iCON Education HYD, 79930 92826, 73309 7282620 May 2026
Which of the following is formed when $\mathrm{SO}_3$ is absorbed by concentrated $\mathrm{H}_2 \mathrm{SO}_4$ ?
A.
$\mathrm{H}_2 \mathrm{S}_2 \mathrm{O}_8$
B.
$\mathrm{H}_2 \mathrm{S}_2 \mathrm{O}_3$
C.
$\mathrm{H}_2 \mathrm{S}_2 \mathrm{O}_7$
D.
$\mathrm{H}_2 \mathrm{S}_2 \mathrm{O}_5$
Correct Answer: C
Explanation:
When $\mathrm{SO}_3$ gas is passed through $\mathrm{H}_2 \mathrm{SO}_4$ solution, it get absorbed in concentrated $\mathrm{H}_2 \mathrm{SO}_4$. The final product formed here is oleum or fuming sulphuric acid.
iCON Education HYD, 79930 92826, 73309 7282620 May 2026
Identify the correct statements about boron.
I. It has high melting point.
II. It has high density.
III. It has high electrical conductivity.
IV. B-10 isotope of it has high ability to absorb neutrons.
A.
I and II only
B.
II and III only
C.
III and IV only
D.
I and IV only
Correct Answer: D
Explanation:
I. Boron has a high melting point of 2352 K . So, statement I is correct.
II. Boron has a low density of $2.37 \mathrm{~gcm}^{-3}$. So, statement II is incorrect.
III. Boron has low electrical conductivity at low temperature. So, statement III is incorrect.
IV. B-10 isotope is a neutron absorber due to the high neutron cross-section of isotope ${ }^{10}$ B. So, statement IV is also correct.
Hence, statements I and IV are correct.
2022
Q303
AP-EAPCET
MCQ
iCON Education HYD, 79930 92826, 73309 7282620 May 2026
Assertion (A) HCl gas is dried by passing through concentrated H$_2$SO$_4$.
Reason (R) HCl gas reacts with NH$_3$ that gives white fumes.
A.
Both A and R are correct and R is the correct explanation of A.
B.
Both A and R are correct and R is not the correct explanation of A.
C.
A is correct but R is incorrect.
D.
A is incorrect but R is correct.
Correct Answer: B
Explanation:
HCl gas is dried by passing through the drying agent i.e. concentrated sulphuric acid. So, assertion is correct.
HCl reacts with $\mathrm{NH}_3$ to give white fumes of ammonium chloride.
$\mathrm{HCl}+\mathrm{NH}_3 \longrightarrow \underset{\text { White fumes }}{\mathrm{NH}_4 \mathrm{Cl}}$
So, reason is also correct. But reason is not a correct explanation of assertion.
2022
Q304
BITSAT
MCQ
iCON Education HYD, 79930 92826, 73309 7282611 Jun 2026
Ge (II) compounds are powerful reducing agents whereas Pb (IV) compounds are strong oxidants. It can be because
A.
Pb is more electropositive than Ge.
B.
ionisation potential of lead is less than that of Ge.
C.
ionic radii of Pb2+ and Pb4+ are larger than that of Ge2+ and Ge4+.
D.
more pronounced inert pair effect in lead has.
Correct Answer: D
Explanation:
Inert pair effect is more pronounced in heavier members like Pb.
Hence, Pb (IV) compounds act as strong oxidising agents and are reduced to more stable Pb (II) compounds.
2022
Q305
BITSAT
MCQ
iCON Education HYD, 79930 92826, 73309 7282611 Jun 2026
The number of 3C$-$2e$-$ bonds present in diborane is
A.
1
B.
2
C.
3
D.
4
Correct Answer: B
Explanation:
The structure of diborane is
In this structure, there are two 3C$-$2e$-$ bonds. 3 atoms, B $-$ H $-$ B share two electrons and form an angular geometry, leading bent bond, also known as banana bond.
2022
Q306
BITSAT
MCQ
iCON Education HYD, 79930 92826, 73309 7282611 Jun 2026
The total number of P$-$OH bonds for pyrophosphoric acid
A.
4
B.
5
C.
6
D.
8
Correct Answer: A
Explanation:
The structure of pyrophosphoric acid is
There are 4 P$-$OH bonds, 1 P$-$O$-$P bond and two P==O bonds.
2021
Q307
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The oxide without nitrogen-nitrogen bond is :
A.
N2O
B.
N2O4
C.
N2O3
D.
N2O5
Correct Answer: D
Explanation:
(a) N $ \equiv $ N+ $-$ O$-$
(b) (c) (d)
2021
Q308
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Which one of the following correctly represents the order of stability of oxides, X2O; (X = halogen)?
A.
Br > Cl > I
B.
Br > I > Cl
C.
Cl > I > Br
D.
I > Cl > Br
Correct Answer: D
Explanation:
Stability of oxides of Halogens is
I > Cl > Br
2021
Q309
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The number of S = O bonds present in sulphurous acid, peroxodisulphuric acid and pyrosulphuric acid, respectively are :
A.
2, 3 and 4
B.
1, 4 and 3
C.
2, 4 and 3
D.
1, 4 and 4
Correct Answer: D
Explanation:
2021
Q310
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Which one of the following is formed (mainly) when red phosphorus is heated in a sealed tube at 803 K?
A.
White phosphorus
B.
Yellow phosphorus
C.
$\beta$-Black phosphorus
D.
$\alpha$-Black phosphorus
Correct Answer: D
Explanation:
When red phosphorus is heated in a sealed tube at 803 K, $\alpha$-Black phosphorus is formed.
2021
Q311
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
In which one of the following molecules strongest back donation of an electron pair from halide to boron is expected?
(d) Orthophosphorous acid : ${H_3}\underline P {O_3}$
(+1)3 + x + (–2)3 = 0
x = +3
2021
Q332
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
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 :
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
Correct Answer: C
Explanation:
A : Due to inert pair effect, Tl is more stable in
+1 oxidation state
Hence TlI3 and CSI3 are isomorphous
R : Electronic configuration of Tl (81) =
[Xe] 4f14 5d10 6s2 6p1
Both A and R are correct but R is not the
correct explanation of A.
2021
Q333
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
Find A, B and C in the following reactions :
$N{H_3} + A + C{O_2} \to {(N{H_4})_2}C{O_3}$
${(N{H_4})_2}C{O_3} + {H_2}O + B \to N{H_4}HC{O_3}$
$N{H_4}HC{O_3} + NaCl \to N{H_4}Cl + C$
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.
$A - {H_2}O;B - C{O_2};C - NaHC{O_3}$
Correct Answer: D
Explanation:
2021
Q334
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
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 :
at room temperature $\alpha $-sulphur (Rhombic) is most stable form.
2021
Q336
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The correct statement about B2H6 is :
A.
Its fragment, BH3, behaves as a Lewis base.
B.
All B$-$H$-$B angles are of 120$^\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.
Correct Answer: D
Explanation:
Terminal B – H bonds are shorter than the
bridging B – H bonds which shows that the
terminal B – H bonds have greater s-character
and less p-character.
2021
Q337
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
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
Correct Answer: A
Explanation:
Al2O3 (aluminium oxide) was leached with alkali to get
Na[Al(OH)4] (X) (sodium aluminate). The solution of sodium
aluminate when passed through gas Y, i.e. carbon dioxide (CO2)
forms Al2O3.xH2O (Z).
Complete reactions are as follows
2021
Q338
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
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.
2021
Q339
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
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
2021
Q340
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
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)
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
Q341
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
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
Q342
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
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
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
Q343
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 7282614 Mar 2026
The reaction of sulphur in alkaline medium is given below:
The values of 'a' is _______.
(Integer answer)
Correct Answer: 12
Explanation:
The two half reaction, one separately are as follows
(b) P in H3PO4 is in its highest oxidation state i.e. + 5 hence, it cannot act as reducing agent. But P in H3PO3 is in oxidation state, + 3 hence, it can act as reducing agent.
(c) H3PO3 contains two $-$OH groups, hence it is a dibasic acid.
(d) Hydrogen attached to P, does not ionise in water. Therefore, options (a), (b) and (d) are correct.
2021
Q345
AP-EAPCET
MCQ
iCON Education HYD, 79930 92826, 73309 7282620 May 2026
$\mathrm{H}_3 \mathrm{BO}_3$ or $\mathrm{B}(\mathrm{OH})_3$ is considered as an acid because its molecule
A.
combines with proton from water molecule
B.
accepts $\mathrm{OH}^{-}$ from water, releasing a proton
C.
contains replaceable $\mathrm{H}^{+}$ ion
D.
can donate proton easily
Correct Answer: B
Explanation:
Boric acid $\left(\mathrm{H}_3 \mathrm{BO}_3\right)$ is an acid because its molecule accepts $\mathrm{OH}^{-}$ from water releasing proton.
iCON Education HYD, 79930 92826, 73309 7282620 May 2026
Xenon best reacts with
A.
neutral atoms
B.
most electronegative elements
C.
most electropositive elements
D.
transition elements
Correct Answer: B
Explanation:
As Xe has fully filled stable electronic configuration, it will react only with most electronegative elements like $\mathrm{O}$ and $\mathrm{F}$. Xenon (Xe) reacts with fluorine to form $\mathrm{XeF}_2, \mathrm{XeF}_4$ and $\mathrm{XeF}_6$.
2021
Q347
AP-EAPCET
MCQ
iCON Education HYD, 79930 92826, 73309 7282620 May 2026
The correct order of acidic character of the following is
Silicon is heated with methyl chloride at high temperature in the presence of $\mathrm{Cu}$- powder at $570 \mathrm{~K}$, methyl substituted chlorosilaner $\mathrm{MeSiCl}_3, \mathrm{Me}_2 \mathrm{SiCl}_2, \mathrm{Me}_3 \mathrm{SiCl}$ and $\mathrm{Me}_4 \mathrm{Si}$ are formed. After formation of $\mathrm{Me}_2 \mathrm{SiCl}_2, \mathrm{H}_2 \mathrm{O}$ is used for hydrolysis to form $\left(\mathrm{CH}_3\right)_2 \mathrm{Si}(\mathrm{OH})_2$ i.e. $Q$. Hence, $Q$ is straight chain polymer