Atomic and Ionic radii Questions in English

(A) The species $S^{2-}, K^{+}, Sc^{3+}, V^{5+},$ and $Mn^{7+}$ are isoelectronic,as they all contain $18$ electrons.
For isoelectronic species,the ionic size is inversely proportional to the effective nuclear charge $(Z_{eff})$.
As the atomic number $(Z)$ increases,the effective nuclear charge increases,which leads to a decrease in ionic size.
Since the atomic numbers are $S(16), K(19), Sc(21), V(23),$ and $Mn(25)$,the effective nuclear charge increases in the order $S^{2-} < K^{+} < Sc^{3+} < V^{5+} < Mn^{7+}$.
Therefore,the ionic size decreases in the order: $S^{2-} > K^{+} > Sc^{3+} > V^{5+} > Mn^{7+}$.

(A) The species $Ca^{2+}$,$Cl^{-}$,$S^{2-}$,and $Ar$ are isoelectronic,meaning they all contain $18$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number,$Z$) increases.
The atomic numbers are: $S$ $(Z=16)$,$Cl$ $(Z=17)$,$Ar$ $(Z=18)$,and $Ca$ $(Z=20)$.
Since the nuclear charge follows the order $S < Cl < Ar < Ca$,the ionic size follows the inverse order: $Ca^{2+} < Ar < Cl^{-} < S^{2-}$.
Comparing the given options with the standard ionic radii: $Ca^{2+} (0.99 \ \mathring{A}) < Ar (1.54 \ \mathring{A}) < Cl^{-} (1.81 \ \mathring{A}) < S^{2-} (1.84 \ \mathring{A})$.
None of the provided options are strictly correct based on the inclusion of $Ar$. However,considering the trend for the ions $Ca^{2+}$,$Cl^{-}$,and $S^{2-}$,the order is $Ca^{2+} < Cl^{-} < S^{2-}$. Option $A$ is the closest logical progression if $Ar$ is excluded or misplaced in the sequence.

(D) Ionic size depends on the number of shells and the effective nuclear charge.
$1$. For $Se^{-2} > Cl^{-} > O^{-2} > Na^{+}$,the order is correct as $Se^{-2}$ and $Cl^{-}$ have $4$ and $3$ shells respectively,while $O^{-2}$ and $Na^{+}$ are isoelectronic with $10$ electrons but $O^{-2}$ has a smaller nuclear charge than $Na^{+}$.
$2$. For $Li^{+} > Mg^{+2} > Al^{+3}$,the order is correct as $Li^{+}$ has $2$ shells,while $Mg^{+2}$ and $Al^{+3}$ have $3$ shells,but $Mg^{+2}$ is larger than $Al^{+3}$ due to lower nuclear charge.
$3$. For $P^{-3} > N^{-3} > Mg^{+2}$,the order is correct as $P^{-3}$ has $3$ shells,$N^{-3}$ has $2$ shells,and $Mg^{+2}$ has $2$ shells with higher nuclear charge than $N^{-3}$.
$4$. For $S^{-2} > Ca^{+2} > K^{+} > Ba^{+2}$,this is incorrect because $Ba^{+2}$ has $5$ shells,making it significantly larger than $S^{-2}$ ($3$ shells),$Ca^{+2}$ ($3$ shells),and $K^{+}$ ($3$ shells).

(C) $1$. For isoelectronic species $(O^{2-}, F^{-}, Na^{+})$,the ionic size decreases as the nuclear charge increases. Thus,the order $O^{2-} > F^{-} > Na^{+}$ is correct.
$2$. For ions in the same group $(Cl^{-}, Br^{-}, I^{-})$,the ionic size increases down the group. Thus,$I^{-} > Br^{-} > Cl^{-}$ is correct.
$3$. In option $C$,$O^{2-}$ and $Cl^{-}$ are not isoelectronic,and $S^{2-}$ is larger than $O^{2-}$ because it belongs to a higher period. The correct order for these ions is $S^{2-} > Cl^{-} > O^{2-}$. Therefore,the order $O^{2-} > Cl^{-} > S^{2-}$ is incorrect.

(D) For set $A$: $F^{-}, Na^{+}, Mg^{+2}$ are isoelectronic species with $10$ electrons. The ionic radius decreases as the nuclear charge $(Z)$ increases. Since $Z$ for $Mg$ is $12$,it has the smallest radius.
For set $B$: In a period,atomic radius decreases from left to right due to increased effective nuclear charge. Thus,$Zn$ has the smallest radius among $Ni, Cu, Zn$.
For set $C$: $N^{-3}$ has $10$ electrons,$Cs^{+}$ has $54$ electrons,and $H^{-}$ has $2$ electrons. $H^{-}$ has the smallest size due to the lowest number of shells and electrons.
For set $D$: $Li$ $(Z=3)$,$He$ $(Z=2)$,$Be^{+2}$ $(Z=4)$. $Be^{+2}$ has the smallest radius due to the highest nuclear charge and loss of electrons.
Therefore,the correct set is $Mg^{+2}, Zn, H^{-}, Be^{+2}$.

(D) For isoelectronic species,the ionic radius decreases as the nuclear charge increases. In option $A$,$B$,and $C$,the ions are isoelectronic or follow the trend of decreasing size with increasing positive charge. In option $D$,the order is incorrect because $Cs^{+}$ ($6^{th}$ period) has a much larger ionic radius than $O^{-2}$ ($2^{nd}$ period). The correct order for the ions in option $D$ is $Cs^{+} > Rb^{+} > Sr^{+2} > O^{-2}$.

(D) Atomic radius generally decreases across a period from left to right due to an increase in effective nuclear charge.
In option $A$,the order should be $Li > Be > B > C$.
In option $B$,the order should be $B > C > N > O$.
In option $C$,the order should be $Na > Mg > Si > P$.
In option $D$,$I$ is in the $5^{th}$ period,while $P$,$S$,and $Cl$ are in the $3^{rd}$ period. Since atomic radius increases down a group,$I$ has a larger radius than the elements of the $3^{rd}$ period. Within the $3^{rd}$ period,the order is $P > S > Cl$. Therefore,the correct order is $I > P > S > Cl$.

(A) The interionic distance $d$ is the sum of the ionic radii: $d = r_{Na^{+}} + r_{F^{-}} = 2.31 \ \mathring{A}$.
Since $F^{-}$ is an anion and $Na^{+}$ is a cation,the radius of the anion is larger than the radius of the cation $(r_{F^{-}} > r_{Na^{+}})$.
Based on periodic trends and experimental values,$r_{Na^{+}} \approx 0.95 \ \mathring{A}$ and $r_{F^{-}} \approx 1.36 \ \mathring{A}$.
Summing these: $0.95 \ \mathring{A} + 1.36 \ \mathring{A} = 2.31 \ \mathring{A}$.

(A) For isoelectronic species,the ionic radius decreases as the nuclear charge $(Z)$ increases.
The atomic numbers are: $S$ $(Z=16)$,$Cl$ $(Z=17)$,$K$ $(Z=19)$,and $Ca$ $(Z=20)$.
Since all these ions have the same number of electrons ($18$ electrons),the effective nuclear charge $(Z_{eff})$ increases with the increase in the number of protons $(Z)$.
As $Z_{eff}$ increases,the attraction between the nucleus and the electrons increases,causing the ionic radius to decrease.
Therefore,the decreasing order of ionic radii is: $S^{2-} (Z=16) > Cl^{-} (Z=17) > K^{+} (Z=19) > Ca^{2+} (Z=20)$.

(B) To determine the ionic radius,consider the following rules:
$1$. For isoelectronic species,the radius decreases as the nuclear charge (atomic number) increases.
$2$. For ions with different electron shells,the radius increases as the number of shells increases.
$K^{+}$ ($18$ electrons,$Z=19$) and $Cl^{-}$ ($18$ electrons,$Z=17$) are isoelectronic. Since $Cl^{-}$ has a lower nuclear charge,$Cl^{-} > K^{+}$.
$Br^{-}$ ($36$ electrons,$Z=35$) has $4$ shells,while $Cl^{-}$ and $K^{+}$ have $3$ shells. Thus,$Br^{-} > Cl^{-} > K^{+}$.
$Te^{2-}$ ($54$ electrons,$Z=52$) has $5$ shells,which is more than $Br^{-}$. Thus,$Te^{2-} > Br^{-} > Cl^{-} > K^{+}$.
The correct order is $Te^{2-} > Br^{-} > Cl^{-} > K^{+}$.

(A) For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
Atomic numbers are: $S = 16$,$Cl = 17$,$K = 19$,$Ca = 20$.
Since all these ions have $18$ electrons,the ionic size depends on the effective nuclear charge.
As the nuclear charge increases,the electrons are pulled more strongly towards the nucleus,resulting in a smaller size.
Thus,the decreasing order of size is: $S^{2-} (16) > Cl^{-} (17) > K^{+} (19) > Ca^{2+} (20)$.

(A) All given species ($Na^{+}$,$Mg^{2+}$,$F^{-}$,$O^{2-}$) are isoelectronic,as each contains $10 \ e^-$.
In an isoelectronic series,the ionic radius decreases as the nuclear charge (atomic number) increases.
The atomic numbers are: $O (8)$,$F (9)$,$Na (11)$,$Mg (12)$.
Therefore,the order of increasing ionic radii is $Mg^{2+} < Na^{+} < F^{-} < O^{2-}$.

(C) In the periodic table,as we move down a group,the number of shells increases,which leads to an increase in the atomic radius.
$C$ $(Z=6)$ and $Ge$ $(Z=32)$ belong to Group $14$. Moving from $C$ to $Ge$ represents moving down the group,hence the atomic radius increases.
For option $A$,ionization enthalpy decreases down the group.
For option $B$,electron gain enthalpy (magnitude) generally decreases down the group.
For option $D$,noble character is a property of the group and does not increase significantly in the manner described.

(A) The ionic radii of species depend on the number of shells and the effective nuclear charge.
For species in the same period,the ionic radius decreases as the atomic number increases (e.g.,$N^{3-} > O^{2-}$).
For species in the same group,the ionic radius increases as we move down the group due to the addition of a new shell (e.g.,$P^{3-} > N^{3-}$ and $S^{2-} > O^{2-}$).
Comparing the given species:
$1$. $O^{2-}$ and $N^{3-}$ are in the second period,so $N^{3-} > O^{2-}$.
$2$. $S^{2-}$ and $P^{3-}$ are in the third period,so $P^{3-} > S^{2-}$.
$3$. Since $S^{2-}$ and $P^{3-}$ have more shells than $O^{2-}$ and $N^{3-}$,they are larger.
Combining these,the order of increasing ionic radii is $O^{2-} < N^{3-} < S^{2-} < P^{3-}$.

(A) All the given ions $(O_2^{2-}, F^-, Li^+, B^{3+})$ are isoelectronic with the neon configuration $(1s^2 2s^2 2p^6)$,having $10$ electrons.
For isoelectronic species,the ionic radius increases as the nuclear charge $(Z)$ decreases.
The atomic numbers are: $O (Z=8)$,$F (Z=9)$,$Li (Z=3)$,$B (Z=5)$.
Since $O_2^{2-}$ has the lowest nuclear charge $(Z=8)$,it exerts the least attraction on the electrons,resulting in the largest ionic radius.

(C) In a group,atomic radii increase down the group due to the addition of new shells. Thus,the order for group $16$ elements is $O < S < Se$.
$As$ (Arsenic) belongs to group $15$ and is located to the left of the group $16$ elements in the periodic table. Across a period from left to right,atomic radii decrease due to an increase in effective nuclear charge. Therefore,$As$ has a larger atomic radius than the elements of group $16$ in the same period.
Comparing the sizes,$O < S < Se$ and $Se < As$. Thus,the overall order of increasing atomic radii is $O < S < Se < As$.

(A) The atomic radius increases down a group and decreases across a period from left to right.
Carbon $(C)$ is in Period $2$,Group $14$.
Sulfur $(S)$ is in Period $3$,Group $16$.
Aluminum $(Al)$ is in Period $3$,Group $13$.
Cesium $(Cs)$ is in Period $6$,Group $1$.
Comparing their positions:
$C$ $(Z=6)$ is the smallest as it is in Period $2$.
$S$ $(Z=16)$ and $Al$ $(Z=13)$ are in Period $3$. Since atomic radius decreases across a period,$Al > S$.
$Cs$ is in Period $6$,which is much larger than the others.
Thus,the order is $C < S < Al < Cs$.

(D) For isoelectronic species,the number of electrons is the same,so the size is determined by the effective 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.
Therefore,the size is inversely proportional to the nuclear charge: $\text{Size} \propto \frac{1}{Z}$.

(A) To determine the ionic radius of isoelectronic or related species,we consider the effective nuclear charge $(Z_{eff})$.
For ions with different electron configurations,the radius generally increases as we move down a group and decreases as we move across a period.
Comparing the given ions:
$O^{2-}$ ($Z=8$,$10$ electrons)
$F^{-}$ ($Z=9$,$10$ electrons)
$Se^{2-}$ ($Z=34$,$36$ electrons)
$Rb^{+}$ ($Z=37$,$36$ electrons)
Among these,$Se^{2-}$ and $Rb^{+}$ are isoelectronic ($36$ electrons). For isoelectronic species,the radius is inversely proportional to the atomic number $(Z)$.
Since $Z_{Se} = 34$ and $Z_{Rb} = 37$,$Se^{2-}$ has a larger radius than $Rb^{+}$.
Comparing $O^{2-}$ and $F^{-}$,they are smaller than $Se^{2-}$ because they belong to lower periods.
Thus,$Se^{2-}$ has the largest radius.

(A) The atomic radius decreases as we move from left to right across a period due to an increase in effective nuclear charge $(Z_{eff})$.
$K$ (Potassium) is in Group $1$,Period $4$.
$Sr$ (Strontium) is in Group $2$,Period $5$.
$Rb$ (Rubidium) is in Group $1$,Period $5$.
$Ba$ (Barium) is in Group $2$,Period $6$.
Comparing these,$K$ has the smallest atomic radius because it is in the $4^{th}$ period,while the others are in the $5^{th}$ or $6^{th}$ periods. As we move down a group,the number of shells increases,leading to a larger atomic size.

(B) The radius ratio is calculated as follows:
$\frac{r_{Rb^{+}}}{r_{I^{-}}} = \frac{1.46}{2.16} = 0.676$
Since the radius ratio lies between $0.414$ and $0.732$,the coordination number is $6$.
Therefore,the geometry of $RbI$ is similar to the rock salt $(NaCl)$ structure.

(D) The ionic radius depends on the effective nuclear charge and the number of shells.
For ions of the same element,the radius decreases as the positive charge increases because the remaining electrons are pulled more strongly by the nucleus.
Comparing $P^{+3}$ and $P^{+5}$,$P^{+3}$ has more electrons than $P^{+5}$,resulting in greater electron-electron repulsion and a larger ionic radius.
Therefore,$P^{+3} > P^{+5}$ is the correct relationship.

(B) For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
$1$. All given species ($Na^{+}$,$Mg^{2+}$,$F^{-}$,$O^{2-}$) have $10$ electrons.
$2$. Their atomic numbers are: $O (8)$,$F (9)$,$Na (11)$,$Mg (12)$.
$3$. As the nuclear charge increases,the attraction of the nucleus for the electrons increases,causing the ionic radius to decrease.
$4$. The order of increasing atomic number is $O^{2-} (8) < F^{-} (9) < Na^{+} (11) < Mg^{2+} (12)$.
$5$. Therefore,the order of increasing ionic radii is $Mg^{2+} < Na^{+} < F^{-} < O^{2-}$.

(A) Across a period in the periodic table,the atomic radius decreases from left to right due to an increase in effective nuclear charge $(Z_{eff})$.
Phosphorus $(P)$ is in Group $15$ and Chlorine $(Cl)$ is in Group $17$ of the same period (Period $3$).
Since $Cl$ is to the right of $P$,the effective nuclear charge of $Cl$ is higher than that of $P$.
Therefore,the covalent radius of $Cl$ $(0.099 \ nm)$ is smaller than the covalent radius of $P$ $(0.11 \ nm)$.

(A) The atomic radius increases down a group due to the addition of new shells. However,for $4d$ and $5d$ transition series elements,the atomic radii are very similar due to the lanthanoid contraction.
$1$. For $Sc$,$Y$,and $La$ (Group $3$): $Sc < Y \approx La$ is correct.
$2$. For $K$,$Rb$,and $Cs$ (Group $1$): The radius increases as $K < Rb < Cs$.
$3$. For $Ge$,$Sn$,and $Pb$ (Group $14$): The radius increases as $Ge < Sn < Pb$.
$4$. For $Ni$,$Pd$,and $Pt$ (Group $10$): $Ni < Pd \approx Pt$ is correct due to lanthanoid contraction between $Pd$ and $Pt$.
Since both $A$ and $D$ represent correct trends,$A$ is the most standard representation for the group $3$ elements.

(A) The ionic radius of an ion depends on the number of electron shells and the effective nuclear charge.
As we move down a group in the periodic table,the number of electron shells increases,leading to an increase in ionic radius.
Comparing the given anions: $H^{-}$ has $1$ shell,$F^{-}$ has $2$ shells,$Cl^{-}$ has $3$ shells,and $Br^{-}$ has $4$ shells.
Since $H^{-}$ has the fewest number of electron shells,it has the smallest ionic radius among the given options.

(D) The ions $Li^{+}$,$Be^{2+}$,and $B^{3+}$ are isoelectronic species,all having the same number of electrons ($2$ electrons).
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
The atomic numbers are: $Li$ $(Z=3)$,$Be$ $(Z=4)$,and $B$ $(Z=5)$.
Since the nuclear charge increases in the order $Li^{+} < Be^{2+} < B^{3+}$,the ionic radii will decrease in the same order.
Therefore,the correct order of ionic radii is $Li^{+} > Be^{2+} > B^{3+}$.
Hence,option $D$ is correct.

(B) The species $Na^{+}$,$Ne$,and $F^{-}$ are isoelectronic,meaning they all contain $10$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (number of protons) increases.
$F^{-}$ has $9$ protons,$Ne$ has $10$ protons,and $Na^{+}$ has $11$ protons.
Since $F^{-}$ has the lowest number of protons $(9)$,the effective nuclear attraction on the electrons is the weakest,resulting in the largest ionic radius.
The order of size is $F^{-} > Ne > Na^{+}$.

(A) To determine the smallest size among $F, F^{-}, O$,and $N$:
$1$. Comparing neutral atoms $(N, O, F)$: In the periodic table,atomic radius decreases from left to right across a period due to an increase in effective nuclear charge. Therefore,the order of size is $N > O > F$.
$2$. Comparing an atom with its anion ($F$ vs $F^{-}$): An anion is always larger than its parent neutral atom because the addition of an electron increases inter-electronic repulsion,which causes the electron cloud to expand. Thus,$F < F^{-}$.
$3$. Combining these observations,$F$ is smaller than $O, N$,and $F^{-}$.
Therefore,$F$ has the smallest size.

(A) In the periodic table,the atomic size decreases from left to right across a period due to an increase in effective nuclear charge.
$P$ (Phosphorus) belongs to group $15$,while $Cl$ (Chlorine) belongs to group $17$.
Since $Cl$ is to the right of $P$ in the same period ($3$rd period),its atomic size and consequently its covalent radius will be smaller than that of $P$.

(B) The given ions are $O^{2-}$,$Mg^{2+}$,and $Al^{3+}$.
All these ions are isoelectronic,meaning they all have $10$ electrons ($O^{2-}: 8+2=10$,$Mg^{2+}: 12-2=10$,$Al^{3+}: 13-3=10$).
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
The atomic numbers are $O=8$,$Mg=12$,and $Al=13$.
Therefore,the decreasing order of size is $O^{2-} > Mg^{2+} > Al^{3+}$.

(B) To determine the order of atomic radii,we look at the positions of the elements in the periodic table:
$1$. $O$ (Oxygen) and $C$ (Carbon) are in the $2^{nd}$ period. In a period,atomic radius decreases from left to right. Thus,$C > O$.
$2$. $S$ (Sulfur) and $Se$ (Selenium) are in the $3^{rd}$ and $4^{th}$ periods respectively. Atomic radius increases down a group. Thus,$Se > S$.
$3$. Comparing periods: Elements in the $2^{nd}$ period $(C, O)$ have smaller atomic radii than those in the $3^{rd}$ and $4^{th}$ periods $(S, Se)$.
$4$. Combining these: $O < C < S < Se$.
Therefore,the correct order of increasing atomic radius is $O < C < S < Se$.

(B) The elements $Na$,$Al$,$Si$,and $P$ are present in the same period ($3^{rd}$ period).
Atomic radius decreases from left to right in a period due to an increase in effective nuclear charge.
Therefore,the order of atomic radius for the given elements is $Na > Al > Si > P$.
Thus,the correct order of increasing radius is $P < Si < Al < Na$.

(B) The size of an atom or ion depends on the number of electrons and the effective nuclear charge.
For a given element,the neutral atom $(Pb)$ has the largest size because it has the most electrons and the least effective nuclear charge per electron.
When electrons are removed to form cations ($Pb^{2+}$ and $Pb^{4+}$),the remaining electrons experience a greater effective nuclear charge,causing the electron cloud to be pulled closer to the nucleus.
Therefore,as the positive charge on the ion increases,the ionic radius decreases.
Thus,the order of size is $Pb > Pb^{2+} > Pb^{4+}$.

(D) For isoelectronic species,the ionic radius decreases as the effective nuclear charge ($Z/e$ ratio) increases.
In option $A$,$Sr^{2+}$,$Rb^{+}$,$Br^{-}$,and $Se^{2-}$ are isoelectronic ($36$ electrons). The order of radii is $Sr^{2+} < Rb^{+} < Br^{-} < Se^{2-}$,which is correct.
In option $B$,$Nb^{5+}$,$Zr^{4+}$,and $Y^{3+}$ are isoelectronic ($36$ electrons). The order of radii is $Nb^{5+} < Zr^{4+} < Y^{3+}$,which is correct.
In option $C$,for the same element,the size decreases as the positive charge increases: $Co > Co^{2+} > Co^{3+} > Co^{4+}$,which is correct.
In option $D$,$Ba^{2+}$ and $Cs^{+}$ belong to the $6^{th}$ period,while $Se^{2-}$ and $As^{3-}$ belong to the $4^{th}$ period. Ions in the $6^{th}$ period are significantly larger than those in the $4^{th}$ period. Thus,the correct order is $As^{3-} < Se^{2-} < Ba^{2+} < Cs^{+}$. Therefore,the given order is incorrect.

(C) The ionic radius of $Li^{+}$ is approximately $76 \ pm$.
Due to the diagonal relationship between $Li$ and $Mg$, the ionic radius of $Mg^{2+}$ $(72 \ pm)$ is very close to that of $Li^{+}$ $(76 \ pm)$.
Therefore, the correct option is $(c)$.

(D) In aqueous solutions,the ionic radius is determined by the extent of hydration. Smaller ions have a higher charge density,leading to greater hydration and a larger hydrated radius.
The order of ionic size for bare ions is $Li^{+} < Na^{+} < K^{+}$.
However,the extent of hydration follows the order $Li^{+} > Na^{+} > K^{+}$.
Therefore,the hydrated ionic radius in aqueous solution follows the order $K^{+}_{(aq)} < Na^{+}_{(aq)} < Li^{+}_{(aq)}$.

(B) The ionic radius depends on the number of shells and the effective nuclear charge.
$1$. Comparing $Na^{+}$ $(2, 8)$ and $Li^{+}$ $(2)$: $Na^{+}$ has two shells while $Li^{+}$ has one,so $Na^{+} > Li^{+}$.
$2$. Comparing $Mg^{2+}$ $(2, 8)$ and $Al^{3+}$ $(2, 8)$: Both have two shells,but $Al^{3+}$ has a higher nuclear charge,so $Mg^{2+} > Al^{3+}$.
$3$. Comparing $Li^{+}$ $(2)$ and $Be^{2+}$ $(2)$: $Be^{2+}$ has a higher nuclear charge,so $Li^{+} > Be^{2+}$.
$4$. Comparing $Na^{+}$ $(2, 8)$ and $Mg^{2+}$ $(2, 8)$: $Mg^{2+}$ has a higher nuclear charge,so $Na^{+} > Mg^{2+}$.
Combining these,the correct order is $Na^{+} > Li^{+} > Mg^{2+} > Al^{3+} > Be^{2+}$.

(C) Isoelectronic species are those that have the same number of electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (number of protons) increases.
In option $C$,both $F^{-}$ and $Na^{+}$ have $10$ electrons.
$F^{-}$ has $9$ protons,while $Na^{+}$ has $11$ protons.
Since $F^{-}$ has fewer protons to attract the same number of electrons,it has a larger ionic radius than $Na^{+}$.
Therefore,$F^{-}$ is larger than $Na^{+}$.

(C) All the given ions ($N^{3-}$,$O^{2-}$,$F^{-}$,$Na^{+}$,$Mg^{2+}$) are isoelectronic species,meaning they all contain $10$ electrons.
For isoelectronic species,the ionic size decreases as the atomic number $(Z)$ increases because the effective nuclear charge increases,pulling the electrons closer to the nucleus.
The atomic numbers are: $N (7)$,$O (8)$,$F (9)$,$Na (11)$,$Mg (12)$.
Since $Mg^{2+}$ has the highest atomic number $(12)$,it has the smallest size,and $N^{3-}$ has the lowest atomic number $(7)$,it has the largest size.
Therefore,the correct order of increasing ionic size is $Mg^{2+} < Na^{+} < F^{-} < O^{2-} < N^{3-}$.

(A) To determine the ionic radius,we consider the number of shells and the effective nuclear charge.
$Al^{3+}$ has $2$ shells $(1s^2 2s^2 2p^6)$.
$Mg^{2+}$ has $2$ shells $(1s^2 2s^2 2p^6)$.
$Li^{+}$ has $1$ shell $(1s^2)$.
$K^{+}$ has $3$ shells $(1s^2 2s^2 2p^6 3s^2 3p^6)$.
Comparing $Al^{3+}$ and $Mg^{2+}$ (isoelectronic,$10$ electrons): $Mg^{2+}$ has a lower nuclear charge $(Z=12)$ than $Al^{3+}$ $(Z=13)$,so $Mg^{2+} > Al^{3+}$.
Comparing all: $Al^{3+}$ $(0.54 \ \mathring{A})$ $< Mg^{2+}$ $(0.72 \ \mathring{A})$ $< Li^{+}$ $(0.76 \ \mathring{A})$ $< K^{+}$ $(1.38 \ \mathring{A})$.
Therefore,the correct order is $Al^{3+} < Mg^{2+} < Li^{+} < K^{+}$.

(C) The atomic radius of a metal is always larger than its ionic radius because the loss of an electron results in a decrease in the number of shells or increased effective nuclear charge on the remaining electrons. Thus,the atomic radius of $K$ $(1.96 \ \mathring{A})$ is greater than the ionic radius of $K^{+}$ $(1.34 \ \mathring{A})$.
Conversely,the atomic radius of a non-metal is smaller than its ionic radius because the gain of an electron increases inter-electronic repulsion,causing the electron cloud to expand. Thus,the atomic radius of $F$ $(0.72 \ \mathring{A})$ is smaller than the ionic radius of $F^{-}$ $(1.34 \ \mathring{A})$.
Therefore,the atomic radii of $K$ and $F$ are $1.96 \ \mathring{A}$ and $0.72 \ \mathring{A}$ respectively.

(C) For lanthanides, the ionic radius decreases as the atomic number increases due to lanthanoid contraction. The correct order is $La^{3+} (103 \text{ pm}) > Eu^{3+} (95 \text{ pm}) > Gd^{3+} (94 \text{ pm}) > Lu^{3+} (86 \text{ pm})$. Thus, option $A$ is correct.
For ions of the same element, ionic size decreases with an increase in positive charge: $V^{2+} > V^{3+} > V^{4+} > V^{5+}$. Thus, option $B$ is correct.
For isoelectronic species, ionic size decreases as the nuclear charge increases. $K^{+} (Z=19)$, $Sc^{3+} (Z=21)$, $V^{5+} (Z=23)$, and $Mn^{7+} (Z=25)$ are isoelectronic $(18 \text{ e}^-)$. The order $K^{+} > Sc^{3+} > V^{5+} > Mn^{7+}$ is correct. Thus, option $D$ is correct.
In option $C$, $Tl^{+}$ $(150 \text{ pm})$ and $In^{+}$ $(132 \text{ pm})$ are larger than $Sn^{2+}$ $(118 \text{ pm})$ and $Sb^{3+}$ $(76 \text{ pm})$. However, the comparison between $Tl^{+}$ and $In^{+}$ is correct, but the overall sequence provided in the options needs to be evaluated. Given the standard trends, option $C$ is the incorrect order as $Tl^{+}$ and $In^{+}$ belong to different periods compared to $Sn^{2+}$ and $Sb^{3+}$.

(C) For $(a)$,the ionic size of isoelectronic species increases as the nuclear charge decreases: $Ca^{2+} (20) < K^{+} (19) < S^{2-} (16) < P^{3-} (15)$. Thus,the given order is incorrect.
For $(b)$,in aqueous solution,the extent of hydration decreases as the size of the cation increases $(Na^{+} > K^{+} > Rb^{+} > Cs^{+})$. Smaller hydrated ions move faster,but electrical conductance depends on the mobility of the hydrated ion. The order of ionic mobility and conductance in aqueous solution is $Cs^{+} > Rb^{+} > K^{+} > Na^{+}$. Thus,the given order is incorrect.
For $(c)$,the extent of hydration is directly proportional to the charge density. $Al^{3+}$ has the highest charge density,followed by $Mg^{2+}$ and $Na^{+}$. Therefore,the hydrated size order is $Al_{(aq.)}^{3+} > Mg_{(aq.)}^{2+} > Na_{(aq.)}^{+}$. This is correct.
For $(d)$,the extent of hydration for anions is $F^{-} > Cl^{-} > Br^{-} > I^{-}$. Consequently,the hydrated size is $F^{-} > Cl^{-} > Br^{-} > I^{-}$. Ionic mobility is inversely proportional to hydrated size,so the order is $I_{(aq.)}^{-} > Br_{(aq.)}^{-} > Cl_{(aq.)}^{-} > F_{(aq.)}^{-}$. Thus,the given order is incorrect.

(A) All the given ions are isoelectronic,meaning they have the same number of electrons ($10$ electrons each).
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
The atomic numbers are: $N (7)$,$O (8)$,$F (9)$,$Na (11)$,$Mg (12)$.
As the positive charge increases,the effective nuclear charge increases,pulling the electrons closer to the nucleus,thus decreasing the ionic radius.
Conversely,as the negative charge increases,the electron-electron repulsion increases,leading to a larger ionic radius.
Therefore,the increasing order of ionic radii is: $Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}$.

(A) The oxidation state of $Cr$ in the given compounds is calculated as follows:
$1. \text{ In } CrCl_3, \text{ the oxidation state of } Cr \text{ is } +3$.
$2. \text{ In } CrF_3, \text{ the oxidation state of } Cr \text{ is } +3$.
$3. \text{ In } CrO_2, \text{ the oxidation state of } Cr \text{ is } +4$.
$4. \text{ In } K_2CrO_4, \text{ the oxidation state of } Cr \text{ is } +6$.
As the ionic radius of an ion decreases with an increase in its positive oxidation state (due to increased effective nuclear charge),the $Cr$ ion with the highest oxidation state will have the minimum ionic radius.
Since $Cr$ has the maximum oxidation state of $+6$ in $K_2CrO_4$,it has the minimum ionic radius among the given compounds.

(A) To determine the ionic radius,we consider the number of shells and the effective nuclear charge.
$1$. $Al^{3+}$ ($10$ electrons,$2$ shells,$Z=13$)
$2$. $Mg^{2+}$ ($10$ electrons,$2$ shells,$Z=12$)
$3$. $Li^{+}$ ($2$ electrons,$1$ shell,$Z=3$)
$4$. $K^{+}$ ($18$ electrons,$3$ shells,$Z=19$)
Comparing these,$Li^{+}$ has the smallest radius due to having only one shell.
Among $Al^{3+}$ and $Mg^{2+}$ (isoelectronic species),$Al^{3+}$ has a higher nuclear charge,so it is smaller.
$K^{+}$ has the largest radius as it has $3$ shells.
Thus,the increasing order is $Al^{3+} < Mg^{2+} < Li^{+} < K^{+}$.

(A) The covalent radii of elements decrease as we move from left to right across a period due to an increase in effective nuclear charge.
For the pair $K$ and $Ca$,$K$ is an alkali metal (Group $1$) and $Ca$ is an alkaline earth metal (Group $2$).
Since $K$ is at the extreme left of the period and $Ca$ is next to it,the decrease in size from $K$ to $Ca$ is significant due to the large difference in effective nuclear charge.
In contrast,$Mn, Fe, Co, Ni,$ and $Cr$ are all $d$-block transition elements where the increase in nuclear charge is balanced by the shielding effect of $d$-electrons,leading to a much smaller variation in atomic radii.
Therefore,the difference in covalent radii is maximum for the pair $K, Ca$.

(A) The smallest cation is the hydrogen ion,$H^{+}$,which consists of only a proton with no electrons.
The smallest anion is the hydride ion,$H^{-}$,which has the smallest number of electrons $(2)$ and the smallest nuclear charge among anions.
Therefore,the correct pair is $H^{+}$ and $H^{-}$.