datavorous/entrance-exam-dataset huggingface dataset:
The output of a two-inputs \( O R \) gate is \( 0 \) only when its:

Options:
A Both inputs are \( 0 \)
B Either input is \( 1 \)
C Both inputs are \( 1 \)
D Either input is \( 0 \)

The correct answer is:
Both inputs are \( 0 \)

Solution:
The OR gate is a two inputs and one output logic gate.It combines the inputs A and B with the output Y following the Boolean expression Y =A+B.And truth table of OR gate is drawn below

Topic: Physics, Semiconductors, JEE Main
The output of an OR gate is connected to both the inputs of a NAND gate. The combination will serve as a:

Options:
A NOT gate
B NOR gate
C AND gate
D OR gate

The correct answer is:
NOR gate

Solution:
$(\overline{A+B})=$ NOR gate When both inputs of NAND gate are connected, it behaves as NOT gate $\mathrm{OR}+\mathrm{NOT}=\mathrm{NOR}$

Topic: Physics, Semiconductors, JEE Main
The output of the given combination gates represents:

Options:
A XOR Gate
B NAND Gate
C AND Gate
D NOR Gate

The correct answer is:
NAND Gate

Solution:
By De Morgan's theorem, we have

Here the output of A and B gives A¯ and B¯. This output is fed as inputs to NAND gate and the output is A·B, when when again fed as input to NAND gate to get output as A·B¯.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The output of the given logic circuit is:

Options:
A AB-
B A-B
C AB+AB¯
D AB-+A-B

The correct answer is:
AB-

Solution:
y= ( A. A.B ¯ ¯ ).( A.B ¯ +B ) ¯
A.A.BÂ¯+(A.B).BÌ
=AâA-+B-
=AB-

Topic: Physics, Semiconductors, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The output voltage of a transfor mer connected to $220 \mathrm{~V}$ line is $1100 \mathrm{~V}$ at 2 Acurrent. Itsefficiency is $100 \%$. The current coming from the line is

Options:
A $20 \mathrm{~A}$
B $10 \mathrm{~A}$
C $11 \mathrm{~A}$
D $22 \mathrm{~A}$

The correct answer is:
$10 \mathrm{~A}$

Solution:
For $100 \%$ efficiency
For $100 \%$ enciency- $V_{s} i_{s}=V_{p} i_{p}$ $1100 \times 2=220 \times i_{p}$ i.e., Primary current, $i_{p}=10 \mathrm{~A}$ PART - II (CHEMISTRY)

Topic: Physics, Alternating Current, JEE Main
The output voltage (taken across the resistance) of a LCR series resonant circuit falls to half its peak value at a frequency of 200 Hz and again reaches the same value at 800 Hz. The bandwidth of this circuit is:

Options:
A 200 Hz
B 2003 Hz
C 400 Hz
D 600 Hz

The correct answer is:
2003 Hz

Solution:
VOutput =VR

=irmsR

=V0RZ=V0RR2+XL−XC2

For peak XL=XC⇒Vpeak =V0
For VOutput =V02
V02=V0RR2+XL-XC2
R2+XL-XC2=4R2
XL-XC=±3R
ωL-1ωC=±3R
ω2LC∓3RωC-1=0

ω=±3RC±3R2C2+4LC2LC
ω1=-3RC+3R2C2+4LC2LC=200×2π
ω2=+3RC+3R2C2+4LC2LC=800×2π
ω2-ω1=600×2π=3RL
Bandwidth =RL=2π×6003
Δf=12πRL=6003=2003

Topic: Physics, Alternating Current, JEE Main
The output waveform of the given logical circuit for the following inputs A and B as shown below, is

Options:
A
B
C
D

The correct answer is:

Solution:
In the given logical circuit, all logic gates are NAND gate.

Here, the inputs A and B are inverted using NOT gate (made using NAND gate), then A¯ and B¯ are applied as inputs in other NAND gate.

So, A.A=A and B.B=B

Output at Y is A.B=A+B.

The above boolean expression is for OR gate.

Truth table of OR gate is

A	B	Y

0	0	0

0	1	1

1	0	1

1	1	1

The output waveform is shown below.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2023 (30 Jan Shift 1)
The output $\mathrm{Y}$ of following circuit for given inputs is :

Options:
A $\Lambda \cdot B(\Lambda+B)$
B 0
C $\overline{\mathrm{A}} \cdot \mathrm{B}$
D $A \cdot B$

The correct answer is:
0

Solution:
By truth table

Topic: Physics, Semiconductors, JEE Main, JEE Main 2024 (08 Apr Shift 1)
The output $(Y)$ of logic circuit given below is 0 only when :

Options:
A $\mathrm{A}=1, \mathrm{~B}=0$
B $\mathrm{A}=0, \mathrm{~B}=1$
C $\mathrm{A}=0, \mathrm{~B}=0$
D $\mathrm{A}=1, \mathrm{~B}=1$

The correct answer is:
$\mathrm{A}=0, \mathrm{~B}=0$

Solution:

Topic: Physics, Semiconductors, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The output Y of the combination of logic gates shown is equal to

Options:
A A
B A ¯
C A + B
D AB

The correct answer is:
A

Solution:
From boolen algebra we have
âA . (A + B) = A

Topic: Physics, Semiconductors, JEE Main
The output Y of the logic circuit shown in figure is best represented as

Options:
A A+B.C
B A+B.C
C A+B.C
D A+B¯.C¯

The correct answer is:
A+B¯.C¯

Solution:
At logic gate-I, the Boolean expression is B.C=Y'
At logic gate-II, the Boolean expression is A+ B.C =Y''
At logic gate-III, the Boolean expression is A+B¯.C¯=Y

Topic: Physics, Semiconductors, JEE Main
The output \( Y \) of the logic circuit shown in the figure is best represented as-

Options:
A \( \bar{A}+\overline{B C} \)
B \( A+\bar{B} \cdot C \)
C \( \overline{A+\overline{B \cdot C}} \)
D \( \overline{A+\bar{B} \cdot C} \)

The correct answer is:
\( \overline{A+\bar{B} \cdot C} \)

Solution:
Boolean expression for first NOT gate,

X=B¯
Boolean expression for AND gate,
Z=X·C=B¯·C

Boolean expression for OR gate,
D=Z+A=B¯·C+A

Boolean expression for second NOT gate,

Y=D¯=B¯·C+A¯=A+B¯·C¯

Topic: Physics, Semiconductors, JEE Main
The overall goal of glycolysis, Krebs cycle and the electron transport system is the formation of:

Options:
A ATP in one large oxidation reaction
B sugars
C nucleic acids
D ATP in small stepwise units.

The correct answer is:
ATP in small stepwise units.

Solution:
Glycolysis, Krebs cycle and electron transport system, they all give ATP which is the main energy currency of our body. Glycolysis takes place in cytoplasm, Krebs cycle occurs in mitochondrial matrix and ETS takes place in the inner membrane of mitochondria. (Ref. to Fig. 25.6)

Topic: Biology, Respiration in Plants, NEET, NEET 2007
The overall stability constant of the complex ion CuNH342+ is 2.1×1013. The overall dissociation constant is y×10-14. Then y is (Nearest integer)

Options:

The correct answer is:
5

Solution:
Dissociation constant =1 stability constant =12.1×1013=100×10-1421

=4.7×10-14

≈5×10-14

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The overlapping of orbitals in benzene is of the type

Options:
A sp-sp
B sp2-sp3
C $\mathrm{sp}^{2}-\mathrm{sp}^{2}$
D $\mathrm{sp}^{3}-\mathrm{sp}^{3}$

The correct answer is:
$\mathrm{sp}^{2}-\mathrm{sp}^{2}$

Solution:
The molecular orbital picture of benzene shows that in it all the six carbon atoms are $\mathrm{sp}^{2}$ hybridised. Out of these three ${s p^{2}}^{2}$ hybrid orbitals of each C atom, two orbitals overlap with $\mathrm{sp}^{2}$ hybrid orbitals of adjacent $C$ atoms to form six $\mathrm{C}-\mathrm{C}$ single bonds. The remaining $\mathrm{sp}^{2}$ orbital of each $C$ atom overlaps with -orbital of each hydrogen atom to form six $\mathrm{C}-\mathrm{H}$ single sigma bonds. Each C atom is now left with one unhybridised -orbital perpendicular to the plane of the ring.

Topic: Chemistry, Chemical Bonding and Molecular Structure, KCET, KCET 2008
The overlapping of orbitals in benzene is of the type

Options:
A sp-sp
B sp2-sp3
C $\mathrm{sp}^{2}-\mathrm{sp}^{2}$
D $\mathrm{sp}^{3}-\mathrm{sp}^{3}$

The correct answer is:
$\mathrm{sp}^{2}-\mathrm{sp}^{2}$

Solution:
The molecular orbital picture of benzene shows that in it all the six carbon atoms are $\mathrm{sp}^{2}$ hybridised. Out of these three ${s p^{2}}^{2}$ hybrid orbitals of each C atom, two orbitals overlap with $\mathrm{sp}^{2}$ hybrid orbitals of adjacent $C$ atoms to form six $\mathrm{C}-\mathrm{C}$ single bonds. The remaining $\mathrm{sp}^{2}$ orbital of each $C$ atom overlaps with -orbital of each hydrogen atom to form six $\mathrm{C}-\mathrm{H}$ single sigma bonds. Each C atom is now left with one unhybridised -orbital perpendicular to the plane of the ring.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The ovule is absent in:

Options:
A Pteridophytes
B Gymnosperm
C Angiosperm
D All

The correct answer is:
Pteridophytes

Solution:
Gymnosperm (gymnos - naked and Sperma - seed) are plants in which ovules are not enclosed by ovary wall and remain exposed both before and after fertilization. The seeds that develop post-fertilization are not covered, are naked. This is a very important character of gymnosperm. In angiosperms, the ovules are present inside the ovary attached to the placenta. Pteridophytes do not have ovules. Meiospores are formed inside sporangia. Depending upon spore differentiation plants are homosporous (similar spores) and heterosporous (with two type of spores), smaller male microspores and larger female megaspores. Sporangia are associated with leaves called sporophylls.

Topic: Biology, Plant Kingdom, NEET
The oxidation number is changed in which of the following case?

Options:
A SO2 gas is passed into Cr2O72-/H+
B Aqueous solution of CrO42- is acidified
C CrO2Cl2 is dissolved in NaOH
D Cr2O72- solution is made alkaline .

The correct answer is:
SO2 gas is passed into Cr2O72-/H+

Solution:
aCr2O72-+6+SO2+4+2H+â2Cr3++3+3SO42-+6+H2O
b2CrO42-+6+2H+âCr2O72-+6+H2O
c   CrO 2 Cl 2 +6  + 2NaOH→Na 2 CrO 4
dCr2O72-+6+2OH-â2CrO42-+6+H2O

Topic: Chemistry, d and f Block Elements, JEE Main
The oxidation number of $A s$ is

Options:
A +2 and +3
B +3 and +5
C +3 and +4
D None of these

The correct answer is:
+3 and +5

Solution:
Arsenic $[A s]$ belongs to group 15 of the periodic
table, hence can exhibit both +3 and +5 oxidation states.

Topic: Chemistry, Redox Reactions, JEE Main
The oxidation number of $\mathrm{C}$ atom in $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ and $\mathrm{CCl}_{4}$ are respectively

Options:
A $-2$ and $-4$
B 0 and $-4$
C 0 and 4
D 2 and 4

The correct answer is:
0 and 4

Solution:
Let the oxidation number of $C$ is $x$
$\mathrm{CH}_{2} \mathrm{Cl}_{2}$
$x+2(+1)+2(-1)=0$
$x+2-2=0$
$x=0$
$\mathrm{CCl}_{4}$
$x+4(-1)=0$
$x-4=0$
$x=4$

Topic: Chemistry, Redox Reactions, JEE Main
The oxidation number of $\mathrm{Cr}$ in $\mathrm{CrO}_{5}$ which has the following structure, is

Options:
A $+4$
B $+5$
C $+3$
D $+6$

The correct answer is:
$+6$

Solution:
In the structure of $\mathrm{CrO}_{5}$, it has $\mathrm{O}$-atoms as peroxide which have oxidation number $=-1$ and one $\mathrm{O}$ as oxide (double bonded $\mathrm{O}$ ) atom with oxidation number $=-2$
Let the oxidation number of $\mathrm{Cr}$ be $x$.
Then, $x+4 \times(-1)+1 \times(-2)=0$
or $\quad x=+6$
$\therefore$ The oxidation number of $\mathrm{Cr}$ in $\mathrm{CrO}_{5}$ is $+6$.

Topic: Chemistry, d and f Block Elements, JEE Main
The oxidation number of Mn in the product of alkaline oxidative fusion of MnO2 is

Options:

The correct answer is:
6

Solution:
The oxidation number of Mn in the product of alkaline oxidative fusion of MnO2 is +6 as follows.
4KOH+2MnO2+O2→2K2MnO4+2+x-8+2H2OPotassium maganate
+2+x-8=0
x=+6.

Topic: Chemistry, d and f Block Elements, JEE Main
The oxidation number of $\mathrm{S}$ in $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8$ is

Options:
A $+2$
B $+4$
C $+6$
D $+7$

The correct answer is:
$+6$

Solution:
\(\mathrm{H}-\mathrm{O}-\underset{\underset{ }{||} \atop \huge \mathrm{O}}{\stackrel{\huge \mathrm{O} \atop ||}{\mathrm{S}}}-\mathrm{O}-\mathrm{O}-\underset{\underset{ }{||} \atop \huge \mathrm{O}}{\stackrel{\huge \mathrm{O} \atop ||}{\mathrm{S}}}-\mathrm{O}-\mathrm{H}\)
\(\because 2\) O-atoms are in peroxide linkage
\(\therefore\) Their oxidation state \(=-1\)
and, oxidation state of remaining \(6 \mathrm{O}\)-atoms \(=-2\)
Oxidation state of \(\mathrm{H}-\) atoms \(=+1\)
Let oxidation state of \(\mathrm{S}\) atom \(=\mathrm{x}\)
\(\mathrm{S}_2 \mathrm{H}_2 \mathrm{O}_6 \mathrm{O}_2\)
\(\therefore 2 \mathrm{x}+2(+1)+6(-2)+2(-1)=0\)
\(2 x+2-12-2=0\)
\(\therefore \mathrm{x}=+6\).

Topic: Chemistry, Redox Reactions, JEE Main
The oxidation number of xenon in $\mathrm{XeOF}_2$ is

Options:
A Zero
B 2
C 4
D 3

The correct answer is:
4

Solution:
Suppose the oxidation state of Xe is
$\begin{aligned} & x+(-2)+2(-1)=0 \\ & \Rightarrow x-2-2=0 \\ & \Rightarrow x=4\end{aligned}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The oxidation of SO2 to SO3 is an exothermic reaction. The yield of SO3 will be maximum if:

Options:
A Temperature is increased and pressure is kept constant
B Temperature is reduced and pressure is increased
C Both temperature and pressure are increased
D Both temperature and pressure are reduced

The correct answer is:
Temperature is reduced and pressure is increased

Solution:
SO 2 + 1 2 O 2 ⇌ SO 3 ; Δ H = - ve ;
Apply Le Chatelier principle.

Topic: Chemistry, Chemical Equilibrium, JEE Main
The oxidation sate of phosphorus in hypophosphoric acid is

Options:

The correct answer is:
4

Solution:
Hypophosphoric acid H4P2O6

The oxidation state of phosphorus is

⇒ 4 × 1 + 2 × x + 6 × -2 = 0

⇒ 4 + 2x - 12 = 0

⇒ 2x - 8 = 0

⇒ 2x = 8

⇒ x = 82 = 4

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2023 (31 Jan Shift 1)
The oxidation state exhibited by the transition metal which is present in Ziegler‐Natta catalyst is

Options:

The correct answer is:
4

Solution:
ZieglerâNatta catalyst is TiCl 4 +Al C 2 H 5 3

Topic: Chemistry, d and f Block Elements, JEE Main
The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is

Options:
A
+4

B
+6

C
+2

D
+3

The correct answer is:
+3

Solution:

Topic: Chemistry, d and f Block Elements, JEE Main
The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is

Options:
A
+4

B
+6

C
+2

D
+3

The correct answer is:
+3

Solution:

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2005
The oxidation state of $\mathrm{Cr}$ in $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ is

Options:
A $4$
B $-6$
C $6$
D $-2$

The correct answer is:
$6$

Solution:
$\begin{aligned} & \mathrm{Cr}_2 \mathrm{O}_7^{2-} \\ & 2 x-2 \times 7=-2 ; \quad 2 x=14-2=12 \\ & x=\frac{12}{2}=+6 .\end{aligned}$

Topic: Chemistry, Redox Reactions, JEE Main
The oxidation state of $\mathrm{Cr}$ in $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}$is

Options:
A
$+3$

B
$+2$

C
$+1$

D
0

The correct answer is:
$+3$

Solution:
$\left(\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right)^{+}$
$X+4 \times 0+2 \times-1=1$
$\mathrm{X}=+3$

Topic: Chemistry, Coordination Compounds, JEE Main
The oxidation state of $\mathrm{Cr}$ in $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}$is

Options:
A
$+3$

B
$+2$

C
$+1$

D
0

The correct answer is:
$+3$

Solution:
$\left(\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right)^{+}$
$X+4 \times 0+2 \times-1=1$
$\mathrm{X}=+3$

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2005
The oxidation state of $\mathrm{Cr}$ in $\mathrm{CrO}_5$ is

Options:
A -6
B +12
C +6
D +4

The correct answer is:
+6

Solution:
$\text {The structure of } \mathrm{CrO}_5 \text { is }$
Oxidation state of $\mathrm{Cr}$ is +6 due to the presence of two peroxide linkages, which can be calculated as. In $\mathrm{CrO}\left(\mathrm{O}_2\right)_2$, let the oxidation state of $\mathrm{Cr}$ be $\mathrm{x}$
$\begin{aligned}
\mathrm{x}+(-1) 4+(-2) & =0 \\
\mathrm{x}-6 & =0 \\
\mathrm{x} & =+6
\end{aligned}$

Topic: Chemistry, Redox Reactions, JEE Main
The oxidation state of iron in $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ is

Options:
A 1
B 4
C 3
D 2

The correct answer is:
2

Solution:
Let the oxidation state of iron in $\mathrm{K}_{4}\left[\mathrm{Fe}\left(\mathrm{CN}_{6}\right]\right.$ is $x$
$\begin{aligned} 4(+1)+x+6(-1) &=0 \\ 4+x-6 &=0 \\ x &=+2 \\ \mathrm{Au}+8 \mathrm{CN}^{-}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} \longrightarrow & \underset{\text { Soluble }}{\longrightarrow} \\ 2\left[\mathrm{Au}(\mathrm{CN})_{2} \Gamma+\mathrm{Zn} \longrightarrow\right.& \mathrm{Au}(s) \downarrow+\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-} \end{aligned}$

Topic: Chemistry, Coordination Compounds, JEE Main
The oxidation state of Mo in its oxo-complex species Mo2O4C2H42H2O22- is

Options:
A +5
B +4
C +3
D +2

The correct answer is:
+3

Solution:
-2=2×x+4×-2+0+0
or 2x=-2+8=+6
or x=+3

Topic: Chemistry, Coordination Compounds, JEE Main
The oxidation state of sulphur in H2S2O7 is

Options:
A +4
B +6
C +5
D +7

The correct answer is:
+4

Solution:
Antibiotics which kill or inhibit a wide range of gram positive and gram negative bacteria are called broad spectrum antibiotics. Among the given options, penicillin is not a broad spectrum antibiotics. It is a narrow spectrum antibiotics i.e. it is effective mainly against specific families of bacteria.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), MHT CET, MHT CET 2019 (Shift 2)
The oxidation states of Cr in [CrH2O)6Cl3,[CrC6H6)2 and K2[Cr(CN)2O)2O2NH3 respectively are

Options:
A +3, +4 and +6
B +3, +2 and +4
C +3, 0 and +6
D +3, 0 and +4

The correct answer is:
+3, 0 and +6

Solution:
[CrH2O)6Cl3:x+6×0+3×-1=0
x=+3
[CrC6H6)2:x+2×0
x=0
K2[CrCN)2O2O2NH3
2×(+1)+x+2×(-1)+2×(-2)+2×(-1)+0=0
2+x-2-4-2=0
x=+6

Topic: Chemistry, Redox Reactions, JEE Main
The oxidation states of Cr in CrH2O6Cl3, CrC6H62 and K2CrCN2O2O2NH3 , respectively, are:

Options:
A +3,Â 0 and +4
B +3,Â +4 and +6
C +3,Â +2 and +4
D +3, 0 and +6

The correct answer is:
+3, 0 and +6

Solution:
(i) CrH2O6Cl3: Hexaaquachromium III chloride
x+6×0+-1×3=0
x=+3
(ii) CrC6H62:bis η6-benzenechromium 0
y+2×0=0
y=0
(iii) K2CrCN2O2O2NH3: Potassium amminedicyanido dioxide peroxide chromate VI
2×1+z+2×-1+2×-2+-2+0=0
z=+6
The oxidation states of Cr in CrH2O6Cl3, CrC6H62 and K2CrCN2O2O2NH3, respectively, are +3, 0 and +6.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2018 (08 Apr)
The oxidation states of iron atoms in compounds A, B and C, respectively, are x, y and z. Then sum of x, y and z is ........... Na4FeCN5NOSA Na4FeO4B Fe2CO9C

Options:

The correct answer is:
6

Solution:
Oxidation state of Fe in Na4FeCN5NOS:

Charge of CN =-1

Charge of NOS=-1

Suppose oxidation state of Fe = x

4+x-5-1=0⇒x=+2

Oxidation state of Fe in Na4FeO4:

4+x-8=0⇒x=+4

Oxidation state of Fe in Fe2CO9 =0 because carbonyl is neutral ligand.

The oxidation states of iron in these compounds will be

A = +2 B = +4 C = 0

The sum of oxidation states will be = 6.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (02 Sep Shift 1)
The oxidation states of nitrogen in NO, NO2, N2O and NO3- are in the order of :

Options:
A NO3->NO2>NO>N2O
B NO2>NO3->NO>N2O
C N2O>NO2>NO>NO3-
D NO>NO2>N2O>NO3-

The correct answer is:
NO3->NO2>NO>N2O

Solution:
The oxidation states of Nitrogen in following molecules are as follows

NO3-→+5

NO2→+4

NO→+2

N2O→+1

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2021 (18 Mar Shift 2)
The oxidation states of $\mathrm{P}$ atom in $\mathrm{POCl}_{3}, \mathrm{H}_{2} \mathrm{PO}_{3}$ and $\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}$, respectively, are

Options:
A $+5,+4,+4$
B $+5,+5,+4$
C $+4,+4,+5$
D $+3,+4,+5$

The correct answer is:
$+5,+4,+4$

Solution:
O. No.
$\begin{array}{l}
\mathrm{POCl}_{3}+5 \\
\mathrm{H}_{2} \mathrm{PO}_{3}+5 \\
\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}+4
\end{array}$

Topic: Chemistry, Redox Reactions, JEE Main
The oxidation states of 'P' in H4P2O7, H4P2O5 and H4P2O6, respectively, are :

Options:
A 7, 5 and 6
B 5, 4 and 3
C 5, 3 and 4
D 6, 4 and 5

The correct answer is:
5, 3 and 4

Solution:
Oxidation state of P in H4P2O7, H4P2O5 and H4P2O6 is 5, 3 & 4 respectively

H4P2O7
2x+4(+1)+7(-2)=0
x=+5
H4P2¯O5
2x+4(+1)+5(-2)=0
x=+3
H4P2¯O6
2x+4(+1)+6(-2)=0
x=+4

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2021 (27 Jul Shift 1)
The oxidation states of transition metal atoms in K2Cr2O7, KMnO4 and K2FeO4, respectively, are x,y and z . The sum of x, y and z is______

Options:

The correct answer is:
19

Solution:
Compound Oxidation state of transition element.

Oxidation states of K =+1 and O=-2

(i) K2Cr2O7 x=+6
(ii) KMnO4 y=+7
(iii) K2FeO4 z=+6
so (x+y+z)=19

Topic: Chemistry, Redox Reactions, JEE Main
The oxidation states of transition metal atoms in K2Cr2O7, KMnO4 and K2FeO4, respectively, are x,y and z . The sum of x, y and z is______

Options:

The correct answer is:
19

Solution:
Compound Oxidation state of transition element.

Oxidation states of K =+1 and O=-2

(i) K2Cr2O7 x=+6
(ii) KMnO4 y=+7
(iii) K2FeO4 z=+6
so (x+y+z)=19

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2020 (02 Sep Shift 2)
The oxide that gives H2O2 most readily on treatment with H2O is :

Options:
A Na2O2
B BaO2.8H2O
C SnO2
D PbO2

The correct answer is:
Na2O2

Solution:
Among the given compounds Na2O2 and BaO2.8H2O are the peroxides. SnO2 and PbO2 are the oxides.

Na2O2+H2O→NaOH+H2O2

Acidified barium peroxide gives hydrogen peroxide.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2021 (27 Aug Shift 2)
The oxide that gives hydrogen peroxide $\left(\mathrm{H}_2 \mathrm{O}_2\right)$ on the treatment with a dilute acid $\left(\mathrm{H}_2 \mathrm{SO}_4\right)$ is

Options:
A $\mathrm{MnO}_2$
B $\mathrm{PbO}_2$
C $\mathrm{Na}_2 \mathrm{O}_2$
D $\mathrm{TiO}_2$

The correct answer is:
$\mathrm{Na}_2 \mathrm{O}_2$

Solution:
$\mathrm{Na}_2 \mathrm{O}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}_2$

Topic: Chemistry, Hydrogen, JEE Main
The oxide which contains an odd electron at the nitrogen atom is

Options:
A N2O
B NO2
C N2O3
D N2O5

The correct answer is:
NO2

Solution:
Structures of the given oxides of nitrogen are as follows:

As we can see only NO2 is having an odd electron on the nitrogen atom.It is the correct answer.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2022 (26 Jun Shift 2)
The oxide, which is best soluble in $\mathrm{H}_2 \mathrm{O}$ is

Options:
A $\mathrm{Ba}(\mathrm{OH})_2$
B $\mathrm{Mg}(\mathrm{OH})_2$
C $\mathrm{Sr}(\mathrm{OH})_2$
D $\mathrm{Ca}(\mathrm{OH})_2$

The correct answer is:
$\mathrm{Ba}(\mathrm{OH})_2$

Solution:
$\mathrm{Ba}(\mathrm{OH})_2 > \mathrm{Sr}(\mathrm{OH})_2 > \mathrm{Ca}(\mathrm{OH})_2 > \mathrm{Mg}(\mathrm{OH})_2$
Solubility decreasing order.

Topic: Chemistry, s Block Elements, JEE Main
The oxide without nitrogen-nitrogen bond is :

Options:
A N2O4
B N2O
C N2O5
D N2O3

The correct answer is:
N2O5

Solution:
Formula	Resonance structure

N2O4

N2O

N2O3

N2O5

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2021 (01 Sep Shift 2)
The oxoacid of phosphorus that is easily obtained from a reaction of alkali and white phosphorus and has two P-H bonds, is

Options:
A Phosphonic acid
B Phosphinic acid
C Hypophosphoric acid
D Pyrophosphorus acid

The correct answer is:
Phosphinic acid

Solution:
White phosphorus is a translucent white waxy solid. It is poisonous, insoluble in water but soluble in carbon disulphide and glows in dark (chemiluminescence). It dissolves in boiling NaOH solution in an inert atmosphere giving PH3.

P4+3NaOH+3H2O→PH3+3NaH2PO2 (sodium hypophosphite)

Sodium hypophosphite is a salt of Phosphinic acid (Hypophosphorous acid).

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2022 (29 Jun Shift 1)
The oxoacid of sulphur that does not contain bond between sulphur atom is:

Options:
A H2S2O7
B H2S2O4
C H2S4O6
D H2S2O3

The correct answer is:
H2S2O7

Solution:
Hence correct option is 1 as there is no bond present between S-S atom. It is also known as oleum.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2019 (10 Apr Shift 1)
The oxoacid of sulphur which contains two sulphur atoms in different oxidation states is

Options:
A pyrosulphurous acid
B hyposulphurous acid
C pyrosulphuric acid
D persulphuric acid

The correct answer is:
pyrosulphurous acid

Solution:
The structure of the given oxoacids of sulphur are as
Thus, the oxidation states of two $S$ atoms in pyrosulphurous acid are different.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main