p-Block Elements
267 Questions
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
The number of ionisable hydrogens present in the product obtained from a reaction of phosphorus trichloride and phosphonic acid is :
A.
3
B.
0
C.
2
D.
1
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Evening Shift
The set that represents the pair of neutral oxides of nitrogen is :
A.
NO and N2O
B.
N2O and NO2
C.
NO and NO2
D.
N2O and N2O3
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Evening Shift
The INCORRECT statement regarding the structure of C60 is :
A.
The five-membered rings are fused only to six-membered rings.
B.
It contains 12 six-membered rings and 24 five-membered rings.
C.
Each carbon atom forms three sigma bonds.
D.
The six-membered rings are fused to both six and five-membered rings.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Morning Shift
Match List - I with List - II :
Choose the correct answer from the options given below :
| List - I Industrial process |
List - II Application |
||
|---|---|---|---|
| (a) | Haber's process | (i) | $HN{O_3}$ synthesis |
| (b) | Ostwald's process | (ii) | Aluminium extraction |
| (c) | Contact process | (iii) | $N{H_3}$ synthesis |
| (d) | Hal-Heroult process | (iv) | ${H_2}S{O_4}$ synthesis |
Choose the correct answer from the options given below :
A.
(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
B.
(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
C.
(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
D.
(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Morning Shift
A group of 15 element, which is a metal and forms a hydride with strongest reducing power among group 15 hydrides. The element is :
A.
As
B.
Sb
C.
Bi
D.
P
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Morning Shift
Match List - I with List - II :
Choose the correct answer from the options given below :
| List - I Name of oxo acid |
List - II Oxidation state of 'P' |
||
|---|---|---|---|
| (a) | Hypophosphorous acid | (i) | +5 |
| (b) | Orthophosphoric acid | (ii) | +4 |
| (c) | Hypophosphoric acid | (iii) | +3 |
| (d) | Orthophosphorous acid | (iv) | +2 |
| (v) | +1 |
Choose the correct answer from the options given below :
A.
(a) - (iv), (b) - (v), (c) - (ii), (d) - (iii)
B.
(a) - (iv), (b) - (i), (c) - (ii), (d) - (iii)
C.
(a) - (v), (b) - (i), (c) - (ii), (d) - (iii)
D.
(a) - (v), (b) - (iv), (c) - (ii), (d) - (iii)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Evening Shift
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : In TlI3, isomorphous to CsI3, the metal is present in +1 oxidation state.
Reason R : Tl metal has fourteen f electrons in its electronic configuration.
In the light of the above statements, choose the most appropriate answer from the options given below :
Assertion A : In TlI3, isomorphous to CsI3, the metal is present in +1 oxidation state.
Reason R : Tl metal has fourteen f electrons in its electronic configuration.
In the light of the above statements, choose the most appropriate answer from the options given below :
A.
A is correct but R is not correct
B.
A is not correct but R is correct
C.
Both A and R are correct but R is NOT the correct explanation of A
D.
Both A and R are correct and R is the correct explanation of A
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Morning Shift
Find A, B and C in the following reactions :
$N{H_3} + A + C{O_2} \to {(N{H_4})_2}C{O_3}$
${(N{H_4})_2}C{O_3} + {H_2}O + B \to N{H_4}HC{O_3}$
$N{H_4}HC{O_3} + NaCl \to N{H_4}Cl + C$
$N{H_3} + A + C{O_2} \to {(N{H_4})_2}C{O_3}$
${(N{H_4})_2}C{O_3} + {H_2}O + B \to N{H_4}HC{O_3}$
$N{H_4}HC{O_3} + NaCl \to N{H_4}Cl + C$
A.
$A - {O_2};B - C{O_2};C - N{a_2}C{O_3}$
B.
$A - {H_2}O;B - {O_2};C - NaHC{O_3}$
C.
$A - {H_2}O;B - {O_2};C - N{a_2}C{O_3}$
D.
$A - {H_2}O;B - C{O_2};C - NaHC{O_3}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Morning Shift
On treating a compound with warm dil. H2SO4, gas X is evolved which turns K2Cr2O7 paper acidified with dil. H2SO4 to a green compound Y. X and Y respectively are :
A.
X = SO2, Y = Cr2O3
B.
X = SO3, Y = Cr2(SO4)3
C.
X = SO3, Y = Cr2O3
D.
X = SO2, Y = Cr2(SO4)3
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
Given below are two statements :
Statement I : $\alpha$ and $\beta$ forms of sulphur can change reversibly between themselves with slow heating or slow cooling.
Statement II : At room temperature the stable crystalline form of sulphur is monoclinic sulphur.
In the light of the above statements, choose the correct answer from the options given below :
Statement I : $\alpha$ and $\beta$ forms of sulphur can change reversibly between themselves with slow heating or slow cooling.
Statement II : At room temperature the stable crystalline form of sulphur is monoclinic sulphur.
In the light of the above statements, choose the correct answer from the options given below :
A.
Statement I is true but Statement II is false.
B.
Statement I is false but Statement II is true.
C.
Both Statement I and Statement II are true.
D.
Both Statement I and Statement II are false.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Morning Shift
The correct statement about B2H6 is :
A.
Its fragment, BH3, behaves as a Lewis base.
B.
All B$-$H$-$B angles are of 120$^\circ$.
C.
The two B$-$H$-$B bonds are not of same length.
D.
Terminal B$-$H bonds have less p-character when compared to bridging bonds.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Morning Shift
Al2O3 was leached with alkali to get X. The solution of X on passing of gas Y, forms Z.
X, Y and Z respectively are :
A.
X = Na[Al(OH)4], Y = CO2, Z = Al2O3.xH2O
B.
X = Al(OH)3, Y = CO2, Z = Al2O3
C.
X = Al(OH)3, Y = SO2, Z = Al2O3.xH2O
D.
X = Na[Al(OH)4], Y = SO2, Z = Al2O3
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Morning Shift
Consider the sulphides HgS, PbS, CuS, Sb2S3, As2S3 and CdS. Number of these sulphides soluble in 50% HNO3 is ___________.
Correct Answer: 4
Explanation:
Pbs, CuS, As2S3, CdS are soluble in 50% HNO3.
HgS, Sb2S3 are insoluble in 50% HNO3
So, answer is 4.
HgS, Sb2S3 are insoluble in 50% HNO3
So, answer is 4.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Morning Shift
The number of halogen/(s) forming halic (V) acid is ___________.
Correct Answer: 3
Explanation:
Except F and At, all other halide can form Halic (V)
acid.
F cannot go in +5 oxidation state.
At is radioactive.
The number of halogen forming halic (V) acid
HClO3
HBrO3
HIO3
So answer is 3
F cannot go in +5 oxidation state.
At is radioactive.
The number of halogen forming halic (V) acid
HClO3
HBrO3
HIO3
So answer is 3
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Evening Shift
A xenon compound 'A' upon partial hydrolysis gives XeO2F2. The number of lone pair of electrons present in compound A is _________. (Round off to the Nearest Integer)
Correct Answer: 19
Explanation:
XeF6 on partial hydrolysis form XeO2F2.
XeF6 + H2O $\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{Hydrolysis}^{Partial}} $ XeO2F2 + 2HF
In XeF6, central atom Xe has one lone pair all 6 fluorine have 3 lone pairs each.
So, total number of lone pair on XeF6 = 1 + (6 $\times$ 3) = 19
XeF6 + H2O $\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{Hydrolysis}^{Partial}} $ XeO2F2 + 2HF
In XeF6, central atom Xe has one lone pair all 6 fluorine have 3 lone pairs each.
So, total number of lone pair on XeF6 = 1 + (6 $\times$ 3) = 19
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Morning Shift
The reaction of white phosphorus on boiling with alkali in inert atmosphere resulted in the formation of product 'A'. The reaction of 1 mol of 'A' with excess of AgNO3 in aqueous medium gives ___________ mol(s) of Ag. (Round off to the Nearest Integer).
Correct Answer: 4
Explanation:
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Evening Shift
Among the following allotropic forms of sulphur, the number of allotropic forms, which will show paramagnetism is _________.
(A) $\alpha$-sulphur
(B) $\beta$-sulphur
(C) S2-form
(A) $\alpha$-sulphur
(B) $\beta$-sulphur
(C) S2-form
Correct Answer: 1
Explanation:
Only S2-form of sulphur is paramagnetic in nature. Because S2 is like O2 i.e. paramagnetic as per molecular orbital theory. It contains unpaired electron. While $\alpha $-sulphur and $\beta $-sulphur are diamagnetic as they do not have unpaired electron.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Morning Shift
The reaction of sulphur in alkaline medium is given below:
The values of 'a' is _______. (Integer answer)
The values of 'a' is _______. (Integer answer)
Correct Answer: 12
Explanation:
The two half reaction, one separately are as follows
${S_8} + 16{e^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }}$ (Reduction)
${S_8} + 12{H_2}O\buildrel {} \over \longrightarrow 4{S_2}O_3^{2 - } + 24{H^ + } + 16{e^ - }$
_____________________________________
$2{S_8} + 12{H_2}O\buildrel {} \over \longrightarrow \mathop {8{S^{2 - }} + 4{S_2}O_3^{2 - }}\limits_{} + 24{H^ + }$
For balancing in basic medium, Add an equal number of OH$-$ that of H+, we get
$2{S_8} + 12{H_2}O + 24O{H^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }} + 4{S_2}O_3^{2 - } + 24{H_2}O$
$2{S_8} + 24O{H^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }} + 4{S_2}O_3^{2 - } + 12{H_2}O$
or ${S_8} + 12O{H^ - }\buildrel {} \over \longrightarrow 4{S^2} + 2{S_2}O_3^{2 - } + 6{H_2}O$ .... (i)
On comparing (i) with
${S_8} + aO{H^ - }(aq)\buildrel {} \over \longrightarrow b{S^2}(aq) + c{S_2}O_3^{2 - } + d{H_2}O$
We get, a = 12; b = 4; c = 2; d = 6
${S_8} + 16{e^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }}$ (Reduction)
${S_8} + 12{H_2}O\buildrel {} \over \longrightarrow 4{S_2}O_3^{2 - } + 24{H^ + } + 16{e^ - }$
_____________________________________
$2{S_8} + 12{H_2}O\buildrel {} \over \longrightarrow \mathop {8{S^{2 - }} + 4{S_2}O_3^{2 - }}\limits_{} + 24{H^ + }$
For balancing in basic medium, Add an equal number of OH$-$ that of H+, we get
$2{S_8} + 12{H_2}O + 24O{H^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }} + 4{S_2}O_3^{2 - } + 24{H_2}O$
$2{S_8} + 24O{H^ - }\buildrel {} \over \longrightarrow 8{S^{2 - }} + 4{S_2}O_3^{2 - } + 12{H_2}O$
or ${S_8} + 12O{H^ - }\buildrel {} \over \longrightarrow 4{S^2} + 2{S_2}O_3^{2 - } + 6{H_2}O$ .... (i)
On comparing (i) with
${S_8} + aO{H^ - }(aq)\buildrel {} \over \longrightarrow b{S^2}(aq) + c{S_2}O_3^{2 - } + d{H_2}O$
We get, a = 12; b = 4; c = 2; d = 6
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
The reaction of NO with N2O4 at 250 K gives :
A.
N2O3
B.
N2O5
C.
N2O
D.
NO2
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
Reaction of an inorganic sulphite X with dilute
H2SO4 generates compound Y. Reaction of Y
with NaOH gives X. Further, the reaction of X
with Y and water affords compound Z. Y and Z,
respectively, are :
A.
S and Na2SO3
B.
SO2 and NaHSO3
C.
SO2 and Na2SO3
D.
SO3 and NaHSO3
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
The correct statement with respect to
dinitrogen is :
A.
N2 is paramagnetic in nature
B.
it can be used as an inert diluent for
reactive chemicals
C.
it can combine with dioxygen at 25oC
D.
liquid dinitrogen is not used in cryosurgery
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
Boron and silicon of very high purity can be obtained through :
A.
Liquation
B.
Electrolytic refining
C.
Zone refining
D.
Vapour phase refining
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
Reaction of ammonia with excess Cl2
gives :
A.
NH4Cl and N2
B.
NH4Cl and HCl
C.
NCl3
and HCl
D.
NCl3
and NH4Cl
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
On heating, lead (II) nitrate gives a brown gas
(A). The gas (A) on cooling changes to a
colourless solid/liquid (B). (B) on heating with
NO changes to a blue solid (C). The oxidation
number of nitrogen in solid (C) is :
A.
+3
B.
+4
C.
+5
D.
+2
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
Aqua regia is used for dissolving noble metals (Au, Pt, etc.). The gas evolved in this process is :
A.
N2O3
B.
N2
C.
N2O5
D.
NO
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
In a molecule of pyrophosphoric acid, the number of P - OH, P = O and P - O - P bonds/moiety(ies)
respectively are :
A.
4, 2 and 0
B.
4, 2 and 1
C.
3, 3 and 3
D.
2, 4 and 1
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
On heating compound (A) gives a gas (B) which
is a constituent of air. This gas when treated
with H2 in the presence of a catalyst gives
another gas (C) which is basic in nature. (A)
should not be :
A.
Pb(NO3)2
B.
(NH4)2Cr2O7
C.
NH4NO2
D.
NaN3
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Evening Slot
The reaction of H3N3B3Cl3 (A) with LiBH4 in
tetrahydrofuran gives inorganic benzene (B).
Further, the reaction of (A) with (C) leads to
H3N3B3(Me)3. Compounds (B) and (C)
respectively, are :
A.
Borazine and MeBr
B.
Diborane and MeMgBr
C.
Boron nitride and MeBr
D.
Borazine and MeMgBr
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
'X' melts at low temperature and is a bad
conductor of electricity in both liquid and solid
state. X is :
A.
Mercury
B.
Carbon tetrachloride
C.
Zinc sulphide
D.
Silicon carbide
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
A metal (A) on heating in nitrogen gas gives
compound B. B on treatment with H2O gives
a colourless gas which when passed through
CuSO4 solution gives a dark blue-violet
coloured solution. A and B respectively, are :
A.
Na and NaNO3
B.
Na and Na3N
C.
Mg and Mg3N2
D.
Mg and Mg(NO3)2
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
White Phosphorus on reaction with concentrated
NaOH solution in an inert atmosphere of CO2
gives phosphine and compound (X). (X) on
acidification with HCl gives compound (Y). The
basicity of compound (Y) is :
A.
2
B.
4
C.
1
D.
3
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
The number of bonds between sulphur and oxygen atoms in
S2O82- and the number of bonds between sulphur
and sulphur atoms in rhombic sulphur, respectively, are :
A.
4 and 8
B.
8 and 6
C.
4 and 6
D.
8 and 8
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
In the following reaction, products (A) and (B) respectively, are :
NaOH + Cl2 $ \to $ (A) + side products
(hot and conc.)
Ca(OH)2 + Cl2 $ \to $ (B) + side products (dry)
NaOH + Cl2 $ \to $ (A) + side products
(hot and conc.)
Ca(OH)2 + Cl2 $ \to $ (B) + side products (dry)
A.
NaOCl and Ca(OCl)2
B.
NaClO3 and Ca(ClO3)2
C.
NaOCl and Ca(ClO3)2
D.
NaClO3 and Ca(OCl)2
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
The C–C bond length is maximum in :
A.
C60
B.
diamond
C.
graphite
D.
C70
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
The basic structural unit of feldspar, zeolites, mica, and asbestos is :
A.
B.
(SiO4)4–
C.
(SiO3)2–
D.
SiO2
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
The noble gas that does not occur in the atmosphere is :
A.
Ne
B.
He
C.
Kr
D.
Ra
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
The number of pentagons in C60 and trigons (triangles) in white phosphorus, respectively, are :
A.
12 and 3
B.
20 and 3
C.
20 and 4
D.
12 and 4
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
The correct order of catenation is :
A.
C > Sn > Si $ \approx $ Ge
B.
Si > Sn > C > Ge
C.
C > Si > Ge $ \approx $ Sn
D.
Ge > Sn > Si > C
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
The correct statements among I to III regarding
group 13 element oxides are,
(I) Boron trioxide is acidic.
(II) Oxides of aluminium and gallium are amphoteric.
(III) Oxides of indium and thalliumare basic.
(I) Boron trioxide is acidic.
(II) Oxides of aluminium and gallium are amphoteric.
(III) Oxides of indium and thalliumare basic.
A.
(II) and (III) only
B.
(I), (II) and (III)
C.
(I) and (III) only
D.
(I) and (II) only
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
The amorphous form of silica is :
A.
kieselguhr
B.
quartz
C.
tridymite
D.
cristobalite
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
C60 an allotrope of carbon contains :
A.
12 hexagons and 20 pentagons.
B.
20 hexagons and 12 pentagons.
C.
16 hexagons and 16 pentagons
D.
18 hexagons and 14 pentagons.
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Morning Slot
Diborane (B2H6) reacts with O2 and H2O independently, then products formed are :
A.
B2O3 and HBO2
B.
B2O3 and H3BO3
C.
HBO2 and H3BO3
D.
HBO2 and HBO3
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Evening Slot
The element that does NOT show catenation is :
A.
Ge
B.
Si
C.
Sn
D.
Pb
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Evening Slot
The element that shows greater ability to form p$\pi $-p$\pi $ multiple bonds, is :
A.
Si
B.
Ge
C.
C
D.
Sn
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Evening Slot
Chlorine on reaction with hot and concentrated sodium hydroxide give :
A.
C$\ell $O$_3^ - $ and C$\ell $O$_2^ - $
B.
C$\ell $$-$ and C$\ell $O$-$
C.
C$\ell $$-$ and C$\ell $O$_2^ - $
D.
C$\ell $$-$ and C$\ell $O$_3^ - $
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Morning Slot
Iodine reacts with concentrated HNO3 to yield Y along with other products. The oxidation state of iodine in Y, is :
A.
3
B.
1
C.
7
D.
5
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Evening Slot
The hydride that is NOT electron deficient :
A.
B2H6
B.
AlH3
C.
GaH3
D.
SiH4
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Evening Slot
The relative stability of +1 oxidation state of group 13 elements follows the order :
A.
Tl < In < Ga < Al
B.
Al < Ga < TI < In
C.
AI < Ga < In < TI
D.
Ga < AI < In < TI
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Morning Slot
The chloride that CANNOT get hydrolysed is :
A.
SnCl4
B.
SiCl4
C.
PbCl4
D.
CCl4
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Evening Slot
Among the following reactions of hydrogen with halogens, the one that requires a catalyst is :
A.
H2 + Br2 $ \to $ 2HBr
B.
H2 + Cl2 $ \to $ 2HCl
C.
H2 + F2 $ \to $ 2HF
D.
H2 + I2 $ \to $ 2HI