### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2004

The formation of the oxide ion O2-(g) requires first an exothermic and then an endothermic step as shown below

O(g) + e- = $O_{(g)}^{-}$ $\Delta$Ho = -142 kJmol-1

$O_{(g)}^{-}$ + e- = $O_{(g)}^{2-}$ $\Delta$Ho = 844 kJmol-1

This because
A
O- ion will tend to resist the addition of another electron
B
Oxygen has high electron affinity
C
Oxygen is more electronegative
D
O- ion has comparatively larger size than oxygen atom

## Explanation

$O$ atom is highly electronegative so will add first electron easily by releasing energy. So it is an exothermic.

After adding first electron $O$ becomes ${O^ - }\,\,$ and size of ${O^ - }\,\,$ becomes slightly more than $O$ atom. Now when a new electron is trying to add into $\,\,{O^ - }$ ion, two forces will be act on new electron $\to$

$\left( 1 \right)\,\,\,\,$ attraction between nucleus and new electron

$\left( 2 \right)\,\,\,\,$ Repulsion between outer most shell electrons and electron.

But repulsion between electrons are more compare to the attraction between nucleus and electron as distance between nucleus and electron is more compare to electrons of outer most shell and new electron.

So, to overcome the repulsion energy should be added.

That is why ${O^ - }\left( g \right) + {e^ - }\buildrel \, \over \longrightarrow {O^{2 - }}\left( g \right)\,\,\,$ is an endothermic reaction.
2

### AIEEE 2004

Among Al2O3, SiO2, P2O3 and SO2 the correct order of acid strength is
A
SO2 < P2O3 < SiO2 < Al2O3
B
Al2O3 < SiO2 < P2O3 < SO2
C
Al2O3 < SiO2 < SO2 < P2O3
D
SiO2 < SO2 < Al2O3 < P2O3

## Explanation

In periodic table from left to right (group $1$ to group $18$ ) acidic strength increases.

$Al$ belongs to group $13,$ $Si$ belongs to group $14,$ $P$ belongs to group $15$ and $S$ belongs to group $16$.

So, $\,\,\,A{l_2}{O_3}$ will be least acidic and $S{o_2}$ will most acidic among then.

So, right order will be

$A{l_2}{O_3} < Si{O_2} < {P_2}{O_3} < S{O_2}$
3

### AIEEE 2004

Which one of the following ions has the highest value of ionic radius?
A
O2-
B
B3+
C
Li+
D
F-

## Explanation

From a neutral atom where a electron is removed then it become cation and when a electron is added to the atom then it becomes anion.

For a atom $x$ when it loose a electron then it becomes cation $x - {e^ - } \to {x^ + }$

Let atomic no. of $x$ is $z$

then no. of proton in ${x^ + } = z$

and no. of electron in ${x^ + } = z - 1$

So the ratio ${z \over e}$ as no. of electrons decreases. And when no of electron decreases then per electron attraction increases from nucleus and radius of cation deceases.

Similarly when atom $x$ becomes ${x^ - }$ by adding a electron $\left( {x + {e^ - } \to {x^ - }} \right)$ then no. of proton in ${x^ - } = z$ and no. of electron in ${x^ - } = z + 1.$

Now the ratio of ${z \over e}\,\,$ in ${x^ - }$ decreases as no. of electron increases. As electron increases in ${x^ - }$ then attraction from nucleus decreases on per electron and the distance between electron and nucleus increases so the radius also increase in anion.

All $4$ ions are from second period and for ${O^{2 - }}\,\,$ and ${F^ - }$ both have $1s,$ $2s$ and $2p$ shell and for $L{i^ + }\,\,$ and ${B^{3 + }}$ both have $1s$ shell.

So, size of ${O^{2 - }}\,\,\,$ and $\,\,\,{F^ - }\,\,\,$ are more then $L{i^ + }\,\,\,\,$

and $\,\,\,\,{B^{3 + }}$ and among ${O^{2 - }}\,\,\,$ and $\,\,\,\,{F^ - }$

For $\,\,\,$ ${O^{2 - }},\,\,{z \over e} = {8 \over {10}} = 0.8$

For $\,\,\,$ ${F^ - },\,\,\,{z \over e} = {9 \over {10}} = 0.9$

as for ${O^{2 - }},\,\,{z \over e}\,\,\,$ is less so per electron attraction from nucleus decreases and radius increases more than ${F^ - }.$
4

### AIEEE 2003

The atomic numbers of Vanadium (V), Chromium (cr), Manganese (Mn) and Iron (Fe), respectively, $23,24,25$ and $26$. Which one of these may be expected to have the higher second ionization enthalpy?
A
Cr
B
Mn
C
Fe
D
V

## Explanation

Ionization enthalpy is the energy required to remove the electron from the outer most orbit.

Electronic configuration :

\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,V\left( {23} \right) = \left[ {Ar} \right]\,\,\,3{d^3}\,\,\,4{S^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Cr\left( {24} \right) = \left[ {Ar} \right]\,\,\,3{d^5}\,\,\,4{S^1} \cr & \,\,\,\,\,\,\,\,\,\,\,Mn\left( {25} \right) = \left[ {Ar} \right]\,\,\,3{d^5}\,\,\,4{S^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Fe\left( {26} \right) = \left[ {Ar} \right]\,\,\,3{d^6}\,\,\,4{S^2} \cr}

After first ionization enthalpy the electronic configuration will be,
\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,V\left( {23} \right) = \left[ {Ar} \right]\,\,\,3{d^3}\,\,\,4{S^1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Cr\left( {24} \right) = \left[ {Ar} \right]\,\,\,3{d^5} \cr & \,\,\,\,\,\,\,\,\,\,\,Mn\left( {25} \right) = \left[ {Ar} \right]\,\,\,3{d^5}\,\,\,4{S^1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Fe\left( {26} \right) = \left[ {Ar} \right]\,\,\,3{d^6}\,\,\,4{S^1} \cr}

Now in 2nd ionization enthalpy one more electron will be removed. And after second ionization enthalpy electronic configuration will be,
\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,V\left( {23} \right) = \left[ {Ar} \right]\,\,\,3{d^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Cr\left( {24} \right) = \left[ {Ar} \right]\,\,\,3{d^4} \cr & \,\,\,\,\,\,\,\,\,\,\,Mn\left( {25} \right) = \left[ {Ar} \right]\,\,\,3{d^5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,Fe\left( {26} \right) = \left[ {Ar} \right]\,\,\,3{d^6} \cr}

For $V(23), Mn (25)$ and $Fe (26)$ electron is removed from $4S$ orbital but for $Cr (24)$ electron is removed from $3d$ orbital.

We know distance of $4S\,\, > \,\,3d$ from the nucleus. So, the attraction on the electrons of $45$ shell is less compared to $3d$ shell by the nucleus. So, the removed of electron will be easier for the electrons of $4S$ shell than $3d$ shell.

So, for $Cr(24)$ second ionization is more than other as electron is removed from $3d$ shell.

And for $Cr(24)$ outer most shell $''3d''$ was half filled after $1st$ ionization enthalpy, so $3d$ shell was stable. So removal of electron will be difficult from stable shell. That is why also $2nd$ ionization enthalpy of $Cr(24)$ is higher compare to other given atoms.