Periodic Table & Periodicity

Across a period (left to right) in the periodic table :

Ionization enthalpy tends to increase

Electron gain enthalpy (in magnitude) tends to increase (more negative values)

Electronegativity tends to increase

Atomic radius tends to decrease

Hence, atomic radius shows the opposite trend compared to the other three properties. Therefore, the correct answer is

(D) Atomic radius.

First ionization enthalpies = 496 kJ/mole

Second ionization enthalpies = 4560 kJ/mol

According to the given information, the difference between first and second ionization enthalpy is very high so Metal belong to 1st group i.e. Monovalent cation.

Metal hydroxide will be of type, MOH.

MOH + HCl $ \to $ MCl + H₂O

MOH + ${1 \over 2}$H₂SO₄ $ \to $ ${1 \over 2}$M₂SO₄ + H₂O

So one mole of HCl required to react with one mole MOH.

And ${1 \over 2}$ mole of H₂SO₄ required to react with one mole MOH.

Non metal oxide $ \to $ acidic.

Metal oxide $ \to $ basic.

Al₂O₃ amphoteric.

1st I.E. of Be > B

In case of Be, electron is removed from 2s orbital which has more penetration power, while in case of B electron is removed from 2p orbital which has less penetration power.

2p electron of B is more shielded from nucleus by the inner electrons than 2s electrons of Be

$ \therefore $ It is easier to remove 2p electron than 2s electron.

Eu²⁺ : [xe] 4f⁷

Ce³⁺ : [xe] 4f¹

Generally in a period Left to Right Atomic radius decrease.

In a Group Top to Bottom Atomic radius increase.

Atomic radius order : Br > Cl > C > O > F

Elecronic configuration of Na = [Ne] 3s¹
Mg = [Ne] 3s²
Al = [Ne] 3s²3p¹
Si = [Ne] 3s²3p²

Correct order is :

Na $<$ Al $<$ Mg $<$ Si

Electronic configuration of

₂₅Mn = [Ar]3d⁵4s²

₂₅Mn²⁺ = [Ar]3d⁵4s⁰

₂₆Fe = [Ar]3d⁶4s²

₂₆Fe²⁺ = [Ar]3d⁶4s⁰

₂₇Co = [Ar]3d⁷4s²

₂₇Co²⁺ = [Ar]3d⁷4s⁰

₂₈Ni = [Ar]3d⁸4s²

₂₈Ni²⁺ = [Ar]3d⁸4s⁰

So third ionisation energy is minimum for Fe.

Electron affinity of second period p-block element is less than third period p-block element due to small size of second period p-block element.

$ \therefore $ Electron affinity order : F $<$ Cl

Down the group electron affinity decreases due to size increases.

$ \therefore $ Electron affinity E.A. order : S $>$ Se

Li $>$ Na

Order of electron gain enthalpy (magnitude) is Cl > F > Br > I

Note: Electron gain enthalpy increases with electro negativity but chlorine has higher electron gain enthalpy than fluorine (exception).

The basic difference in approach between Mendeleev's periodic law and Modern periodic.

law is as follows :

The Mendeleev's periodic law is based on atomic weight while modern periodic law is based on atomic number. i.e. according to Mendeleev's periodic law.

If the element are arranged in order of increasing atomic weights their properties vary in definite manner from member to member of the series, but return more or less nearly to same value at certain fixed points in the series, while according to modern periodic law.

Physical and chemical properties of the elements are periodic function of their atomic numbers.

Hence, option (b) is the correct answer.

A diagonal relationship is said to exist between certain pairs of diagonally adjacent elements in the 2nd and 3rd periods.

(a) Li shows diagonal relationship with Mg. Due to comparable :

Electronegativities

$ \mathrm{Li}=1.0 \Rightarrow \mathrm{Mg}=1.20 $

(b) Atomic size, i.e. atomic radii (in pm)

$ \mathrm{H}=152 \Rightarrow \mathrm{Mg}=160 $

(c) Atomic volume

$ \begin{aligned} & \mathrm{Li}=12.97 \mathrm{~mL} \Rightarrow \mathrm{Mg} \\ = & 13.97 \mathrm{~mL} \end{aligned} $

Thus, Li-resembles Mg in many respects.

Hence, option (a) is the correct answer.

The electronic configuration of $\mathrm{Be}^{2+}$ is $\left[{ }_2 \mathrm{He}\right], \mathrm{Na}^{+}$is $\left[{ }_{10} \mathrm{Ne}\right]$, $\mathrm{Cl}^{-}$is $\left[{ }_{18} \mathrm{Ar}\right]$ and $\mathrm{Br}^{-}$is $\left[{ }_{36} \mathrm{Kr}\right]$.

We know, site of group 18 elements follow the order

$ \mathrm{He}<\mathrm{Ne}<\mathrm{Ar}<\mathrm{Kr}<\mathrm{Xe} . $

So, the order of ionic radii of the given ions will be

$ \mathrm{Br}^{-}>\mathrm{Cl}^{-}>\mathrm{Na}^{+}>\mathrm{Be}^{2+} $

The maximum number of elements for sixth period of the periodic table are 32, i.e. from atomic number $(Z)=55$ to atomic number $(Z) =86 .(Z)=55$ stands for caesium and $( Z)=86$ stands for radon. Hence, option (c) is the correct answer.

Higher the electrostatic attraction between nucleus and valence electron, smaller will be the size of the atom/ion. Electrostatic attraction is given by $Z_{\text {eff }}=Z-S$ where, $Z=$ the number of protons in the nucleus of an atom or ion (the atomic number) and $S=$ shielding from core electrons. In case of $\mathrm{Al}^{3+}$ and $\mathrm{Mg}^{2+}$, they are isoelectronic species and thus have same number of electrons as $1 s^2, 2 s^2, 2 p^6$ isoelectric series of atoms and ions with different numbers of protons (and thus different nuclear attraction, gives) the relative ionic sizes of each atom or ion with respect to atomic number. Atomic number, $Z_{\mathrm{Al}}=13$ and $Z_{\mathrm{Mg}}=12$. Thus, $\mathrm{Al}^{3+}$ has lower ionic radii than $\mathrm{Mg}^{2+}$ due to higher nuclear charge. Hence, A is true, R is true and R is the correct explanation for A .

Maximum difference is between the fifth and fourth ionisation energy. (Removal of the 5th electron requires almost more than double the energy required to remove the 4th electron.) This implies that the first four electrons are removed from the valence shell. Hence, number of valence electrons in the atom ' $X$ ' will be four.

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.

Here Z of K = 15, Na = 11, H = 1, Li = 3.

In K, because of more number of protons (high atomic number) the 2s electron experiences a higher effective nuclear charge and is closer to nucleus, thus having less energy

Electronic configuration of Potassium (K) :

1s²2s²2p⁶3s²3p⁶4s¹

After first ionisation enthalpy its configuration becomes :

1s²2s²2p⁶3s²3p⁶

which is a inert gas configuration. To remove an electron from inert gas configuration requires very high energy.

So Potassium (K) have high difference in the first ionisation and the second ionisation energy as it achieve stable noble gas configuration after first ionisation.

	Cl-	Ar	Ca+2
Protons	17	18	20
Electrons	18	18	18

Nuclear charge means number of protons present in the nucleus. Here maximum number of protons present in the Ca⁺² ion, so the attraction towards those 18 electrons wll be most by the nucleus of the Ca⁺² ion. That is why size of the Ca⁺² ion wll be least because of highest nuclear charge.

Electronic configuration of

Sm = [Xe] 4f⁶ 6s²

$ \therefore $ Electronic configuration of

Sm⁺³ = [Xe] 4f⁵

Sm⁺³ shows yellow colour due to partially filled f orbital and f-f transition.

Electronic configuration of

La⁺³ = [Xe] 4f⁰

Lu⁺³ = [Xe] 4f¹⁴

Gd⁺³ = [Xe] 4f⁷

La⁺³, Lu⁺³ and Gd⁺³ are colourless due to absence of f-f transition.

Electronic configuration of element Og (118) is
[Rn] 5f¹⁴6d ¹⁰7s ²7p⁶, where Og is oganesson.

[Og₁₁₈] 8s² is configuration for Z = 120,

As per the configuration it is in II^nd group. Thus, element with Z = 120 will be an alkaline earth metal.

Electronegativity increases from left to right in a period and decreases down the group. The correct orders are

Si > Al, Ga < Ge, Te < Se and P < S.

Atomic radii increase by moving down the group and decrease across a period.

Hence, the correct order of atomic radii is : C < S < Al < Cs.

E.N. of Al = (1.5) $ \cong $ Be (1.5)

Second electron gain enthalpy is always positive for every element.
O^$-$_(g) + e^$-$ $ \to $ O^$-$2_(g) ; $\Delta $H = positive

Electronic configuration of Thallium (Tl) is
= [Xe]4f¹⁴ 5d¹⁰ 6s² 6p¹

Here because of poor sheilding of 4f and 5d orbital on 6s orbital, electrons of 6s orbital is more tightly held by the nucleus and perticipate less in bond formation. This is called as innert pair effect. And that is why Tl become stable in +1 oxidation state.

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.