Periodic Table & Periodicity

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.

Electron affinity means tendency of gaining an electron by an atom.

In a period from left ot right the electron affinity increases and in a group it decreases from top to bottom.

So according to this theory Fluorine(F) should have most electron affinity. But 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.

Note : In entire periodic table Cl have highest electron affinity.

Here all of them are isoelectric. For isoelectric anion size is more than cation. Among two anions which anion has more negative charge will have more radius and among two cation which cation has less positive charge will have more radius.

In option (D) electron is removed from 4p subshell for 1^st ionization enthalpy, but in all other options electron is removed from 3p subshell. As 3p subshell is closer to nucleus than 4p, then attraction to the electrons in 3p subshell is more compared to the electrons in 4p subhell. That is why removal of electrons from 4p subshell is easy compared to 3p subshell. So among all the options, option (D) will have least ionization enthalpy.

Among (A), (B), (C) options, (C) has half filled 3p subshell so it is more stable compare to the others and you have to provide more energy to remove electrons from stable 3p subshell. That is why electron configuration [Ne] 3s² 3p³ has highest ionization enthalpy.

Na, H, Ba and Sr produce Peroxide.

K, Rb and Cs produces Superoxide.

So, SiO₂ will be oxide.

From the table you can see ionization enthalpy of A is greater than B in all the cases.

So, B comes below A in the group as in a group from top to bottom ionization enthalpy decreases.

After 1^st ionization enthalpy each element become cation by removing a electron. In A⁺ and B⁺, after removing one electron from each element, effective nuclear charge increases as per-electron attraction increase by the neuclers. So, the removal of next electron will be more difficult, therefore more energy is required for 2^nd ionization energy. From the table you can see 2^nd ionization energy is more than first ionization energy for both elements.

But you can see 3^rd ionization enthalpy is so much higher than 2^nd ionization enthalpy. It means after 2^nd ionization enthalpy outermost shell is empty and in 3^rd ionization enthalpy from a new shell electron is removed, as the new shell is closer to the neucleus so the attraction by the nucleus to the electrons of this shell is more and to remove a electron from this shell you have to provide very high energy, that is why 3^rd ionization enthalpy is so high.

If A and B are from group 1 then after 1^st ionization enthalpy outermost shell will be empty and 2^nd electron will be removed from inner shell, so 2^nd ionization will be so high. But from table you can see 2^nd ionization energy is around double of 1^st ionization energy, it is not so much high than 1^st onization energy. So A and B can't be from group 1.

A and B are from group 2 as in group 2 element, outermost shell has 2 electron and after 2^nd ionization enthalpy outermost shell will be empty and in 3^rd ionization enethalpy, third electron will removed from innershell so 3^rd ionization enthalpy will be very high.

(A) Group 17 elements are halogens; they exist in all three states of matter at room temperature.

(B) Except for the lightest element (boron), all other Group 13 elements are relatively electropositive and are thus metals.

(C) Group 15 elements have half-filled electronic configuration, thus have high values of ionization enthalpy, while Group 16 elements achieve halffilled configuration on loss of one electron and therefore have lower ionisation enthalpy.

(i) In Group 15 elements, due to the inert pair affect the stability of +5 oxidation state decreases down the group.

Here all of them are isoelectric and for isoelectric species size of anion increases as negative charge increases.

So, correct order is :

N^3– > O^2– > F^–

$ \therefore $ Radius of N^3– = 1.71 Å

and Radius of O^2– = 1.40 Å

and Radius of F^– = 1.36 Å

Ionization potential of Na means energy required to convert of Na to Na⁺ ion.

Na $ \to $ Na⁺ + e^-          IE = 5.1 ev

Electron gain enthalpy of Na⁺ means energy required to convert Na⁺ ion to Na.

Na $ \to $ Na⁺ + e^-          $\Delta {H_{eq}}$ = - 5.1 ev

According to Lavoisier and Laplace law of thermochemistry, when we change the direction of reaction then the sign of energy also changes.

Those elements are present in periodic table like this -

	Group 2		Group 16		Group 18
Period 3	Ca	..........	S	.........	Ar
Period 4			Se
Period 5	Ba

In periodic table,

1. From left to right ionization energy increases.

2. From top to bottom ionization energy decreases.

$ \therefore $ Ba < Ca and Se < S and S < Ar

$ \therefore $ Correct order is -

Ba < Ca < Se < S < Ar

Iso-electronic ions have same number of electrons. So, for iso-electronic ions, number of electrons = constant.
$ \therefore $ $\sigma $(Slaten's Constant) = Constant. As $\sigma $ depends on number of electrons. If a element's number of electrons increases then that element's $\sigma $ increases.

Also we know, Z_effective = Z - $\sigma $

As for iso-electronic ions, $\sigma $(Slaten's Constant) = Constant. So Z_effective depend on only value of Z. If Z of an ion increases then Z_effective also increases and if Z of an ion decreases then Z_effective also decreases. And when Z_effective increases then nuclear attraction towards outermost electrons increase and size of ion decreases. Similarly when Z_effective decrease then nuclear attraction towards outermost electrons decreases and size of ion increases.

$ \therefore $ Size $ \propto $ ${1 \over {{Z_{eff}}\,or\,Z}}$

	Ca2+	K+	Cl-	S2-
Z	20	19	17	16

$ \therefore $ Ionic radius order

Ca²⁺ < K⁺ < Cl^– < S^2–

You should know that,

(1)$\,\,\,$ In a period, from left to right the acidic strength increases that mean basic nature decreases.

(2)$\,\,\,$ In a group, from top to bottom the basic nature increases that means acidic nature decreases.

Na, Mg, Al and K are present in periodic table like this : -

	Group 1	Group 2		Group 13
Period 3	Na	Mg	.....	Al
Period 4	K

So, you can see Na, Mg and Al are present in same period so the order of basic nature among Na, Mg and Al is Na > Mg > Al.

And Na and K are from same group so basic nature of K > Na.

So the correct order of oxides

Al₂O₃ < MgO < Na₂O < K₂O

Electronic configuration of Gd(64) : [Xe]4f⁷5s²5p⁶5d¹6s²

So the outer electronic configuration will be : 4f⁷5d¹6s²

Shortcut Method : Gd belongs to 4f series which is also called lanthanide series. In this lanthanide series for Cerium (Ce), Gadolinium (Gd) and Lutetium (Lu) outer d orbital has 1 electron and for all other lanthanide series elements outer d orbital has 0 electron. From this you can easily say option (C) is correct.

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

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

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, $N{a^ + } > L{i^ + } > M{g^{ + 2}} > B{e^{ + 2}}$

$(A)\,\,\,$ 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 $B < C < O < N$

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.

The correct order of ionisation enthalpies is $F > P > S > B$

NOTE : On moving along a period ionization enthalpy increases from left to right and decreases from top to bottom in a group. But this trend breaks up in case of atom having fully or half filled stable orbitals.

In this case $P$ has a stable half filled electronic configuration hence its ionisation enthalpy is greater in comparison to $S.$ For B(5) electronic configuration is = 1s²2s²2p¹ and P(15) electronic configuration is = 1s²2s²2p⁶3s²3p³. As P has half filled valence shell so removing electron will be harder compare to B. Hence by checking all options the correct possible order is

B < S < P < F

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.

$\left( a \right)\,\,\,\,\,$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

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

$O$ 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.

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 ${F^ - }.$

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

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

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.

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.

This can't be $f$ - block elements as $f$ block means only lanthanides and actinides but we know lanthanide contraction happens on the elements before lantanidine elements also. So, radius also decrease for those element when increase the atomic number.

Radioactive Series starts from atomic number $84$ ( Polonium ) and affer that all are radioactive. But we know before $84$ atomic number also the atomic size decrease with increase in atomic number. So, only radioactive series don't show this characteristic.

So, correct answer should be high atomic mass.

According to modified modern periodic law, physical and chemical properties of the elements are periodic functions of their atomic numbers.

Those are the elements of lanthanides series.


In lanthanide series the extra electron is added in $4f$ subshell and we know in periodic table from left to right the effective nuclear charge increases as electron is added so the size of elements will decreases.

So, the correct order is $Ce > Sm > Tb > Lu.$

Here $C{e^{ + 3}},\,\,P{m^{ + 3}},\,\,Y{b^{ + 3}}$ are from lanthanide series.

$L{a^{ + 3}}$ atomic number is $57.$ This is a $d$ block element but all the property of $La$ is similar to lanthanides. It is present before $Ce.$

As we know in lanthanides the radius of elements decreases from left to right of periodic table.

So, order will be $L{a^{ + 3}} > C{e^{ + 3}} > P{m^{ + 3}} > Y{b^{ + 3}}$