### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2010

The correct sequence which shows decreasing order of the ionic radii of the elements is
A
Al3+ > Mg2+ > Na+ > F- > O2-
B
Na+ > Mg2+ > Al3+ > O2– > F
C
Na+ > F > Mg2+ > O2– > Al3+
D
O2– > F > Na+ > Mg2+ > Al3+

## Explanation

As you can see all of them are isoelectronic and for isoelectric elements which ion has more positive charge will have lesser size. And which ion has more negative charge that ion's size will be more.

So, the correct order is

${O^{2 - }} > {F^ - } > N{a^ + } > m{g^{2 + }} > A{l^{3 + }}$
2

### AIEEE 2009

The set representing the correct order of ionic radius is :
A
$N{a^ + } > L{i^ + } > M{g^{2 + }} > B{e^{2 + }}$
B
$L{i^ + } > N{a^ + } > M{g^{2 + }} > B{e^{2 + }}$
C
$M{g^{2 + }} > B{e^{2 + }} > L{i^ + } > N{a^ + }$
D
$L{i^ + } > B{e^{2 + }} > N{a^ + } > M{g^{2 + }}$

## Explanation

In periodic table $Li, Na, Mg, Be$ are present in following ways :-

Group  1 Group  2
Li Be
Na Mg

Note :

$(1)\,\,\,$ In periodic table from left to right the size of element decreases.

$(2)\,\,\,$ In periodic table, from top to bottom size of element increases.

So, $B{e^{ + 2}}$ will be the smallest in size.

Among $L{i^ + }$ and $N{a^ + },$ size of $N{a^ + } > L{i^ + }$ and among $B{e^{ + 2}}$ and $M{g^{ + 2}},$ size of $M{g^{ + 2}} > B{e^{ + 2}}$.

Note : Size of $L{i^ + } > M{g^{ + 2}},$ this is a exception case.

So, the correct order will be, $N{a^ + } > L{i^ + } > M{g^{ + 2}} > B{e^{ + 2}}$
3

### 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
4

### 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 $B < C < O < N$