datavorous/entrance-exam-dataset huggingface dataset:
Which of the following cell organelles is responsible for the maintenance of turgidity of a plant cell?

Options:
A Mitochondria
B Vacuoles
C Plastids
D Ribosomes

The correct answer is:
Vacuoles

Solution:
Vacuoles are large organelles found in plant cells. Vacuoles have a membrane, tonoplast that surrounds it. This organelle help plant cell with storage of plant products and also water. Since it can store water, it provides structural support to the cell and also regulates the osmotic expansion and contraction.

Topic: Biology, Transport in Plants, NEET
Which of the following character is wrong regarding brown algae?

Options:
A Presence of chlorophyll 'a' and 'b'
B They remain attached to the surface by holdfast
C Presence of chlorophyll 'a' and 'c'
D Presence of fucoxanthin which gives brown colour

The correct answer is:
Presence of chlorophyll 'a' and 'b'

Solution:
Chlorophyll 'a' and 'b' are the main pigments of green algae, chlorophyll 'a', and chlorophyll 'c' is the main pigments present in brown algae. In addition, they have xanthophyll, fucoxanthin which gives them a brown color. Their body is differentiated into holdfast by which they remain attached to the surface, they have stem-like cylindrical stipe and leaf-like blade.

Topic: Biology, Plant Kingdom, NEET
Which of the following characteristics is mainly responsible for the diversification of insects on land?

Options:
A Segmentation
B Bilateral symmetry
C Exoskeleton
D Eyes

The correct answer is:
Exoskeleton

Solution:
Exoskeleton plays a role in defence from the prey and is also an important characteristic in the diversification of many species, e.g., insects having chitin as the form of an exoskeleton.

Topic: Biology, Animal Kingdom, NEET, NEET 2015 (Phase 1)
Which of the following characteristics is mainly responsible for the diversification of insects on land?

Options:
A Segmentation
B Bilateral symmetry
C Exoskeleton
D Eyes

The correct answer is:
Exoskeleton

Solution:
Exoskeleton plays a role in defense from the prey and is also an important characteristic in the diversification of many species, e.g., insects having chitin as the form of an exoskeleton.

Topic: Biology, Animal Kingdom, NEET
Which of the following characteristics of the transition metals is associated with their catalytic activity?

Options:
A Color of hydrated ions
B Diamagnetic behavior
C Paramagnetic behavior
D Variable oxidation

The correct answer is:
Variable oxidation

Solution:
Catalytic activity is the increase in the rate of a specified chemical reaction caused by catalyst under specified conditions. Catalysis at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst. This has the effect of increasing the concentration of the reactants at the catalyst surface. Also, because the transition metal ions can change their oxidation states, they become more effective as catalysts.

Transition elements exhibit variable valency to form intermediate compounds. Hence they act as a good catalyst. The catalytic activity of the transition metals and their compounds is described to their ability to adopt multiple oxidation states and their complexing ability.

Topic: Chemistry, d and f Block Elements, JEE Main
Which of the following chemical reactions depicts the oxidizing behaviour of $\mathrm{H}_2 \mathrm{SO}_4$ ?

Options:
A
$2 \mathrm{HI}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{I}_2+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}$

B
$\mathrm{Ca}(\mathrm{OH})_2+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{CaSO}_4+2 \mathrm{H}_2 \mathrm{O}$

C
$\mathrm{NaCl}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{NaHSO}_4+\mathrm{HCl}$

D
$2 \mathrm{PCl}_5+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow 2 \mathrm{POCl}_3+2 \mathrm{HCl}+\mathrm{SO}_2 \mathrm{Cl}_2$

The correct answer is:
$2 \mathrm{HI}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{I}_2+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}$

Solution:
No Solution Available

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2006
Which of the following chemical reactions represents Hall-Heroult Process?

Options:
A Cr2O3+2Al→Al2O3+2Cr
B 2Al2O3+3C→4Al+3CO2
C FeO+CO→Fe+CO2
D 2AuCN2-aq+Zns→2Aus+ZnCN42-

The correct answer is:
2Al2O3+3C→4Al+3CO2

Solution:
The Hall–Héroult process is the major industrial process for smelting aluminium. It involves dissolving aluminium oxide (alumina) (obtained most often from bauxite, aluminium's chief ore, through the Bayer process) in molten cryolite, and electrolyzing the molten salt bath, typically in a purpose-built cell.

In this process carbon electrodes are used.

The following reactions take place

2Al2O3+3C→4Al+3CO2

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2022 (24 Jun Shift 2)
Which of the following chemicals can be added for sweetening of food items at cooking temperature and does not provide calories?

Options:
A Sucrose
B Glucose
C Aspartame
D Sucralose

The correct answer is:
Sucralose

Solution:
Sucralose is a sweetening agent with zero calorific value and can be used at cooking temperature. It is a derivative of sucrose (Aspartame is also sweetening agent but is decomposed at cooking temperature).

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Which of the following circuits correctly represents the following truth table ?

Options:
A

B

C

D

The correct answer is:

Solution:
For circuit 1
$A \cdot B=\overline{Y+\bar{A}}=C$
A B $\overline{Y+\bar{A}}=C$
$\begin{array}{lllll}0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 & 0\end{array}$

Topic: Physics, Semiconductors, JEE Main, JEE Main 2013 (25 Apr Online)
Which of the following circuits is reverse - biased ?

Options:
A
B
C
D

The correct answer is:

Solution:
Reverse biasing in a PN-junction occurs when an external voltage is applied in such a way that the positive terminal of the voltage source is connected to the N-type semiconductor, and the negative terminal is connected to the P-type semiconductor.

In other words, if the potential of the P-side is less than that of the N-side of the junction, the diode is said to be in the reverse bias condition.

Amongst the given configurations, this is the correct one which indicates the reverse biasing of the diode.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Which of the following circular rods, (given radius $r$ and length $l$ ) each made of the same material and whose ends are maintained at the same temperature will conduct most heat?

Options:
A $r=2 r_0 ; l=2 l_0$
B $r=2 r_0 ; l=l_0$
C $r=r_0 ; l=l_0$
D $r=r_0 ; l=2 l_0$

The correct answer is:
$r=2 r_0 ; l=l_0$

Solution:
Heat conduction through a rod is given by
$H=\frac{\Delta Q}{\Delta t}=K A\left(\frac{T_1-T_2}{l}\right)$
$\Rightarrow \quad H \propto \frac{r^2}{l}$ ...(i)
(a) When $r=2 r_0 ; l=2 l_0$
$\begin{aligned} & H \propto \frac{\left(2 r_0\right)^2}{2 l_0} \\ \Rightarrow \quad & H \propto \frac{2 r_0{ }^2}{l_0}\end{aligned}$
(b) When $r=2 r_0 ; l=l_0$
$\begin{aligned} & H \propto \frac{\left(2 r_0\right)^2}{l_0} \\ \Rightarrow \quad & H \propto \frac{4 r_0^2}{l_0}\end{aligned}$
(c) When $r=r_0 ; l=l_0$
$H \propto \frac{r_0^2}{l_0}$
(d) When $r=r_0 ; l=2 l_0$
$H \propto \frac{r_0^2}{2 l_0}$
It is obvious that heat conduction will be more in case (b).

Topic: Physics, Thermal Properties of Matter, JIPMER, JIPMER 2008
Which of the following circular rods, (given radius $r$ and length $l$) each made of the same material and whose ends are maintained at the same temperature will conduct most heat?

Options:
A $r=2 r_0, l=2 l_0$
B $r=2 r_0, l=l_0$
C $r=r_0, l=l_0$
D $r=r_0, l=2 l_0$

The correct answer is:
$r=2 r_0, l=l_0$

Solution:
Heat conduction through a rod is rate of change of heat $\left(\frac{\Delta Q}{\Delta t}\right)$.
$\therefore H=\frac{\Delta Q}{\Delta t}=K A\left(\frac{T_1-T_2}{l}\right)$
$\Rightarrow H \propto \frac{r^2}{l}...(i)$
(a) When
$\begin{aligned} r & =2 r_0, l=2 l_0 \\ H & \propto \frac{\left(2 r_0\right)^2}{2l_0} \Rightarrow H \propto \frac{2 r_0^2}{l_0}\end{aligned}$
(b) When
$\begin{aligned} r & =2 r_0, l=l_0 \\ H & \propto \frac{\left(2 r_0\right)^2}{ l_0} \Rightarrow H \propto \frac{4 r_0^2}{l_0}\end{aligned}$
(c) When
$\begin{gathered}r=r_0, l=l_0 \\ H \propto \frac{r_0^2}{l_0}\end{gathered}$
(d) When
$\begin{aligned} r & =r_0, l=2 l_0 \\ H & \propto \frac{r_0^2}{2 l_0}\end{aligned}$
It is obvious that heat conduction will be more in case (b).

Topic: Physics, Thermal Properties of Matter, JIPMER, JIPMER 2007
Which of the following colligative properties can provide molar mass of proteins (or polymers or colloids) with greater precision

Options:
A Relative lowering of vapour pressure
B Elevation of boiling point
C Depression in freezing point
D Osmotic pressure

The correct answer is:
Osmotic pressure

Solution:
Osmotic pressure method is especially suitable for the determination of molecular masses of macromolecules such as protein \& polymer because for these substances the value of other colligative properties such as elevation in boiling point or depression in freezing point are too small to be measured on the other hand osmotic pressure of such substances are measurable.

Topic: Chemistry, Solutions, JEE Main
Which of the following combination of reagents does not undergo redox reaction in aqueous medium?

Options:
A SnCl2+HgCl2
B CuSO4+KCN
C Pb(CH3COO)2+KI
D Ag2O+SO2

The correct answer is:
Pb(CH3COO)2+KI

Solution:
Here the reaction between Pb(CH3COO)2+KI is not a redox reaction while rest other are redox reactions
Pb(CH 3 COO) 2 +KI → Non-redox reaction Ion exchange reaction PbI 2 ↓+  CH 3 COOK
SnCl 2 + HgCl 2 → Redox Hg↓ +  SnCl 4
CuSO 4 +KCN→ K 2 SO 4 + Cu(CN) 2 ↓ → Intramolecular redox Spontaneous (R.T.) Cu(CN)↓ +  (CN) 2 ↑
Ag 2 O +  SO 2 → Rediox Ag +  SO 3 ↑

Topic: Chemistry, Redox Reactions, JEE Main
Which of the following combination of statements is true regarding the interpretation of the atomic orbitals?
(A) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum.
(B) For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number.
(C) According to wave mechanics, the ground state angular momentum is equal to h2π .
(D) The plot of ψ Vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value.

Options:
A (B),Â (C)
B (A),Â (B)
C (A), (C)
D (A), (D)

The correct answer is:
(A), (D)

Solution:
Statements (B) and (C) are incorrect.
For a given value of the principal quantum number, the size of the orbit is directly proportional to the azimuthal quantum number and not inversely proportional.
The formula for angular momentum is h2πl(l+1)
If the electron is in ground state, n will be 1 and the only azimuthal quantum number value can be 0. If l = 0, angular quantum momentum will also be 0. The plot of ψ Vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2019 (09 Jan Shift 2)
Which of the following combination will produce H2 gas?

Options:
A Fe metal and conc. HNO3
B Cu metal and conc. HNO3
C Au metal and NaCN(aq) in the presence of air
D Zn metal and NaOH(aq)

The correct answer is:
Zn metal and NaOH(aq)

Solution:
Fe+conc.HNO3→Passivity
Cu+conc.HNO3→NO2
Au+NaCN+O2→[Au(CN)2]-
Zn+NaOH→Na2ZnO2+H2

Topic: Chemistry, s Block Elements, JEE Main
Which of the following combinations has the dimension of electrical resistance (ϵ0 is the permittivity of vacuum and μ0 is the permeability of vacuum)?

Options:
A ϵ0μ0
B ϵ0μ0
C μ0ϵ0
D μ0ϵ0

The correct answer is:
μ0ϵ0

Solution:
Resistance=VoltAmpere=ML2T-3A-1A=ML2T-3A-2
We know, μ0=MLT-2A-2    ε0=M-1L-3T4A2
Hence,
μ0ε0=μ0ε0=MLT-2A-2M-1L-3T4A21/2
=M2L4T-6A-41/2=ML2T-3A-2

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2019 (12 Apr Shift 1)
Which of the following combinations is required for chemiosmosis?

Options:
A proton pump, electron gradient, NADP synthase
B membrane, proton pump, proton gradient, ATP synthase
C membrane, proton pump, proton gradient, NADP synthase
D proton pump, electron gradient, ATP synthase

The correct answer is:
membrane, proton pump, proton gradient, ATP synthase

Solution:
ATP synthesis according to chemiosmosis hypothesis takes place in both chloroplasts (photophosphorylation) and mitochondria (oxidative phosphorylation). Both these processes need the involvement of:

A membrane (thylakoid membrane or mitochondrial membrane, respectively).

Proton pumps to maintain the pH.

Proton gradient to move the proton across the membrane.

ATP synthase enzyme which is present in F0-F1 particles in mitochondria or CF0-CF1 particles in plastids.

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2023 (All India)
Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication?

Options:
A \( R=20 \Omega, L=1.5 \mathrm{H}, C=35 \mu \mathrm{F} \)
B \( R=25 \Omega, L=2.5 H, C=45 \mu F \)
C \( R=15 \Omega, L=3.5 H, C=30 \mu F \)
D \( R=25 \Omega, L=1.5 \mathrm{H}, C=45 \mu \mathrm{F} \)

The correct answer is:
\( R=15 \Omega, L=3.5 H, C=30 \mu F \)

Solution:
Better tuning means low bandwidth =RL

Topic: Physics, Alternating Current, JEE Main
Which of the following combinations would not result in the formation of a buffer solution?

Options:
A $\mathrm{NH}_{3}+\mathrm{HCl}$
B $\mathrm{NH}_{4} \mathrm{Cl}+\mathrm{NH}_{3}$
C $\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaCl}$
D $\mathrm{NaOH}+\mathrm{CH}_{3} \mathrm{COOH}$

The correct answer is:
$\mathrm{NaOH}+\mathrm{CH}_{3} \mathrm{COOH}$

Solution:
Combination of $\mathrm{NaOH}$ and $\mathrm{CH}_{3} \mathrm{COOH}$ is the mixture of alkali and acetic acid. Therefore this combination can not be buffer forming solution.

Topic: Chemistry, Ionic Equilibrium, JEE Main
Which of the following complex compounds will exhibit highest paramagnetic behaviour?
(At. no. : $\mathrm{Ti}=22, \mathrm{Cr}=24$, $\mathrm{Co}=27, \mathrm{Zn}=30$ )

Options:
A $\left[\mathrm{Ti}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
B $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
C $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
D $\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_6\right]^{2+}$

The correct answer is:
$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$

Solution:
(A) Electronic configuration of $\mathrm{Ti}^{3+}$ in $\left[\mathrm{Ti}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
(B) Electronic configuration of $\mathrm{Cr}^{3+}$ in $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
(C) Electronic configuration of $\mathrm{Co}^{3+}$ in $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ :
In the presence of strong field ligand $\mathrm{NH}_3$, pairing of electrons takes place and thus, octahedral complex, $\quad\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ is diamagnetic.
$$
\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}
$$
$$
\left[\begin{array}{l}
\text { inner orbital or } \\
\text { low spin complex }
\end{array}\right]\left(6 \mathrm{NH}_3 \text { molecules }\right)
$$
(D) Electronic configuration of $\mathrm{Zn}^{2+}$ in $\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_6\right]^{2+}$ :
$\therefore\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_6\right]^{2+}$ is an outer orbital complex and is diamagnetic.

Topic: Chemistry, Coordination Compounds, NEET, NEET 2011 (Mains)
Which of the following complex has the highest value of Δ t

Options:
A CoCl42-
B CoBr 4 2−
C CoI42-
D CoNCS42-

The correct answer is:
CoNCS42-

Solution:
NCS- is the strongest ligand among Cl − , Br − and I −

Topic: Chemistry, Coordination Compounds, JEE Main
Which of the following complex has zero magnetic moment (spin only)?

Options:
A NiNH36Cl2
B Na3[FeF6]
C CrH2O6SO4
D K4[FeCN6]

The correct answer is:
K4[FeCN6]

Solution:
Ferrocyanide ion FeCN64- is diamagnetic in nature hence K4[FeCN6] complex has zero magnetic moment.
Fe atom 3d 6   4s 2   4p 0
Fe (II)
d 2   sp 3 hybridisation

Topic: Chemistry, Coordination Compounds, JEE Main
Which of the following complex ions has electrons that are symmetrically filled in both t2g and eg orbitals?

Options:
A CoNH362+
B MnCN64-
C CoF63-
D FeF63-

The correct answer is:
FeF63-

Solution:
Metal ion having either d5 or d10 configuration can be filled symmetrically in both t2g and eg orbitals. Co+2 has d7 configuration, Co+3 has d6 configuration. Mn+2 has d5 configuration but CN- is strong field ligand and it will pair the electrons due to which electronic configuration is t2g5eg0.

As F is weak field ligand hence back pairing of electrons do not take place or in case of weak field ligand pairing energy of electrons (P )is greater than crystal field splitting energy in octahedral complex ∆o i.e.P>∆o , hence electrons will go in eg orbitals, instead of back paired in t2g orbitals.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2015 (11 Apr Online)
Which of the following complex ions is expected to absorb visible light ?
(At no. $\mathrm{Zn}=30, \mathrm{Sc}=21, \mathrm{Ti}=22, \mathrm{Cr}=24$ )

Options:
A $\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_6\right]^{2+}$
B $\left[\mathrm{Sc}\left(\mathrm{H}_2 \mathrm{O}\right)_3\left(\mathrm{NH}_3\right)_3\right]^{3+}$
C $\left[\mathrm{Ti}(\mathrm{en})_2\left(\mathrm{NH}_3\right)_2\right]^{4+}$
D $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$

The correct answer is:
$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$

Solution:
$\mathrm{Cr}^{3+} \rightarrow \mathrm{Is}^2, 2 \mathrm{~s}^2 2 \mathrm{p}^6, 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^3$
Unpaired electron shows colour so absorb visible light.

Topic: Chemistry, d and f Block Elements, NEET, NEET 2009 (Mains)
Which of the following complex ions will exhibit optical isomerism? (en = 1, 2-diamine ethane).

Options:
A
$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]^{+}$

B
$\left[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2\right]^{+}$

C
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^{+}$

D
$\left[\mathrm{Zn}(\mathrm{en})_2\right]^{2+}$

The correct answer is:
$\left[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2\right]^{+}$

Solution:
Non superimposable mirror images are called optical isomers and may be described as "chiral'. They are also called enantiomers and rotate plane polarised light in opposite directions.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2012 (12 May Online)
Which of the following complex is homoleptic?

Options:
A Ni(CN)42-
B NiNH32Cl2
C FeNH34Cl2+
D CoNH34Cl2+

The correct answer is:
Ni(CN)42-

Solution:
Homoleptic complexes are compounds in which all the ligands bound to the central metal atom or ion are identical.

Homoleptic complexes have only one type of ligand.

Ni(CN)42- has only one type of ligand.

Hence, option A is the answer.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (01 Feb Shift 1)
Which of the following complex is octahedral, diamagnetic and the most stable?

Options:
A Na3[CoCl6]
B NiNH36Cl2
C K3Co(CN)6
D CoH2O6Cl2

The correct answer is:
K3Co(CN)6

Solution:
The coordination compound K3[Co(CN)6]contains a cobalt(III) ion, which has a d6 electronic configuration. The cyanide ligands in this complex are indeed strong field ligands, which means they will cause a large splitting of the d orbitals. So here cobalt is in +3 state having 3d6 configuration and also cyanide ligand which is a strong field ligand so, it is most stable and will have octahedral shape. The given compounds have the following nature:

(1) Na3[CoCl6] – Paramagnetic

(2) NiNH36Cl2 – Paramagnetic

(3) K3Co(CN)6– Diamagnetic

(4) CoH2O6Cl2 – Paramagnetic

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Which of the following complex species is not expected to exhibit optical isomerism?

Options:
A Co NH 3 3 Cl 3
B Co en NH 3 2 Cl 2 +
C Co en 3 3 +
D Co en 2 Cl 2 +

The correct answer is:
Co NH 3 3 Cl 3

Solution:
Co NH 3 3 Cl 3
MA 3 B 3 type compounds exhibit facial as well as meridional isomerism. Since both isomers have a plane of symmetry they do not exhibit optical isomerism.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2013 (07 Apr)
Which of the following complex species is not expected to exhibit optical isomerism?

Options:
A Co NH 3 3 Cl 3
B Co en NH 3 2 Cl 2 +
C Co en 3 3 +
D Co en 2 Cl 2 +

The correct answer is:
Co NH 3 3 Cl 3

Solution:
Co NH 3 3 Cl 3
MA 3 B 3 type compounds exhibit facial as well as meridional isomerism. Since both isomers have a plane of symmetry they do not exhibit optical isomerism.

Topic: Chemistry, Coordination Compounds, JEE Main
Which of the following complex will show geometrical as well as optical isomerism (en = ethylene diamine)

Options:
A $\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2$
B $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right) \mathrm{Cl}_4\right]$
C $\left[P t(e n)_3\right]^{4+}$
D $\left[\mathrm{Pt}(e n)_2\right] \mathrm{Cl}_2$

The correct answer is:
$\left[\mathrm{Pt}(e n)_2\right] \mathrm{Cl}_2$

Solution:
$\left[\mathrm{Pt}(e n)_2\right] \mathrm{Cl}_2$

Topic: Chemistry, Coordination Compounds, JEE Main
Which of the following complexes exhibits the highest paramagnetic behaviour?
Where gly = glycine, en = ethylenediamine and bpy = bipyridyl moities). (At. number $\mathrm{Ti}$ $=22, \mathrm{~V}=23, \mathrm{Fe}=26, \mathrm{Co}=27$ )

Options:
A $\left[\mathrm{Ti}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
B $\left[\mathrm{V} \text { (gly) } 2(\mathrm{OH})_2\left(\mathrm{NH}_3\right)_2\right]^{+}$
C $\left[\mathrm{Fe}(\mathrm{en})(\mathrm{bpy})\left(\mathrm{NH}_3\right)_2\right]^{2+}$
D $\left[\mathrm{Co}(\mathrm{OX})_2(\mathrm{OH})_2\right]^{-}$

The correct answer is:
$\left[\mathrm{Co}(\mathrm{OX})_2(\mathrm{OH})_2\right]^{-}$

Solution:
No Solution Available

Topic: Chemistry, Coordination Compounds, NEET, NEET 2008 (Mains)
Which of the following complexes has a square planar geometry

Options:
A $\mathrm{Ag}\left(\mathrm{NH}_3\right)_2^{+}$
B $\mathrm{Cu}(\mathrm{en})_{2+}^{2+}$
C $\left[\mathrm{MnCl}_4\right]^{2-}$
D $\mathrm{Ni}(\mathrm{CO})_4$

The correct answer is:
$\mathrm{Cu}(\mathrm{en})_{2+}^{2+}$

Solution:
Copper complexes usually involve co-ordination number of four and are thus square planar in shape.

Topic: Chemistry, Coordination Compounds, JEE Main
Which of the following complexes has the highest stability constant at 298 K

Options:
A $\left[\mathrm{CdCl}_4\right]^{2-}$
B $\left[\mathrm{CdBr}_4\right]^{2-}$
C $\left[\mathrm{Cdl}_4\right]^{2-}$
D $\left[\mathrm{Cd}(\mathrm{CN})_4\right]^{2-}$

The correct answer is:
$\left[\mathrm{Cd}(\mathrm{CN})_4\right]^{2-}$

Solution:
The cyano and ammine complexes are far more stable than those formed by halide ions. This is due to the fact that $\mathrm{NH}_3$ and $\mathrm{CN}^{-}$are strong lewis bases.

Topic: Chemistry, Coordination Compounds, JEE Main
Which of the following complexes show six coordination number

Options:
A $\left[\mathrm{Zn}(\mathrm{CN})_4\right]^{2-}$
B $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$
C $\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{2-}$
D $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_4\right]^{2+}$

The correct answer is:
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$

Solution:
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$

Topic: Chemistry, Coordination Compounds, JEE Main
Which of the following complexes will exhibit maximum attraction to an applied magnetic field?

Options:
A [Ni(H2O)6] 2+
B [Co(en)3] 3+
C [Zn(H2O)6] 2+
D [Co(H2O)6] 2+

The correct answer is:
[Co(H2O)6] 2+

Solution:
Complex having maximum number of unpaired electrons will exhibit maximum attraction to applied magnetic field.

ZnH2O62+ →d10 system, t62g eg4, 0 unpaired electrons
CoH2O62+→d7 system, t52g eg2, 3 unpaired electrons
Co(en)33+→d6 system, t62g eg0, 0 unpaired electrons
NiH2O62+→d8 system, t62g eg2, 2 unpaired electrons.

So Option D is correct.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (13 Apr Shift 2)