datavorous/entrance-exam-dataset huggingface dataset:
The correct order for acidity of the following hydroxyl compound is

(A) CH3OH

(B) CH33COH

(C)

(D)

(E)

Choose the correct answer from the options given below:

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

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

Solution:
The compounds A & B are alcohols, and rest of the compounds are phenols. Phenols are more acidic than alcohols. Among the A &B, methyl alcohol is more acidic than tertiary butyl alcohol as alkyl groups decreases the acidic nature by electron releasing tendency. Among the phenols, the phenol with electron withdrawing group is more acidic. NO2(Electron withdrawing group) group will increase the acidity, OCH3(Electron releasing group) group at para position will decrease the acidity.

So, Acidic strength E>C>D>A>B

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (10 Apr Shift 2)
The correct order for the wavelength of absorption in the visible region is

Options:
A $\left[\mathrm{Ni}\left(\mathrm{NO}_2\right)_6\right]^{4-} \lt \left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+} \lt \left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
B $\left[\mathrm{Ni}\left(\mathrm{NO}_2\right)_6\right]^{4-} \lt \left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \lt \left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}$
C $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \lt \left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+} \lt \left[\mathrm{Ni}\left(\mathrm{NO}_2\right)_6\right]^{4-}$
D $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+} \lt \left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \lt \left[\mathrm{Ni}\left(\mathrm{NO}_2\right)_6\right]^{4-}$

The correct answer is:
$\left[\mathrm{Ni}\left(\mathrm{NO}_2\right)_6\right]^{4-} \lt \left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+} \lt \left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

Solution:
The absorption of energy or the observation of colour in a complex transition compound depends on the charge of the metal ion and the nature of the ligands attached. The same metal ion with different ligands shows different absorption depending upon the type of ligand. The presence of weak field ligands make the central metal ion to absorb low energies i.e., of higher wavelength. The field strength of ligands can be obtained from spectrochemical series. i.e.
(weak field) $\mathrm{I}^{-} \lt \mathrm{Br}^{-} \lt \mathrm{S}^{2-} \lt \mathrm{Cl}^{-} \lt \mathrm{NO}_3^{-} \lt \mathrm{F}^{-} \lt \mathrm{OH}^{-} \lt \mathrm{H}_2 \mathrm{O} \lt \mathrm{NH}_3 \lt \mathrm{NO}_2^{-} \lt \mathrm{CN}^{-} \lt \mathrm{CO}$ (strong field)

Topic: Chemistry, Coordination Compounds, JEE Main
The correct order in aqueous medium of basic strength in case of methyl substituted amines is:

Options:
A Me2NH>MeNH2>Me3N>NH3
B Me2NH>Me3N>MeNH2>NH3
C NH3>Me3N>MeNH2>Me2NH
D Me3N>Me2NH>MeNH2>NH3

The correct answer is:
Me2NH>MeNH2>Me3N>NH3

Solution:
In aqueous medium basic strength is dependent on electron density on nitrogen as well as solvation of cation formed after accepting H+. In protic solvent like water, the order of basic strength is

Me2NH>MeNH2>Me3N>NH3.

This happens beacuse in case of Me2NH there will be optimum (+I) and solvation while in MeNH2 has more significant hydration than Me3N.

Topic: Chemistry, Amines, JEE Main, JEE Main 2023 (25 Jan Shift 1)
The correct order of acid character of the following compounds is :

Options:
A IV>III>II>I
B II>III>IV>I
C III>II>I>IV
D I>II>III>IV

The correct answer is:
II>III>IV>I

Solution:
Presence of electron withdrawing groups such as nitro group enhance the acidic strength of phenol and the presence of electron releasing groups such as alkyl group decreases the acidic strength of phenol.

2>3>4>1

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The correct order of acid strength is:

Options:
A $\mathrm{HClO}_4 < \mathrm{HClO}_3 < \mathrm{HClO}_2 < \mathrm{HClO}$
B $\mathrm{HClO} < \mathrm{HClO}_2 < \mathrm{HClO}_3 < \mathrm{HClO}_4$
C $\mathrm{HClO}_4 < \mathrm{HClO}^2 < \mathrm{HClO}_2 < \mathrm{HClO}_3$
D $\mathrm{HClO}_2 < \mathrm{HClO}_3 < \mathrm{HClO}_4 < \mathrm{HClO}$

The correct answer is:
$\mathrm{HClO} < \mathrm{HClO}_2 < \mathrm{HClO}_3 < \mathrm{HClO}_4$

Solution:
Oxygen is more electronegative than chlorine. With an increase in the number of $\mathrm{O}$ atoms attached to $\mathrm{Cl}$, more electrons are pulled away from $\mathrm{O}-\mathrm{H}$ bond and weaker becomes the $\mathrm{O}-\mathrm{H}$ bond. This increases the acid strength. The decreasing order of acidic strength is $\mathrm{HClO} < \mathrm{HClO}_2 < $ $\mathrm{HClO}_3 < \mathrm{HClO}_4$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The correct order of acid strength of the following substituted phenols in water at $28^{\circ} \mathrm{C}$ is

Options:
A $p$-nitrophenol $ < $ p-fluorophenol $ < $ p-chlorophenol
B $p$-chlorophenol $ < $ p-fluorophenol $ < $ p-nitrophenol
C p-fluorophenol $ < $ p-chlorophenol $ < $ p-nitrophenol
D $p$-flurophenol $ < $ p-nitrophenol $ < p$-chlorophenol

The correct answer is:
p-fluorophenol $ < $ p-chlorophenol $ < $ p-nitrophenol

Solution:
The order of electron withdrawing tendency from benzene ring i.e..
$$
-F < -C l < -N O_{2}
$$
$\therefore$ Correct order of acidic strength of substituted phenols will be

Topic: Chemistry, General Organic Chemistry, JEE Main
The correct order of acidic strength of the following is
(I) Oxalic acid HOOC-COOH
(II) Malonic acid HOOC-CH2-COOH
(III) Succinic acid HOOC-CH22-COOH
(IV) Glutaric acid HOOC-CH23-COOH

Options:
A I > II > III > IV
B I > III> II > IV
C IV> III > II > I
D IV> II > III > I

The correct answer is:
I > II > III > IV

Solution:
Acidic strength of dicarboxylic acid is based on the +I effect of alkyl group. As +I increases acidic nature decreases.
Compounds with formula HOOC-CH2n-COOH are called dicarboxylic acid.
Acidic strength order is:
Oxalic acid > Malonic acid > Succinic acid > Glutaric acid

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The correct order of acidity for the following compounds is:

Options:
A $II < I V < I < I$
B $II < I < III < IV$
C $II < III < IV < I$
D $III < II < I < IV$

The correct answer is:
$II < I < III < IV$

Solution:
Hint:
Compound IV is a carboxylic acid whereas others are alcohols. So IV is most acidic, amongst the alcohols. Presence of $-\mathrm{NO}_{2}$ increases acidic strength while the presence of $-\mathrm{CH}_{3}$ lowers. Amongst I and III, more acidic is $\mathrm{I}$ as $-\mathrm{NO}_{2}$ exerts it's $-\mathrm{R}$ influence in para position.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The correct order of acidity for the following compounds is

Options:
A I > II > III > IV
B III > I > II > IV
C III > IV > II > I
D I > III > IV > II

The correct answer is:
I > II > III > IV

Solution:
In I and II, the conjugate base is satbilized by intra-molecular H-bonding.

Topic: Chemistry, General Organic Chemistry, JEE Main
The correct order of acidity of the following is

Options:
A III>II>I
B II >III >I
C I>II>III
D III>I>II

The correct answer is:
I>II>III

Solution:
The electron withdrawing groups increases the acidic strength of phenols and the electron releasing groups decreases the acidic strength of phenols. The methyl group and methoxy groups are electron releasing groups at para position. Hence, both p-cresol and p-methoxy phenol are less acidic than phenol. Methoxy group is electron releasing group by +M effect and methyl group is hyperconjugation. Hence, p-methoxy pheniol is less acidic than p-cresol. The correct acidic strength order is

Topic: Chemistry, General Organic Chemistry, JEE Main
The correct order of atomic radii is:

Options:
A Ce>Eu>Ho>N
B N>Ce>Eu>Ho
C Ho>N>Eu>Ce
D Eu>Ce>Ho>N

The correct answer is:
Eu>Ce>Ho>N

Solution:
As the number of electrons are increased it increases the radius of the element. As the number of protons are increased the radius is decreased.

As we can see:

Radius of :

Eu=199 pmCe=183 pmHo=176 pmN=170 pm

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The correct order of basicity is

Options:
A
B
C
D

The correct answer is:

Solution:
CH3N¨H2 (Aliphatic amines) Most basic due to +I effect of methyl group. Methoxy group is electron donating group and increase electron density by (+M) effect on NH2 group.
-NO2 group withdraws electron density from N of -NH2 and decrease basic strength.

CH3N¨H2 Most basic because NH2 connected with sp3 C while in i,ii&iii Connected with sp2 C & basic strength ∝ +M/+I
basic strength ∝ 1 −M ∝ 1 −I
NO2 exert -M so least basic while -OCH3 exert +M so more basic.
M effect is a stronger effect than the I effect.

Aliphatic amines are more basic than aromatic amines because in aliphatic amines lone pair of electrons on nitrogen is localized and easy to donate, while in aromatic amines lone pair of electrons on nitrogen atom is delocalized and participate in resonance.

Topic: Chemistry, Amines, JEE Main, JEE Main 2015 (10 Apr Online)
The correct order of basicity is

Options:
A
B
C
D

The correct answer is:

Solution:
CH3NÂ¨H2 (Aliphatic amines) Most basic due to +I effect of methyl group. Methoxy group provides electron density (+M)- NH2
-NO2 group with draws electron density from N of -NH2

CH3NÂ¨H2 Most basic because NH2 connected with sp3 C while in i,ii&iii Connected with
sp2 C & basic strength â +M/+I
basic strength â 1 âM â 1 âI
NO2 exert -M so least basic while -OCH3 exert +M so more basic
M effect is a stronger effect than the effect

Topic: Chemistry, General Organic Chemistry, JEE Main
The correct order of basicity of amines in water is :

Options:
A $\left(\mathrm{CH}_3\right)_2 \mathrm{NH}\gt\left(\mathrm{CH}_3\right)_3 \mathrm{~N}\gt\mathrm{CH}_3 \mathrm{NH}_2$
B $\mathrm{CH}_3 \mathrm{NH}_2\gt\left(\mathrm{CH}_3\right)_2 \mathrm{NH}\gt\left(\mathrm{CH}_3\right)_3 \mathrm{~N}$
C $\left(\mathrm{CH}_3\right)_3 \mathrm{~N}\gt\left(\mathrm{CH}_3\right)_2 \mathrm{NH}\gt\mathrm{CH}_3 \mathrm{NH}_2$
D $\left(\mathrm{CH}_3\right)_3 \mathrm{~N}\gt\mathrm{CH}_3 \mathrm{NH}_2\gt\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$

The correct answer is:
$\left(\mathrm{CH}_3\right)_2 \mathrm{NH}\gt\left(\mathrm{CH}_3\right)_3 \mathrm{~N}\gt\mathrm{CH}_3 \mathrm{NH}_2$

Solution:
Basicity of amines increase with increase in number of $-\mathrm{CH}_3$ groups (or any group which cause $+/$ effect), due to increase in electron density on $N$ atom. As a rule, the basicity of $t$-amine should be more than that of s-amine, but actually it is found to be lesser than s-amines. This is due to stearic hinderence of bulkier alkyl groups, which decreases the availability of lone pair of electron on the $N$ atom of the amino group. Hence the correct order of basicity is :
$\left(\mathrm{CH}_3\right)_2 \mathrm{NH}\gt\left(\mathrm{CH}_3\right)_3 \mathrm{~N}\gt\mathrm{CH}_3 \mathrm{NH}_2$

Topic: Chemistry, Amines, JEE Main
The correct order of basicity of oxides of vanadium is

Options:
A V2O3>V2O4>V2O5
B V2O3>V2O5>V2O4
C V2O5>V2O4>V2O3
D V2O4>V2O3>V2O5

The correct answer is:
V2O3>V2O4>V2O5

Solution:
V2O3: In, V2O3, the oxidation state of V is as follows

2(x) + 3(-2) = 0

⇒ 2x - 6 = 0

⇒ x = 62 = 3

The oxidation state of V in V2O3 is +3. The vanadium is present in its lowest oxidation state. The vanadium tends to donate the pair of electrons. Hence, it is basic oxide.

V2O5: In V2O5, the oxidation state of V is as follows

2(x) + 5(-2) = 0

⇒ 2(x) - 10 = 0

⇒x = 102 = 5

The oxidation state of V in V2O5 is +5. The Vanadium tends to donate and accept the pair of electrons. It is amphoteric oxide.

V2O4: In V2O4, the oxidation state of V is as follows

2(x) + 4(-2) = 0

⇒ 2(x) - 8 = 0

⇒ x = 82 = 4

The oxidation state of V in V2O4 is +4. The Vanadium is present in its other low oxidation state. The Vanadium tends to donate and accept the pair of electrons. It is amphoteric oxide and less basic than V2O3, more basic than V2O5.

So, the correct order of basicity of oxides of vanadium is V2O3>V2O4>V2O5.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2023 (31 Jan Shift 1)
The correct order of basicity of the following compounds is

Options:
A $1 < 2 < 3 < 4$
B $1 < 2 < 4 < 3$
C $2 < 1 < 3 < 4$
D $4 < 3 < 2 < 1$

The correct answer is:
$2 < 1 < 3 < 4$

Solution:
The correct order is $2 < 1 < 3 < 4$. Since, more the number of $2^{\circ}$ - NH- group, more is the basic nature.
(1) Thus, 3 and 4 are more basic than 1 and 2 .
(ii) Between 3 and 4,4 has one two - $\mathrm{NH}_{2}$ group while 3 has only one $-\mathrm{NH}_{2}$ group. $\therefore 4$ is more basic than 3
(iii) Between 1 and 2,1 has one $-\mathrm{NH}_{2}$ group with no resonance, thus is more basic than 2.

Topic: Chemistry, Amines, JEE Main
The correct order of boiling point is

Options:
A T2<D2<H2
B n-pentane < neo-pentane
C Xe<Ar<He
D m-nitrophenol > o-nitrophenol

The correct answer is:
m-nitrophenol > o-nitrophenol

Solution:
Here option D is incorrect while rest are correct as follows
AT2>D2>H2 [BP ∝ mol. wt.]
B n-pentane > neo-pentane [BP ∝ VWf ∝ contact area ∝1Branching ]
CXe>Ar>He [BP ∝ mol. wt.]
D

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The correct order of bond angles (smallest first) in $\mathrm{H}_2 \mathrm{~S}, \mathrm{NH}_3, \mathrm{BF}_3$ and $\mathrm{SiH}_4$ is

Options:
A
$\mathrm{H}_2 \mathrm{~S} < \mathrm{SiH}_4 < \mathrm{NH}_3 < \mathrm{BF}_3$
B
$\mathrm{H}_2 \mathrm{~S} < \mathrm{NH}_3 < \mathrm{BF}_3 < \mathrm{SiH}_4$
C
$\mathrm{H}_2 \mathrm{~S} < \mathrm{NH}_3 < \mathrm{SiH}_4 < \mathrm{BF}_3$
D
$\mathrm{NH}_3 < \mathrm{H}_2 \mathrm{~S} < \mathrm{SiH}_4 < \mathrm{BF}_3$

The correct answer is:
$\mathrm{H}_2 \mathrm{~S} < \mathrm{NH}_3 < \mathrm{SiH}_4 < \mathrm{BF}_3$

Solution:
$\mathrm{H}_2 \mathrm{~S} \longrightarrow \mathrm{sp}^3$
$\mathrm{NH}_3 \longrightarrow \mathrm{sp}^3$
$\mathrm{BF}_3 \longrightarrow \mathrm{sp}^2$
$\mathrm{SiH}_4 \longrightarrow \mathrm{sp}^3$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2004
The correct order of bond dissociation energy among $\mathrm{N}_2, \mathrm{O}_2, \mathrm{O}_2^{-}$is shown in which of the following arrangements?

Options:
A
$\mathrm{N}_2>\mathrm{O}_2^{-}>\mathrm{O}_2$

B
$\mathrm{O}_2^{-}>\mathrm{O}_2>\mathrm{N}_2$

C
$\mathrm{N}_2>\mathrm{O}_2>\mathrm{O}_2^{-}$

D
$\mathrm{O}_2>\mathrm{O}_2^{-}>\mathrm{N}_2$

The correct answer is:
$\mathrm{N}_2>\mathrm{O}_2>\mathrm{O}_2^{-}$

Solution:
The bond order of $\mathrm{N}_2, \mathrm{O}_2$, and $\mathrm{O}_2^{-}$are respectively 3,2 and $1.5$
Since higher bond order implies higher bond dissociation energy hence the correct order will be
$$
\mathrm{N}_2>\mathrm{O}_2>\mathrm{O}_2^{-}
$$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2014 (11 Apr Online)
The correct order of bond dissociation enthalpy of halogens is:

Options:
A F2>Cl2>Br2>I2
B Cl2>F2>Br2>I2
C I2>Br2>Cl2>F2
D Cl2>Br2>F2>I2

The correct answer is:
Cl2>Br2>F2>I2

Solution:
Correct order of bond dissociation enthalpy of halogens is Cl2>Br2>F2>I2.

Due to inter electronic repulsions F-F bond becomes weak and easily broken.

Bond dissociation energy of halogen family decreases down the group as the size of atom increases.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2021 (25 Feb Shift 2)
The correct order of bond enthalpy (kJ mol–1) is:

Options:
A Si-Si>C-C>Sn-Sn>Ge-Ge
B Si-Si>C-C>Ge-Ge>Sn-Sn
C C-C>Si-Si>Sn-Sn>Ge-Ge
D C-C>Si-Si>Ge-Ge>Sn-Sn

The correct answer is:
C-C>Si-Si>Ge-Ge>Sn-Sn

Solution:
As bond length increases, bond energy decreases.

Bond	Bond energy (kJ mol-1)

C - C	348

Si - Si	297

Ge - Ge	260

Sn - Sn	240

So, bond enthalpy order C - C > Si - Si > Ge - Ge > Sn - Sn

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2023 (01 Feb Shift 2)
The correct order of bond orders of C22-, N22- and O22- is

Options:
A C22->N22->O22-
B O22->N22->O22-
C N22->O22->C22-
D C22->O22->N22-

The correct answer is:
C22->N22->O22-

Solution:
Bond order =Nb-Na2

NB= Number of electrons in bonding molecular orbitals.

NA·B= Number of electrons in anti bonding molecular orbitals.

C22- has 14 electrons, it is isoelectronic with N2 where Bond order =3,

N22- has 16 electrons is isoelectronic with O2 whose bond order =2

O22- has 18 electrons and is isoelectronic with F2 whose bond order =1

So, the correct order of Bond order is

C22->N22->O22-.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2022 (24 Jun Shift 2)
The correct order of Cl-O bond lengths in ClO-,ClO2-,ClO3- and ClO4- is

Options:
A ClO-<ClO2-<ClO3-<ClO4-
B ClO4-<ClO3-<ClO2-<ClO-
C ClO3-<ClO4->ClO2->ClO-
D ClO4-=ClO3-=ClO2-=ClO-

The correct answer is:
ClO4-<ClO3-<ClO2-<ClO-

Solution:
The correct order of Cl-O bond lengths in ClO-,ClO2-,ClO3- and ClO4- is as follows
ClO4-<ClO3-<ClO2-<ClO-
Bond length ∝1Bond order
or Bond length ∝1Extent of delocalization

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The correct order of C-O bond length among CO,CO32-,CO2 is

Options:
A CO<CO32-<CO2
B CO32-<CO2<CO
C CO<CO2<CO32-
D CO2<CO32-<CO

The correct answer is:
CO<CO2<CO32-

Solution:
More single bond character in resonance hybrid, more is the bond length. Hence the increasing bond length is CO<CO2<CO32-

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The correct order of chemical reactivity with water according to electrochemical series

Options:
A $\mathrm{K} \gt \mathrm{Mg} \gt \mathrm{Zn} \gt \mathrm{Cu}$
B $\mathrm{Mg} \gt \mathrm{Zn} \gt \mathrm{Cu} \gt \mathrm{K}$
C $\mathrm{K} \gt \mathrm{Zn} \gt \mathrm{Mg} \gt \mathrm{Cu}$
D $\mathrm{Cu} \gt \mathrm{Zn} \gt \mathrm{Mg} \gt \mathrm{K}$

The correct answer is:
$\mathrm{K} \gt \mathrm{Mg} \gt \mathrm{Zn} \gt \mathrm{Cu}$

Solution:
The standard reduction potential of, $\mathrm{K}^{+} \mathrm{Mg}^{2+}, \mathrm{Zn}^{+2} \mathrm{Cu}^{2+}$ increase in this order.

Topic: Chemistry, Electrochemistry, JEE Main
The correct order of conductivity of ions in water is:

Options:
A Na+>K+>Rb+>Cs+
B Cs+>Rb+>K+>Na+
C K+>Na+>Cs+>Rb+
D Rb+>Na+>K+>Li+

The correct answer is:
Cs+>Rb+>K+>Na+

Solution:
→Li+Na+K+Rb+Cs+Hydration energy↑

As the size of gaseous ion decreases, it gets more hydrated in water and hence, the size of aqueous ion increases. When this bulky ion move in solution, it experience greater resistance and hence lower conductivity.

Size of gaseous ion: Cs+>Rb+>K+>Na+

Size of aqueous ion: Cs+<Rb+<K+<Na+

Conductivity: Cs+>Rb+>K+>Na+

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The correct order of decreasing acid strength of trichloroacetic acid $(A)$, trifluoroacetic acid $(B)$, acetic acid $(C)$ and formic acid $(D)$ is

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

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

Solution:
If an electron withdrawing group ( $-I$ showing group) is present, it makes the removal of proton more easy by stabilising the remaining carboxylate ion and thus, makes the acid more acidic.
The order of acidity of given compounds is

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2012 (Screening)
The correct order of decreasing acidic nature of oxides

Options:
A Li2O>BeO>CO2>B2O3>N2O3
B CO2>N2O3>B2O3>Li2O>BeO
C CO2>BeO>Li2O>B2O3>N2O3
D N2O3>CO2>B2O3>BeO>Li2O

The correct answer is:
N2O3>CO2>B2O3>BeO>Li2O

Solution:
A binary compound of oxygen with another element is called oxide

Oxides can be simple (e.g., MgO, Al2O3 ) or mixed (Pb3O4 ,Fe3O4 ). Simple oxides can be classified on the basis of their acidic, basic or amphoteric character. An oxide that combines with water to give an acid is termed acidic oxide (e.g., SO2 ,Cl2O7,CO2 , N2O5 ). For example, SO2 combines with water to give H2SO3 , an acid.

The oxides which give a base with water are known as basic oxides (e.g., Na2O, CaO, BaO). For example, CaO combines with water to give Ca(OH)2, a base.

Some metallic oxides exhibit a dual behaviour. They show characteristics of both acidic as well as basic oxides. Such oxides are known as amphoteric oxides. They react with acids as well as alkalies. For example, Al2O3 reacts with acids as well as alkalies.

There are some oxides which are neither acidic nor basic. Such oxides are known as neutral oxides. Examples of neutral oxides are CO, NO and N2O.

Group 1	Group 2	Group 13	Group 14	Group 15

Period 2	Li	Be	B	C	N

According to periodic trend, the electronegativity of the period 2 elements increases in the order :

Li < Be < B < C < N

Electronegativty of the central atom of the oxide is direclty proportional acidic nature of the oxide.

Therefore, the acidic character of their oxides decreases in the order as follows :

N2O3 > CO2 > B2O3 > BeO > Li2O

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The correct order of decreasing second ionisation enthalpy of Ti (22), V (23), Cr
(24) and $\mathrm{Mn}(25)$ is

Options:
A $\mathrm{Ti}>\mathrm{V}>\mathrm{Cr}>\mathrm{Mn}$
B $\mathrm{Cr}>\mathrm{Mn}>\mathrm{V}>\mathrm{Ti}$
C $\mathrm{V}>\mathrm{Mn}>\mathrm{Cr}>\mathrm{Ti}$
D $\mathrm{Mn}>\mathrm{Cr}>\mathrm{Ti}>\mathrm{V}$

The correct answer is:
$\mathrm{Cr}>\mathrm{Mn}>\mathrm{V}>\mathrm{Ti}$

Solution:
$$
\mathrm{Cr}(24) \rightarrow[\mathrm{Ar}] 3 d^5 4 s^1
$$
After removing one electron from chromium, the resulting structure becomes more stable. Hence $\mathrm{Cr}$ has higher second ionisation enthalpy.
Thus, the correct order is
$$
\mathrm{Cr}>\mathrm{Mn}>\mathrm{V}>\mathrm{Ti}
$$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET, NEET 2008 (Mains)
The correct order of density is

Options:
A Be>Mg>Ca>Sr
B Sr>Ca>Mg>Be
C Sr>Be>Mg>Ca
D Be>Sr>Mg>Ca

The correct answer is:
Sr>Be>Mg>Ca

Solution:
The size of alkali metals increases down the group, so volume also shows increment, and since volume is inversely proportional to the density but in the case Sr and Ba, increase in the volume is less compared to increasing mass.

So, down the group density decreases first and then increases.

Therefore, the correct order of density is as follows:

'Ca<Mg<Be<Sr

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2022 (25 Jul Shift 2)
The correct order of $\mathrm{E}_{\mathrm{SR}^2 / \mathrm{M}}^0$ values with negative sign for the four successive elements $\mathrm{Cr}, \mathrm{Mn}, \mathrm{Fe}$ and $\mathrm{Co}$ is

Options:
A
$\mathrm{Mn}>\mathrm{Cr}>\mathrm{Fe}>\mathrm{Co}$

B
$\mathrm{Cr}>\mathrm{Fe}>\mathrm{Mn}>\mathrm{Co}$

C
$\mathrm{Fe}>\mathrm{Mn}>\mathrm{Cr}>\mathrm{Co}$

D
$\mathrm{Cr}>\mathrm{Mn}>\mathrm{Fe}>\mathrm{Co}$

The correct answer is:
$\mathrm{Mn}>\mathrm{Cr}>\mathrm{Fe}>\mathrm{Co}$

Solution:
No Solution Available

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2010
The correct order of electron affinity is:

Options:
A
$\mathrm{O}>\mathrm{F}>\mathrm{Cl}$

B
$\mathrm{F}>\mathrm{O}>\mathrm{Cl}$

C
F $>\mathrm{Cl}>\mathrm{O}$

D
$\mathrm{Cl}>\mathrm{F}>\mathrm{O}$

The correct answer is:
$\mathrm{Cl}>\mathrm{F}>\mathrm{O}$

Solution:
On moving from left to right across a period, the electron affinity becomes more negative. On moving from top to bottom in a group, the electron affinity becomes less negative.
Chlorine has more negative electron affinity than fluorine, Because adding an electron to fluorine ( $2 p$ orbital) causes greater repulsion than adding an electron to chlorine $(3 p$ orbital) which is larger in size.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The correct order of electron affinity of $B, C, N, O$ is

Options:
A $O \gt C \gt N \gt B$
B $B \gt N \gt C \gt 0$
C $O \gt C \gt B \gt N$
D $O \gt B \gt C \gt N$

The correct answer is:
$O \gt C \gt B \gt N$

Solution:
$O \gt C \gt B \gt N$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main