Periodic Table & Periodicity

The correct order of atomic radii is given in (ii) and (iv).

The remaining are incorrect. Their correct form are $\mathrm{Li}>\mathrm{Cl}>\mathrm{F}$,

$\mathrm{Eu}>\mathrm{Sm}>\mathrm{Tm}$

The correct match is A-IV, B-I, C-II, D-V.

Let's determine the block for each element provided:

(A) Cd (Cadmium)

Cadmium (atomic number 48) is located in Group 12, Period 5 of the periodic table.

Its electron configuration is [Kr] 4d¹⁰ 5s².

Elements in Groups 3 through 12, where the d-orbitals are being filled, are known as d-block elements. Cadmium fits this description.

So, (A) Cd matches with (III) d-block.

(B) Eu (Europium)

Europium (atomic number 63) is a lanthanide element.

Lanthanides and actinides are typically placed below the main body of the periodic table and comprise the f-block elements, where the f-orbitals are being filled.

Its electron configuration is [Xe] 4f⁷ 6s².

So, (B) Eu matches with (I) f-block.

(C) Se (Selenium)

Selenium (atomic number 34) is located in Group 16, Period 4 of the periodic table.

Its electron configuration is [Ar] 3d¹⁰ 4s² 4p⁴.

Elements in Groups 13 through 18, where the p-orbitals are being filled, are known as p-block elements. Selenium fits this description.

So, (C) Se matches with (IV) p-block.

(D) Ba (Barium)

Barium (atomic number 56) is located in Group 2, Period 6 of the periodic table.

Its electron configuration is [Xe] 6s².

Elements in Group 1 (alkali metals) and Group 2 (alkaline earth metals), where the s-orbitals are being filled, are known as s-block elements. Barium fits this description.

So, (D) Ba matches with (II) s-block.

Combining the matches:

A - III

B - I

C - IV

D - II

The given atomic number 78 corresponds to the element platinum (Pt).

The electronic configuration of platinum is

$ \operatorname{Pt}(Z=78)=[X e] 4 f^{14} 5 d^9 6 s^1 $

From the electronic configuration, the highest principal quantum number is 6 (from $6 s^1$ ), so the period number $x=6$ For $d$-block elements, the group number is obtained by adding the electrons that is present in the $(n-1) d$ and $n s$-orbitals.

Here, $5 d^9 6 s^1, 10$ electrons

Hence, the group number $y=10$

Therefore, the sum $x+y=6+10=16$.

The first ionisation enthalpy generally increases across a period due to increasing effective nuclear charge. However, exceptions to this trend occur when a stable electronic configuration is involved.

In the second period (Li to Ne), two such exceptions occur.

1. Beryllium (Be) The electronic configuration is $1 s^2 2 s^2$. The first ionisation enthalpy of beryllium is higher than that of boron, the succeeding element. It is also higher than that of proceeding Ind element is lithium. This is due to the stable, completely filled 2 s subshell, which makes it more difficult to remove an electron.

2.Nitrogen ( N ) The electronic configuration is $1 s^2 2 s^2 2 y^3$. The first ionisation enthalpy of nitrogen is higher than that of oxygen, the succeeding element. It is also higher than the proceeding element is carbon. This is because of the extra stability associated with the half-filled $2 p$ subshell, which resists the removal of an electron.

Therefore, the two elements $X$ and $Y$ are beryllium ( Be ) and nitrogen ( N ).

Let's determine the electron gain enthalpies for the given elements based on periodic trends and known values. Electron gain enthalpy ($\Delta_{eg} H$) is the energy change when an electron is added to a neutral gaseous atom. A more negative value indicates a greater tendency to accept an electron (more exothermic process).

General Trends:

1. Group 17 (Halogens) have the most negative electron gain enthalpies. Among halogens, chlorine (Cl) generally has a more negative electron gain enthalpy than fluorine (F) because F's small size leads to significant interelectronic repulsion when an incoming electron is added, making it less favorable than Cl.

   • Chlorine (Cl): Expect a very large negative value.

   • Fluorine (F): Expect a large negative value, but slightly less negative than Cl.

2. Group 16 (Chalcogens) also have negative electron gain enthalpies. Similar to halogens, sulfur (S) generally has a more negative electron gain enthalpy than oxygen (O) due to O's small size and resulting interelectronic repulsion.

   • Sulfur (S): Expect a negative value.

   • Oxygen (O): Expect a negative value, but less negative than S.

Let's examine the provided values in List-II:

I. -200 kJmol⁻¹

II. -349 kJmol⁻¹

III. -141 kJmol⁻¹

IV. -328 kJmol⁻¹

V. +48 kJmol⁻¹ (This is a positive value, meaning energy is required to add an electron, which is characteristic of elements like noble gases or alkaline earth metals, but not O, F, Cl, S for the first electron gain).

Now let's match the elements:

1. Chlorine (Cl): It is known to have the most negative electron gain enthalpy among all elements. Among the given negative values, -349 kJmol⁻¹ is the most negative.

   • So, C (Cl) $\rightarrow$ II (-349 kJmol⁻¹).

2. Fluorine (F): It has a very large negative electron gain enthalpy, but due to its smaller size and interelectronic repulsion, it's slightly less exothermic than Cl. The second most negative value is -328 kJmol⁻¹.

   • So, B (F) $\rightarrow$ IV (-328 kJmol⁻¹).

3. Sulfur (S): Belonging to Group 16, its electron gain enthalpy is exothermic, and it should be more negative than Oxygen. Comparing the remaining negative values (-200 and -141), -200 kJmol⁻¹ is more negative.

   • So, D (S) $\rightarrow$ I (-200 kJmol⁻¹).

4. Oxygen (O): Also in Group 16, its small size leads to interelectronic repulsion, making its electron gain enthalpy less negative (less exothermic) than that of Sulfur. The remaining negative value is -141 kJmol⁻¹.

   • So, A (O) $\rightarrow$ III (-141 kJmol⁻¹).

Let's summarize the matching:

• A (O) $\rightarrow$ III (-141 kJmol⁻¹)

• B (F) $\rightarrow$ IV (-328 kJmol⁻¹)

• C (Cl) $\rightarrow$ II (-349 kJmol⁻¹)

• D (S) $\rightarrow$ I (-200 kJmol⁻¹)

Comparing this with the given options:

Option C: A-III, B-IV, C-II, D-I

This matches our derived assignments.

The final answer is $\boxed{\text{Option C}}$

Among the given elements, element II is most reactive metals.

The reason is that, it has low 1st ionisation enthalpy, high 2nd ionisation enthalpy and low electron gain enthalpy.

From periodic table,

$E_1$ will be Gallium with $\mathrm{Z}_1=31$

$E_2$ will be Bismuth with $Z_2=83$

$Z_2-Z_1=83-31=52$

Element $X=\mathrm{Cl}, Y=\mathrm{S}, Z=\mathrm{I}$

More the value of electron gain enthalpy, greater is the tendency to gain electron.

The correct order of increasing atomic radius is $\mathrm{F}<\mathrm{N}<\mathrm{P}<\mathrm{Si}<\mathrm{Na}$.

It is because atomic radius decreases across a period. While it increases down the group.

(i) Elements with atomic number 13, $11,9,7$ and 16 are aluminium, sodium, fluorine, nitrogen and sulphur respectively. Thus, largest ions is $\mathrm{S}^{2-}$, sulphide ion while smallest ion will be $\mathrm{Al}^{3+}$ ion. So, elements $X$ and $Y$ are $E$ and $A$ respectively.

The correct order of atomic radii in group 13 elements is, $\mathrm{Tl}>\mathrm{In}>\mathrm{Al}>\mathrm{Ga}$.

Usually radii increases down the group, due to addition of shell, but Ga has smaller radii than Al due to poor shielding effect of $d$-electrons.

Amphoteric oxide:

$\mathrm{SnO}, \mathrm{PbO}, \mathrm{SnO}_2, \mathrm{PbO}_2, \mathrm{Ga}_2 \mathrm{O}_3$

Acidic oxide : $\mathrm{CO}_2, \mathrm{~B}_2 \mathrm{O}_3$

Neutral : CO

The electronic configurations of the given elements are as follows:

Na (11): [Ne] 3s¹

Mg (12): [Ne] 3s²

Al (13): [Ne] 3s² 3p¹

Si (14): [Ne] 3s² 3p²

The ionization energy for Mg is greater than that for Na due to its fully filled 3s orbitals, which offer greater stability. In contrast, Al has a lower ionization energy than Mg because it has one electron more than a stable configuration, making it easier to remove. However, Al's ionization energy is still less than that of Si because Si has a higher effective nuclear charge, which increases its ionization enthalpy.

Thus, the correct order of their first ionization enthalpies is:

$ \text{Na} < \text{Mg} > \text{Al} < \text{Si} $

Among the options provided, only option (A) $ 575 \, \text{kJ/mol} $ conforms to this order.

$ \mathrm{Kr}, \mathrm{Xe}, \mathrm{Rn} $ has positive electron gain enthalpies. All of above gases are noble gases. All the occupied subshell of noble gases are completely filled. Therefore, the additional electron has to be placed in an orbital of next higher shell. We can also say that energy has to be supplied to add an additional electron and hence, electron gain enthalpy of all noble gases are positive.

A is not correct but R is correct.

The correct form of A is $ \mathrm{Na}^{+} $has smaller size than $ \mathrm{F}^{-} $because after loosing an electron Na will has less electron cloud and more effective nuclear charge whereas F after gaining an electron will have less effective nuclear charge than the parent atom.

Hence, $ \mathrm{F}^{-} $size increases.

$ \mathrm{Na}^{+} $has smaller size than $ \mathrm{F}^{-} $. It is because after losing an electron Na will have more effective nuclear charge whereas $ F $ after gaining an electron will have less effective nuclear charge than the parent atom so the size increases.

In the given species the ratio of $s$-electrons and $p$-electrons are same in $\mathrm{K}^{+}, \mathrm{Cr}^{3+}$.

$\mathrm{K}^{+}$(18 electrons) : $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 \quad \ldots \text { (i) }$

$ \begin{gathered} s \text {-electrons }=6 ; p \text {-electron }=12 \\\\ s: p=6: 12 \text { or } 1: 2 \quad \ldots \text { (ii) } \end{gathered} $

Similarly,

$\mathrm{Cr}^{3+}$ (21 electrons) :

$ \begin{gathered} 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^3 \quad \ldots \text { (iii) } \\\\ s \text {-electrons }=6 ; p \text {-electron }=12 \\\\ s: p=6: 12 \text { or } 1: 2 \quad \ldots \text { (iv) } \end{gathered} $

Option D is correct:

A - II: Ra (Radium) belongs to the s-block as it is an alkaline earth metal.

B - I: Uuq (Ununquadium) is now known as Fl (Flerovium) and belongs to the p-block, as it is a part of the post-transition metals.

C - IV: Ds (Darmstadtium) belongs to the d-block, as it is a transition metal.

D - III: Fm (Fermium) belongs to the f-block, as it is an actinide element.

These elements are classified based on their electron configurations, which determine their position in the periodic table's blocks.

Electronic configuration of $P$

$ \Rightarrow \quad \begin{aligned} & 1 s^2 2 s^2 3 p^6 3 s^2 3 p^3 \\\\ & s_{\text {electron }}=6, p_{\text {electron }}=9 \\\\ & s: p \Rightarrow 6: 9 \text { or } 2: 3 \quad \ldots \text { (i) } \end{aligned} $

Electronic configuration of Ca

$ \begin{aligned} & 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 \\\\ & \Rightarrow \quad s_{\text {electron }}=8, p_{\text {electron }}=12 \\\\ & \quad s: p \Rightarrow 8: 12 \text { or } 2: 3 \quad \ldots \text { (ii) } \end{aligned} $

Hence, P and Ca have same ratio of $S_{\text {electron }}$ and $p_{\text {electron }}$.

Electron gain enthalpy generally becomes more negative across a period as we move from left to right. Within a group, electron gain enthalpy becomes less negative down the group. However adding an electron to the $2 p$-orbital leads to greater repulsion than adding an electron to the larger $3 p$-orbital. Hence, most electronegative enthalpy is chlorine and least electron gain enthalpy is oxygen.

Assertion (A) is true but Reason (R) is false.

Nitrogen has more ionisation enthalpy than oxygen due to half-filled $p$-orbital, which is more stable.

The atomic radius of gallium is less than that of aluminium. This is due to poor shielding effect of $d$-electrons which leads to a greater attraction of the outermost electron, in turn leads to smaller size of gallium.

The correct match is A-IV, B-III, C-I, D-II.

${ }_{56} \mathrm{Ba}$ is an alkaline earth metal,

So, group number = 2

Block $=s$-block.

For, $s$-block element, period number can be calculated for its outermost electronic configuration.

$ { }_{56} \mathrm{Ba}=[\mathrm{Xe}] 6 s^2 $

Hence, period number $=6$

(ii) ${ }_{50} \mathrm{Sn}=[\mathrm{Kr}] 4 d^{10} 5 s^2 5 p^2$

Its a $p$-block element.

Hence, period number = 5

Group number $=10+4=14$

(iii) ${ }_{27} \mathrm{Co}=[\mathrm{Ar}] 3 d^7 4 s^2$

Its a $d$-block element.

Hence, period number = 4

Group number $=7+2=9$

(iv) ${ }_{58} \mathrm{Ce}=[\mathrm{Xe}] 4 f^1 5 d^1 6 s^2$

Its a $f$-block element.

Hence, period number = 6

Group number = 3

Metallic character generally increases as the ionisation enthalpy decreases. Thus, it is highest for alkali metals and lowest for halogen or we can say that it decreases on moving from left to right in the period.

Thus, the order is, $\mathrm{Sn}<\mathrm{Cd}<\mathrm{Zr}<\mathrm{Sr}$ (14th group) (12th group) (4th group) (2nd group)

The correct order of second ionisation enthalpy is oxygen $>$ fluorine > nitrogen > carbon.

Electronic configuration of elements after removal of first electron,

$ \begin{aligned} & \mathrm{O}^{+}=1 s^2, 2 s^2, 2 p^3 \\ & \mathrm{~F}^{+}=1 s^2, 2 s^2, 2 p^4 \\ & \mathrm{~N}^{+}=1 s^2, 2 s^2, 2 p^2 \\ & \mathrm{C}^{+}=1 s^2, 2 s^2, 2 p^1 \end{aligned} $

The second I.E. of oxygen is more, as it acquires a half-filled stable electronic configuration. So, require a lot of energy to remove electron from oxygen. Carbon has least I.E. as it will remove one electron to attain a stable fully-filled configuration.

The electron gain enthalpy $(E A)$ increases along a period and decreases down the group. So, elements of group? 17 i.e., Cl and F have more (-ve) ${ }^{\text {EA }}$ than 16th group element i.e., $s$ has more (-ve) EA than 15 th group element P. However, EA of F is less than Cl due to small size of fluorine. Therefore, correct order of electron affinity is $\mathrm{Cl}>\mathrm{F}>\mathrm{S}>\mathrm{P}$.

$ \begin{array}{llll} \hline \text { Elements } & \begin{array}{l} \text { Boiling } \\ \text { point }\left(T_2(\mathrm{~K})\right) \end{array} & \begin{array}{l} \text { Melting } \\ \text { point }\left(T_1(\mathrm{~K})\right) \end{array} & \begin{array}{l} T_2-T_1 \\ =T(\mathrm{~K}) \end{array} \\ \hline \mathrm{Al} & 2740 & 933.5 & 1806.5 \\ \hline \mathrm{Ga} & 2478 & 302.92 & 2175.08 \\ \hline \mathrm{In} & 2350 & 429.78 & 1920.22 \\ \hline \mathrm{~B} & 4275 & 2365 & 1910 \\ \hline \end{array} $

From above table, it clearly seen that, $T_2-T_1$ is maximum for Ga .

Let's analyze each statement one by one:

Statement (I): "The order of first ionisation enthalpy of first three elements of 3rd period is $\mathrm{Mg}>\mathrm{Al}>\mathrm{Na}.$"

The first three elements in the 3rd period are sodium (Na), magnesium (Mg), and aluminium (Al). The experimental ionisation energies are approximately:

Na: 496 kJ/mol

Mg: 738 kJ/mol

Al: 578 kJ/mol

Since magnesium has the highest ionisation energy, followed by aluminium and then sodium, the order $\mathrm{Mg}>\mathrm{Al}>\mathrm{Na}$ is correct.

Statement (II): "The element with electronegativity of 3.5 is chlorine."

On the commonly used Pauli electronegativity scale, the values are approximately:

Fluorine: 4.0

Oxygen: 3.5

Chlorine: 3.16

Hence, an electronegativity of 3.5 corresponds more nearly to oxygen, not chlorine. This statement is incorrect.

Statement (III): "The order of sizes of ions $\mathrm{Mg}^{2+}, \mathrm{Na}^{+}, \mathrm{F}^{-}, \mathrm{O}^{2-}$ is $\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}^{-}<\mathrm{O}^{2-}.$"

All these ions are isoelectronic (they each have 10 electrons). Their sizes depend on the nuclear charge. More protons pull the electrons closer, giving a smaller ionic radius.

$\mathrm{Mg}^{2+}$ (12 protons) is the smallest,

followed by $\mathrm{Na}^{+}$ (11 protons),

then $\mathrm{F}^{-}$ (9 protons),

and finally $\mathrm{O}^{2-}$ (8 protons) is the largest.

Thus, the ordering given is correct.

Statement (IV): "The IUPAC name of the element with atomic number 106 is bohrium."

According to IUPAC, the element with atomic number 106 is called seaborgium (Sg). Bohrium (Bh) is element 107. Therefore, this statement is incorrect.

Since only Statements (I) and (III) are correct, the answer is:

Option C: I and III only.

Statements given in II and III are correct. Whille those in I and IV are incorrect. It is because Au and Pt both are soluble in aqua regia. Also, the correct stability order of oxides of halogens are $\mathrm{I}>\mathrm{Cl}>\mathrm{Br}$.

The size of the ion depends on number of shells, more are the number of shells, larger is the size of anion. When number of shells are same (like in $\mathrm{O}^{2-}$ and $\mathrm{F}^{-}$), more negative ion will have larger size. Therefore, the correct decreasing order of ionic radius will be

$ \mathrm{I}^{-}>\mathrm{Se}^{2-}>\mathrm{Br}^{-}>\mathrm{O}^{2-}>\mathrm{F}^{-} $

There is a sudden rise observed in fourth ionisation energy after 3rd ionisation energy. The element might have achieved inert gas configuration after the removal of 3rd electron and hence there is large energy required for removal of electron from a stable configuration. Thus, element belongs to thirteenth group of periodic table. The option possible from given options is boron which in all has 5 electrons.

∴ Correct option is boron (B).

(A) is true because the ionic radii goes on decreasing on moving left to right across a period as the nuclear charge goes on increasing from left to right. Thus, nucleus will hold the newly added electron more tightly. Hence ionic radii of alkaline earth metals is smaller than alkali metals.

$(\mathrm{R})$ is false because as the number of electron increases from left to right in a period and the atomic number also increases by a value of one. Therefore, alkaline earth metals have higher nuclear charge than the alkali metals.

Isoelectronic species means species having same number of electrons.

(a) $\mathrm{O}^{2-}=10$ electrons

(b) $\mathrm{F}^{-}=10$ electrons

(c) $\mathrm{Na}^{+}=10$ electrons

(d) $\mathrm{Mg}^{2+}=10$ electrons

∴ All are isoelectronic species.

Here all these species are isoelectronic species, i.e. possess 10 electrons.

The species which has maximum negative charge will have maximum ionic radii and species which have maximum positive charge will have minimum ionic radii. Thus order of ionic radii will be

$ \mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+} $

The correct order of ionic radii is $\mathrm{K}^{+}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}$.

In this $\mathrm{Na}^{+}, \mathrm{Mg}^{2+}$ and $\mathrm{Al}^{3+}$ are isoelectronic species so in these size decreases with rising atomic number because on rising atomic number $Z_{\text {eff }}$ increases ( $\mathrm{Z}_{\mathrm{eff}}=$ effective nuclear charge).

Also, $\mathrm{K}^{+}$has the highest ionic radii as it belongs to fourth period and has highest number of shells.

Metallic character decreases along a period due to decrease in atomic radius. As effective nuclear charge between electrons and nucleus increases. So, atomic radius decreases and hence, ability to loose electron increase and metallic character decreases.

On moving down the group, from S to Se electron gain enthalpy decreases. Oxygen has an unexpectedly low value of electron gain enthalpy. This is because of smaller size of oxygen atom due to which incoming electron faces more electronic repulsions which decreases its electron affinity. So, the correct order is $\mathrm{O}>\mathrm{Se} >\mathrm{S}$.

$ \left.\begin{array}{l} \mathrm{Se} \rightarrow-195.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{~S} \rightarrow-208.8 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{O} \rightarrow-141.4 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{array}\right\} \begin{aligned} & \text { electron gain } \\ & \text { enthalpy } \end{aligned} $

The acidity of oxides increases on moving from left to right in the periodic table. Oxides of group 1 are highly basic while those of group 17 are highly acidic. Therefore, the correct order of decreasing acidic nature of given oxides is as follows

$ \mathrm{N}_2 \mathrm{O}_3>\mathrm{CO}_2>\mathrm{B}_2 \mathrm{O}_3>\mathrm{BeO}>\mathrm{Li}_2 \mathrm{O} $

Helium (He) is the second-most abundant element in the universe, after hydrogen, and accounts for about 25 per cent of the elements in the universe. Most of the helium in the universe was created in the Big Bang, but it is also

obtained as the product of hydrogen fusion in stars. The Darmstadtium is a chemical element with the symbol Ds and atomic number 110 . The very small atomic radius of osmium results in a small metal-metal separation. This small atomic separation alongwith osmium's relatively high atomic number gives rise to osmium's highest density.

In group 1A, francium is more electropositive than ceasium but, not stable as it is highly radioactive (maximum half-life of only 22 minutes).

The electronic configuration of elements are as follows

$ \begin{aligned} & \mathrm{Li}=1 s^2 2 s^1 ; \quad \mathrm{Be}=1 s^2 2 s^2 \\ & \mathrm{~B}=1 s^2 2 s^2 2 p^1 ; \mathrm{N}=1 s^2 2 s^2 2 p^3 \end{aligned} $

All these element belong to second period. The ionisation energy generally increases as you move from left to right across a period. The first ionisation energy of beryllium is greater than that of boron since the order of attraction of electrons towards nucleus is $2 s>2 p$. The

beryllium has a stable complete electronic configuration $1 s^2 2 s^2$ whereas boron has the electronic configuration $1 s^2 2 s^2 2 p^1$ which needs lesser energy than that of beryllium to remove the valence electron. Therefore, the order of first ionisation energy is $\mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{N}$.

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.