Atomic and Ionic radii Questions in English

(C) The covalent radius decreases across a period from left to right due to an increase in effective nuclear charge.
In the second period,the order of atomic radii is $B > C > N$.
$Si$ belongs to the third period and has a much larger radius than elements of the second period.
Therefore,$N$ has the minimum covalent radius among the given options.

(C) As we move down the group $(Group \ 17)$,the number of electron shells increases,which leads to an increase in the atomic and ionic radius.
Therefore,the correct option is $(C)$.

(A) As we move down a group in the periodic table,the number of shells increases,which leads to an increase in the atomic radius.
As we move from left to right across a period,the effective nuclear charge increases while the number of shells remains the same,which leads to a decrease in the atomic radius.

(D) The size of an ionic species decreases with an increase in the effective nuclear charge for isoelectronic species or across a period.
Among the given ions,$Na^{+}$,$Mg^{2+}$,and $Al^{3+}$ are isoelectronic (all have $10$ electrons).
For isoelectronic species,the ionic radius decreases as the nuclear charge $(Z)$ increases.
Since the atomic number of $Al$ is $13$,$Mg$ is $12$,and $Na$ is $11$,the nuclear charge is highest for $Al^{3+}$.
Therefore,$Al^{3+}$ is the smallest among the given ions.

(C) All the given ions $(C^{4-}, N^{3-}, O^{2-}, F^{-})$ are isoelectronic species,as they all contain $10$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
The atomic numbers are: $C = 6, N = 7, O = 8, F = 9$.
Since $F^{-}$ has the highest nuclear charge $(Z = 9)$,it exerts the strongest attraction on the electrons,resulting in the smallest ionic radius.

(A) The size of an ion is smaller than its corresponding neutral atom because the removal of electrons leads to a decrease in the number of shells or an increase in the effective nuclear charge per electron.
For the same element,the size decreases as the positive charge increases.
Therefore,the order of size is $Al > Al^{+} > Al^{2+} > Al^{3+}$.
Thus,$Al$ has the largest size.

(A) All the given species are isoelectronic,meaning they all have $18$ electrons.
$Cl^{-}: 17 \text{ protons}, 18 \text{ electrons}$
$Ar: 18 \text{ protons}, 18 \text{ electrons}$
$K^{+}: 19 \text{ protons}, 18 \text{ electrons}$
$Ca^{2+}: 20 \text{ protons}, 18 \text{ electrons}$
For isoelectronic species,the ionic radius decreases as the nuclear charge (number of protons) increases,because the electrons are pulled more strongly towards the nucleus.
Since $Cl^{-}$ has the lowest number of protons $(17)$,it experiences the weakest effective nuclear charge,resulting in the largest ionic radius.

(D) The ionic radius decreases as the effective nuclear charge increases for ions with the same number of electrons (isoelectronic species) or across a period.
$Li^{+}$ $(Z=3)$ has $2$ electrons.
$Be^{2+}$ $(Z=4)$ has $2$ electrons.
Since $Be^{2+}$ has a higher nuclear charge $(Z=4)$ than $Li^{+}$ $(Z=3)$,it exerts a stronger pull on the electrons,resulting in a smaller ionic radius.
Therefore,$Be^{2+}$ is the smallest.

(A) $1$. In a period,atomic radius decreases from left to right. Therefore,the atomic radius of $O > F$.
$2$. $O^{2-}$ and $F^{-}$ are isoelectronic species,both having $10$ electrons.
$3$. For isoelectronic species,the ionic radius increases as the nuclear charge (atomic number) decreases.
$4$. Since the atomic number of $O$ $(Z=8)$ is less than $F$ $(Z=9)$,the ionic radius of $O^{2-} > F^{-}$.
$5$. Combining these,the overall order is $O^{2-} > F^{-} > O > F$.

(B) The species $Na^{+}$,$Mg^{2+}$,$F^{-}$,and $Cl^{-}$ are not all isoelectronic.
$Na^{+}$ $(10 \ e^{-})$,$Mg^{2+}$ $(10 \ e^{-})$,and $F^{-}$ $(10 \ e^{-})$ are isoelectronic,while $Cl^{-}$ $(18 \ e^{-})$ is larger due to an extra shell.
Among the isoelectronic species ($Na^{+}$,$Mg^{2+}$,$F^{-}$),the size decreases as the nuclear charge increases.
$Mg^{2+}$ has the highest atomic number $(Z=12)$ compared to $Na^{+}$ $(Z=11)$ and $F^{-}$ $(Z=9)$.
Therefore,$Mg^{2+}$ has the strongest attraction for electrons,resulting in the smallest ionic radius.

(B) All the given ions are isoelectronic,meaning they all have $18$ electrons ($Na^{+}$ has $10$ electrons,so let us compare the isoelectronic series $S^{2-}, Cl^{-}, K^{+}, Ca^{2+}$).
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
Atomic numbers are: $S$ $(16)$,$Cl$ $(17)$,$Na$ $(11)$,$F$ $(9)$.
$S^{2-}$ has the lowest nuclear charge $(Z = 16)$ among the given options $S^{2-}$ and $Cl^{-}$,therefore it has the largest ionic radius.

(D) As we move down a group in the periodic table,the number of shells increases,which leads to an increase in the distance between the nucleus and the outermost electrons. Consequently,the atomic radius or the size of the atom increases progressively down a group. Other properties like electronegativity,electron affinity,and ionization potential generally decrease down a group.

(A) The atomic radius of fluorine is determined by its covalent radius,which is approximately $0.72 \ \mathring{A}$.
Neon is a noble gas and does not form covalent bonds under normal conditions,so its atomic radius is measured as the Van der Waal's radius,which is approximately $1.60 \ \mathring{A}$.
Since the Van der Waal's radius is always larger than the covalent radius,the atomic radius of neon is significantly larger than that of fluorine.
Therefore,the values are $0.72 \ \mathring{A}$ and $1.60 \ \mathring{A}$ respectively.

(D) The ionic radius increases as we move down a group in the periodic table due to the addition of new electron shells.
Comparing the halide ions $(F^{-}, Cl^{-}, Br^{-}, I^{-})$,$I^{-}$ belongs to the $5^{th}$ period,while $F^{-}$ belongs to the $2^{nd}$ period,$Cl^{-}$ to the $3^{rd}$ period,and $Br^{-}$ to the $4^{th}$ period.
Since $I^{-}$ has the highest number of shells,it has the largest ionic radius.
Therefore,the correct option is $(D)$.

(D) The correct answer is $(D)$.
In the periodic table,as we move from left to right across a period,the atomic radius decreases due to an increase in effective nuclear charge.
All the given elements ($Mg$,$Al$,$Si$,and $P$) belong to the $3^{rd}$ period.
Their positions are: $Mg$ (Group $2$),$Al$ (Group $13$),$Si$ (Group $14$),and $P$ (Group $15$).
Since $Mg$ is the leftmost element among these in the $3^{rd}$ period,it has the largest atomic radius.

(A) The ionic radius is inversely proportional to the $Z/e$ ratio (where $Z$ is the nuclear charge and $e$ is the number of electrons).
For the given ions:
$O^{2-}$: $Z=8, e=10$,$Z/e = 0.8$
$F^{-}$: $Z=9, e=10$,$Z/e = 0.9$
$Li^{+}$: $Z=3, e=2$,$Z/e = 1.5$
$B^{3+}$: $Z=5, e=2$,$Z/e = 2.5$
Since $O^{2-}$ has the lowest $Z/e$ ratio,it has the largest ionic radius.

(A) The correct answer is $(A)$.
As we move down a group in the periodic table,the principal quantum number $(n)$ increases,which means the number of electron shells increases.
This leads to an increase in the distance between the nucleus and the outermost valence electrons,resulting in a continuous increase in the atomic radius.

(D) . $Na^{+} < F^{-} < O^{2-} < N^{3-}$
All these ions are isoelectronic,meaning they contain the same number of electrons ($10$ electrons).
For isoelectronic species,the ionic radius decreases as the atomic number $(Z)$ increases because the effective nuclear charge increases.
Since the atomic numbers are $N(7)$,$O(8)$,$F(9)$,and $Na(11)$,the effective nuclear charge is highest for $Na^{+}$,resulting in the smallest ionic size.

(D) The size of an atom or ion depends on the number of electrons and the effective nuclear charge.
For the same element,the size order is $I^{-} > I > I^{+}$.
$I^{-}$ has $54$ electrons,$I$ has $53$ electrons,and $I^{+}$ has $52$ electrons.
As the number of electrons decreases for the same nuclear charge,the effective nuclear charge per electron increases,leading to a smaller size.
Therefore,the correct order is $I^{-} > I > I^{+}$.

(D) To determine the largest radius,we compare the electronic configurations and nuclear charges of the given species:
$1$. $Na$ $(Z=11)$: Electronic configuration is $1s^2 2s^2 2p^6 3s^1$. It has $3$ shells.
$2$. $Na^{+}$ $(Z=11)$: Electronic configuration is $1s^2 2s^2 2p^6$. It has $2$ shells.
$3$. $F$ $(Z=9)$: Electronic configuration is $1s^2 2s^2 2p^5$. It has $2$ shells.
$4$. $F^{-}$ $(Z=9)$: Electronic configuration is $1s^2 2s^2 2p^6$. It has $2$ shells.
Comparing these,$Na$ has the largest radius because it has electrons in the $3rd$ shell $(n=3)$,whereas all other species have electrons only up to the $2nd$ shell $(n=2)$.

(A) The correct answer is $A$. As we move from $Na$ to $Cl$ across the third period,the atomic number increases,which leads to an increase in the effective nuclear charge $(Z_{eff})$. This stronger pull of the nucleus on the valence electrons causes the atomic radius to decrease continuously.

(A) The species given are $Mg^{2+}$,$Na^{+}$,$F^{-}$,and $Al$.
$1$. $Al$ is a neutral atom,while $Mg^{2+}$,$Na^{+}$,and $F^{-}$ are ions.
$2$. Comparing the ionic radii of isoelectronic species ($Mg^{2+}$,$Na^{+}$,$F^{-}$): These all have $10$ electrons. The ionic radius decreases as the nuclear charge $(Z)$ increases. The nuclear charges are $F^{-} (Z=9)$,$Na^{+} (Z=11)$,and $Mg^{2+} (Z=12)$. Thus,the order of size is $Mg^{2+} < Na^{+} < F^{-}$.
$3$. The neutral atom $Al$ $(Z=13)$ has a larger atomic radius than these cations. However,comparing $Al$ to the others,the correct order of size is $Mg^{2+} < Na^{+} < F^{-} < Al$.

(B) The correct option is $B$.
$K^{+}$ and $F^{-}$ are isoelectronic species,both having $18$ electrons $(1s^2 2s^2 2p^6 3s^2 3p^6)$.
However,$K^{+}$ has $Z = 19$ (protons) and $F^{-}$ has $Z = 9$ (protons).
Since $K^{+}$ has a higher nuclear charge,its ionic radius is smaller than $F^{-}$.
However,the question asks for the atomic radius of $K$ (which is $K$ atom) compared to the ionic radius of $F^{-}$.
The atomic radius of $K$ ($19$ electrons,$4$ shells) is significantly larger than the ionic radius of $F^{-}$ ($10$ electrons,$2$ shells).

(D) All the given species ($Na^{+}$,$F^{-}$,$Ne$,$O^{2-}$) are isoelectronic,meaning they all contain $10$ electrons.
For isoelectronic species,the ionic radius increases as the nuclear charge (number of protons) decreases.
The number of protons in these species are: $Na^{+} = 11$,$Ne = 10$,$F^{-} = 9$,and $O^{2-} = 8$.
Since $O^{2-}$ has the lowest nuclear charge $(8)$,it has the weakest attraction for the electrons,resulting in the maximum size.

(A) The given ions $N^{3-}$,$O^{2-}$,$F^{-}$,and $Na^{+}$ are isoelectronic species,as each contains $10$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (number of protons) increases.
The number of protons in these ions are: $N$ $(7)$,$O$ $(8)$,$F$ $(9)$,and $Na$ $(11)$.
Since the nuclear charge increases from $N^{3-}$ to $Na^{+}$,the ionic radius decreases in the order: $N^{3-} > O^{2-} > F^{-} > Na^{+}$.

(A) On moving down a group,the atomic number increases.
As we move down,new electron shells are added to the atom,which increases the distance between the nucleus and the outermost electrons.
Consequently,both atomic and ionic radii increase.

(C) The correct order is $Be > C > F > Ne$.
Atomic size decreases across a period from left to right due to an increase in effective nuclear charge.
$Be$ (Group $2$),$C$ (Group $14$),$F$ (Group $17$),and $Ne$ (Group $18$) all belong to the $2^{nd}$ period.
Therefore,the atomic size decreases as we move from $Be$ to $Ne$.

(C) Atomic radii and ionic radii are extremely small,typically on the order of $10^{-10} \ m$.
Therefore,they are commonly expressed in units of $\mathring{A}$ $(\mathring{A})$,where $1 \ \mathring{A} = 10^{-10} \ m$.

(B) Atomic radius decreases from left to right within a period due to an increase in effective nuclear charge.
Conversely,moving from right to left across a period,the effective nuclear charge decreases,and the shielding effect remains relatively constant,causing the atomic radius to increase.
Therefore,the atomic radii in the periodic table among elements from right to left increases.

(D) The given ions are $K^{+}$,$Ca^{2+}$,$Ti^{3+}$,and $Ti^{4+}$.
These ions are isoelectronic,meaning they all have the same number of electrons ($18$ electrons).
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
The atomic numbers are: $K (19)$,$Ca (20)$,and $Ti (22)$.
Since $Ti^{4+}$ has the highest nuclear charge $(Z = 22)$ among the given ions,it exerts the strongest attraction on the electrons,resulting in the smallest ionic radius.
Therefore,the correct option is $(D)$.

(D) The ions $Na^{+}$,$Mg^{2+}$,$Al^{3+}$,and $Si^{4+}$ 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) increases.
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,pulling the electron cloud closer to the nucleus.
Therefore,the order of ionic radii is $Na^{+} > Mg^{2+} > Al^{3+} > Si^{4+}$.

(A) The ions $N^{3-}$,$O^{2-}$,and $F^{-}$ are isoelectronic species,meaning they all contain $10$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
The atomic numbers are: $N = 7$,$O = 8$,and $F = 9$.
Since the nuclear charge increases in the order $N < O < F$,the ionic radii decrease in the order $N^{3-} > O^{2-} > F^{-}$.
Therefore,the correct order is $N^{3-} > O^{2-} > F^{-}$.

(D) . Potassium $(K)$ is a metal that loses an electron to form a cation $(K^{+})$. The removal of an electron decreases the electron-electron repulsion and increases the effective nuclear charge,resulting in an ionic radius that is smaller than the atomic radius $(K > K^{+})$.
Bromine $(Br)$ is a non-metal that gains an electron to form an anion $(Br^{-})$. The addition of an electron increases the electron-electron repulsion,resulting in an ionic radius that is larger than the atomic radius $(Br < Br^{-})$.
Therefore,for potassium,the atomic radius $>$ ionic radius,and for bromine,the atomic radius $ < $ ionic radius.

(A) The size of an ion depends on the effective nuclear charge and the number of electrons.
$1$. $A$ cation $(O_2^+)$ is formed by the loss of an electron,which increases the effective nuclear charge per electron,resulting in a smaller size compared to the parent molecule $(O_2)$.
$2$. Anions ($O_2^-$ and $O_2^{2-}$) are formed by the gain of electrons,which increases inter-electronic repulsion and decreases the effective nuclear charge,resulting in a larger size compared to the parent molecule.
$3$. Therefore,the order of size is $O_2^{2-} > O_2^- > O_2 > O_2^+$.
$4$. Thus,$O_2^+$ is the smallest ion.

(B) For isoelectronic species,the ionic radius increases as the nuclear charge (atomic number) decreases.
In the series $F^{-}$,$O^{2-}$,and $N^{3-}$,all have $10$ electrons.
The atomic numbers are $F (9)$,$O (8)$,and $N (7)$.
Since the nuclear charge decreases from $F$ to $N$,the attraction on the electrons decreases,and the ionic radius increases.
Therefore,the correct order is $F^{-} < O^{2-} < N^{3-}$.

(B) The ionic radii depend on the number of shells and the effective nuclear charge.
$K^{+}$ has $4$ shells,$Na^{+}$ has $3$ shells,$Li^{+}$ has $2$ shells,and $F^{-}$ has $2$ shells.
Comparing ions with the same number of shells: $F^{-}$ $(Z=9)$ has a lower effective nuclear charge than $Li^{+}$ $(Z=3)$,so $F^{-} > Li^{+}$.
Comparing ions with different shells: The order is $K^{+} (4 \text{ shells}) > Na^{+} (3 \text{ shells}) > F^{-} (2 \text{ shells}) > Li^{+} (2 \text{ shells})$.
Thus,the correct order is $K^{+} > Na^{+} > F^{-} > Li^{+}$.

(A) To determine the size,we compare the atomic/ionic radii of the given species:
$1$. $Li^+$ ($Z=3$,$2$ electrons): It is a cation with a $1s^2$ configuration. The removal of an electron leads to a decrease in the number of shells and increased effective nuclear charge,making it very small.
$2$. $H$ ($Z=1$,$1$ electron): It has a $1s^1$ configuration.
$3$. $Li$ ($Z=3$,$3$ electrons): It has a $1s^2 2s^1$ configuration.
$4$. $He$ ($Z=2$,$2$ electrons): It has a $1s^2$ configuration.
Comparing these,$Li^+$ has the smallest radius because it has lost its valence shell and has a high effective nuclear charge relative to the number of electrons. The order of size is $Li^+ < He < H < Li$.

(A) All the given ions $(C^{4-}, N^{3-}, O^{2-}, Mg^{2+})$ are isoelectronic species,meaning they all have $10$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (number of protons) increases.
The number of protons in these ions are: $C = 6, N = 7, O = 8, Mg = 12$.
Since $C^{4-}$ has the lowest nuclear charge $(Z = 6)$,it has the weakest attraction for the electrons,resulting in the largest ionic radius.

(A) The ionic radius of elements increases as we move down a group in the periodic table due to the addition of new electron shells.
Since $Li^{+}$,$Na^{+}$,$K^{+}$,and $Cs^{+}$ all belong to Group $1$ of the periodic table,the ionic radius increases in the order: $Li^{+} < Na^{+} < K^{+} < Cs^{+}$.
Therefore,$Cs^{+}$ has the largest ionic radius.

(A) The correct statement is that the $X^{-}$ ion is larger in size than the $X$ atom.
This occurs because,in an anion $(X^{-})$,the addition of an electron increases the electron-electron repulsion while the nuclear charge remains the same.
Consequently,the effective nuclear charge per electron decreases,causing the electron cloud to expand and the ionic radius to become larger than the atomic radius of the neutral atom $X$.

(D) For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number,$Z$) increases because the effective nuclear attraction on the electrons 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 size follows the order of increasing atomic number: $S^{2-} (Z=16) > Cl^{-} (Z=17) > K^{+} (Z=19) > Ca^{2+} (Z=20)$.
Therefore,the correct decreasing order is $S^{2-} > Cl^{-} > K^{+} > Ca^{2+}$.

(B) An isoelectronic series is a group of ions that all have the same number of electrons. For example,one isoelectronic series could include $O^{2-}, F^{-}, Na^{+}, Mg^{2+}$. These all have $10$ electrons. The number of protons,however,increases as the atomic number increases,which leads to an increase in nuclear charge. As the nuclear charge increases,the effective nuclear charge experienced by the electrons also increases. Consequently,the nucleus attracts the electrons more strongly,pulling them closer. Therefore,the ionic radii of species in an isoelectronic series decrease as the nuclear charge increases.

(A) The covalent radii decrease as we move from left to right across a period due to an increase in effective nuclear charge.
For the pair $K$ $(19)$ and $Ca$ $(20)$,both are $s$-block elements where the shielding effect is less effective,leading to a significant decrease in atomic size.
In contrast,for transition elements like $Mn, Fe, Co, Ni,$ and $Cr$,the electrons are added to the penultimate $(n-1)d$ subshell.
This $(n-1)d$ subshell provides a screening effect that offsets the increase in nuclear charge,resulting in a very small change in atomic radii across the transition series.
Therefore,the difference in covalent radii is maximum for the pair $K$ and $Ca$.

(A) To determine the smallest radius among the given ions,we analyze their electronic configurations and nuclear charges:
$1$. The ions $O^{2-}$,$F^{-}$,$Li^{+}$,and $Be^{2+}$ are isoelectronic,all having $10$ electrons (configuration $1s^2 2s^2 2p^6$).
$2$. For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number,$Z$) increases.
$3$. The atomic numbers are: $O (Z=8)$,$F (Z=9)$,$Li (Z=3)$,and $Be (Z=4)$.
$4$. Among these,$Be^{2+}$ has the highest nuclear charge $(Z=4)$,which exerts the strongest attraction on the electrons,resulting in the smallest ionic radius.

(B) Atomic radius generally increases down a group and decreases across a period from left to right.
$Na$ $(Z=11)$ and $Mg$ $(Z=12)$ belong to the $3^{rd}$ period,while $K$ $(Z=19)$ and $Rb$ $(Z=37)$ belong to the $4^{th}$ and $5^{th}$ periods respectively.
Across the $3^{rd}$ period,atomic radius decreases from $Na$ to $Mg$,so $Mg < Na$.
Down Group $1$,atomic radius increases as $Na < K < Rb$.
Combining these trends,the order is $Mg < Na < K < Rb$.

(C) For isoelectronic species,the ionic radius decreases as the effective nuclear charge $(Z_{eff})$ increases.
$N^{3-}$ $(Z=7)$,$O^{2-}$ $(Z=8)$,and $F^{-}$ $(Z=9)$ all have $10$ electrons.
As the atomic number $(Z)$ increases from $7$ to $9$,the attraction of the nucleus for the electrons increases,causing the ionic radius to decrease.
The ionic radii are:
$N^{3-} = 1.71 \ \mathring{A}$
$O^{2-} = 1.40 \ \mathring{A}$
$F^{-} = 1.36 \ \mathring{A}$
Therefore,the correct order is $1.71, 1.40, 1.36$.

(B) $Al^{3+}$ and $Mg^{2+}$ have an equal number of electrons ($10$ electrons) and are,therefore,isoelectronic.
For isoelectronic species,the ionic radius depends upon the effective nuclear charge $(Z_{eff})$.
Ionic radius is inversely proportional to the effective nuclear charge.
Since $Al$ $(Z=13)$ has a higher atomic number than $Mg$ $(Z=12)$,$Al^{3+}$ experiences a greater effective nuclear charge than $Mg^{2+}$.
Consequently,the electrons in $Al^{3+}$ are pulled more strongly towards the nucleus,resulting in a smaller ionic radius.

(C) During the conversion of a neutral atom to a cation,the size decreases because after the removal of one or more $e^-$:
$(i)$ The effective nuclear charge per electron increases.
$(ii)$ The outermost shell may be completely removed,leading to a reduction in the number of electron shells.

(B) As the atomic number increases within a group,the number of electron shells increases.
Consequently,the distance between the nucleus and the outermost shell increases,leading to an increase in the atomic radius.

(A) All the given species are isoelectronic,each containing $10 \ e^{-}$.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
$Mg^{2+}$ $(Z=12)$,$Na^{+}$ $(Z=11)$,$Ne$ $(Z=10)$,$F^{-}$ $(Z=9)$,$O^{2-}$ $(Z=8)$.
As the nuclear charge increases,the electrons are pulled more strongly towards the nucleus,resulting in a smaller radius.
Therefore,the increasing order of radii is $Mg^{2+} < Na^{+} < Ne < F^{-} < O^{2-}$.