datavorous/entrance-exam-dataset huggingface dataset:
Which of the following complex ions absorbs the light of minimum wavelength?

Options:
A CoH2O63+
B CoF63-
C CoCN63-
D CoNH363+

The correct answer is:
CoCN63-

Solution:
Wavelength of light absorb â1stability of complex
C N â is strong field ligand
So complex is more stable

Topic: Chemistry, Coordination Compounds, NEET
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 $\left[\mathrm{Sc}\left(\mathrm{H}_2 \mathrm{O}\right)_3\left(\mathrm{NH}_3\right)_3\right]^{3+}$
B $\left[\mathrm{Ti}(\mathrm{en})_2\left(\mathrm{NH}_3\right)_2\right]^{4+}$
C $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
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:
Key Idea Only those transition metal complexes are expected to absorb visible light, in which d-subshell is incomplete (ie, has unpaired electron) and excitation of electron from a lower energy arbital to higher energy orbital is possible.
(a) In $\left[\mathrm{Sc}\left(\mathrm{H}_2 \mathrm{O}\right)_3\left(\mathrm{NH}_3\right)_3\right]^{3+}, \mathrm{Sc}$ is present as $\mathrm{Sc}^{3+}$.
$$
\mathrm{Sc}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^0, 4 \mathrm{~s}^0
$$
Since, in this complex excitation of electron is not possible, it will not absorb visible light.
(b) In $\left[\mathrm{Ti}(\mathrm{en})_2\left(\mathrm{NH}_3\right)_2\right]^{4+}, \mathrm{Ti}$ is present as $\mathrm{Ti}^{4+}$.
$$
\mathrm{Ti}^{4+}=[\mathrm{Ar}] 3 \mathrm{~d}^0, 4 \mathrm{~s}^0
$$
Hence, it will not absorb visible light.
(c) In $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}, \mathrm{Cr}$ is present as $\mathrm{Cr}^{3+}$.
$$
\mathrm{Cr}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^3, 4 \mathrm{~s}^0
$$
Since, this complex has three unpaired electrons, excitation of electrons is possible and thus, it is expected that this complex will absorb visible light.
(d) In $\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_6\right]^{2+}, \mathrm{Zn}$ is present as $\mathrm{Zn}^{2+}$.
$$
\begin{gathered}
\mathrm{Zn}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{10}, 4 \mathrm{~s}^0 \\
{\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_6\right]^{2+}=[\mathrm{Ar}] 3 d^{10}}
\end{gathered}
$$
Hence, this complex will not absorb visible light.

Topic: Chemistry, Coordination Compounds, NEET, NEET 2009 (Screening)
Which of the following complex ions is not expected to absorb visible light?

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

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

Solution:
Key Idea For the absorption of visible light, presence of unpaired d-electrons is the necessity.
(a) In $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}, \mathrm{Ni}$ is present as $\mathrm{Ni}^{2+}$.
$$
\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0
$$
(Pairing occurs because $\mathrm{CN}^{-}$is a strong field ligand).
Since, in $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$, no unpaired electron is present in d-orbitals, it does not absorb visible light.
(b) In $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}, \mathrm{Cr}$ is present as $\mathrm{Cr}^{3+}$. $\mathrm{Cr}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^3 4 \mathrm{~s}^0$ (Three unpaired electrons)
(c) In $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}, \mathrm{Fe}$ is present as $\mathrm{Fe}^{2+}$.
$\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^6 4 \mathrm{~s}^0$ (Four unpaired electrons)
(d) In $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$, $\mathrm{Ni}$ is present as $\mathrm{Ni}^{2+}$.
$\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0$ (Two unpaired electrons)
The complexes given in option (b), (c), (d) have unpaired electrons, thus absorb visible light.

Topic: Chemistry, d and f Block Elements, NEET, NEET 2010 (Screening)
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, NEET
Which of the following complex will give white precipitate with barium chloride solution?

Options:
A [Cr(NH3)5Cl]SO4
B [Cr(NH3)SO4]Cl
C [Co(NH3)6]Br3
D None of these

The correct answer is:
[Cr(NH3)5Cl]SO4

Solution:
[Cr(NH3)5Cl]SO4 ionise as [Cr(NH3)5Cl]2+ and SO42- and its aqueous solution will give test for SO42- ions.
BaCl2(aq)+SO42-→BaSO4+2Cl-

Topic: Chemistry, Coordination Compounds, NEET
Which of the following complexes are not correctly matched with hybridisation of their central metal ion?

1. [NiCO4] ; sp3

2. NiCO42- ; sp3

3. CoF63- ; d2sp3

4. FeCN63- ; sp3d2

Select the correct answer using the codes given below

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

The correct answer is:
2, 3 and 4

Solution:
Apart from [Ni(CO)4], all other options have wrong hybridizations. Nickel has the configuration of [Ar]3d84s2. Now,CO is a strong field ligand, so it will pair the electrons to give [Ar]3d104s0. Since inner d orbitals are not available, 4s and 4p orbitals will be used to give sp3 hybridized orbitals. So, the geometry will be tetrahedral
Complex Hybridisation
[NiCO4] sp3
NiCO42- dsp2
CoF63- sp3d2
FeCN63- d2sp3

Topic: Chemistry, Coordination Compounds, NEET
Which of the following complexes exhibits the highest paramagnetic behaviour?
where gly = glycine, en = ethylenediamine and bpy = bipyridyl moities
(At no: $\mathrm{Ti}=22, \mathrm{~V}=23, \mathrm{Fe}=26, \mathrm{Co}=27$ )

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

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

Solution:
Key Idea : Greater is the number of unpaired electrons, larger is the paramagnetism.
$\begin{aligned}
& {\left[\mathrm{V}(\mathrm{gly})_2(\mathrm{OH})_2\left(\mathrm{NH}_3\right)_2\right]^{+}} \\
& \mathrm{V}_{23}=[\mathrm{Ar}] 4 s^2, 3 d^3
\end{aligned}$
Oxidation state of $\mathrm{V}$ in $\left[\mathrm{V}(\mathrm{gly})_2(\mathrm{OH})_2\left(\mathrm{NH}_3\right)_2\right]^{+}$ is
$\begin{gathered}
x+(-1) \times 2+(-1) \times 2+(0) \times 2=+1 \\
x=+5 \\
\mathrm{~V}^{5+}=[\mathrm{Ar}] 3 d^0
\end{gathered}$
(No unpaired electron in option (A))
So, option (B),
$\begin{aligned}
& {\left[\mathrm{Fe}\left(\text { en) }(\text { bpy })\left(\mathrm{NH}_3\right)_2\right]^{2+}\right.} \\
& \mathrm{Fe}_{26}=[\mathrm{Ar}] 4 s^2, 3 d^6
\end{aligned}$
Oxidation state of $\mathrm{Fe}$ in $\left[\mathrm{Fe}(\mathrm{en}) \text { (bpy) }\left(\mathrm{NH}_3\right)_2\right]^{2+}$ is
$\begin{gathered}
x+(0)+(0)+(0) \times 2=+2 \\
x=+2 \\
\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 d^6
\end{gathered}$
But en, bpy and $\mathrm{NH}_3$ all are strong field ligands so pairing occurs, thus no unpaired electrons.
So, option (C),
$\begin{gathered}
{\left[\mathrm{Co}(\mathrm{OX})_2(\mathrm{OH}]_2\right]^{-}} \\
\mathrm{Co}_{27}=[\mathrm{Ar}] 4 s^2, 3 d^7
\end{gathered}$
Oxidation state of $\mathrm{Co}$ in $\left[\mathrm{Co}(\mathrm{OX})_2(\mathrm{OH})_2\right]^{-}$is
$x+(-2) \times 2+(-1) \times 2=-1$
$\begin{aligned}
& x-6=-1 \\
& x=+5 \\
& \mathrm{Co}^{5+}=[\mathrm{Ar}] 3 d^4 \\
& \text{(4 paired electrons)} \\
& \text{Now, option (D),} \\
& {\left[\mathrm{Ti}\left(\mathrm{NH}_3\right)_6\right]^{3+}} \\
& \mathrm{Ti}_{22}=[\mathrm{Ar}] 4 s^2, 3 d^2
\end{aligned}$
$\mathrm{Ti}_{22}=[\mathrm{Ar}] 4 s^2, 3 d^2$
Oxidation state of Ti in $\left[\mathrm{Ti}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ is +3, thus it contains 1 unpaired electron.
Hence, $\quad\left[\mathrm{Co}(\mathrm{OX})_2(\mathrm{OH})_2\right]^{-}$has highest paramagnetic behaviour.

Topic: Chemistry, Coordination Compounds, NEET, NEET 2008 (Screening)
Which of the following complexes has minimum magnitude of Δ0 ?

Options:
A [Cr(CN)6]3-
B [Co(NH3)6]3+
C [CoCl6]3-
D [Cr(H2O)6]3+

The correct answer is:
[CoCl6]3-

Solution:
[Cr(CN)6]3-
Cr3+:d3:t32ge0g=-1.2 Î0
[Co(NH3)6]3+
Co3+:d6:t62ge0g=-2.4 Î0
[CoCl6]3-
Co3+:d6:t42ge2g=-0.4 Î0
[Cr(H2O)6]3+
Cr3+:d3:t32ge0g=-1.2 Î0
The minimum magnitude of Î0 is for [CoCl6]3-

Topic: Chemistry, Coordination Compounds, NEET
Which of the following complexes is used as an anticancer agent?

Options:
A mer-CoNH33Cl
B Cis-Pt Cl2NH32
C cis-K2PtÂ Cl2Â Br2
D Na2CoCl4

The correct answer is:
Cis-Pt Cl2NH32

Solution:
Cis-Pt Cl2NH32 known as cis platin is used as an anticancer agent.

Topic: Chemistry, Coordination Compounds, NEET, NEET 2014
Which of the following complexes show Geometric Isomerism in Coordination compounds?

Options:
A Homoleptic Complexes
B Heteroleptic Complexes
C Both of them
D None of them

The correct answer is:
Heteroleptic Complexes

Solution:
Correct Option is : (B)
Heteroleptic Complexes

Topic: Chemistry, Coordination Compounds, NEET
Which of the following components of phosphorylation are not found in the stroma lamellae?

Options:
A PS II complex
B PS I complex
C NADP reductase
D Both (A) and (C)

The correct answer is:
Both (A) and (C)

Solution:
When only PS I is functional, the electron is circulated within the photosystem and the phosphorylation occurs due to cyclic flow of electrons. A possible location where this could be happening is in the stroma lamellae. While the membrane or lamellae of the grana have both PS I and PS II the stroma lamellae membranes lack PS II as well as NADP reductase enzyme. The excited electron does not pass on to NADP+ but is cycled back to the PS I complex through the electron transport chain. The cyclic flow hence results only in the synthesis of ATP, but not of NADPH + H+.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Which of the following components provides sticky character to the bacterial cell?

Options:
A A layer made up of murein
B A fluid mosaic layer made up of carbohydrates
C A phospholipid layer
D A rich glycoprotein layer

The correct answer is:
A rich glycoprotein layer

Solution:
Sticky character of the bacteria is due to glycocalyx or slime layer. This layer is rich in glycoproteins. The capsule acts as a virulence factor as it enhances the ability of bacteria to cause disease by preventing phagocytosis.

Topic: Biology, Biological Classification, NEET, NEET 2017
Which of the following components provides sticky character to the bacterial cell?

Options:
A A layer made up of murein
B A fluid mosaic layer made up of carbohydrates
C A phospholipid layer
D A rich glycoprotein layer

The correct answer is:
A rich glycoprotein layer

Solution:
Sticky character of the bacteria is due to glycocalyx or slime layer. This layer is rich in glycoproteins. The capsule acts as a virulence factor as it enhances the ability of bacteria to cause disease by preventing phagocytosis.

Topic: Biology, Biological Classification, NEET
Which of the following compound can form Zwitter ion?

Options:
A
B
C
D

The correct answer is:

Solution:
Here can form Zwitter ion as it is α -amino acid.

Topic: Chemistry, Biomolecules, NEET
Which of the following compound does not de-colourise the Baeyer's reagent?

Options:
A CH3-C≡C-CH3
B CH2=CH-CH=CH-CH=CH2
C
D H2C=C=CH2

The correct answer is:

Solution:
Benzene is an aromatic cyclic compound.

Benzene is not a pure unsaturated compound due to its de-localised double bonds.

It is reasonance stabilised and it cannot be oxidised by KMnO4,

hence benzene does not de-colourise the Baeyer's reagent.

Topic: Chemistry, Hydrocarbons, NEET
Which of the following compound does not liberate oxygen gas on warming with concentrated H2SO4?

Options:
A SO3
B PbO2
C MnO2
D CrO5

The correct answer is:
SO3

Solution:
SO3 on warming with concentrated H2SO4 give Oleum and not oxygen. Rest all given here gives O2 on warming with concentrated sulphuric acid.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Which of the following compound does not produce oxyacid of central atom on hydrolysis?

Options:
A BF3
B NCl3
C SF4
D PCl5

The correct answer is:
NCl3

Solution:
NCl3 does not produce oxyacid of central atom on hydrolysis as in NCl3 molecule, H2O molecule attacks on less electronegative Cl-atom rather than central N-atom.
NCl3+3H2O→NH3+3HOCl

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Which of the following compound does undergo aldol condensation

Options:
A HCHO
B
C CCl3.CHO
D CH3.CH2.CHO

The correct answer is:
CH3.CH2.CHO

Solution:
βαCH3-CH2-CHO
aldehydes having α-H atom can participate in aldol condensation. The H- atom attached to α carbon atom are called α- hydrogen.

Topic: Chemistry, Aldehydes and Ketones, NEET
Which of the following compound gives benzoic acid when it reacts with hot/alkaline KMnO 4 followed by acidification.

Options:
A
B
C
D All of these

The correct answer is:
All of these

Solution:
Since all of the above compounds have alpha hydrogen, Hence they will react with KMn O 4 to give benzoic acid.
At benzylic C if H,C,O are present then this group convert into -COOH group by the oxidation by alkaline KMnO4.

Topic: Chemistry, General Organic Chemistry, NEET
Which of the following compound is likely to show both Frenkel and Schottky defects in its crystalline form?

Options:
A KBr
B AgBr
C ZnS
D CsCl

The correct answer is:
AgBr

Solution:
AgBr show both Frenkel and Schottky defect. KBr and CsCl show Schottky defect while ZnS shows Frenkel defect.
Highly ionic solids with more coordination number show Schottky defect while Less ionic solids with low coordination number show Frenkel defect.

Topic: Chemistry, Solid State, NEET
Which of the following compound is most rapidly hydrolysed by SN1 mechanism?

Options:
A CH3CH=CHCl
B ClCH2CH=CH2
C
D (C6H5)3CCl

The correct answer is:
(C6H5)3CCl

Solution:
C6H5CCl→SN1C6H5C⊕
Rate of SN1 rection ∝ stability of carbocation
Stability of carbocation
(1) CH3-CH=CH+ +ve Csp2
(2) C⊕H2-CH=CH2 +ve change in configuration
3 +ve change is more delocalized in benzene ring
(4) highly stable due to +ve charge is more delocalised
Stability of carbocation: - 4>3>2>1
Higher rate of SN1 reaction in (4) Ph3C-Cl

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Which of the following compound is most reactive in electrophilic aromatic substitution?

Options:
A
B
C
D

The correct answer is:

Solution:
Electron withdrawing groups decreases the reactivity of benzene and electron releasing groups increases the acidic nature.

+R effect of -OH group enhances the reactivity in aromatic electrophilic substitution reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET, NEET 2020 (Phase 2)
Which of the following compound is (S)-4-chloro-1-methylcyclohexene?

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, General Organic Chemistry, NEET
Which of the following compound is used in visualisation of DNA fragments in gel electrophoresis?

Options:
A

Hexachlorobenzene

B

Silver bromide

C

Ethyl chloride

D

Ethidium bromide

The correct answer is:

Ethidium bromide

Solution:
The separated DNA fragments can be visualised only after staining the DNA with a compound known as ethidium bromide followed by exposure to UV radiation.

Topic: Biology, Biotechnology - Principles and Processes, NEET
Which of the following compound on heating does not produce metal oxide?

Options:
A MgCl2.6H2O
B K2Cr2O7
C K2CO3
D Cu(NO3)2

The correct answer is:
K2CO3

Solution:
Here K2CO3 on heating does not produce its oxide as it is thermally stable
MgCl2.6H2OâÎMgO+2HCl+5H2O
2K2Cr2O7âÎ2K2CrO4+Cr2O3+32O2
K2CO3âÎ does not decompose but melts
Cu ( NO 3 ) 2 → Δ CuO+2 NO 2 + 1 2 O 2

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Which of the following compound on oxidative ozonolysis give malonic acid as only product?

Options:
A CH 2  = CH− CH 2 −CH =  CH 2
B CH 2  = CH−CH =  CH 2
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, NEET
Which of the following compound present in the blood is associated with immunity response?

Options:
A Gamma globulin protein
B Albumin protein
C Alpha globulin protein
D Both (A) and (C)

The correct answer is:
Gamma globulin protein

Solution:
Gamma globulin plasma protein of blood helps in the synthesis of antibodies or immunoglobulin molecules, which binds and destroy the harmful antigens, therefore provide antibody-mediated immunity.

Topic: Biology, Human Health and Diseases, NEET
Which of the following compound will undergo tautomerism?

Options:
A
B
C
D

The correct answer is:

Solution:
This compound has α -H on saturated α -carbon atom so it can show tautomerism.
In rest other compounds there is no α -H-atom on saturated α -Carbon so no tautomerism is possible in them.

Topic: Chemistry, General Organic Chemistry, NEET
Which of the following compounds (a) and (b) will not react with a mixture of NaBr and H2SO4. Explain why?

Options:
A Compound (a) will not react with a mixture of NaBr and H2SO4, C—O bond is more stable because of OH group.
B Compound (b) will not react with a mixture of NaBr and H2SO4, C—O bond is more stable because of resonance.
C Both compounds will not react with a mixture of NaBr and H2SO4
D Both compounds will react with a mixture of NaBr and H2SO4

The correct answer is:
Compound (b) will not react with a mixture of NaBr and H2SO4, C—O bond is more stable because of resonance.

Solution:
Correct Option is : (B)
Compound (b) will not react with a mixture of NaBr and H2SO4, C—O bond is more stable because of resonance.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Which of the following compounds are aromatic according to Huckel’s rule?

Options:
A A,D and E only
B B,D,E and F(only if it is a planar molecule)
C B,C and E only
D A,B and C only

The correct answer is:
B,D,E and F(only if it is a planar molecule)

Solution:
Correct Option is : (B)
B,D,E and F(only if it is a planar molecule)

Topic: Chemistry, Hydrocarbons, NEET
Which of the following compounds are not ionic?

Options:
A MgCl2
B CaCl2
C BeCl2
D SrCl2

The correct answer is:
BeCl2

Solution:
Correct Option is : 3
BeCl2

Topic: Chemistry, s Block Elements, NEET
Which of the following compounds are soluble in organic compounds?

Options:
A BaCl2
B CaCl2
C MgCl2
D BeCl2

The correct answer is:
BeCl2

Solution:
Correct Option is : 4
BeCl2

Topic: Chemistry, s Block Elements, NEET
Which of the following compounds can be used during the free radical mechanism for the polymerisation of ethene?

Options:
A Acetyl peroxide
B 1,3 butadiene
C Novolac
D Formaldehyde

The correct answer is:
Acetyl peroxide

Solution:
Correct Option is : (A)
Acetyl peroxide

Topic: Chemistry, Polymers, NEET
Which of the following compounds can form a zwitter ion?

Options:
A Benzoic acid
B Acetanilide
C Aniline
D Glycine

The correct answer is:
Glycine

Solution:
The molecule which forms zwitter ion is glycine.

Every zwitterion has an isoelectric point (pI). The isoelectric point is the pH at which a zwitterion is uncharged. The pH can affect the charge of a molecule by introducing protons (H+). The amino group of an amino acid is a very effective proton acceptor, so is considered to be basic.

Topic: Chemistry, Biomolecules, NEET, NEET 2018
Which of the following compounds can form a zwitter ion?

Options:
A Benzoic acid
B Acetanilide
C Aniline
D Glycine

The correct answer is:
Glycine

Solution:
Glycine contains acidic group –COOH and basic group –NH2 so
Benzoic acid, acetanilide and aniline do not form zwitter ion.

Topic: Chemistry, Biomolecules, NEET
Which of the following compounds can form H-bonding with each other?

Options:
A CH3COOH and H2O
B Phenol and CH4
C C H 3 F and acetone
D PH3 and HF

The correct answer is:
CH3COOH and H2O

Solution:
Here acetic acid and water can have hydrogen bonding as follows

In options B, C and D there is no scope of hydrogen bonding.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Which of the following compounds cannot be prepared singly by the Wurtz reaction?

Options:
A C2H6
B CH32 CHCH3
C CH3CH2CH2CH3
D All of the above can be prepared

The correct answer is:
CH32 CHCH3

Solution:
If two different alkyl halides R1-X and R2-X are used, a mtextxture of three alkanes is obtained which is difficult to separate.

Topic: Chemistry, Hydrocarbons, NEET
Which of the following compounds contain an aromatic ring?

Options:
A Valeraldehyde
B Pthaldehyde
C Mesityl oxide
D Acrolein

The correct answer is:
Pthaldehyde

Solution:
Correct Option is : 2
Pthaldehyde

Topic: Chemistry, Aldehydes and Ketones, NEET
Which of the following compounds contain(s) no covalent bond(s)?
KCl,PH3,O2,B2H6,H2SO4

Options:
A KCl,B2H6
B KCl,B2H6,PH3
C KCl,H2SO4
D KCl

The correct answer is:
KCl

Solution:
KCl contains only ionic bond between K+ and Cl- ions (K+Cl-) while other structures have covalent bonds as follows

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Which of the following compounds does not have a carboxyl functional group?

Options:
A Picric acid

B Ethanoic acid

C Aspirin

D Benzoic acid

The correct answer is:
Picric acid

Solution:
Picric acid
Thus, Picric acid does not have a -COOH group. Other molecules have carboxyl groups.
Ethanoic acid
Aspirin
Benzoic Acid

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Which of the following compounds doesnâ€™t show ferrimagnetism?

Options:
A Fe3O4
B Fe2O3
C ZnFe2O4
D MgFe2O4

The correct answer is:
Fe2O3

Solution:
Correct Option is : 2
Fe2O3

Topic: Chemistry, Solid State, NEET
Which of the following compounds form pairs of metamers?

Options:
A V and VI , V and VII , VI and VII form pairs of metamers
B I and VI , V and VII , VI and VII form pairs of metamers
C V and VI , IV and VII , VI and VII form pairs of metamers
D V and VI , V and VII , V and III form pairs of metamers

The correct answer is:
V and VI , V and VII , VI and VII form pairs of metamers

Solution:
Correct Option is : (A)
V and VI , V and VII , VI and VII form pairs of metamers

Topic: Chemistry, General Organic Chemistry, NEET
Which of the following compounds forms a hydrate?

Options:
A Na2CO3
B K2CO3
C MgCO3
D CaCO3

The correct answer is:
MgCO3

Solution:
MgCO3 forms a hydrate. Na2CO3, K2CO3, and CaCO3 do not form hydrates.

Topic: Chemistry, s Block Elements, NEET
Which of the following compound(s) give N2 on heating?

Options:
A NH4NO2
B NH4NO3
C NaN3
D Both (A) and (C)

The correct answer is:
Both (A) and (C)

Solution:
Dinitrogen is a colourless, tasteless, odourless gas and it can be formed by the decomposition of the nitrogenous compounds like nitrates, nitrites, and azides.

Decomposition of the ammonium nitrite gives the dinitrogen and water. This is one of the laboratory preparation of the dinitrogen.

1NH4NO2â¶ÎN2+2H2O

Thermal decomposition of the ammonium nitrate gives the nitrous oxide gas. It is also known as the laughing gas.

2 NH4NO3â¶ÎN2O+2H2O

Thermal decomposition of the sodium azide gives the dinitrogen gas and sodium metal. This heating should be the absence of oxygen. This method is one of the laboratory preparation of the dinitrogen.

3 2NaN3â¶Î3N2+2Na

Hence, Only ammonium nitrite and sodium azide will be able to produce dinitrogen gas on thermal heating.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Which of the following compounds has dipole moment non-zero?

Options:
A CCl4
B CHCl3
C CO2
D BF3

The correct answer is:
CHCl3

Solution:
Correct Option is : 2
CHCl3

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET