datavorous/entrance-exam-dataset huggingface dataset:
In which of the following arrangements, the sequence is not strictly according to the property written against it?

Options:
A CO2 < SiO2 < SnO2 < PbO2 : increasing oxidising power
B HF < HCl < HBr < HI : increasing acid strength
C B < C < O < N : increasing first ionisation energy
D NH3 < PH3 < AsH3 < SbH3 : increasing basic strength

The correct answer is:
NH3 < PH3 < AsH3 < SbH3 : increasing basic strength

Solution:
Drago's Rule :-
In case of NH3, PH3, - - - bond angle decrease ,NH3 ( â107 ° ) & PH 3 ( â90 ° ) So, %s character of lp in NH3 is less & %s charactor of lp top to bottom increase so basic charactor decrease.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In which of the following arrangements, the sequence is not strictly according to the property written against it ?

Options:
A
$\mathrm{CO}_2 < \mathrm{SiO}_2 < \mathrm{SnO}_2 < \mathrm{PbO}_2$ : increasing oxidising power
B
$\mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}$ : increasing acid strength
C
$\mathrm{NH}_3 < \mathrm{PH}_3 < \mathrm{AsH}_3 < \mathrm{SbH}_3$ : increasing basic strength
D
$\mathrm{B} < \mathrm{C} < \mathrm{O} < \mathrm{N}$ : increasing first ionization enthalpy.

The correct answer is:
$\mathrm{NH}_3 < \mathrm{PH}_3 < \mathrm{AsH}_3 < \mathrm{SbH}_3$ : increasing basic strength

Solution:
Correct basic strength is $\mathrm{NH}_3>\mathrm{PH}_3>\mathrm{AsH}_3>\mathrm{BiH}_3$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2009
In which of the following arrangements, the sequence is not strictly according to the property written against it?

Options:
A
$\mathrm{CO}_2 < \mathrm{SiO}_2 < \mathrm{SnO}_2 < \mathrm{PbO}_2$ : increasing oxidising power
B
$\mathrm{NH}_3 < \mathrm{PH}_3 < \mathrm{AsH}_3 < \mathrm{SbH}_3$ : increasing basic strength
C
$\mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}$ : increasing acid strength
D
$\mathrm{B} < \mathrm{C} < \mathrm{O} < \mathrm{N}$ : increasing first ionisation enthalpy.

The correct answer is:
$\mathrm{NH}_3 < \mathrm{PH}_3 < \mathrm{AsH}_3 < \mathrm{SbH}_3$ : increasing basic strength

Solution:
Correct order of increasing basic strength is $\mathrm{NH}_3>\mathrm{PH}_3>\mathrm{AsH}_3>\mathrm{SbH}_3>\mathrm{BiH}_3$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2012 (07 May Online)
In which of the following case, does the reaction go fartiest to completion if

Options:
A \( \mathrm{K}_{\mathrm{c}}=10^{2} \)
B \( \mathrm{K}_{\mathrm{c}}=10 \)
C \( \mathrm{K}_{\mathrm{c}}=10^{-2} \)
D \( \mathrm{K}_{\mathrm{c}}=1 \)

The correct answer is:
\( \mathrm{K}_{\mathrm{c}}=10^{2} \)

Solution:
Higher the value of equilibrium constant, higher the reaction rate.

Hence, KC = 102 It gives the fastest reaction among the all.

Topic: Chemistry, Chemical Equilibrium, JEE Main
In which of the following case, mice will remain alive in Griffith's transforming experiments?

Options:
A Heat killed S - strain injected into mice.
B DNA from heat killed S - strain + R - strain injected into mice.
C S - strain injected into mice.
D More than one option is correct

The correct answer is:
Heat killed S - strain injected into mice.

Solution:
Transforming Principle (by Frederick Griffith, 1928)
- Frederick Griffith has done experiments on bacteria Streptococcus pneumonia. This bacteria causes pneumonia.
- Griffith has observed the change in the physical form of the bacteria during the experiments.
- Streptococcus pneumonia produces 2 types of colonies :
1. Smooth and shiny colonies (S - strain or virulent strain) - Bacteria possess mucous
(polysaccharide) coat.
2. Rough colonies (R - strain) - Bacteria lacks mucous coat.

Step 1	S âstrain injected into the mice	Mice die

Step 2	R âstrain injected into the mice	Mice live

Step 3	Heat killed S âstrain injected into the mice	Mice live

Step 4	Heat killed S âstrain + R âstrain injected into the mice	Mice die

Topic: Biology, Molecular Basis of Inheritance, NEET
In which of the following column I is incorrectly matched with column II?

Options:
A

Column I (Cell wall of organism)	Column II (Composition)
Plants	Cellulose + hemicellulose

B

Column I (Cell wall of organism)	Column II (Composition)
Algae	Cellulose + Galactans

C

Column I (Cell wall of organism)	Column II (Composition)

Bacteria	Cellulose + Galactans

D

Column I (Cell wall of organism)	Column II (Composition)
Fungi	Chitin

The correct answer is:

Column I (Cell wall of organism)	Column II (Composition)

Bacteria	Cellulose + Galactans

Solution:
Organism	Composition

Bacteria	Peptidoglycans (N-acetyl muramic acid + N-acetyl glucosmaine)

Fungi	Chitin (also known as fungal cellulose)

Plants	Cellulose + Hemicellulose + Pectins + Proteins

Algae	Cellulose + Galactans + Mannans + Minerals like calcium carbonate

Topic: Biology, Cell - The Unit of Life, NEET
In which of the following compound transition metal has zero oxidation state

Options:
A $\mathrm{CrO}_5$
B $\mathrm{NH}_2 \cdot \mathrm{NH}_2$
C $\mathrm{NOClO}_4$
D $\left[\mathrm{Fe}(\mathrm{CO})_5\right]$

The correct answer is:
$\left[\mathrm{Fe}(\mathrm{CO})_5\right]$

Solution:
In $[F e(C O) 5]$, transition metal $F e$ has zero oxidation state.

Topic: Chemistry, Redox Reactions, JEE Main
In which of the following compounds B-F bond length is shortest?

Options:
A $\mathrm{BF}_{4^{-}}$
B $\mathrm{BF}_3 \leftarrow \mathrm{NH}_3$
C $\mathrm{BF}_3$
D $\mathrm{BF}_3 \leftarrow \mathrm{N}\left(\mathrm{CH}_3\right)_3$

The correct answer is:
$\mathrm{BF}_3$

Solution:
In other case $\mathrm{B}$ is $\mathrm{sp}^3$ hybridised (tetrahedral molecule or ion) and the possibility for $\pi$ bonding no longer exists.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In which of the following compounds deoxygenation is possible when heated with Zn

Options:
A A, B and C
B A, C and D
C B and D
D B and C

The correct answer is:
B and D

Solution:
The compound A is benzyl alcohol. It can not give the deoxygenation with zinc dust because there must be phenol. Deoxygenation s the loss of oxygen atom. So, the phenol can be converts into the benzene by the zinc dust.

Reaction :-

Hence, the compounds B and D deoxygenation is possible when heated with Zn.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In which of the following compounds does hydrogen bonding occur

Options:
A $\mathrm{SiH}_4$
B $L i H$
C $\mathrm{HI}$
D $\mathrm{NH}_3$

The correct answer is:
$\mathrm{NH}_3$

Solution:
Only $\mathrm{NH}_3$ forms $\mathrm{H}$-bonds

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In which of the following compounds hydrolysis takes place through SN1 and SN2 mechanisms reactivity?

Options:
A NF3,NCl3
B P4O10,SiCl4
C SF4,TeF6
D SiCl4,SiF4

The correct answer is:
NF3,NCl3

Solution:
In case of NF3,NCl3 hydrolysis takes place through SN1 and SN2 mechanisms reactivity respectively
NF3+H2O→SN1
NCl3+H2O→SN2
P4O10→SNAE
SiCl4→SN2
SF4,TeF6→ Both SN2
SiCl4,SiF4→ Both SN2

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
In which of the following condition, the genetic equilibrium will not be hampered for a species?

Options:
A Sexual selection
B Random mating
C Mutation
D Gene flow

The correct answer is:
Random mating

Solution:
Gene frequencies and genotype ratios in a randomly-breeding population remain constant from generation to generation.
This is known as the HardyâWeinberg law in honor of the two researchers who first realized the significance of the binomial expansion to population genetics and hence to evolution.
Evolution involves changes in the gene pool. A population in HardyâWeinberg equilibrium shows no change.
The mutation is any random sudden heritable change occurring in the genetic material. Sexual selection refers to the selection of a mate by an organism.

Topic: Biology, Evolution, NEET
In which of the following conditions, a reaction will definitely be spontaneous:

Options:
A Exothermic and increasing disorder
B Exothermic and decreasing disorder
C Endothermic and increasing disorder
D Endothermic and decreasing disorder

The correct answer is:
Exothermic and increasing disorder

Solution:
For a spontaneous reaction, ∆G<0.

As per the Gibbs-Helmholtz equation,

∆G=∆H-T∆S

When ∆H is negative, i.e., the reaction is exothermic and ∆S is positive, i.e., the disorder is increasing, the value of ∆G will definitely be negative and the process will be spontaneous for all temperatures.

Topic: Chemistry, Thermodynamics (C), JEE Main
In which of the following coordination entities the magnitude of $\Delta \mathrm{O}$ (CFSE in octahedral field) will be maximum (at. no. $\mathrm{Co}=27$ )?

Options:
A $\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$
B $\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$
C $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
D $\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$

The correct answer is:
$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$

Solution:
No Solution Available

Topic: Chemistry, Coordination Compounds, NEET, NEET 2008 (Mains)
In which of the following crystals alternate tetrahedral voids are occupied ?

Options:
A NaCl
B ZnS
C CaF2
D Na2O

The correct answer is:
ZnS

Solution:
In ZnS structure, sulphides occupy all the lattice points while Zn2+ ions are present in alternative tetrahedral voids.
NaCl has FCC structure and Na+ atom occupied all octahedral hole.
CaF2 has FCC structure and all T-holes will occupied by guest.
Na2O has FCC sturcture and all T-holes will occupied by guest.

Topic: Chemistry, Solid State, JEE Main
In which of the following devices the bending of bimetallic strip due to thermal expansion is not used?

Options:
A Refrigerator
B Fire alarm system
C Electric iron
D Telephone

The correct answer is:
Telephone

Solution:
The telephone needs no bimetallic strip for its working. A bimetallic strip works on the principle of thermal expansion and it is used in electrical devices to protect them through excess current flow.

Topic: Physics, Thermal Properties of Matter, JEE Main
In which of the following exothermic reactions, the heat liberated per mole is the highest ?

Options:
A
$\mathrm{CaO}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_2$

B
$\mathrm{SrO}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Sr}(\mathrm{OH})_2$

C
$\mathrm{BaO}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Ba}(\mathrm{OH})_2$

D
$\mathrm{MgO}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Mg}(\mathrm{OH})_2$

The correct answer is:
$\mathrm{CaO}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_2$

Solution:
No Solution Available

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2013 (25 Apr Online)
In which of the following ionization processes the bond energy has increased and also the magnetic behaviour has changed from paramagnetic to diamagnetic ?

Options:
A
$\mathrm{NO} \rightarrow \mathrm{NO}^{+}$

B
$\mathrm{N}_2 \rightarrow \mathrm{N}_2^{+}$

C
$\mathrm{C}_2 \rightarrow \mathrm{C}_2^{+}$

D
$\mathrm{O}_2 \rightarrow \mathrm{O}_2^{+}$

The correct answer is:
$\mathrm{NO} \rightarrow \mathrm{NO}^{+}$

Solution:
For $\mathrm{NO}$
Total no. of electrons $=15$
$\mathrm{B.O}=2.5$
Mag. Behaviour $=$ Paramagnetic
For $\mathrm{NO}^{+}$
Total no. of electrons $=14$
$\mathrm{B.O}$=3$
Mag. Behaviour $=$ Diamagnetic

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2013 (09 Apr Online)
In which of the following ionization processes, the bond order has increased and the magnetic behaviour has changed?

Options:
A
$\mathrm{C}_2 \longrightarrow \mathrm{C}_2^{+}$

B
$\mathrm{NO} \longrightarrow \mathrm{NO}^{+}$

C
$\mathrm{O}_2 \longrightarrow \mathrm{O}_2^{+}$

D
$\mathrm{N}_2 \longrightarrow \mathrm{N}_2^{+}$

The correct answer is:
$\mathrm{NO} \longrightarrow \mathrm{NO}^{+}$

Solution:
In $\mathrm{C}_2-\mathrm{C}_2{ }^{+}$electron is removed from bonding molecular orbital so bond order decreases. In $\mathrm{NO}$ $\longrightarrow \mathrm{NO}^{+}$, electron is removed from anti bonding molecular orbital so bond order increases and nature changes from paramagnetic to diamagnetic.
Hence, (B) is correct.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2007
In which of the following is sulphur a component?

Options:
A Vitamins
B Enzymes
C Cell walls
D None of the above

The correct answer is:
Vitamins

Solution:
Sulphur is a main constituent of several vitamins (thiamine, biotin, Coenzyme A).

Topic: Biology, Mineral Nutrition, NEET
In which of the following lanthanides, oxidation state \( +2 \) is the most stable?

Options:
A \( \mathrm{Ce} \)
B Eu
C \( \mathrm{Tb} \)
D Dy

The correct answer is:
Eu

Solution:
Europium is a lanthanide with an atomic number 63. Its electronic configuration is[Xe]4f76s2.

In its +2 Oxidation state, it will lose 2 electrons from its outermost shell, which gives the final oxidation state as [Xe]4f7. In this final oxidation state, Eu+2 possesses half-filled 4f- orbital which contributes to the stability of ion.

Topic: Chemistry, d and f Block Elements, JEE Main
In which of the following minerals, Al is present?
1. Fluorspar
2. Mica
3. Feldspar
4. Cryolite

Options:
A 1, 2, 3
B 1, 2, 4
C 2, 3, 4
D 1, 2, 3, 4

The correct answer is:
2, 3, 4

Solution:
Aluminium is not present in Fluorspar but present in all others.
Cryolite: Na3AlF6
Mica: KAlSi3O10(OH)2
Feldspar: KAlSi3O8
Fluorspar: CaF2 .

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
In which of the following Mn has highest oxidation state ?

Options:
A K2MnO4
B MnO2
C KMnO4
D Mn3O4

The correct answer is:
KMnO4

Solution:
It is +7 in KMnO4
In other:
K2MnO4 +6,
MnO2 +4,
Mn3O4 +8/3

Topic: Chemistry, d and f Block Elements, JEE Main
In which of the following molecular species σ -dative bond is present?

Options:
A BF 3
B Be2Cl4
C NH 3
D BH 3

The correct answer is:
Be2Cl4

Solution:
In Be2Cl4 octet of Be was not complete. So, it shows bridge bonding.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In which of the following molecules / ions $\mathrm{BF}_3$, $\mathrm{NO}_2^{-}, \mathrm{NH}_2^{-}$and $\mathrm{H}_2 \mathrm{O}$, the central atom is $\mathrm{sp}^2$ hybridized?

Options:
A $\mathrm{BF}_3$ and $\mathrm{NO}_2^{-}$
B $\mathrm{NO}_2^{-}$and $\mathrm{NH}_2$
C $\mathrm{NH}_2^{-}$and $\mathrm{H}_2 \mathrm{O}$,
D $\mathrm{NO}_2^{-}$and $\mathrm{H}_2 \mathrm{O}$

The correct answer is:
$\mathrm{BF}_3$ and $\mathrm{NO}_2^{-}$

Solution:
Both $\mathrm{BF}_3$ and $\mathrm{NO}_2^{-}$is $\mathrm{sp}^2$ hybridized.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2009 (Mains)
In which of the following molecules the central atom does not have $\mathrm{sp}^3$ hybridisation?

Options:
A $\mathrm{CH}_4$
B $\mathrm{SF}_4$
C $\mathrm{BF}_4^{-}$
D $\mathrm{NH}_4^{+}$

The correct answer is:
$\mathrm{SF}_4$

Solution:
When the number of hybrid orbitals, $H$ is 4 , the hybridisation is $\mathrm{sp}^3$.
$\mathrm{H}=\frac{1}{2}[\mathrm{~V}+\mathrm{M}-\mathrm{C}+\mathrm{A}]$
where, $V=$ number of valence electrons of central atom
$\begin{aligned}
M & =\text { number of monovalent atoms } \\
C & =\text { total positive charge } \\
A & =\text { negative charge }
\end{aligned}$
(a) For $\mathrm{CH}_4$,
$\mathrm{H}=\frac{1}{2}[4+4-0+0]=4 \text {, thus } \mathrm{sp}^3$
hybridisation
(b) For $\mathrm{SF}_4$,
$\mathrm{H}=\frac{1}{2}[6+4-0+0]=5 \text {, thus } \mathrm{sp}^3 \mathrm{~d}$
hybridisation
(c) For $\mathrm{BF}_4^{-}$,
$\mathrm{H}=\frac{1}{2}[3+4-0+1]=4, \text { thus } \mathrm{sp}^3$
hybridisation
(d) For $\mathrm{NH}_4^{+}$, $H=\frac{1}{2}[5+4-1+0]=4$, thus $s p^3$ hybridisation Thus, only in $\mathrm{SF}_4$, the central atom does not have $\mathrm{sp}^3$ hybridisation

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2010 (Mains)
In which of the following molecules, the number of possible ∠ XAX angles is maximum in the anionic part of their solid state? [A : Central atom; X : Surrounding atom]

Options:
A PBr5
B N2O5
C PCl5
D Cl2O6

The correct answer is:
PCl5

Solution:
PBr 5 (s)→ PBr 4 + +  Br −
N 2 O 5 (s)→ NO 2 +  +  NO 3 − ( sp 2 )
PCl 5 (s)→ PCl 4 +  +  PCl 6 − ( sp 3 d 2 )
Cl 2 O 6 (s)→ ClO 2 +  +  ClO 4 − ( sp 3 )

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In which of the following molecules/ions are all the bonds not equal?

Options:
A
$\mathrm{SF}_4$

B
$\mathrm{SiF}_4$

C
$\mathrm{XeF}_4$

D
$\mathrm{BF}_4^{-}$

The correct answer is:
$\mathrm{SF}_4$

Solution:
No Solution Available

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2006
In which of the following molecules/ions is the bond angle largest?

Options:
A H 2 O
B SCl2
C ICl 2 −
D SnCl2

The correct answer is:
ICl 2 −

Solution:
Here I is sp3d hybridized with structure.
The bond angle is 180° .

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In which of the following octahedral complex species the magnitude of $\Delta_0$ will be maximum?

Options:
A
$\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

B
$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$

C
$\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$

D
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$

The correct answer is:
$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$

Solution:
Crystal field splitting depends upon the nature of ligand. The nature of ligand $\Delta$ decreases as shown below
$\mathrm{C}_2 \mathrm{O}_4^{--} < \mathrm{H}_2 \mathrm{O} < \mathrm{NH}_3 < \mathrm{CN}^{-}$
hence the crystal field splitting will be maximum for $\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2013 (25 Apr Online)
In which of the following octahedral complexes of Co (at. no. 27), will the magnitude of $\Delta_{\circ}$ be the highest?

Options:
A
$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$

B
$\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$

C
$\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$

D
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$

The correct answer is:
$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$

Solution:
$\mathrm{CN}^{\Theta}$ is stronger ligand hence $\Delta_{\circ}$ is highest.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2008
In which of the following order the given complex ions are arranged correctly with respect to their decreasing spin only magnetic moment?
(i) FeF63-

(ii) CoNH363+

(iii) NiCl42-

(iv) CuNH342+

Options:
A ii>i>iii>iv
B iii>iv>ii>i
C i>iii>iv>ii
D ii>iii>i>iv

The correct answer is:
i>iii>iv>ii

Solution:
Complex
(i) FeF63-Fe3+⇒[Ar]3 d5, F- is WFL

n=5, μ=35 B.M

(ii)CoNH363+⇒Co3+Ar3 d6,NH3 is SFL

n=0, μ=0 B.M

(iii) NiCl42-Ni2+⇒Ar3 d8,Cl-→WFL

n=2, μ=8 B.M

(iv) CuNH342+Cu2+⇒Ar3 d9,NH3→SFL

n=1, μ=3 B.M

Thus correct order of spin only magnetic moment is

(i)>(iii)>(iv)>(ii)

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (25 Feb Shift 2)
In which of the following, ortho/para substitution by an electrophile is very facile?

Options:
A Nitrobenzene
B Phenol
C Benzoic acid
D Acetophenone

The correct answer is:
Phenol

Solution:
- OH group is ortho/para directing and makes the benzene ring more active towards electrophilic substitution.
Thus, ortho/para substitution in phenol by an electrophile is very facile.

Topic: Chemistry, General Organic Chemistry, JEE Main
In which of the following oxyanions the oxidation state of central atom is not same as that of its group number in periodic table?

Options:
A MnO4-
B Cr2O72-
C VO43-
D FeO42-

The correct answer is:
FeO42-

Solution:
A MnO4-=+7;Ar183d54s2=7
B Cr2O72-=+6;Ar183d54s1=6
C VO43-=+5;Ar183d34s2=5
D FeO42-=+6;Ar183d64s2=8

Topic: Chemistry, d and f Block Elements, JEE Main
In which of the following pairs A is more stable than B?

Options:
A
B
C
D

The correct answer is:

Solution:
Ph 3 C •  ​>  CH 3 C •
In Ph 3 C • e − undergoes in delocalization while CH 3 C • is stable due to the +I and hyperconjugation effect of − CH 3 group, because resonance is a predominant factor so the order is above one.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2014 (09 Apr Online)
In which of the following pairs A is more stable than B?

Options:
A
B
C
D

The correct answer is:

Solution:
Ph 3 C â¢ ââ>â CH 3 C â¢
In Ph 3 C â¢ e â undergo in delocalization while CH 3 C â¢ is stable due to the +I effect of − CH 3 group, because resonance is a predominant factor so order is above

Topic: Chemistry, General Organic Chemistry, JEE Main
In which of the following pairs, electron gain enthalpies of constituent elements are nearly the same or identical?

(A) Rb and Cs

(B) Na and K

(C) Ar and Kr

(D) I and At

Choose the correct answer from the options given below

Options:
A A & B only
B B & C only
C A & C only
D C & D only

The correct answer is:
A & C only

Solution:
Rb & Cs have nearly same electron gain enthalpy Their values are nearly =-46 kJ/mol

Ar & Kr also have same ΔHe.g.. Value is +96 kJ/mol.

Na and K have different values (-53 and -48 kJ/mol) . And I and At has values of -295 and -270 kJ/mol.

As we move down the group value of electron gain enthalpy decreases, with few exceptions.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2022 (28 Jul Shift 1)
In which of the following pairs molecules have bond order three and are isoelectronics

Options:
A $\mathrm{CN}^{-}, \mathrm{CO}$
B $\mathrm{NO}^{+}, \mathrm{CO}^{\circ}$
C $\mathrm{CN}^{-}, \mathrm{O}_2^{+}$
D $\mathrm{CO}, \mathrm{O}_2^{+}$

The correct answer is:
$\mathrm{CN}^{-}, \mathrm{CO}$

Solution:
$C N^{-}=14$ electrons ; $\mathrm{CO}=14$ electrons
B.O. $=\frac{1}{2}[10-4]=\frac{6}{2}=3$.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In which of the following pairs of molecules/ions, both the species are not likely to exist?

Options:
A H 2 2 + , He 2
B H 2 - , He 2 2 +
C H 2 + , He 2 2 -
D H 2 - , He 2 2 -

The correct answer is:
H 2 2 + , He 2

Solution:
(i) H 2 2 + ⇒ B . O = 0 He 2 ⇒ B . O = 2 - 2 2 = 0 }
(ii) H2-⇒B.O=2-12=0.5He22+⇒B.O=1
(iii) H 2 + = B . O = 1 - 0 2 = 0 . 5 He 2 2 - = B . O = 4 - 2 2 = 1 . 0
(iv) H 2 - ⇒ B . O = 2 - 1 2 = 0 . 5 He 2 2 - ⇒ B . O = 4 - 2 2 = 1 . 0
As both H22+ and He2 have zero bond order and hence do not exist.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2013 (07 Apr)
In which of the following pairs, the outermost electronic configuration will be the same?

Options:
A Cr+and Mn2+
B Ni2+ and Cu+
C Fe2+ and Co+
D V2+ and Cr+

The correct answer is:
Cr+and Mn2+

Solution:
Option 1: Mn+2Ar3 d5,Cr+Ar3 d5

Cr is expected to have the highest second ionization enthalpy as after removal of 4s electrons, the electrons will be removed from d-orbital. Cr+ have stable half-filled d-orbital with maximum spin multiplicity.

Option -2 Ni+2Ar3 d8,Cu+Ar3 d10

Option -3 Fe+2Ar3 d6,Co+Ar3 d74 s1

Option -4 V+2Ar3 d3,Cr+Ar3 d5

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (25 Feb Shift 1)
In which of the following pairs the two species are not isostructural?

Options:
A
$\mathrm{CO}_3^{2-}$ and $\mathrm{NO}_3^{-}$

B
$\mathrm{PCl}_4^{+}$and $\mathrm{SiCl}_4$

C
$\mathrm{PF}_5$ and $\mathrm{BrF}_5$

D
$\mathrm{AlF}_6^{3-}$ and $\mathrm{SF}_6$

The correct answer is:
$\mathrm{PF}_5$ and $\mathrm{BrF}_5$

Solution:
(1) $\mathrm{CO}_3^{2-}$ \& $\mathrm{NO}_3^{-} \rightarrow \mathrm{Sp}^2$ hybridized, Trigonal planar
(2) $\mathrm{PCl}_4^{+}$\& $\mathrm{SiCl}_4 \rightarrow \mathrm{Sp}^3$ hybridized, Tetrahedral
(3) $\mathrm{PF}_5 \rightarrow \mathrm{Sp}^3 \mathrm{d}$ hybridized, Trigonal bipyramidal
$\mathrm{BrF}_5 \rightarrow \mathrm{Sp}^3 \mathrm{d}^2$ hybridized, square pyramidal
(4) $\mathrm{AlF}_6^{3-}$ \& $\mathrm{SF}_6 \rightarrow \mathrm{Sp}^3 \mathrm{d}^2$ hybridized, octahedral

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2012 (Offline)