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Periodic Table & Periodicity

JEE Chemistry · 270 questions · Page 5 of 27 · Click an option or "Show Solution" to reveal answer

Q41
Which one of the following ions has the highest value of ionic radius?
A O2-
B B3+
C Li+
D F-
Correct Answer
Option A
Solution

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

xx

when it loose a electron then it becomes cation

x−e−→x+x - {e^ - } \to {x^ + }

Let atomic no. of

xx

is

zz

then no. of proton in

x+=z{x^ + } = z

and no. of electron in

x+=z−1{x^ + } = z - 1

So the ratio

ze{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

xx

becomes

x−{x^ - }

by adding a electron

(x+e−→x−)\left( {x + {e^ - } \to {x^ - }} \right)

then no. of proton in

x−=z{x^ - } = z

and no. of electron in

x−=z+1.{x^ - } = z + 1.

Now the ratio of

ze  {z \over e}\,\,

in

x−{x^ - }

decreases as no. of electron increases. As electron increases in

x−{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

44

ions are from second period and for

O2−  {O^{2 - }}\,\,

and

F−{F^ - }

both have

1s,1s,
2s2s

and

2p2p

shell and for

Li+  L{i^ + }\,\,

and

B3+{B^{3 + }}

both have

1s1s

shell. So, size of

O2−   {O^{2 - }}\,\,\,

and

   F−   \,\,\,{F^ - }\,\,\,

are more then

Li+    L{i^ + }\,\,\,\,

and

    B3+\,\,\,\,{B^{3 + }}

and among

O2−   {O^{2 - }}\,\,\,

and

    F−\,\,\,\,{F^ - }

For

   \,\,\,
O2−,  ze=810=0.8{O^{2 - }},\,\,{z \over e} = {8 \over {10}} = 0.8

For

   \,\,\,
F−,   ze=910=0.9{F^ - },\,\,\,{z \over e} = {9 \over {10}} = 0.9

as for

O2−,  ze   {O^{2 - }},\,\,{z \over e}\,\,\,

is less so per electron attraction from nucleus decreases and radius increases more than

F−.{F^ - }.
Q42
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
Correct Answer
Option B
Solution

In periodic table from left to right (group

11

to group

1818

) acidic strength increases.

AlAl

belongs to group

13,13,
SiSi

belongs to group

14,14,
PP

belongs to group

1515

and

SS

belongs to group

1616

. So,

   Al2O3\,\,\,A{l_2}{O_3}

will be least acidic and

So2S{o_2}

will most acidic among then. So, right order will be

Al2O3<SiO2<P2O3<SO2A{l_2}{O_3} < Si{O_2} < {P_2}{O_3} < S{O_2}
Q43
In general, the properties that decrease and increase down a group in the periodic table, respectively, are :
A Atomic Radius and Electronegativity
B Electron Gain Enthalpy and Electronegativity.
C Electronegativity and Atomic Radius.
D Electronegativity and Electron Gain Enthalpy.
Correct Answer
Option C
Solution

Electronegativity decreases down the group because the increased number of energy levels puts the outer electrons very far away from the pull of nucleus.

Atomic radius increases down the group because the number of energy levels increases when you move down the group.

Each subsequent energy level is further from the molecules than the last.

That is why atomic radius increases down the group.

Q44
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)−O_{(g)}^{-} Δ\Delta Ho = -142 kJmol-1 O(g)−O_{(g)}^{-} + e- = O(g)2−O_{(g)}^{2-} Δ\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
Correct Answer
Option A
Solution
OO

atom is highly electronegative so will add first electron easily by releasing energy.

So it is an exothermic.

After adding first electron

OO

becomes

O−  {O^ - }\,\,

and size of

O−  {O^ - }\,\,

becomes slightly more than

OO

atom. Now when a new electron is trying to add into

  O−\,\,{O^ - }

ion, two forces will be act on new electron →\to

(1)    \left( 1 \right)\,\,\,\,

attraction between nucleus and new electron

(2)    \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−(g)+e−⟶ O2−(g)   {O^ - }\left( g \right) + {e^ - }\overset{\,}\longrightarrow {O^{2 - }}\left( g \right)\,\,\,

is an endothermic reaction.

Q45
Which of the following oxides is amphoteric in character?
A CaO
B CO2
C SiO2
D SnO2
Correct Answer
Option D
Solution

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

(1)    Zn\left( 1 \right)\,\,\,\,Zn
(2)    Be\left( 2 \right)\,\,\,\,Be
(3)    Al\left( 3 \right)\,\,\,\,Al
(4)    B\left( 4 \right)\,\,\,\,B
(5)    Cr\left( 5 \right)\,\,\,\,Cr
(6)    Ga\left( 6 \right)\,\,\,\,Ga
(7)    Pb\left( 7 \right)\,\,\,\,Pb
(8)    Sn\left( 8 \right)\,\,\,\,Sn

Here

Zn,Be,Al,Pb   Zn,Be,Al,Pb\,\,\,

and

 \,
SnSn

are mostly amphoteric.

B,CrB, Cr

and

GaGa

are less amphoteric.

Q46
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
Correct Answer
Option C
Solution
(a)     \left( a \right)\,\,\,\,\,

It is True. Here

Li,Na,KLi, Na, K

and

RbRb

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.

(b)     \left( b \right)\,\,\,\,\,

It is True. In entire periodic table

ClCl

(chlorine) has the highest, and in a group it decreases from top to bottom. So the correct order is

I<Br<F<Cl.{\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.

(c)     \left( c \right)\,\,\,\,\,

It is False. Electronic configuration of

B(5)   =   1s22s22p1B\left( 5 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^1}
C(6)   =   1s22s22p2C\left( 6 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^2}
N(7)   =   1s22s22p3N\left( 7 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^3}
O(8)   =   1s22s22p4O\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(1s2 2s2 2p3)N\left( {1{s^2}\,2{s^2}\,2{p^3}} \right)

has stable half filled

2p2p

subshell so it is more stable than

O.O.

So, correct order is

B<C<O<NB < C < O < N
(d)     \left( d \right)\,\,\,\,\,

It is True.

Al3+,Mg2+,Na+  A{l^{3 + }},M{g^{2 + }},N{a^ + }\,\,

and

  F−\,\,{F^ - }

are isoelectric. So all have

1010

electrons. So,

  Ze  \,\,{Z \over e}\,\,

of

Al3+=1310=1.3A{l^{3 + }} = {{13} \over {10}} = 1.3
  Ze  \,\,{Z \over e}\,\,

of

Mg2+=1210=1.2M{g^{2 + }} = {{12} \over {10}} = 1.2
  Ze  \,\,{Z \over e}\,\,

of

Na+=1110=1.1N{a^ + } = {{11} \over {10}} = 1.1
  Ze  \,\,{Z \over e}\,\,

of

F−=910=0.9{F^ - } = {9 \over {10}} = 0.9

We know

ze{z \over e}

means

zz

no of protons in nucleus is pulling

ee

no. of electrons present in the orbital. So for

Al3+, 13A{l^{3 + }},\,13

protons is pulling

1010

electron so the size of

Al3+A{l^{3 + }}

will decrease most. That is why more

ze{z \over e}

means less size of ion. So, the correct order is

Al3+<Mg2+<Na+<F−A{l^{3 + }} < M{g^{2 + }} < N{a^ + } < {F^ - }
Q47
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
Correct Answer
Option D
Solution

For alkali metals on going down the group, number of electron shells increase, so forces between the oppositely charged electron and nucleus is less so electron can be removed easily, therefore, reactivity increases.

But for halogens they react by abstracting electron from less electronegative atoms and the attraction for additional electron is stronger when there are fewer filled electron shells between the valence electrons and atomic nucleus.

Q48
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
Correct Answer
Option C
Solution
(A)   (A)\,\,\,

Option A is true. Here in

PbO2Pb{O_2}

oxidation number of

PbPb

is

+4,+4,

due to inert pair effect the oxidation number

+4+4

is not stable for

PbPb

, So, to become stable oxide

+2+2
PbPb

will oxidize other and reduced to

Pb+2.P{b^{ + 2.}}

So,

PbO2Pb{O_2}

has the highest oxidizing power.

(B)    (B)\,\,\,\,

Option B is true. Bond length from top to bottom in periodic table increases, So the bond length order is

HI>HBr>HCl>HFHI > HBr > HCl > HF

The compound which can give

H+{H^ + }

ion easily that compound has higher acidic strength. As the bond length between

HH

and

II

in

HIHI

is longest then it will be carier to break the bond between

HH

and

II

. So

HIHI

will have highest acidic strength.

(C)    (C)\,\,\,\,

Option C is false. The structure of those

44

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

22

nd period so in

NN

new electron is added in the second period and as size of

NN

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

NH3.N{H_3}.

On the other hand

pp

is a

33

rd block, As is a

44

th block and

SbSb

is a

55

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

PH3,AsH3P{H_3},As{H_3}

and

SbH3Sb{H_3}

will have less ability to donate the electron. So the correct order is

SbH3<AsH3<PH3<NH3Sb{H_3} < As{H_3} < P{H_3} < N{H_3}

(D) Option (D) is true Electronic configuration of

B(5)   =   1s22s22p1B\left( 5 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^1}
C(6)   =   1s22s22p2C\left( 6 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^2}
N(7)   =   1s22s22p3N\left( 7 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^3}
O(8)   =   1s22s22p4O\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(1s2 2s2 2p3)N\left( {1{s^2}\,2{s^2}\,2{p^3}} \right)

has stable half filled

2p2p

subshell so it is more stable than

O.O.

So, correct order is

B<C<O<NB < C < O < N
Q49
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+
Correct Answer
Option D
Solution

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

O2−>F−>Na+>mg2+>Al3+{O^{2 - }} > {F^ - } > N{a^ + } > m{g^{2 + }} > A{l^{3 + }}
Q50
In comparison to boron, berylium has :
A lesser nuclear charge and greater first ionisation enthalpy
B greater nuclear charge and greater first ionisation enthalpy
C greater nuclear charge and lesser first ionisation enthalpy
D lesser nuclear charge and lesser first ionisation ethalpy
Correct Answer
Option A
Solution

Electronic configuration of Be(4) = 1s22s2 Electronic configuration of B(5) = 1s22s22p1 Removing electron from outer most shell of Be is harder compare to B, as for Be 2s orbital is full filled so it is stable therefore required more ionization enthalpy to remove an electron.

We know in a period from left to righ effective nuclear charge increases, so B will have more nuclear charge.

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