datavorous/entrance-exam-dataset huggingface dataset:
The correct bond order in the following species is:

Options:
A $\mathrm{O}_{2}^{2+}<\mathrm{O}_{2}^{-}<\mathrm{O}_{2}^{+}$
B $\mathrm{O}_{2}^{+}<\mathrm{O}_{2}^{-}<\mathrm{O}_{2}^{2+}$
C $\mathrm{O}_{2}^{-}<\mathrm{O}_{2}^{+}<\mathrm{O}_{2}^{2+}$
D $\mathrm{O}_{2}^{2+}<\mathrm{O}_{2}^{+}<\mathrm{O}_{2}^{-}$

The correct answer is:
$\mathrm{O}_{2}^{-}<\mathrm{O}_{2}^{+}<\mathrm{O}_{2}^{2+}$

Solution:
$\mathrm{O}_{2}^{+}$ ion - Total number of electrons
$(16-1)=15$
E.C.:
$$
\begin{array}{r}
\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}^{2} \\
\pi 2 p_{x}^{2}=\pi 2 p_{y}^{2} \pi^{*} 2 p_{x}^{1}=\pi^{*} 2 p_{y} \\
\text { Bond order }=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{10-5}{2}=\frac{5}{2}=2 \frac{1}{2}
\end{array}
$$
$\mathrm{O}_{2}^{-}$ (Super oxide ion): Total number of
electrons $(16+1)=17$
E.C.:
$$
\begin{array}{l}
\text { \sigma1s }^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}^{2} \\
\pi 2 p_{x}^{2}=\pi 2 p_{y}^{2} \pi^{*} 2 p_{x}^{2}=\pi^{*} 2 p_{y}^{1} \\
\text { Bond order }=\frac{\left(\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)}{2}=\frac{10-7}{2}=\frac{3}{2}=1 \frac{1}{2}
\end{array}
$$
$\mathrm{O}_{2}^{2+}$ ion: Total number of electrons $=(16-2)=14$
E.C.
$$
\begin{array}{l}
\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_{x}^{2}=\pi 2 p_{y}^{2} \sigma 2 p_{z}^{2} \\
\text { Bond order }=\frac{\left(N_{b}-N_{a}\right)}{2}=\frac{10-4}{2}=\frac{6}{2}=3
\end{array}
$$
So bond order: $\mathrm{O}_{2}^{-}<\mathrm{O}_{2}^{+}<\mathrm{O}_{2}^{2+}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The correct bond order in the following species is

Options:
A O22+<O2+<O2-
B O2+<O2-<O22+
C O2-<O2+<O22+
D O22+<O2-<O2+

The correct answer is:
O2-<O2+<O22+

Solution:
According to molecular orbital theory (MOT)

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The correct code for stability, of oxidation states for given cations is
I. Pb2+>Pb4+,Tl+<Tl3+ II. Bi3+<Sb3+,Sn2+<Sn4+
III. Pb2+>Pb4+,Bi3+>Bi5+ IV. Tl3+<In3+,Sn2+>Sn4+
V. Sn2+<Pb2+,Sn4+>Pb4+ VI. Sn2+<Pb2+,Sn4+<Pb4+

Options:
A V and VI
B I, III and VI
C III and V
D II and IV

The correct answer is:
III and V

Solution:
Due to inert effect the stability of lower oxidation state gradually increases while stability of higher oxidation state gradually decreases down the group in elements of group 13th to 15th. So correct order are
III. Pb2+>Pb4+,Bi3+>Bi5+
IV. Sn2+<Pb2+,Sn4+<Pb4+

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The correct combination is

Options:
A NiCl42-- Square planar; Ni(CN)42-- paramagnetic
B Ni(CN)42-- tetrahedral; Ni(CO)42-- paramagnetic
C NiCl42-- paramagnetic; Ni(CO)4- tetrahedral
D NiCl42-- diamagnetic; Ni(CO)4- square-planar

The correct answer is:
NiCl42-- paramagnetic; Ni(CO)4- tetrahedral

Solution:
[NiCl4]−2
Tetrahedral
Paramagnetic
[Ni(CO)4]
Tetrahedral
Diamagnetic
Weak field ligand (WFL)
Strong field ligand(SFL)

NiCl42-	NiCO4

Paramagnetic (2 unpaired electrons)	Sp3

Ni2+→Ar 3d8, 4s0, 4p0

Cl- (WFL) (No pairing)	NiO→Ar3d8, 4s2, 4p0

CO is SFL Ar3d10, 4s0, 4p0 (tetrahedral)

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2018 (15 Apr)
The correct combination is :

Options:
A
$\left[\mathrm{NiCl}_4\right]^{2-}$ - square-planar; $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ - paramagnetic

B
$\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$-tetrahedral; $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ - paramagnetic

C
$\left[\mathrm{NiCl}_4\right]^{2-}$- paramagnetic; $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ - tetrahedral

D
$\left[\mathrm{NiCl}_4\right]^{2-}$- dimagnetic; $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ -square-planar

The correct answer is:
$\left[\mathrm{NiCl}_4\right]^{2-}$- paramagnetic; $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ - tetrahedral

Solution:
$\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is square planar, diamagnetic $(0$ unpaired electrons) with $d s p^2$ hybridisation.
$\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ - is tetrahedral,diamagnetic ( $0$ unpaired electrons) with $s p^3$ hybridisation.
$\left[\mathrm{NiCl}_4\right]^{2-}$ is tetrahedral, paramagnetic ($2$ unpaired electrons) with $s p^3$ hybridisation.
Hence, the option (c) is the correct answer.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The correct decreasing order for metallic character is

Options:
A Na>Mg>Be>Si>P
B P>Si>Be>Mg>Na
C Si>P>Be>Na>Mg
D Be>Na>Mg>Si>P

The correct answer is:
Na>Mg>Be>Si>P

Solution:
A metal is a material that, when freshly prepared, polished, or fractured, shows a lustrous appearance, and conducts electricity and heat relatively well. Metals are typically malleable (they can be hammered into thin sheets) or ductile (can be drawn into wires)

On moving top to bottom of the periodic table metallic character increases and on moving left to right metallic character decreases.

Metallic character is directly proportional to the atomic size.

The correct order of atomic size of the given elements is Na>Mg>Be>Si>P.

Hence, correct order of metallic character is Na>Mg>Be>Si>P

All these elements are form 3rd period except beryllium which is from second period

Na is an alkali metal

Mg and Be are alkaline earth metals,

Phosphorous belongs to a group which is commonly called as pnicogens.

Si belongs to 14th group

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2022 (28 Jul Shift 2)
The correct decreasing order of bond angles is

Options:
A ClF 3 > PF 3 > NF 3 > BF 3
B BF 3 > PF 3 > NF 3 > ClF 3
C BF 3 > ClF 3 > PF 3 > NF 3
D BF 3 > NF 3 > PF 3 > ClF 3

The correct answer is:
BF 3 > NF 3 > PF 3 > ClF 3

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The correct decreasing order of densities of the following compounds is :

Options:
A (D)>(C)>(B)>(A)
B (C)>(D)>(A)>(B)
C (C)>(B)>(A)>(D)
D (A)>(B)>(C)>(D)

The correct answer is:
(D)>(C)>(B)>(A)

Solution:
The density order

Density is directly proportional to molecular mass.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (25 Jul Shift 2)
The correct decreasing order of electropositive character among the following elements is :

Fe, Sc, Rb, Br, Te, F, Ca

Options:
A Fe > Sc > Rb > Br > Te > F > Ca
B Ca > Rb > Sc > Fe > Te > F > Br
C Rb > Ca > Sc > Fe > Br > Te > F
D Rb > Ca > Sc > Fe > Te > Br > F

The correct answer is:
Rb > Ca > Sc > Fe > Te > Br > F

Solution:
(i) Alkali and alkaline earth metals are most electro-positive. Alkali metals are more electropositive than alkaline earth metals.
(ii) In d-block series, the elements near alkaline earth metals are more electropositive than rest of the members.
(iii) Metalloids are less electropositive than metals.
(iv) Halogens (Non-metals) are least electropositive. In halogen group, electropositive character increases as atomic number increases.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The correct decreasing order of electropositive character among the following elements is

Options:
A \( \mathrm{Fe}>\mathrm{Sc}>\mathrm{Rb}>\mathrm{Br}>\mathrm{Te}>\mathrm{F}>\mathrm{Ca} \)
B \( \mathrm{Ca}>\mathrm{Rb}>\mathrm{Sc}>\mathrm{Fe}>\mathrm{Te}>\mathrm{F}>\mathrm{Br} \)
C \( \mathrm{Rb}>\mathrm{Ca}>\mathrm{Sc}>\mathrm{Fe}>\mathrm{Br}>\mathrm{Te}>\mathrm{F} \)
D \( \mathrm{Rb}>\mathrm{Ca}>\mathrm{Sc}>\mathrm{Fe}>\mathrm{Te}>\mathrm{Br}>\mathrm{F} \)

The correct answer is:
\( \mathrm{Rb}>\mathrm{Ca}>\mathrm{Sc}>\mathrm{Fe}>\mathrm{Te}>\mathrm{Br}>\mathrm{F} \)

Solution:
Metallic character(electropositive) is inversely proportional to ionization energy.

Alkali and alkaline earth metals are most electropositive. Alkali metals are more electropositive than alkaline earth metals.

In d-block elements , the elements near the alkaline earth metals are more electropositive than rest of the members. Metalloids are less electropositive than metals. Halogens (nonmetals) are least electropositive. In halogen group electropositive character increases down the group.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The correct decreasing order of energy, for the orbitals having, following set of quantum numbers:

(A) n=3,1=0, m=0

(B) n=4,l=0, m=0

(C) n=3,l=1, m=0

(D) n=3,1=2, m=1

Options:
A D>B>C>A
B B>D>C>A
C C>B>D>A
D B>C>D>A

The correct answer is:
D>B>C>A

Solution:
(A) n+l=3+0=3⇒3s

(B) n+l=4+0=4⇒4s

(C) n+l=3+1=4⇒3p

(D) n+l=3+2=5⇒3d

Higher the n+l value, higher the energy of the orbital and if two orbitals have same n+l value, then the orbital with higher n value has higher energy.

Thus: D>B>C>A.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2022 (27 Jul Shift 2)
The correct decreasing order of negative electron gain enthalpy for $\mathrm{C}, \mathrm{Ca}, \mathrm{Al}, \mathrm{F}$ and $\mathrm{O}$ is

Options:
A $\mathrm{F}>\mathrm{O}>\mathrm{C}>\mathrm{Al}>\mathrm{Ca}$
B $\mathrm{Ca}>\mathrm{Al}>\mathrm{O}>\mathrm{F}>\mathrm{C}$
C $\mathrm{Al}>\mathrm{F}>\mathrm{Ca}>\mathrm{C}>\mathrm{O}$
D $\mathrm{F}>\mathrm{C}>\mathrm{O}>\mathrm{Ca}>\mathrm{Al}$

The correct answer is:
$\mathrm{F}>\mathrm{O}>\mathrm{C}>\mathrm{Al}>\mathrm{Ca}$

Solution:
Electron gain enthalpy becomes less negative as we move down in a group white it becomes more negative along the period.
$\therefore$ Correct order is
$\mathrm{F}>\mathrm{O}>\mathrm{C}>\mathrm{Al}>\mathrm{Ca}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The correct decreasing order of priority for the functional groups of organic compounds in the IUPAC system of nomenclature is

Options:
A
$-\mathrm{COOH},-\mathrm{SO}_3 \mathrm{H},-\mathrm{CONH}_2,-\mathrm{CHO}$

B
$-\mathrm{SO}_3 \mathrm{H},-\mathrm{COOH},-\mathrm{CONH}_2,-\mathrm{CHO}$

C
$-\mathrm{CHO},-\mathrm{COOH},-\mathrm{SO}_3 \mathrm{H},-\mathrm{CONH}_2$

D
$-\mathrm{CONH}_2,-\mathrm{CHO},-\mathrm{SO}_3 \mathrm{H},-\mathrm{COOH}$

The correct answer is:
$-\mathrm{SO}_3 \mathrm{H},-\mathrm{COOH},-\mathrm{CONH}_2,-\mathrm{CHO}$

Solution:
$$
-\mathrm{SO}_3 \mathrm{H},-\mathrm{COOH},-\mathrm{CONH}_2,-\mathrm{CHO}
$$

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2008
The correct decreasing order of priority of functional groups in naming an organic compound as per IUPAC system of nomenclature is

Options:
A -COOH>-CONH2>-COCl>-CHO
B -SO3H>-COCl>-CONH2>-CN
C

D COOH>--COOR>-CONH2>-COCl

The correct answer is:
-SO3H>-COCl>-CONH2>-CN

Solution:
In the case of polyfunctional compounds, one of the functional groups is chosen as the principal functional group and the compound is then named on that basis. The remaining functional groups, which are subordinate functional groups, are named as substituents using the appropriate prefixes. The choice of principal functional group is made on the basis of order of preference. The order of decreasing priority for some functional groups is:

-COOH, –SO3H, -COOR (R=alkyl group), COCl, -CONH2, -CN, -HC=O, >C=O, -OH, -NH2, >C=C<, -C≡C-.

For example, the IUPAC name of

HOCH2(CH2)3CH2COCH3 is 7-hydroxyheptan-2-one.

The correct decreasing order of priority of given functional groups is

-SO3H>-COCl>-CONH2>-CN

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2022 (26 Jul Shift 2)
The correct decreasing order of the basic strength is

Options:
A HS->CH3->NH2->CN-
B HS->CN->NH2->CH3-
C CH3->NH2->CN->HS-
D NH2->CN->CH3->HS-

The correct answer is:
CH3->NH2->CN->HS-

Solution:
According to the question, Conjugate acids and bases are part of the Bronsted-Lowry theory of acids and bases.

According to this theory, the species that donates a hydrogen cation or proton in a reaction is a conjugate acid, while the remaining portion or the one that accepts a proton or hydrogen is the conjugate base.

The conjugate base may be recognised as an anion.

Conjugated base of weak acid is worked as strong base.

CH3- + H+→CH4 (W.A)CN-+H+→HCNNH2-+H+→NH3HS-+H+ →H2S (S.A)

Topic: Chemistry, Ionic Equilibrium, TS EAMCET, TS EAMCET 2021 (05 Aug Shift 2)
The correct dimensional formula for impulse is given by

Options:
A $\mathrm{ML}^{2} \mathrm{T}^{-2}$
B MLT $^{-1}$
C $\mathrm{ML}^{2} \mathrm{T}^{-1}$
D $\mathrm{MLT}^{-2}$

The correct answer is:
MLT $^{-1}$

Solution:
As we know that,
Impulse, $I=$ force $(F) \times$ small time interval
$\begin{array}{ll}
I=m a \times t & {[\because F=m a]} \\
I=[\mathrm{M}]\left[\mathrm{LT}^{-2}\right] \times[\mathrm{T}] \\
I=\left[\mathrm{MLT}^{-1}\right]
\end{array}$
Hence, the correct dimensional formula for impulse is given by $\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-1} 1 .\right.$

Topic: Physics, Units and Dimensions, JEE Main
The correct dimensions of the coefficient of viscosity η are

Options:
A [M L -1 T - 2 ]
B [M L T - 1 ]
C [ M L - 1 T - 1 ]
D [ M L - 2 T - 2 ]

The correct answer is:
[ M L - 1 T - 1 ]

Solution:
Viscous force is given as, F⁡=6πηrv.

∴η=F6πrv

[η]=[F][r][v]

[η]=[MLT-2][L][LT-1]

Topic: Physics, Units and Dimensions, JEE Main
The correct evaluation of $\int_0^\pi\left|\sin ^4 x\right| d x$ is

Options:
A $\frac{8 \pi}{3}$
B $\frac{2 \pi}{3}$
C $\frac{4 \pi}{3}$
D $\frac{3 \pi}{8}$

The correct answer is:
$\frac{3 \pi}{8}$

Solution:
$\int_0^\pi\left|\sin ^4 x\right| d x=2 \int_0^{\pi / 2} \sin ^4 x d x$
Applying gamma function,
$2 \int_0^{\pi / 2} \sin ^4 x d x=2 \frac{\Gamma(5 / 2) \cdot \Gamma(1 / 2)}{2 \Gamma \Gamma(6 / 2)}=\frac{3 \pi}{8}$

Topic: Mathematics, Definite Integration, JEE Main
The correct figure that shows, schematically, the wave pattern produced by the superposition of two waves of frequencies 9 Hz and 11 Hz, is

Options:
A
B
C
D

The correct answer is:

Solution:
Beat frequency =11-9=2 Hz
No.of maxima per second is 2.
Time period of beat =12 s.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The correct floral formula of soybean is

Options:
A
B
C
D

The correct answer is:

Solution:
Soybean or Glycine max is a member of familyFabaceae, sub-family-Papilionidae. Its floral formula is
$\% \not \phi^\pi K_{(5)} C_{1+2+(2)} A_{(9)+1} G_1 .$

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2010 (Mains)
The correct graph representation of potential along the branch PQRS is

Options:
A
B
C
D

The correct answer is:

Solution:
As we move from P to S we are moving from high potential to low potential. Since P is connected to +ve terminal of battery it should be at a positive potential. Q is grounded hence it should be at zero potential. R and S should be at a negative potential.

Topic: Physics, Current Electricity, JEE Main
The correct graph representing the variation of total energy $\left.{ }^{(} E_t\right)$ kinetic energy $\left(E_k\right)$ and potential energy $(U)$ of a satellite with its distance from the centre of earth is

Options:
A

B

C

D

The correct answer is:

Solution:
$U=\frac{-G M m}{r}, K=\frac{G M m}{2 r}$ and $E=\frac{-G M m}{2 r}$
For a satellite $U, K$ and $E$ varies with $r$ and also $U$ and $E$ remains negative whereas $\mathrm{K}$ remain always positive.

Topic: Physics, Gravitation, JEE Main
The correct graph which shows the variation of paramagnetic susceptibility S with temperature T is

Options:
A

B

C

D

The correct answer is:

Solution:
Paramagnetic susceptibility is inversely proportional to temperature.
S∝1T . The graph is a hyperbola

Topic: Physics, Magnetic Properties of Matter, JEE Main
The correct group of halide ions which can be oxidised by oxygen in acidic medium is

Options:
A Br– and I– only
B Br– only
C I– only
D Cl– , Br– and I– only

The correct answer is:
I– only

Solution:
Fluorine oxidises water to oxygen whereas chlorine & bromine react with water to form corresponding HX & HOX acids.

2F2+2H2O→4HF+O2Cl2+H2O→HCl+HOCl

The reaction of I2 with water is nonspontaneous. In fact I– can be oxidised by oxygen in acidic medium. It is why the I– in nature is not present in that much amount as other halides are present in nature.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2023 (13 Apr Shift 2)
The correct hybridization of Fe in complex K 3 [Fe(CN) 6 ] is

Options:
A dsp3
B sp3
C sp3d2
D d2sp3

The correct answer is:
d2sp3

Solution:
K3[Fe(CN)6]
Electronic configuration of Fe=[Ar]4s23d6
Electronic configuration of Fe+3=[Ar]3d5
Number of ligand (coordination numbr) =6
Nature of ligand is strong field.
Hybridization of Fe is d2sp3 .

Topic: Chemistry, Coordination Compounds, JEE Main
The correct increasing order for bond angles among \(\mathrm{BF}_3, \mathrm{PF}_3\) and \(\mathrm{ClF}_3\) is :

Options:
A \(\mathrm{BF}_3 < \mathrm{PF}_3 < \mathrm{ClF}_3\)
B \(\mathrm{ClF}_3 < \mathrm{PF}_3 < \mathrm{BF}_3\)
C \(\mathrm{PF}_3 < \mathrm{BF}_3 < \mathrm{ClF}_3\)
D \(\mathrm{BF}_3=\mathrm{PF}_3 < \mathrm{ClF}_3\)

The correct answer is:
\(\mathrm{ClF}_3 < \mathrm{PF}_3 < \mathrm{BF}_3\)

Solution:
Order of bond angle is $\mathrm{ClF}_3 < \mathrm{PF}_3 < \mathrm{BF}_3$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (09 Apr Shift 2)
The correct increasing order of basic character of Ce(OH)3,Gd(OH)3 and Nd(OH)3 is

Options:
A Ce(OH)3<Nd(OH)3<Gd(OH)3
B Gd(OH)3<Ce(OH)3<Nd(OH)3
C Gd(OH)3<Nd(OH)3<Ce(OH)3
D Ce(OH)3<Gd(OH)3<Nd(OH)3

The correct answer is:
Gd(OH)3<Nd(OH)3<Ce(OH)3

Solution:
Due to lanthanide contraction, as the size of lanthanide ions decreases from left to right La3+to Lu3+, the covalent character of the hydroxides increases and hence, the basic strength decreases.

So, increasing order of basic characters: Gd(OH)3 <Nd(OH)3 <Ce(OH)3

Topic: Chemistry, d and f Block Elements, JEE Main
The correct increasing order of electronegativity for \( \mathrm{Sn}, \mathrm{Sn}^{2+} \) and \( \mathrm{Sn}^{4+} \) species is:

Options:
A \( \mathrm{Sn} < \mathrm{Sn}^{2+} < \mathrm{Sn}^{4+} \)
B \( \mathrm{Sn}^{4+} < \mathrm{Sn}^{2+} < \mathrm{Sn} \)
C \( \mathrm{Sn}=\mathrm{Sn}^{2+}=\mathrm{Sn}^{4} \)
D \( \mathrm{Sn} < \mathrm{Sn}^{2+}=\mathrm{Sn}^{4+} \)

The correct answer is:
\( \mathrm{Sn} < \mathrm{Sn}^{2+} < \mathrm{Sn}^{4+} \)

Solution:
Higher the oxidation state, greater is the electronegativity.

So, order will be Sn<Sn2+<Sn4+

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The correct increasing order of extent of hydrolysis is

Options:
A CCl4<MgCl2<AlCl3<SiCl4<PCl5
B CCl4<AlCl3<MgCl2<PCl5<SiCl4
C CCl4<SiCl4<PCl5<AlCl3<MgCl2
D CCl4<PCl5<SiCl4<AlCl3<MgCl2

The correct answer is:
CCl4<MgCl2<AlCl3<SiCl4<PCl5

Solution:
Extent of hydrolysis is proportional to the amount of positive formation charge present on the atomic site prone to attack of H2O molecules.
CCl4<MgCl2<AlCl3<SiCl4<PCl5

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The correct increasing order of the ionic radii is

Options:
A Cl-<Ca2+<K+<S2-
B K+<S2-<Ca2+<Cl-
C S2-<Cl-<Ca2+<K+
D Ca2+<K+<Cl-<S2-

The correct answer is:
Ca2+<K+<Cl-<S2-

Solution:
For isoelectronic species, as nuclear charge increases radius decreases. Greater the positive charge, the lesser the size of the ion. Greater the negative charge, the larger the size of an ion.

Ca2+< K+ < Cl- < S2-p 20 19 17 16e 18 18 18 18

So, the correct option is (D).

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2023 (31 Jan Shift 1)
The correct IUPAC name of following compound is

Options:
A 4-Aminomethyl-3-hydroxycyclohex-5-ene-1-carboxylic acid
B 2-Aminomethyl-5-carboxycyclohex-3-en-1-ol
C 4-Aminomethyl-5-hydroxycyclohex-2-ene-1-carboxylic acid
D 3-Hydroxy-4-aminomethylcyclohex-5-en-1-oic acid

The correct answer is:
4-Aminomethyl-5-hydroxycyclohex-2-ene-1-carboxylic acid

Solution:
4-Aminomethyl-5-hydroxycyclohex-2-ene-1-carboxylic acid

Topic: Chemistry, General Organic Chemistry, JEE Main
The correct IUPAC name of K2MnO4 is:

Options:
A Potassium tetraoxopermanganate (VI)
B Potassium tetraoxidomanganate (VI)
C Dipotassium tetraoxidomanganate (VII)
D Potassium tetraoxidomanganese (VI)

The correct answer is:
Potassium tetraoxidomanganate (VI)

Solution:
The structure of the IUPAC name is

name of cationic part + Name of anionic part.

Metal present in the anionic complex, so 'ate', must be added at the end of the metal name.

Oxidation state of Mn

Mn+4-2=-2 =+6

The structure of anionic complex name is

ligand names + metal name(oxidation state)

K2MnO4

2+x–8=0

⇒x=+6

Oxidation state of Mn=+6

IUPAC Name is:

Potassium tetraoxidomanganate VI

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (29 Jan Shift 2)
The correct IUPAC name of \(\left[\mathrm{PtBr}_2\left(\mathrm{PMe}_3\right)_2\right]\) is :

Options:
A dibromodi(trimethylphosphine)platinum(II)
B bis(trimethylphosphine)dibromoplatinum(II)
C dibromobis(trimethylphosphine)platinum(II)
D bis[bromo(trimethylphosphine)]platinum(II)

The correct answer is:
dibromobis(trimethylphosphine)platinum(II)

Solution:
Dibromo bis(trimethylphosphine) platinum (II)

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (06 Apr Shift 2)
The correct IUPAC name of the compound is

Options:
A 4-Bromo-5-chloro-3-iodohexane
B 3-Bromo-2-chloro-4-iodohexane
C 3-Bromo-4-iodo -2-chlorohexane
D 2-Bromo-3-bromo-4-iodohexane

The correct answer is:
3-Bromo-2-chloro-4-iodohexane

Solution:
The decreasing order of priority of prefix in numbering the carbon chain of an organic compound is
Bromo > Chloro > Iodo

Topic: Chemistry, General Organic Chemistry, JEE Main
The correct IUPAC name of the following compound is:

Options:
A \( 5-(1,2 \)-dimethylpropyl \( )-6-(2,2- \) dimethylpropyl \( )-3 \)-ethyl-8-methyldecane
B \( 5 \)-cyclopropyl-4-ethyl-4-isopropyl-2, \( 2,3,3 \) tetra methylhexane
C bis- \( 4,5-(1,1- \) dimethylpropyl \( )-2,2,6,7- \) tetramethyl-loctane
D \( 5-(1,2- \) dichloroethyl \( )-5-(2,2 \)-dichloroethyl \( )-7 \)-ethyl- \( 2 \)-methylnonane

The correct answer is:
\( 5 \)-cyclopropyl-4-ethyl-4-isopropyl-2, \( 2,3,3 \) tetra methylhexane

Solution:
Numbering should be started from that side where substituent get least possible number possible. Numbering started from left most side of hydrocarbon.

Priority of cyclopropyl is more than that of methyl group. Using above name of compound will be

5-cyclopropyl-4-ethyl-4-isopropyl-2,2,3,3 tetra methylhexane

Topic: Chemistry, General Organic Chemistry, JEE Main
The correct IUPAC name of the following compound

is :

Options:
A 4 - methyl - 3 - ethylhexane
B 3 - ethyl - 4 - methylhexane
C 4 - ethyl - 3 - methylhexane
D 3, 4 - ethylmethylhexane

The correct answer is:
3 - ethyl - 4 - methylhexane

Solution:
â
â3 - ethyl - 4 - methylhexane

Longest chain - 6 carbon

Substituent present -ethyl and methyl, since ethyl comes alphabeticaly first, hence the numbering will be such that ethyl group get lowest possible number.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2014 (19 Apr Online)
The correct IUPAC name of the following compound is:

Options:
A 5-chloro-4-methyl-1-nitrobenzene.
B 2-chloro-1-methyl-4-nitrobenzene.
C 2-methyl-5-nitro-1-chlorobenzene.
D 3-chloro-4-methyl-1-nitrobenzene.

The correct answer is:
2-chloro-1-methyl-4-nitrobenzene.

Solution:
Priority,
2-chloro-1-methyl-4-nitrobenzene

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The correct IUPAC name of the following compound is :

Options:
A 1-Bromo-1-ethyl-2-fluoro-2-iodo-1-nitroethane.
B 3-Bromo-4-fluoro-4-iodo-3-nitrobutane.
C 2-Bromo-1-fluoro-1-iodo-2-nitrobutane.
D 1-Fluoro-1-iodo-2-bromo-2-ethyl-2-nitroethane.

The correct answer is:
2-Bromo-1-fluoro-1-iodo-2-nitrobutane.

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main