Atomic and Ionic radii Questions in English

(D) Across a period from left to right, the effective nuclear charge $(Z_{eff})$ increases, which leads to a decrease in atomic radius $(A.R.)$.
For the elements $Be, B, C, N, O$ (all in period $2$), the order of atomic radii is $Be > B > C > N > O$.
The corresponding values in $pm$ are $111, 88, 77, 74, 66$ respectively.
Thus, the correct sequence for $Be, O, C, N, B$ is $111, 66, 88, 77, 74$.

(B) The size ratio of an anion to its corresponding neutral atom is determined by the increase in electron-electron repulsion upon adding an electron to the valence shell.
For $H$, the addition of an electron to the $1s$ orbital doubles the electron cloud size, resulting in a very high ratio.
Comparing the ratios:
$1$. For $H^{-}/H$, the radius increases from $37 \text{ pm}$ to $208 \text{ pm}$ (ratio $\approx 5.6$).
$2$. For $O^{-}/O$, the radius increases from $66 \text{ pm}$ to $140 \text{ pm}$ (ratio $\approx 2.1$).
$3$. For $F^{-}/F$, the radius increases from $64 \text{ pm}$ to $133 \text{ pm}$ (ratio $\approx 2.0$).
$4$. For $Na^{-}/Na$, the ratio is significantly smaller as the electron enters a new shell but the nuclear charge is high.
Thus, the maximum ratio is observed for $H^{-}/H$.

(D) In the periodic table,atomic radius decreases across a period from left to right due to an increase in effective nuclear charge.
Oxygen $(O)$ is in Group $16$ and Period $2$.
Fluorine $(F)$ is to the right of oxygen in the same period,so its atomic size is smaller than oxygen.
Neon $(Ne)$ is a noble gas in the same period,but its atomic radius is generally considered larger than oxygen due to the van der Waals radius,but in terms of covalent radius,it is not typically compared directly.
However,looking at the options provided,$F$ is smaller,and $Ne$ is a noble gas.
Actually,moving down a group increases size. Since no element from the same group below oxygen is provided,and $F$ is smaller,the correct answer is none of these.

(B) The ionic radius increases as we move down a group in the periodic table due to the addition of new electron shells.
$Na^{+}$ and $Mg^{2+}$ belong to the $3^{rd}$ period,while $Ca^{2+}$ belongs to the $4^{th}$ period and $Cs^{+}$ belongs to the $6^{th}$ period.
Since $Cs^{+}$ has the highest number of electron shells (principal quantum number $n=6$),it has the largest ionic radius among the given options.

(D) The given ions $N^{3-}$,$O^{2-}$,$F^{-}$,and $Na^{+}$ are isoelectronic species,meaning they all have the same number of electrons ($10$ electrons).
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number,$Z$) increases.
The atomic numbers are: $N (Z=7)$,$O (Z=8)$,$F (Z=9)$,and $Na (Z=11)$.
Since $Na^{+}$ has the highest nuclear charge $(Z=11)$,it exerts the strongest attraction on the electrons,resulting in the smallest ionic radius.

(D) Atomic radius generally increases down a group due to the addition of a new electron shell.
In the case of $N$ and $P$,both belong to Group $15$.
$N$ is in the $2^{nd}$ period and $P$ is in the $3^{rd}$ period.
Since $P$ is below $N$ in the group,the atomic radius of $P$ is larger than that of $N$.
For other options:
$Br$ ($4^{th}$ period) is larger than $Cl$ ($3^{rd}$ period).
$Na$ ($1^{st}$ group) is larger than $Mg$ ($2^{nd}$ group) due to decreasing size across a period.
$Sr$ ($5^{th}$ period) is larger than $Ca$ ($4^{th}$ period).

(B) $1$. For a given element,the ionic radius follows the order: $Anion > Neutral \ atom > Cation$. Thus,$I^{-} > I > I^{+}$ is correct.
$2$. For isoelectronic species or ions of the same element,the radius decreases as the positive charge increases. $N^{3-}$ has a larger radius than $N^{5+}$ because $N^{3-}$ has more electrons and greater electron-electron repulsion,while $N^{5+}$ has fewer electrons and higher effective nuclear charge. Therefore,the order $N^{3-} < N^{5+}$ is incorrect.
$3$. Similarly,$P^{5+} < P^{3+}$ is correct as the cation with higher positive charge is smaller.
$4$. Across a period from left to right,atomic radius decreases due to an increase in effective nuclear charge. Thus,$Li > Be > B$ is correct.
$5$. The incorrect order is $N^{3-} < N^{5+}$.

(A) The atomic radius generally decreases across a period from left to right due to an increase in effective nuclear charge.
$F$ (Fluorine) is a halogen with atomic number $9$,and $Ne$ (Neon) is a noble gas with atomic number $10$.
For noble gases,the atomic radius is defined as the van der Waals radius,which is significantly larger than the covalent radius of other elements in the same period.
The covalent radius of $F$ is approximately $0.72 \ \mathring{A}$.
The van der Waals radius of $Ne$ is approximately $1.60 \ \mathring{A}$.
Therefore,the values are $0.72 \ \mathring{A}$ and $1.60 \ \mathring{A}$ respectively.

(C) The ionic radii of the given ions are as follows:
$Li^{+}$: $60\, pm$
$Na^{+}$: $95\, pm$
$Br^{-}$: $195\, pm$
$I^{-}$: $216\, pm$
Matching these values:
$(i) Li^{+} \rightarrow (r) 60\, pm$
$(ii) Na^{+} \rightarrow (s) 95\, pm$
$(iii) Br^{-} \rightarrow (q) 195\, pm$
$(iv) I^{-} \rightarrow (p) 216\, pm$
Therefore, the correct matching is $i-r, ii-s, iii-q, iv-p$.

(C) The atomic radius is defined based on the type of bonding between atoms.
$1$. $Van \ der \ Waals$ radius is the distance between two non-bonded atoms in adjacent molecules,which is determined by weak $Van \ der \ Waals$ forces.
$2$. Covalent radius is half the distance between the nuclei of two atoms bonded by a single covalent bond. Since covalent bonds are much stronger and involve electron sharing,the internuclear distance is shorter than in $Van \ der \ Waals$ interactions. Thus,$Van \ der \ Waals$ radius > covalent radius.
$3$. Metallic radius is half the internuclear distance between two adjacent metal atoms in a metallic lattice. Metallic bonds are stronger than $Van \ der \ Waals$ forces,making metallic radii smaller than $Van \ der \ Waals$ radii.
Therefore,both statements are correct.

(A) The given ions $P^{3-}$,$S^{2-}$,and $Cl^{-}$ are isoelectronic species,as they all have $18$ electrons.
For isoelectronic species,the ionic radius increases as the nuclear charge (atomic number) decreases.
The atomic numbers are: $P = 15$,$S = 16$,and $Cl = 17$.
Since the nuclear charge follows the order $P < S < Cl$,the ionic radius follows the inverse order: $P^{3-} > S^{2-} > Cl^{-}$.
Therefore,the correct increasing order is $Cl^{-} < S^{2-} < P^{3-}$.

(A) Atomic radius generally increases down a group and decreases across a period from left to right.
$1$. Comparing elements in the same period ($2^{nd}$ period): $C > O > F$.
$2$. Comparing elements in the same group ($17^{th}$ group): $Br > Cl > F$.
$3$. Combining these trends: $F$ is the smallest (top-right),followed by $O$ and $C$. $Cl$ and $Br$ are larger than elements of the $2^{nd}$ period.
$4$. The correct order is $F < O < C < Cl < Br$.

(C) For isoelectronic species (species having the same number of electrons),the ionic radius depends on the effective nuclear charge $(Z_{eff})$.
As the positive charge on a cation increases,the number of protons remains the same while the number of electrons decreases,leading to a higher $Z_{eff}$ and a smaller ionic radius.
Conversely,as the negative charge on an anion increases,the number of electrons increases relative to the number of protons,leading to a lower $Z_{eff}$ and a larger ionic radius.
Therefore,for a cation,a smaller positive charge means a larger ionic radius,not a smaller one.
Thus,statement $C$ is incorrect.

(C) For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
All the given ions ($Na^+$,$Mg^{2+}$,$Al^{3+}$,$Si^{4+}$) have $10$ electrons.
The atomic numbers are: $Na (11)$,$Mg (12)$,$Al (13)$,and $Si (14)$.
As the nuclear charge increases,the electrostatic attraction between the nucleus and the electrons increases,causing the ionic radius to decrease.
Therefore,the order of ionic radii is $Na^+ > Mg^{2+} > Al^{3+} > Si^{4+}$.

(B) Atomic radii increase as we move down the group due to the addition of an extra energy shell.
Although the nuclear charge also increases,the effect of the addition of a new energy shell predominates.
The increase in atomic radius is most significant when moving from $Na$ ($Z=11$,$3s^1$) to $K$ ($Z=19$,$4s^1$) because the transition from the $n=3$ shell to the $n=4$ shell involves a substantial increase in shielding and distance from the nucleus compared to subsequent transitions in the group.
Therefore,the largest difference in radii is observed for the pair $Na, K$.

$(D)$ The atomic radius of $Al$ is $143 \ pm$ and that of $Ga$ is $135 \ pm$.
Thus, the atomic radius of $Ga$ is actually smaller than that of $Al$.
This occurs because, in $Ga$, the $3d^{10}$ electrons provide poor shielding of the nuclear charge, leading to a higher effective nuclear charge which pulls the valence shell closer to the nucleus.
Since the Assertion is false and the Reason is true, the correct option is $D$.

(B) Atomic radius decreases across a period from left to right and increases down a group.
For the given elements:
$1.$ In the second period,the order of atomic radii is $C > O > F$.
$2.$ In the halogen group,the order is $F < Cl < Br$.
Combining these trends,the atomic radii follow the order: $F < O < C < Cl < Br$.
Thus,the increasing order is $c < b < a < d < e$.

(A) Atomic radii decrease across a period from left to right. Thus,$Mg$ is larger than $Al$.
Cations are always smaller than their parent neutral atoms due to the loss of electrons and increased effective nuclear charge. Therefore,$Mg > Mg^{2+}$ and $Al > Al^{3+}$.
Comparing $Mg^{2+}$ and $Al^{3+}$,both have $10$ electrons (isoelectronic). $Al^{3+}$ has a higher nuclear charge $(Z = 13)$ compared to $Mg^{2+}$ $(Z = 12)$,resulting in a stronger pull on the electrons,making $Al^{3+}$ smaller than $Mg^{2+}$.
Comparing all,$Mg$ is the largest and $Al^{3+}$ is the smallest.

Atomic radius is the distance from the center of the nucleus to the outermost shell of electrons in an atom. It measures the size of an atom.
For metals,it is the metallic radius,defined as half the internuclear distance between two adjacent metal atoms in a metallic crystal. For example,the internuclear distance in solid copper is $256 \, pm$,so the metallic radius is $\frac{256}{2} \, pm = 128 \, pm$.
For nonmetals,it is the covalent radius,defined as half the distance between the nuclei of two atoms bonded by a single covalent bond. For example,the distance between two chlorine atoms in a $Cl_2$ molecule is $198 \, pm$,so the covalent radius is $\frac{198}{2} \, pm = 99 \, pm$.
Ionic radius is the effective distance from the center of the nucleus to the outer shell of an ion. Cations are smaller than their parent atoms due to the loss of electrons and increased effective nuclear charge (e.g.,$Na^+$ is $95 \, pm$ vs $Na$ is $186 \, pm$). Anions are larger than their parent atoms due to increased electron-electron repulsion and decreased effective nuclear charge (e.g.,$F^-$ is $136 \, pm$ vs $F$ is $64 \, pm$).

(N/A) Atomic radius generally decreases from left to right across a period. This is because within a period,the outer electrons are present in the same valence shell and the atomic number increases from left to right,resulting in an increased effective nuclear charge. As a result,the attraction of electrons to the nucleus increases.
On the other hand,the atomic radius generally increases down a group. This is because down a group,the principal quantum number $(n)$ increases,which results in an increase of the distance between the nucleus and valence electrons.

(N/A) Each of the given species (ions) has the same number of electrons ($10$ electrons). Hence,the given species are isoelectronic.
$(b)$ The ionic radii of isoelectronic species increase with a decrease in the magnitude of nuclear charge.
The arrangement of the given species in order of their increasing nuclear charge is as follows:
$N^{3-} < O^{2-} < F^{-} < Na^{+} < Mg^{2+} < Al^{3+}$
Nuclear charge $= +7, +8, +9, +11, +12, +13$
Therefore,the arrangement of the given species in order of their increasing ionic radii is as follows:
$Al^{3+} < Mg^{2+} < Na^{+} < F^{-} < O^{2-} < N^{3-}$

(N/A) cation has fewer electrons than its parent atom,while its nuclear charge remains the same. As a result,the attraction of electrons to the nucleus is stronger in a cation than in its parent atom,leading to a smaller size.
On the other hand,an anion has more electrons than its parent atom,resulting in increased electron-electron repulsion and a decrease in the effective nuclear charge. As a result,the distance between the valence electrons and the nucleus increases,making the anion larger in radius than its parent atom.

(N/A) When an atom gains an electron,it forms an anion. The addition of an electron increases the interelectronic repulsion,while the nuclear charge remains constant. This causes the electron cloud to expand,leading to an increase in the ionic radius compared to the atomic radius.
$(b)$ When an atom loses an electron,it forms a cation. The removal of an electron decreases the interelectronic repulsion,while the nuclear charge remains constant. This results in a stronger pull by the nucleus on the remaining electrons,leading to a decrease in the ionic radius compared to the atomic radius.

(A) Isoelectronic species have the same number of electrons but different nuclear charges $(Z).$
The size of isoelectronic species is primarily determined by the magnitude of the nuclear charge $(Z).$
As the nuclear charge $(Z)$ increases,the force of attraction between the nucleus and the electrons increases,which pulls the electron cloud closer to the nucleus,resulting in a decrease in ionic/atomic size.
For the given species:
$F^{-} (Z=9), Ne (Z=10), Na^{+} (Z=11)$
Since the nuclear charge increases from $F^{-}$ to $Na^{+},$ the size decreases in the order: $F^{-} > Ne > Na^{+}.$
Thus,the size is affected by the nuclear charge $(Z).$

(N/A) The atomic radius decreases as we move from left to right in a period. This is because, as we move along a period, the nuclear charge increases while the electrons are added to the same principal energy shell. This leads to an increase in the effective nuclear charge, which pulls the electrons closer to the nucleus, thereby decreasing the atomic size.
Within a group, the atomic radius increases as we move down the group. This is because a new principal energy shell is added at each succeeding element, which increases the distance between the nucleus and the outermost electrons, outweighing the effect of increased nuclear charge.

Atom (Period $II$)	$Li, Be, B, C, N, O, F$
Atomic radius $(pm)$	$152, 111, 88, 77, 74, 66, 64$
Atom (Period $III$)	$Na, Mg, Al, Si, P, S, Cl$
Atomic radius $(pm)$	$186, 160, 143, 117, 110, 104, 99$

(N/A) There is no direct method to measure the absolute size of an individual atom.
The size of an atom is extremely small,approximately $1.2 \ \mathring{A}$ (i.e.,$1.2 \times 10^{-10} \ m$).
Since the electron cloud surrounding the nucleus does not have a sharp,well-defined boundary,the determination of an exact atomic size is not possible.
Consequently,there is no practical way to measure the size of an isolated atom.
Instead,atomic radius is defined as covalent or metallic radius,which is determined experimentally using techniques such as $X$-ray diffraction or spectroscopic methods.

(N/A) There is no direct practical way to measure the size or volume of an individual atom because the electron cloud does not have a sharp boundary.
However,the size of an atom can be estimated by measuring the distance between atoms in a bonded state. The three common methods are:
$1$. Metallic radius: Used for metals,defined as half the internuclear distance between two adjacent metal atoms in a metallic lattice.
$2$. Covalent radius: Used for non-metals,defined as half the distance between the nuclei of two atoms bonded by a single covalent bond.
$3$. Van der Waals radius: Used for noble gases and non-bonded atoms,defined as half the distance between the nuclei of two identical non-bonded atoms in a solid state.

(N/A) Covalent Radius: It is defined as one-half of the distance between the nuclei of two identical atoms bonded by a single covalent bond in a molecule.
Example: The bond length of $Cl-Cl$ in a $Cl_{2}$ molecule is $198 \ pm$. Therefore, the covalent radius of chlorine is $\frac{198}{2} = 99 \ pm$.
Ionic Radius: It is defined as the effective distance from the center of the nucleus of an ion up to which it has an influence on its electron cloud.
Example: In an ionic crystal like $NaCl$, the distance between the nuclei of $Na^{+}$ and $Cl^{-}$ ions is $276 \ pm$. If the ionic radius of $Cl^{-}$ is $181 \ pm$, then the ionic radius of $Na^{+}$ is $276 - 181 = 95 \ pm$.

(N/A) Atomic radius:
- $\rightarrow$ Atomic radius is defined as one-half of the distance between the nuclei of two covalently bonded atoms of the same element in a molecule.
- In the case of metals,the atomic radius is referred to as the metallic radius,which corresponds to one-half of the distance between two adjacent atoms in a metallic crystal lattice.
Ionic radius:
- $\rightarrow$ Ionic radius refers to the effective size of an ion (cation or anion).
- It represents the distance from the nucleus of the ion to the point where it exerts an influence in an ionic bond.
- The size of a cation is always smaller than its parent atom due to the loss of electrons and increased effective nuclear charge,whereas the size of an anion is always larger than its parent atom due to increased electron-electron repulsion.

(N/A) Radius of positive ion $(cation)$: An atom loses electrons to form a positive ion. The size of the positive ion is smaller than its parent atom.
This is because the positive ion has fewer electrons than the parent atom, while the nuclear charge remains the same. Consequently, the remaining electrons experience a greater effective nuclear charge, pulling them closer to the nucleus.
Example: The radius of a sodium atom is $186 \ pm$, whereas the radius of a $Na^{+}$ ion is $95 \ pm$.
Radius of negative ion $(anion)$: When an atom gains an electron, it becomes a negative ion. The size of the negative ion is larger than its parent atom.
This is because when one or more electrons are added to the parent atom, the inter-electronic repulsion increases, and the effective nuclear charge per electron decreases, causing the electron cloud to expand.
Example: The radius of a fluorine atom is $95 \ pm$, whereas the radius of a fluoride ion $(F^{-})$ is $136 \ pm$.

(N/A) All these ions have the same number $(10)$ of electrons. Therefore,these are called isoelectronic species.
$(b)$ Since the number of electrons is the same,the ionic size decreases with an increase in nuclear charge. Therefore,the ions can be arranged in increasing order of ionic radii as $Al^{3+} < Mg^{2+} < Na^{+} < F^{-} < O^{2-} < N^{3-}$.

(N/A) The atomic radius is defined based on the nature of the bond,such as metallic radius or covalent radius,which is half the internuclear distance between bonded atoms.
$(a)$ When an atom gains an electron,it forms a negative ion (anion). The radius of an anion is always larger than that of its parent neutral atom. This is because the addition of one or more electrons increases the electron-electron repulsion,while the effective nuclear charge $(Z_{eff})$ decreases,causing the electron cloud to expand.
$(b)$ When an atom loses an electron,it forms a positive ion (cation). The radius of a cation is always smaller than that of its parent neutral atom. This is because the loss of electrons results in a higher effective nuclear charge per remaining electron,pulling the electron cloud closer to the nucleus.
In summary,for isoelectronic species: $\text{radius of cation} < \text{radius of neutral atom} < \text{radius of anion}$.

(A) The ions $K^{+}$,$Cl^{-}$,$S^{2-}$,and $Ca^{2+}$ are isoelectronic,meaning they all contain $18$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (number of protons) increases.
$S^{2-}$ has $16$ protons.
$Cl^{-}$ has $17$ protons.
$K^{+}$ has $19$ protons.
$Ca^{2+}$ has $20$ protons.
As the number of protons increases,the effective nuclear charge increases,which pulls the electron cloud closer to the nucleus,resulting in a smaller ionic radius.
Therefore,the decreasing order of ionic radii is: $S^{2-} > Cl^{-} > K^{+} > Ca^{2+}$.

(A) The atomic radius of $F$ is measured as a covalent radius, whereas the atomic radius of $Ne$ is measured as a van der Waals' radius.
By definition, the van der Waals' radius is always larger than the covalent radius for a given element.
Therefore, the atomic radius of $F$ is smaller than the atomic radius of $Ne$ ($F = 72 \ pm$, $Ne = 160 \ pm$).
The correct order is $F < Ne$.

(N/A) For the formation of a cation, there is a loss of an electron, which means its radius decreases. Due to the decrease in the number of electrons, the nuclear pull per electron increases.
As a result, the radius of the cation decreases as the effective nuclear charge increases. For example, the ionic radius of $Na^{+}$ is smaller than that of the parent atom $Na$.
$Na \longrightarrow Na^{+} + 1e^{-}$
Electrons: $11, 10$
Nuclear charge: $11, 11$
Ionic size: $186 \text{ pm}, 95 \text{ pm}$
$(1 \text{ pm} = 10^{-12} \text{ m})$

(N/A) Covalent radius:
- The covalent radius is defined as one-half of the distance between the nuclei of two identical atoms bonded by a single covalent bond in a molecule.
- For a heteronuclear molecule $AB$, the bond length $d_{AB}$ is given by $d_{AB} = r_A + r_B$, where $r_A$ and $r_B$ are the covalent radii of atoms $A$ and $B$ respectively.
- Example: The bond length in $Cl_2$ is $198 \ pm$, so the covalent radius of $Cl$ is $198 \div 2 = 99 \ pm$.
van der Waals radius:
- The van der Waals radius represents the overall size of an atom, including its valence shell, in a non-bonded state.
- It is defined as one-half of the distance between the nuclei of two identical atoms belonging to neighboring molecules of an element in the solid state.

(A) In a group,as we move from top to bottom,the number of shells increases,which leads to an increase in the atomic and ionic size.
Therefore,the ionic radius increases with an increase in the atomic number.
The correct order of ionic radius for group $1$ elements is:
$Li^{+} < Na^{+} < K^{+} < Rb^{+} < Cs^{+} < Fr^{+}$

(B) Although $Ga$ has one shell more than $Al$,its atomic radius is slightly smaller than that of $Al$.
This is due to the poor shielding effect of the $3d$-electrons.
The $d$-electrons do not shield the valence electrons effectively from the nuclear charge.
Consequently,the effective nuclear charge experienced by the valence electrons in $Ga$ is significantly higher than in $Al$,which pulls the valence shell closer to the nucleus.

(N/A) $1$. Across a period from left to right,the effective nuclear charge increases,which causes the atomic and ionic radii to decrease. Since $_{11}Na$ is to the left of $_{12}Mg$ in the same period,$_{11}Na$ has a larger radius than $_{12}Mg$.
$2$. Down a group,the number of shells increases,leading to an increase in atomic and ionic radii. Since $_{20}Ca$ is in the same group as $_{12}Mg$ but is located below it,$_{20}Ca$ has a larger radius than $_{12}Mg$.

(B) Across a period from left to right,the atomic radius decreases.
This is because the nuclear charge increases as the number of protons increases,while the electrons are added to the same principal energy shell.
The increased nuclear attraction pulls the electron cloud closer to the nucleus,resulting in a smaller atomic radius.

(A) As we move down a group in the periodic table,a new principal energy shell $(n)$ is added for each successive element.
This increases the distance between the nucleus and the outermost valence electrons.
Although the nuclear charge increases,the effect of the additional shell (shielding effect) outweighs the increase in nuclear charge,resulting in an increase in the atomic radius.

(A) $1$. $Covalent \ radius$: It is defined as one-half of the distance between the nuclei of two identical atoms bonded by a single covalent bond in a molecule. For example,in $Cl_2$ molecule,the distance between two $Cl$ nuclei is $198 \ pm$,so the covalent radius is $99 \ pm$.
$2$. $Metallic \ radius$: It is defined as one-half of the internuclear distance between two adjacent metal atoms in a metallic crystal lattice. For example,the distance between two adjacent $Cu$ atoms in a metallic crystal is $256 \ pm$,so the metallic radius is $128 \ pm$.

(N/A) The atomic radius is measured using $X$-ray diffraction or spectroscopic methods. These techniques allow for the determination of the internuclear distance between bonded atoms,from which the atomic radius can be calculated.

(A) Atomic radius decreases across a period from left to right and increases down a group.
$(i)$ These elements belong to the $2^{nd}$ period. The order of increasing atomic radius is: $F < O < N < C < B < Be < Li$.
$(ii)$ These elements belong to the $3^{rd}$ period. The order of increasing atomic radius is: $Cl < S < P < Si < Al < Mg < Na$.
$(iii)$ These elements belong to Group $1$. The order of increasing atomic radius is: $Li < Na < K < Rb < Cs$.
$(iv)$ These elements belong to Group $17$. The order of increasing atomic radius is: $F < Cl < Br < I < At$.

(A) The atomic radius increases down a group and decreases across a period from left to right.
$Na$ $(Group \ 1, Period \ 3)$ has the largest radius.
$Li$ $(Group \ 1, Period \ 2)$ is smaller than $Na$.
$S$ $(Group \ 16, Period \ 3)$ is smaller than $Li$ because it is to the right of $Na$ in the same period.
$O$ $(Group \ 16, Period \ 2)$ is the smallest as it is at the top right of the given elements.
Thus,the decreasing order is $Na > Li > S > O$.

(B) In reality,atomic radius is defined based on the nature of the element:
$1$. In metals,it is the $ \text{metallic radius} $.
$2$. In non-metals,it is the $ \text{covalent radius} $.
$3$. In noble gases,it is the $ \text{van der Waals radius} $.

(A) Chlorine is a non-metal, so its atomic radius is defined as the covalent radius, which is half of the bond length of the $Cl-Cl$ bond in the $Cl_2$ molecule.
$\text{Atomic radius} = \frac{\text{Bond length}}{2}$
$\text{Atomic radius} = \frac{198 \ pm}{2} = 99 \ pm$

(C) Noble gases have the largest atomic radius.
This is because the atomic radius of noble gases is measured as the $Van der Waals$ radius.
Since noble gases exist as monoatomic gases,their $Van der Waals$ radius is significantly larger than the covalent or metallic radii of other elements.

(B) $Mg^{2+}$ has a smaller size than $Mg$.
This is because $Mg^{2+}$ is a cation formed by the loss of two electrons from the neutral $Mg$ atom.
As the number of electrons decreases while the nuclear charge remains the same,the effective nuclear charge per electron increases,causing the remaining electrons to be pulled more strongly towards the nucleus,resulting in a smaller ionic radius.

(B) $Al^{3+}$ has a smaller radius. This is because $Al$ is the parent atom of $Al^{3+}$,and $Al^{3+}$ is formed by the removal of $3$ electrons from $Al$. Consequently,the effective nuclear charge increases in $Al^{3+}$,which results in a smaller size compared to the neutral $Al$ atom.