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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2005

MCQ (Single Correct Answer)
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

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

AIEEE 2005

MCQ (Single Correct Answer)
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.
3

AIEEE 2004

MCQ (Single Correct Answer)
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.
4

AIEEE 2004

MCQ (Single Correct Answer)
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

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

Questions Asked from Periodic Table & Periodicity

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