JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2009

In which of the following arrangements, the sequence is not strictly according to the property written against it
A
CO2 < SiO2 < SnO2 < PbO2 : increasing oxidising power
B
HF < HCl < HBr, HI : increasing acid strength
C
NH3 < PH3 < AsH3 < SbH3 : increasing basic strength
D
B < C < O < N : increasing first ionization enthalpy

Explanation

$(A)\,\,\,$ Option A is true.

Here in $Pb{O_2}$ oxidation number of $Pb$ is $+4,$ due to inert pair effect the oxidation number $+4$ is not stable for $Pb$, So, to become stable oxide $+2$ $Pb$ will oxidize other and reduced to $P{b^{ + 2.}}$ So, $Pb{O_2}$ has the highest oxidizing power.

$(B)\,\,\,\,$ Option B is true.

Bond length from top to bottom in periodic table increases, So the bond length order is $HI > HBr > HCl > HF$

The compound which can give ${H^ + }$ ion easily that compound has higher acidic strength.

As the bond length between $H$ and $I$ in $HI$ is longest then it will be carier to break the bond between $H$ and $I$. So $HI$ will have highest acidic strength.

$(C)\,\,\,\,$ Option C is false.

The structure of those $4$ molecules are

Due to the lone pair electron all of them are basic nature. The molecule which can easily donate the lone pair electron will have more basic strength.

Here, Nitrogen (N) is a element of $2$nd period so in $N$ new electron is added in the second period and as size of $N$ is small so electron density is more on the outer shell, so electron become unstable due to repulsion with each other. That is why electron pair is easily removable from $N{H_3}.$

On the other hand $p$ is a $3$rd block, As is a $4$th block and $Sb$ is a $5$th block elements. So, their size is bigger. So, electron density on the outer shell is less and their repulsion also decreases. So, outer shell electrons are more stable and $P{H_3},As{H_3}$ and $Sb{H_3}$ will have less ability to donate the electron.

So the correct order is $Sb{H_3} < As{H_3} < P{H_3} < N{H_3}$

(D) Option (D) is true

Electronic configuration of

$B\left( 5 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^1}$

$C\left( 6 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^2}$

$N\left( 7 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^3}$

$O\left( 8 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^4}$

In general in periodic table from left to right effective nuclear change increase and it becomes more difficult to remove electron from an atom. That is why ionization enthalpy increases.

But for those atoms which has half filled or full filled outer most shell, those atoms will be more stable and more energy is needed to remove electron from those atoms. Here $N\left( {1{s^2}\,2{s^2}\,2{p^3}} \right)$ has stable half filled $2p$ subshell so it is more stable than $O.$

So, correct order is $B < C < O < N$
2

AIEEE 2006

Following statements regarding the periodic trends of chemical reactivity of the alkali metals and the halogens are given. Which of these statements gives the correct picture?
A
The reactivity decreases in the alkali metals but increases in the halogens with increase in atomic number down the group
B
In both the alkali metals and the halogens the chemical reactivity decreases with increase in atomic number down the group
C
Chemical reactivity increases with increase in atomic number down the group in both the alkali metals and halogens
D
In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic number down the group
3

AIEEE 2005

In which of the following arrangements the order is NOT according to the property indicated against it?
A
Li < Na < K < Rb Increasing metallic radius
B
I < Br < F < Cl Increasing electron gain enthalpy (with negative sign)
C
B < C < N < O Increasing first ionization enthalpy
D
Al3+ < Mg2+ < Na+ < F- Increasing ionic size

Explanation

$\left( a \right)\,\,\,\,\,$It is True.

Here $Li, Na, K$ and $Rb$ all are belongs to the same group, so their effective nuclear charge is same but in a group from top to bottom the no. of shells increase in a atom. So the radius of atom increases.

$\left( b \right)\,\,\,\,\,$ It is True.

In entire periodic table $Cl$ (chlorine) has the highest, and in a group it decreases from top to bottom.

So the correct order is ${\rm I} < Br < F < Cl.$

Note : Among F and Cl, when an electron is added to the F atom, electron comes to the 2p orbital and for Cl atom electron is added in 3p orbital. As 2p orbital is closer to the nucleus than 3p orbital as 3p orbital is larger in size, so when a new electron comes to 2p orbital then it will face a strong repulsion force by the nucleus than if electron comes to 3p orbital.

So, F have lesser tendency of gaining electron than Cl.

$\left( c \right)\,\,\,\,\,$ It is False.

Electronic configuration of

$B\left( 5 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^1}$

$C\left( 6 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^2}$

$N\left( 7 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^3}$

$O\left( 8 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^4}$

In general in periodic table from left to right effective nuclear change increase and it becomes more difficult to remove electron from an atom. That is why ionization enthalpy increases.

But for those atoms which has half filled or full filled outer most shell, those atoms will be more stable and more energy is needed to remove electron from those atoms. Here $N\left( {1{s^2}\,2{s^2}\,2{p^3}} \right)$ has stable half filled $2p$ subshell so it is more stable than $O.$

So, correct order is $B < C < O < N$

$\left( d \right)\,\,\,\,\,$ It is True.

$A{l^{3 + }},M{g^{2 + }},N{a^ + }\,\,$ and

$\,\,{F^ - }$ are isoelectric.

So all have $10$ electrons. So,

$\,\,{Z \over e}\,\,$ of $A{l^{3 + }} = {{13} \over {10}} = 1.3$

$\,\,{Z \over e}\,\,$ of $M{g^{2 + }} = {{12} \over {10}} = 1.2$

$\,\,{Z \over e}\,\,$ of $N{a^ + } = {{11} \over {10}} = 1.1$

$\,\,{Z \over e}\,\,$ of ${F^ - } = {9 \over {10}} = 0.9$

We know ${z \over e}$ means $z$ no of protons in nucleus is pulling $e$ no. of electrons present in the orbital.

So for $A{l^{3 + }},\,13$ protons is pulling $10$ electron so the size of $A{l^{3 + }}$ will decrease most. That is why more ${z \over e}$ means less size of ion.

So, the correct order is

$A{l^{3 + }} < M{g^{2 + }} < N{a^ + } < {F^ - }$
4

AIEEE 2005

Which of the following oxides is amphoteric in character?
A
CaO
B
CO2
C
SiO2
D
SnO2

Explanation

These elements are called amphoteric which reacts with both acid and base.

Following elements are amphoteric

$\left( 1 \right)\,\,\,\,Zn$

$\left( 2 \right)\,\,\,\,Be$

$\left( 3 \right)\,\,\,\,Al$

$\left( 4 \right)\,\,\,\,B$

$\left( 5 \right)\,\,\,\,Cr$

$\left( 6 \right)\,\,\,\,Ga$

$\left( 7 \right)\,\,\,\,Pb$

$\left( 8 \right)\,\,\,\,Sn$

Here $Zn,Be,Al,Pb\,\,\,$ and $\,$ $Sn$ are mostly amphoteric.

$B, Cr$ and $Ga$ are less amphoteric.