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.

Pearl white is:

Answer»

BiOCl
SbOCl
NOCl
AsOCl

Answer :A
Discussion
2.

Pearl ash and caustic potash are chemically:

Answer»

`K_2CO_3` and NAOH
`KOH and K_2CO_3`
`Na_2CO_3 and KOH`
`Na_2CO_3 and NaOH`

ANSWER :A
Discussion
3.

[PdCl_(2)(PM e_(3))_(2)] is a diamagnetic complex of Pd(II). How many total isomers are possible of analogous paramagnetic complex of Ni(II)?

Answer»

zero
1
2
3

Solution :`[NiCl_(2)(PM e_(3))_(2)]`
`Ni^(2+)` in `[NiCl_(2)(PM e_(3))_(2)]`
Discussion
4.

PDI for natural polymers is generally close to:

Answer»

Zero
100
1
10

Answer :C
Discussion
5.

[PdCl_(2)(PM e_(3))_(2)] is a diamagnetic complex of Pd (II). How many unpaired electrons are presennt in analogous complex of Ni (II)?

Answer»

Zero
1
2
3

Answer :C
Discussion
6.

[PdBr_4]^(2-) is a sqaure planar complex. The hybridisation of Pd^(2+)is

Answer»

`sp^3`
`dsp^2`
`sp^3 d`
`sp^3 d^2`

ANSWER :B
Discussion
7.

Pd has exceptional outer electronic configuration as 4d^(10) 5 s^(0). It belongs to

Answer»

`4^(th)` period, group 11
`5^(th)` period, group 10
`6^(th)` period, group 9
`3^(rd)` period, group 16

Solution :The EXCEPTIONAL outer ELECTRONIC configuration of Pd suggests that, in principle shell its configuration should have been `4d^8 5s^2`. As such it lies in `5^(th)` period and group 10.
Discussion
8.

Pd can adsorb 900 times its volume of hydrogen. This is called:

Answer»

Absorption
Adsorption
Occlusion
Both (B) and (C)

ANSWER :D
Discussion
9.

PCl_(5) +SO_(2) rarr A+B A overset(3H_(2)O)rarrHCl C overset("Red hot")rarrD +H_(2)O Compound 'D'is

Answer»

`P_4O_6`
`P_4O_(10)`
`SO_3`
`SO_2`

Solution :`PCl_(5)+SO_(2)rarrunderset(A)(PCl_3)+underset(B)(SO_2Cl_2)underset(A)(" "PCl_3)+3H_2Orarrunderset(C)(H_3PO_3)+3HCl`
`underset(C)(2H_3PO_3)overset("RED HOT")rarrunderset(P_4O_6)underset(darr"Dimerises")(P_2O_3)+3H_2O" "2PCl_(5)hArrPCl_4^(+)+PCl_6^(-)`
Discussion
10.

PCl_5was found to dissociate to the extent of 42% at 227^@Cand 1 atm. Find the equilibrium constant at (i) constant pressure, and (ii) constant volume.

Answer»

SOLUTION :0.214, 0.005
Discussion
11.

PCl_(5) +SO_(2) rarr A+B A overset(3H_(2)O)rarrHCl C overset("Red hot")rarrD +H_(2)O Compound 'B' is

Answer»

`SOCl_2`
`SO_2Cl_2`
`POCl_3`
`PCl_3`

Solution :`PCl_(5)+SO_(2)rarrunderset(A)(PCl_3)+underset(B)(SO_2Cl_2)underset(A)(" "PCl_3)+3H_2Orarrunderset(C)(H_3PO_3)+3HCl`
`underset(C)(2H_3PO_3)OVERSET("Red hot")rarrunderset(P_4O_6)underset(DARR"Dimerises")(P_2O_3)+3H_2O" "2PCl_(5)hArrPCl_4^(+)+PCl_6^(-)`
Discussion
12.

PCl_(5) reactws with ethanol to form chloroethane. However, with phenol, it does not give chlorobenzene but gives triphenylphoshate. Explain.

Answer»

SOLUTION :To begin with both ethanol and phenol react with `PCl_(5)` to give the corresponding chlorophosphate ESTERS.

Since in ethanol, C-O bond has only single bond CHARACTER, therefore, nucleophilic ATTACK by `Cl^(-)` ion occurs easily on chlorophosphate ester (I) to give chloroethane.

however, in case of phenol, C-O bond has some double bond character due to resonance and hence difficult to break. instead nucleophilic attack by `C_(6)H_(5)OH` occurs twice on the chlorophosphate ester (II) to form intermediate (III). further, nucleophilic attack by `C_(6)H_(5)OH` on III does not occur due to steric hindrance. Instead (III) undergoes hydrolysis by water to form triphenylphosphate.
.
Discussion
13.

PCl_(5) reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous NH_3 solution. Write the reactions involved to explain what happens.

Answer»

Solution :`PCl_5` will oxidise AG to AgCl. AgCl forms a soluble COMPLEX with ammonia when added in EXCESS.
`PCl_(5) + 2Ag to 2AgCl + PCl_(3)`
`underset("WHITE")(AgCl) + 2NH_(3) to underset("Soluble complex")([Ag(NH_(3))_(2)]Cl)`
Discussion
14.

PCl_5 reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous NH_3 solution. Write the reactions involved to explain what happens.

Answer»

Solution :`PCl_5 + 2AG to 2AgCl + PCl_5`
`AGCL + 2NH_3 (AQ) to underset("Soluble COMPLEX")([Ag(NH_3)_2]^(+) Cl^(-)`
Discussion
15.

PCl_5 reacts with propanone, to give:

Answer»

Gem-dichloride
Vic - dichloride
Propanal
propane chloride

Answer :A
Discussion
16.

Pcl_(5) reacts with a compoundcontaining

Answer»

`-SO_(3)` GROUP
`-OH` group
`NO_(3)` group
`-NO` group

Solution :`PCl_(5)` is used in organic chemistry to replace the -OH group by -Cl and carbonylic OXYGEN by `(-Cl_(2))` .
Discussion
17.

PCl_(5), PCl_(3) and Cl_(2) are at equilibrium at 500 K in a closed container and their concentrations are 0.8xx10^(-3) mol L^(-1), 1.2xx10^(-3)mol L^(-1) respectively. The value of K_(c ) for the reactionPCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)will be

Answer»

`1.8xx10^(-3) mol L^(-1)`
`1.8xx10^(3)`
`1.8xx10^(-3)L mol^(-1)`
`0.55xx10^(4)`

Solution :`K_(c )=([PCl_(3)][Cl_(2)])/([PCl_(5)])=((1.2xx10^(-3))xx(1.2xx10^(-3)))/((0.8xx10^(-3)))`
`=1.8xx10^(-3) mol L^(-1)`.
Discussion
18.

PCI_(5) on hydrolysis gives

Answer»

`H_(3)PO_(3)`
`H_(3)PO_(4)`
`H_(3)PO_(2)`
`H_(3)PO_(5)`

Answer :2
Discussion
19.

PCl_5 molecule has

Answer»

THREE FOLD AXIS of symmetry
Two fold axis of symmetry
Both
None of these

Solution :
Discussion
20.

PCl_(5) molecule has the geometry :

Answer»

TRIGONAL bipyramidal
OCTAHEDRAL
SQUARE planar
Square bipyramidal

Answer :A
Discussion
21.

PCl_5 is prepared by the action of Cl_2 on:

Answer»

`P_2O_3`
`P_2O_5`
`H_3PO_3`
`PCl_3`

ANSWER :D
Discussion
22.

PCl_5 is kept in well stoppered bottle because:

Answer»

It is HIGHLY volatile
It REACTS with oxygen
It reacts READILY with moisture
It is explosive

Answer :C
Discussion
23.

PCl_(5) is ionic in nature in the solid state. Give reasons.

Answer»

Solution :`PCl_(5)` conducts electricity in the molten state. This MEANS that in the solid state it CONSISTS of ions.
`2PCl_(5)rarr [PCl_(4)]^(+)[PCl_(6)]^(-)`.
On melting, these ions BECOMES free to move and hence `PCl_(3)` conducts electricity in the molten state.
Discussion
24.

PCl_(5) is an example of a molecule having sp^(3)d-hybridisation. Three out of the orbitals in sp_(x)p_(y)-hybridisation while remaining two have p_(x)d_(x^(2))-hybridisation. If P-atom is attached to substitutuents differ in electronegativity, as in PCl_(x)F_(5-x), then it is has been experimently observed that the more electronegativity substituent occupies the axial position of t.b.p geometry. Q. The incorrect statement regarding PCl_(2)F_(2) molecule will be:

Answer»

given compound is polar
both AXIAL POSITION occupied by F-atoms
both Cl atoms PRESENT in EQUATORIAL position
one Cl atom present at axial and other Cl atom is present at equatorial position ofgeometry

Answer :D
Discussion
25.

PCl_(5) is an example of a molecule having sp^(3)d-hybridisation. Three out of the orbitals in sp_(x)p_(y)-hybridisation while remaining two have p_(x)d_(x^(2))-hybridisation. If P-atom is attached to substitutuents differ in electronegativity, as in PCl_(x)F_(5-x), then it is has been experimently observed that the more electronegativity substituent occupies the axial position of t.b.p geometry. Q.The correct statement is :

Answer»

in `CH_(2)F_(2)` the `F-C-F` BOND angle is larger `109^(@)28'`
in `CH_(2)F_(2)` the C-F bond has more than 25% c-character
in `CH_(2)F_(2)` the `H-C-H`bond angle is larger `109^(@)28'`
in `CH_(2)F_(2)` the C-H` bond has more than 25% c-character

Answer :C
Discussion
26.

PCl_(5) in solid state exists as PCl_(4)^(+) and PCl_(6)^(-) and also in some solvents it undergoes dissociation as 2PCl_(5) hArr PCl_(4)^(+) +PCl_(6)^(-) The geometry and hybridisation of PCl_(6)^(-) is

Answer»

Octahedral , `sp^3`
Octahedral , `sp^3d^2`
TRIGONAL bipyramid `sp^3d`
SEE - saw , `sp^3 d `

SOLUTION :`[PCl_3]^(-)` - No. of HYBRIDE orbitals `=1/2 [5+6-1]=6`
Discussion
27.

PCl_(5) in solid state exists as PCl_(4)^(+) and PCl_(6)^(-) and also in some solvents it undergoes dissociation as 2PCl_(5) hArr PCl_(4)^(+) +PCl_(6)^(-) The geometry and hydridisation of PCl_(4)^(+) is

Answer»

Tetrahderal , `sp^3`
OCTAHEDRAL , `sp^(3) d^(2)`
LINEAR , sp
Angular , `sp^2`

Solution :`[PCl_4]^(+)` - No of HYBRID orbitals `=1/2 [5+4-1]=4`
Discussion
28.

PCl_(5) exists but NCl_(5) does not due to

Answer»

INERTNESS of `N_(2)`
`NCl_(5)` is unstable
larger size of N
non-availability of vacant d-orbitals

Answer :A::B::C::D
Discussion
29.

PCl_5 exists but NCl_3 does not because:

Answer»

NITROGEN has no VACANT .d. ORBITALS
`NCl_5` is UNSTABLE
`N_2` is inert
None of these

Answer :A
Discussion
30.

PCl_(5) exist but NCl_(5) doesnot because :-

Answer»

NITROGEN ATOM is much smaller than P
`NCl_(5)`is unstable
Nitrogen has no VACANT d-orbitals
Nitrogen is HIGHLY inert

Answer :C
Discussion
31.

PCl_5 does not react with:

Answer»

`CH_3COOH`
`C_2H_5NH_2`
`C_6H_5OH`
`H_2SO_4`

ANSWER :B
Discussion
32.

PCl_(5) causes cleavage of ether linkage R-O-R' forming RCl, R'Cl and POCl_(3), C_(5)H_(12)O on reaction with PCl_(5) forms 2 - chloropropane and 1 - chloroethane as main compound. Thus, C_(5)H_(12)O is named as

Answer»

1 - ethoxypropane
2 - ethoxypropane
1 - ETHYL propane
2 - ethylpropane

Answer :B
Discussion
33.

PCl_5 acts as an oxidising agent. Justify .

Answer»

Solution : `PCI_5` acts as an oxidising agent because its oxidation NUMBER is DECREASED in the reaction as under :
`2AG + PCl_5 to 2Ag CL + PCl_3`
Discussion
34.

PCl_3 under goes hydrolysis to produce an oxoacid.It has formula In solid PCl, exist as

Answer»

`HPO_(3)`
`H_3PO_(3)`
`H_(3)PO_(4)`
`H_(3)PO_2`

ANSWER :B
Discussion
35.

PCl_3 reacts with organic compounds containing_____group.

Answer»

`-SO_3`
`-OH`
`-NO_3`
`-NO`

ANSWER :B
Discussion
36.

PCl_(3) on reacting with HCOOH gives

Answer»

Chlorobenzene
Acetyl CHLORIDE
Methyl chloride
FORMYL chloride

ANSWER :D
Discussion
37.

PCl_3 dissociates into PCl_3 and Cl_2 . If the total pressure of the system in equilibrium is p at a density, d, and temperature, T, show thatalpha = (pM)/(dRT) - 1where alphais the degree of dissociation, M is the relative molar mass of PCl_3. If the vapour density has the value 62 when the temperature is 230^@C, what is the value of p/d?

Answer»

SOLUTION :`0.333L ATM G^(-1)`
Discussion
38.

PCI_(3) on hydrolysis gives

Answer»

`H_(3)PO_(2)`
`H_(3)PO_(3)`
`HCl`
Both (2) and (3)

ANSWER :4
Discussion
39.

PCl_3 fumes in moisture because

Answer»

it UNDERGOES HYDROLYSIS giving FUMES of HCl
it has 5 chlorine atoms
P in `PCl_3` is `sp^3` hybridised
Chlorine atom is REPLACED by HYDROXYL group

Answer :A
Discussion
40.

PCI_5(g) at an initial pressure of 1 atm is placed in a closed vessel containing chlorine at a pressure of 0.2 atm. if PCI_5 dissociation to the extent of 20% then match the following : {:("","List-I","List_II",""),((P),"Partial pressure of" PCI_5,(1),0.4 atm),((Q),"Partial pressure of" PCI_3,(2),0.8 atm),((R ),"Partial pressure of"CI_2,(3),0.10 atm),((S),"Equilibrium constant" ("In terms of partial pressures" ) "for dissociation of "PCI_5,(4),0.2 atm):}

Answer»

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

ANSWER :B
Discussion
41.

PCI_5 reacts with propanone, to give

Answer»

VIE dichloride
Propanal
Propane-chloride
Gem-dichloride

Solution :`CH_3COCH_3 + PCl_5 to CH_3 - C Cl_2 - CH_3 -POCl_3`
Discussion
42.

PCI_(5) is less stable. Why?

Answer»

SOLUTION :`PCI_(5)` molecule has trigonal bipyramidal structure. The three EQUITORIAL bonds are equivalent.
The `P-Cl` bond length in equitorial POSITION is 202 pm. The twoaxial bonds (240 pm) are longer.
The axial bonds suffer more repulsion as compared to theequatorial bonds.
HENCE, `PCI_(5)` is less stable.
Discussion
43.

PCI_(5) is known but NCI_(5) is not known. Or Nitrogen does not form pentahalide.

Answer»

Solution :ELECTRONIC configuration of P is `1 s^(2)2 s^(2) 2 s^(6) 3 s^(2)" "3 P_(x)^(1) 3P_(y)^(1)3P_(z)^(1)3d^(0)`. Thus, P has emptly 3d orbitals to

which the 3S ELECTRON can be excited to have five half-filled orbitals needed for formation of `PCI_(5)`. Thus, `PCl_(5)`, is known.
In contrast, electronic configuration of N is `1s^(2)2s^(2)2P_(x)^(1)2P_(y)^(1)2P_(z)^(1)`. Since the valence shell of nitrogen has n = 2, therefore, it cannot have d-orbitals. However, if one of the 2s electrons is excited to 3s orbital, five

half-filled orbitals needed to form `NCI_(5)`can still be obtained. But such an excitation is thermodynamically not favourable since the energy needed for excitation is more than the energy expected to be generated during the formation of TWO additional P-CI bonds. Therefore, nitrogen doesnot form `NCI_(5)`. In other words, `NCI_(5)` is UNKNOWN.
Discussion
44.

PCI_(5) is ionic in nature in the solid state. Give reasons. Or Solid phosphorus pentachloride behaves as an ionic compound. Explain.

Answer»

SOLUTION :`PCI_(5)` conducts electricity in the MOLTEN state. This MEANS that in the SOLID state, it exists as `[PCI_(4)]^(+) [PCI_(6)]^(-)` in which the cation is tetrahedral and the anion is octahedral
`2 PCI_(5) rarr [PCI_(4)]^(+) [PCI_(6)]^(-)`
On MELTING, these ions become free to move and hence `PCI_(5)` conducts electricity in the molten state.
Discussion
45.

PCl_(5)+Cl^(-) to PCl_(6)^(-). The wrong statement regarding the above equation is

Answer»

Hybridisation of P CHANGES from `sp^3d` to `sp^3d^2`
Oxidation NUMBER of P changes +5 to +6
Covalency P changes from 5 to 6
Here `PCI_5` is a LEWIS acid

Answer :B
Discussion
46.

PCI_(5) exists but NCI_(5)does not . Explain .

Answer»

SOLUTION :of 3D sub-shell.
Discussion
47.

PCl_(3) is covalent. It fumes in moisture and its aqueous solution is electrical conductor. Why?

Answer»

Solution :`PCl_(3)` hydrolyses in moisture (or in water) giving fumes of HYDROGEN CHLORIDE.
`PCl_(3)+3H_(2)OrarrH_(3)PO_(3)+3HCl`
In aqueous SOLUTIONS, HCI ionises and HENCE acts as ELECTRICAL conductor.
Discussion
48.

PbSO_(4) is soluble in (a) _____ due to formation of (b) _____.

Answer»


Answer :a.`CH_(3)COONH_(4)` b. `(CH_(3)COO)_(4)Pb`
Discussion
49.

PbS can be separated from ZnS by electrostatic separation method. The property utilized in this method is

Answer»

PBS is a GOOD CONDUCTOR and ZnS is a poor conductor of ELECTRICITY
PbS is a BAD conductor and ZnS is a good conductor of electricity
both PbS and ZnS are good conductors
both PbS and ZnS are bad conductors

Solution :`PbS` is good conductor and ZnS is poor conductor of electricity.
Discussion
50.

PBr_(5) in solid state exists as :

Answer»

`[PBr_(6)]^(-)[PBr_(4)]^(+)`
`[PBr_(4)]^(+)[Br]^(-)`
`[PBr_(6)]^(+)[Br_(4)]^(-)`
`(PBr_(5))_(2).`

Answer :B
Discussion