hexagonal close packed lattice

Answer»

SOLUTION :HEXAGONAL CLOSE PACKED LATTICE.
Discussion
7.

Packing fraction is related to :

Answer»

NUCLEAR spin
electron spin
mass NUMBER
atomic number.

ANSWER :C
Discussion
8.

Packing efficiency by arrangement of atoms in two dimensional hexagonal close packing is

Answer»

60.43
65.78
59.78
68.76

Solution :
let radius of the sphere=r
AREA occupied by sphere in hexagonal close packing
`pir^(2)+6XX((1)/(6)xxpir^(2))=2pir^(2)`
Area of hexagonal`=6xx[(SQRT(sqrt(3))/(4)xx(2r)^(2)]`
`=6xx(sqrt(3))/(4)xx4r^(2)`
`=6sqrt(3)xxr^(2)`
% occupied by `=(2pir^(2))/(6xxsqrt(3)xxr^(2))xx100`
`=(2xx3.14)/(6xxsqrt(3))xx100=60.43%`
Discussion
9.

Packing efficiency (Packing Fraction)

Answer»

Solution :It is the percentage of total SPACE in the tunit cell OCCUPIED by the constituent particles like ATOMS, ions or MOLECULES of the crystal.
Discussion
10.

P_(A)=320 torr, whenX_(A)=x torr If X_(A)=x+0.1,P_(A)=440 then P_(A)^(0)=200x, find x.

Answer»


ANSWER :6
Discussion
11.

P_(A)^(0)=10 torr, 1g of B is dissolved in 20g of A, its vapour pressure reduced to 9.0 torr . If mwt of A is 200, M.wt of B is 30x is

Answer»


ANSWER :3
Discussion
12.

P_A and P_B are the vapour pressure of pure liquid components. A and B, respectively of an ideal binary solution. If x_A represents the mole fraction of component A, the total pressure of the solution will be

Answer»

`P_B+x_A (P_B-P_A)`
`P_A+x_A (P_A-P_B)`
`P_A+x_A(P_B-P_A)`
`P_B+x_A (P_A-P_B)`

ANSWER :D
Discussion
13.

p_(A) and p_(B) are the vapour pressure of pure liquid components, A and B, respectively of an ideal binary solution. If x_(A) represents the mole fraction of component A, the total pressure of the solution will be

Answer»

<P>`p_(A) + x_(A)(p_(B)-p_(A))`
`P_(A)+ x_(A)(p_(A)-p_(B))`
`p_(B)+x_(A)(p_(B)-p_(A))`
`p_(B)+x_(A)(p_(A)-p_(B))`

Solution :According to Raoult.s law
`p=x_(A)p_(A)+x_(B)p_(B)`
For binary solution, `x_(A)+x_(B)=1` or `[x_(B)=1-x_(A)]`
`p=x_(A)p_(A)+(1-x_(A))p_(B)`
`=x_(A)p_(A)+p_(B)-x_(A)p_(B)`
or `=p_(B)+x_(A)(p_(A)-p_(B))`
Discussion
14.

P_(A)and P_(B) are the vapour pressure of pure liquid components ,Aand B respectively of an ideal binary solution,If x_(A) represents the mole fraction of component A, the total pressure of the solution will be

Answer»

`p_(A)+x_(A)(P_(B)-P_(A))`
`p_(A)+x_(A)(p_(A)-P_(B))`
`p_(B)+x_(A)(p_(B)-p_(A))`
`p_(B)+x_(A))p_(A)-p_(B))`

SOLUTION :According to Raoult's LAW,
`P=P_(A)x_(A)+P_(B)x_(B)`
For BINARY solutions
`x_(A)+x_(B)=1,x_(B)=1-x_(A)`
` P=P_(A)x_(A)+P_(B)(1-x_(A))`
`P=P_(B)+x_(A)(P_(A)-P_(B))`
Discussion
15.

P_(4)S_(3) +8O_(2) to P_(4)O_(10)+3SO_(2) Calculate mass of P_(4)S_(3) is required at least 9.6 gm of each product.

Answer»


ANSWER :11 GM
Discussion
16.

P_(4)("white")+SOCl_(2)rarrX("halide of P")+Y("oxide")+Z("halide of S") Incorrect statement among the following is :

Answer»

X on hydrolysis gives a dibasic acid
Y turns acidified `K_(2)Cr_(2)O_(7)` solution to green
Z has angular structure
Hydrolysis of Z is a disproportionation reaction

Answer :C
Discussion
17.

P_(4)overset(Ba(OH)_(2))underset(Delta)(to)underset("salt")(X) + underset("gas")(Y) uarr Xoverset("dil." H_2SO_4)rarrZdarr+underset("acid")WWhich is incorrect about Y?

Answer»

It usually IGNITES due to PRESENCE of `P_2H_4`
It is LESS soluble than `NH_3` in water
It has `107^(@) 28^(1)` bond angles
It can be obtained by HYDROLYSIS of `Ca_(3)P_2`

Solution :`P_(4)+3Ba(OH)_2+6H_(2)Orarr3Baunderset(X)((H_2PO)_2)+underset(Y)(2PH_3uarr)Baunderset(X)(H_2PO_2)+underset("dil")(H_2SO_(4))rarrunderset(Z)(BaSO_4darr)+underset(W)(H_3PO_2)`
Y is `PH_3` - less soluble than `NH_3` in water as it is weak base
`-Ca_(3)P_(2)+6H_(2)Orarr3Ca(OH)_2+2PH_(3)`
`-PH_3` contains impurity diphosphine `(P_2H_4)` - which catches fire readily and ignites `PH_3`
Discussion
18.

P_(4)overset(Ba(OH)_(2))underset(Delta)(to)underset("salt")(X) + underset("gas")(Y) uarr Xoverset("dil." H_2SO_4)rarrZdarr+underset("acid")W Which is correct in view of 'W' ?

Answer»

It's shape is planar
It is a diabasic acid
It can be a reducing agent
It has three 'P - OH' linkages

Solution :`P_(4)+3Ba(OH)_2+6H_(2)Orarr3Baunderset(X)((H_2PO)_2)+UNDERSET(Y)(2PH_3uarr)Baunderset(X)(H_2PO_2)+underset("DIL")(H_2SO_(4))rarrunderset(Z)(BaSO_4darr)+underset(W)(H_3PO_2)`
Y is `PH_3` is WEAKER base than `NH_3`
Discussion
19.

P_(4)overset(Ba(OH)_(2))underset(Delta)(to)underset("salt")(X) + underset("gas")(Y) uarr Xoverset("dil." H_2SO_4)rarrZdarr+underset("acid")W Which statment is correct ?

Answer»

The anion of salt 'X' cannot have a LEWIS base
Gas 'Y' is a better than `NH_3`
W' is tribasic acid
OXIDATION state of 'P' in 'X' is + 3

SOLUTION :`P_(4)+3Ba(OH)_2+6H_(2)Orarr3Baunderset(X)((H_2PO)_2)+underset(Y)(2PH_3uarr)Baunderset(X)(H_2PO_2)+underset("dil")(H_2SO_(4))rarrunderset(Z)(BaSO_4darr)+underset(W)(H_3PO_2)`
`H_(2)PO_(2)^(-)` Anion of SALF .X. No Lone pairs on phosphorous - so not a lewi.s base
Discussion
20.

P_4O_6 reacts with water according to equation P_4O_6 + 6H_2O to4H_3PO_3. Calculate the volume of 0.1 M NaOH solution required to neutralise the acid formed by dissolving 1.1 g of P_(4)O_(6)in H_2O .

Answer»

Solution :MOL. Mass of `P_(4)O_(6) = 220` g/mol
Moles of `P_(4)O_(6)` dissolved in `H_(2)O = (1.1)/220`
`underset(1 mol)(P_(4)O_(6)) + underset(6 "mol")(6H_(2)O) to underset(4 "mol")(4H_(3)PO_(3))`
`therefore` Moles of `H_(3)PO_(3)` produced `=4.4/(220)`
`underset(1 mol)(H_(3)PO_(3)) + 2NaOHto Na_(2)HPO_(3) + 2H_(2)O`
`therefore (4.4)/(220) mol`
Moles of NaOH consumed `=(2 xx 4.4)/220 = 8.8/(220)`
`therefore` Volume of NaOH REQUIRED
`=("Moles of NaOH" xx 1000)/("Molarity of NaOH")`
`=(8.8 xx 1000)/(220 xx 0.1)`
= 400 mL
`V_(NaOH) = 400 mL`
Discussion
21.

P_(4)O_(6) reacts with water according to equation, P_(4)O_(6) + 6 H_(2)O rarr 4 H_(3)PO_(3). Calculate the volume of 0.1 M NaOH solution required to neutralise the acid formed by dissolving 1.1 g of P_(4)O_(6) in H_(2)O.

Answer»

<P>

Solution :The chemical equations for the reactions involved are :
`{:(""P_(4)O_(6) + 6H_(2)O rarr 4 H_(3)PO_(4)""("Hydrolysis reaction")),(ul(H_(3)PO_(3) + 2 NaOH rarr [4 Na_(2)HPO_(3)+ 2H_(2)O]xx 4) " "("Neutralisation reaction")),(" "underset("= 220 g")underset(4 xx 31 + 6 xx 16)(P_(4)O_(6))+underset("= 320 g")underset(8 xx 40)(8 NaOH) rarr 4 Na_(2)HPO_(3)+2H_(2)O""("Overall reaction")):}`
Now 220 g of `P_(4)O_(6)` require NaOH for neutralization = 320 g
`therefore` 1.1 g of `P_(4)O_(6)` will require NaOH `= (320)/(220) xx 1.1 = 1.6 g`
Now 1000 ML of 0.1 MHCl contain NaOH `= 40 xx 0.1 = 4 g`
In other words, 4 g of NaOH are present in 0.1 M NaOH = 1000 mL
`therefore` 1.6 g of NaOH will be present in 0.1 M NaOH `=(1000)/(4) xx 1.6 =400 mL = 0.4 L`
Discussion
22.

P_4O_(6) reacts with water according to equation P_4O_6 + 6H_2O to 4H_3PO_3 . Calculate the volume of 0.1 M NaOH solution required to neutralise the acid formed by dissolving 1.1 g of P_4O_6 in H_2O.

Answer»

SOLUTION :
No. of moles in `P_4O_6=(1.1)/(220)`
1 MOL `P_4O_6` requires = 8 mol NAOH
`(1.1)/(220)` mol `P_4O_6` required = `8 XX (1.1)/(220)` mol NaOH
`=(8.8)/(220) =0.04 mol` NaOH
Molarity of NaOH solution is 0.1 M
0.1 mol of NaOH is present in =1 litre
0.04 mol NaOH is present in `=(1)/(0.1) xx 0.04 `litre = 400 mL
Discussion
23.

P_(4)O_(6) reacts with cold water to give

Answer»

`H_(3)PO_(3)`
`H_(4)P_(2)O_(7)`
`HPO_(3)`
`H_(3)PO_(4)`

Solution :`H_(3)PO_(3)`
Discussion
24.

P_(4)O_(6) is the anhydride of the following

Answer»

`H_3PO_2`
`H_3PO_3`
`H_3PO_4`
`H_3PO_5`

ANSWER :B
Discussion
25.

P_4O_6 and P_4O_10are anhydrides of :

Answer»

`H_3PO_3 and H_3PO_4`
`H_3PO_4 and H_4P_2O_7`
`H_3PO_3 and H_4P_2O_7`
`H_3PO_2 and H_3PO_3` respectively

Answer :A
Discussion
26.

P_(4)O_(10) is the anhydride of the following

Answer»

`H_3PO_2`
`H_3PO_3`
`H_3PO_4`
`H_3PO_5`

ANSWER :C
Discussion
27.

P_(4)O_(10) is not used to dry NH_(3) gas because

Answer»

`P_(4)O_(10)` is not a drying agent
`P_(4)O_(10)` reacts with moisture in `NH_(3)`
`P_(4)O_(10)` is basic and `NH_(3)` is acidic
none of these

Solution :`P_(4)O_(10)` being acidic & `NH_(3)` is basic, so they react with each other in presence of moisture giving ammonium phosphate. Hence `P_(4)O_(10)` can't be used for drying of `NH_(3)`
`P_(4)O_(10) + 12NH_(3) + 6H_(2)O rarr 4(NH_(4))_(3) PO_(4)`
Discussion
28.

P_(4)O_(10) is not used to dry NH_(3) gas because:

Answer»

It is very costly
It is not a dehydration agent.
It reacts with MOISTURE to GIVE an acid which reacts with `NH_(3)` (base)
It is BASIC while `NH_(3)` is acidic.

Answer :C
Discussion
29.

P_4O_10 is an anhydride of which compound ?

Answer»

`H_(3)PO_(2)`
`H_(3)PO_(3)`
`H_(3)PO_(4)`
`H_(4)P_(2)O_(7)`

Answer :C
Discussion
30.

P_(4)O_(10) is an anhydride of :

Answer»

`H_(3)PO_(2)`
`H_(3)PO_(3)`
`H_(3)PO_(4)`
`H_(4)P_(2)O_(7)`

SOLUTION :`4H_(3)PO_(4) to P_(4)O_(10)+6H_(2)O`
Discussion
31.

P_4O_10 has short and long P-O bonds. The number of short P-O bonds in this compound is:

Answer»

1
2
3
4

Answer :D
Discussion
32.

P_(4)+8SOCl_(2)to X +4SO_(2)+YIn this equation mention formula of X and Y

Answer»

`X=PCl_(5(S)), Y=S_((S))`
`X=PCl_(5(S)), Y=SCl_(2(S))`
`X=PCl_(3(l)),Y=S_(2)Cl_(2(G))`
`X=PCl_(3(l)), Y=SCl_(2(S))`

Answer :C
Discussion
33.

P_4 O_(10) is the anhydride of

Answer»

`H_(3)PO_(2)`
`H_(3)PO_(3)`
`H_(3)PO_(4)`
`H_(4)P_(2)O_(7)`

SOLUTION :`P_(4)O_(10) +6H_(2)O to underset("Orthophosphoric ACID")(4H_(3)PO_(4))`
Discussion
34.

P^(3-) ion has the valence shell electronic configuration

Answer»

`3s^(2)3p^(3)`
`3s^(2)3p^(6)`
`2S^(2)2P^(3)`
`4s^(2)4p^(3)`

Solution :`P^(3-)(Z=15)`and total electrons =18 `1s^(2)2s^(2)3s^(2)3p^(6)`
Discussion
35.

P_2O_5 when treated with cold water gives:

Answer»

Orthophosphoric acid
Metaphosphoric acid
Pyrophosphoric acid
Hypophosphoric acid

Answer :B
Discussion
36.

P_2O_5 is used extensively as a:

Answer»

DEHYDRATING agent
Catalytic agent
Reducing agent
Preservative

Answer :A
Discussion
37.

P_(2)O_(5) is anhydride of

Answer»

`H_(3)PO_(3)`
`HPO_(3)`
`H_(3)PO_(4)`
`HCLO`

Answer :B
Discussion
38.

P_2O_5 can dimerise but N_2O_5 cannot. Explain with structure.

Answer»

Solution :Nitrogen on account of small size and high. electronegativity has a strong tendency to form multiple bonds (`ppi-ppi`) with oxygen and form oxide in open chain structure. HOWEVER DUE to larger and diffused p-orbital, phosphorus is unable to construct such `pi`-bonds and prefers to form single bonds in order to gain STABILITY. As a result, `N_2O_5` does not undergo DIMERISATION whereas `P_2O_5` dimerises to form `P_4O_(10)`
For structural detail
Discussion
39.

P_(2)H_(4) can be romoved from phosphine containing traces of it

Answer»

by passing impure `PH_(3)` gas through a freezing mixture
by passing the impure `PH_(3)` gas through HI and then its treatment with KOH (aq)
by both (a) and (B)
by none of these

Solution :Both the methods (a) and (b) can used.
When passed through freezing miture the `P_(2)H_(4)` present condenses and pure `PH_(3)` is OBTAINED.
When passed through `HI, PH_(3)` is obsorbed FORMING `PH_(4)I.PH_(4)I` when treated with KOH(aq) yields pure phosphine.
`PH_(3)+HI to PH_(4)I`
`PH_(4)I+KOH(aq) to KI+H_(2)O+PH_(3) uparrow`
Discussion
40.

P_2 O_5 is used extensively as a/an

Answer»

<P>REDUCING agent
Oxidizing agent
Dehydrating agent
PRESERVATIVE

Solution :Generally `P_(2)O_(5)` is USED as a dehydrating agent.
Discussion
41.

{:((P),1/2A_2+3/2B_2hArrD_2,(1),1//4),((Q),2D_2hArrA_2+3B_2,(2),64),((R ),A_2+3B_2hArr2D_2,(3),1//64),((S),D_2hArrB_2+1/2C,(4),8):}

Answer»

<P>`{:(P,Q,R,S),(4,3,2,1):}`
`{:(P,Q,R,S),(2,3,1,4):}`
`{:(P,Q,R,S),(4,2,1,3):}`
`{:(P,Q,R,S),(2,4,1,3):}`

ANSWER :A
Discussion
42.

P_1:Chloride of A imparts brick red colour to the flame.Select the correct statement P_2:Shape of D is trigonal pyramidal.P_3: E does not decompose on heating.P_4: is used to remove temporary hardness of water...

Answer»

`P_2,P_3,P_4`
`P_1,P_2,P_4`
`P_1,P_3,P_4`
`P_1,P_2,P_3`

ANSWER :B
Discussion
43.

P-V plots for two gases during an adiabatic process are given in the figure: Plot A and plot B should correspond to : (Assume ideal behaviour)

Answer»

He and `O_2`
`SO_2 and AR`
`O_2` and He
Both (A) and (B)

SOLUTION :`GAMMA` for `O_2=1.44 gamma` for He =1.66
Discussion
44.

P-Vplots for three gases (assuming ideal behaviour and similar condition) for reversible adiabatic compression are given in the figure below : Plots X,Y and Z should correspond to respectively :

Answer»

`CO_2,Cl_2` and Ne
`SO_2, N_2O` and He
He, `N_2 and O_3`
`NH_3,H_2S and AR`

Solution :`SO_2(GAMMA =1.33), N_2O(gamma=1.4),He (gamma=1.67)`
As `gamma` increases, for COMPRESSION graph RISES up.
Discussion
45.

p-type semi-conductors are made by mixing silicon with impurities of-

Answer»

germanium
boron
arsenic
ANTIMONY

ANSWER :B
Discussion
46.

p-Toluidine overset(Br_(2)-H_(2)O)to A. A in the above sequence is

Answer»




SOLUTION :
Discussion
47.

P+Q rarr Anisol e overset(HI)rarrR+S Correct statement among the following is

Answer»

P and Q are `C_(6)H_(5)` Ona and `C_(2)H_(5)` CI
R and S are `C_(6)H_(5)I` and `CH_(3)` OH
R and S are `C_(6)H_(5)` OH and `CH_(3))_(3)` CI
P and Q are `C_(6)H_(5)C1` and `CH_(3)` Ona

Answer :C
Discussion
48.

P + Q toAnisoleoverset(HI)(to)R + SCorrect statement among the following is

Answer»

<P>P and Q are `C_6H_5ONa` and `C_2H_5Cl`
R and S are `C_6H_5I` and `CH_3OH`
R and S are `C_6H_5OH` and `CH_3I`
P and Q are `C_6H_5Cl` are `CH_3ONa`

ANSWER :C
Discussion
49.

p pi - d pi bonding is not formed between oxygen and

Answer»

<P>Phosphours in `P_(4)O_(10)`
Nitrogen in `N_(2)O_(5)`
Chlorine in `HClO_(4)`
Chlorine in anionic part of solid `Cl_(2)O_(6)`

Answer :B
Discussion
50.

(P), (P) will be :

Answer»




ANSWER :A
Discussion