Atomic and Ionic radii Questions in English

(D) Chloride ion $(Cl^-)$ and potassium ion $(K^+)$ are isoelectronic,meaning they have the same number of electrons ($18$ electrons each).
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
The atomic number of $Cl$ is $17$,while the atomic number of $K$ is $19$.
Since $K^+$ has a higher nuclear charge $(Z = 19)$ than $Cl^-$ $(Z = 17)$,the electrons in $K^+$ are more strongly attracted by the nucleus.
Therefore,the ionic radius of $Cl^-$ is larger than that of $K^+$.

(C) The ionic radius increases 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 $Cs^{+}$ belongs to the $6^{th}$ period.
Since $Cs^{+}$ has the highest principal quantum number $(n=6)$ among the given ions,it has the largest ionic radius.

(A) The elements $Li$,$Na$,and $K$ belong to Group $1$ of the periodic table.
As we move down a group,the number of shells increases,which leads to an increase in the ionic radius.
Therefore,the order of ionic radii for these ions is $K^{+} > Na^{+} > Li^{+}$.

(D) All the given ions ($Na^{+}$,$Mg^{2+}$,$Al^{3+}$,$Si^{4+}$) are isoelectronic,meaning they all have $10$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
The atomic numbers are: $Na$ $(Z=11)$,$Mg$ $(Z=12)$,$Al$ $(Z=13)$,and $Si$ $(Z=14)$.
Since $Si^{4+}$ has the highest nuclear charge $(Z=14)$,it exerts the strongest attraction on the electrons,resulting in the smallest ionic size.
Therefore,the correct option is $(D)$.

(D) All the given ions are isoelectronic,meaning they all have $10$ electrons.
$F^{-} = 9 + 1 = 10$ electrons
$O^{2-} = 8 + 2 = 10$ electrons
$Al^{3+} = 13 - 3 = 10$ electrons
$N^{3-} = 7 + 3 = 10$ electrons
For isoelectronic species,the ionic radius increases as the nuclear charge $(Z)$ decreases.
The atomic numbers are: $N (7)$,$O (8)$,$F (9)$,$Al (13)$.
Since $N^{3-}$ has the lowest nuclear charge $(Z = 7)$,the electrostatic attraction between the nucleus and the electron cloud is the weakest,resulting in the largest ionic size.

(C) The ionic radius $(r)$ is inversely proportional to the effective nuclear charge $(Z_{eff})$.
According to the relationship $r \propto \frac{1}{Z_{eff}}$,as the effective nuclear charge increases,the electrons are pulled more strongly towards the nucleus,resulting in a decrease in the ionic radius.
Therefore,the correct option is $(C)$.

(D) Across a period,the size of an atom decreases with an increase in atomic number due to an increase in the effective nuclear charge.
Conversely,properties like electron affinity,ionization potential,and electronegativity generally increase across a period.

(D) The ionic radius depends on the effective nuclear charge and the number of shells.
$(A)$ $Ti^{4+}$ $(Z=22)$ and $Mn^{7+}$ $(Z=25)$ are isoelectronic ($18$ electrons). As nuclear charge increases,size decreases,so $Ti^{4+} > Mn^{7+}$.
$(B)$ $^{35}Cl^{-}$ and $^{37}Cl^{-}$ are isotopes. Isotopes have the same number of electrons and protons,so their ionic radii are equal.
$(C)$ $K^{+}$ and $Cl^{-}$ are isoelectronic ($18$ electrons). $K^{+}$ has a higher nuclear charge $(Z=19)$ than $Cl^{-}$ $(Z=17)$,so $K^{+} < Cl^{-}$.
$(D)$ $P^{3+}$ and $P^{5+}$ are ions of the same element. $P^{5+}$ has lost more electrons than $P^{3+}$,resulting in a higher effective nuclear charge per electron,making $P^{3+} > P^{5+}$.
Thus,the correct statement is $(D)$.

(D) The ionic radii of isoelectronic species decrease as the nuclear charge increases.
For the given ions:
$Na^{+}$ ($10$ electrons,$Z=11$)
$Mg^{+2}$ ($10$ electrons,$Z=12$)
$Al^{+3}$ ($10$ electrons,$Z=13$)
$Ca^{+2}$ ($18$ electrons,$Z=20$)
Comparing the isoelectronic series $(Na^{+}, Mg^{+2}, Al^{+3})$,the ionic radius decreases as the atomic number increases.
Thus,$Al^{+3}$ is the smallest among the isoelectronic ions.
$Ca^{+2}$ has a larger shell $(n=3)$ compared to the others $(n=2)$,making it significantly larger.
The order of ionic radii is $Ca^{+2} > Na^{+} > Mg^{+2} > Al^{+3}$.
Therefore,$Al^{+3}$ is the smallest cation.

(D) As we move across a period from left to right,the effective nuclear charge increases,which pulls the electrons closer to the nucleus,resulting in a decrease in atomic size.

Element	$Mg$	$Al$	$Si$	$P$
Atomic radius $(\mathring{A})$	$1.60$	$1.43$	$1.32$	$1.28$

(A) The atomic radius of elements increases down the group in the periodic table due to the addition of new shells.
The order of atomic radii for the halogen group is $F < Cl < Br < I$.
Therefore,$F$ (Fluorine) is the smallest atom among the given options.

(A) All the given ions ($Na^{+}$,$Mg^{2+}$,$Al^{3+}$,$Si^{4+}$) 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.
Since $Na^{+}$ has the lowest atomic number $(Z = 11)$ among the given ions,it experiences the least effective nuclear attraction on its electrons.
Therefore,$Na^{+}$ has the largest ionic radius.

(B) The ionic radius of cations increases as we move down a group in the periodic table due to the addition of new electron shells.
$Al^{3+}$,$Mg^{2+}$,and $Na^{+}$ belong to the $3^{rd}$ period,while $Ba^{2+}$ belongs to the $6^{th}$ period.
Since $Ba^{2+}$ is in the $6^{th}$ period,it has the largest number of shells among the given ions,making it the biggest ion.

(B) The radius ratio is calculated as: $\frac{r^+}{r^-} = \frac{1.46}{2.16} \approx 0.675$.
Since the value lies in the range $0.414 - 0.732$,the coordination number is $6$.
Therefore,the crystal structure is of the $NaCl$ type.

(C) Isoelectronic species are those that have the same number of electrons.
For option $C$:
$N^{3-}$ has $7 + 3 = 10$ electrons.
$F^{-}$ has $9 + 1 = 10$ electrons.
$Na^{+}$ has $11 - 1 = 10$ electrons.
Since all have $10$ electrons,they are isoelectronic.

(D) Isoelectronic species are those that have the same number of electrons.
For option $D$:
$K^{+}$: $19 - 1 = 18$ electrons.
$Ca^{2+}$: $20 - 2 = 18$ electrons.
$Sc^{3+}$: $21 - 3 = 18$ electrons.
$Cl^{-}$: $17 + 1 = 18$ electrons.
Since all these ions have $18$ electrons,they are isoelectronic.

(A) Across a period,the atomic radius decreases from left to right due to an increase in effective nuclear charge.
Down a group,the atomic radius increases due to the addition of new shells.
Comparing the elements:
$C$ and $O$ and $F$ are in the $2^{nd}$ period. Their order of size is $F < O < C$.
$Cl$ and $Br$ are in the $3^{rd}$ and $4^{th}$ periods respectively,so they are larger than the $2^{nd}$ period elements.
The order is $F < O < C < Cl < Br$.

(A) The given ions ${N^{3-}}, {O^{2-}}, {F^-},$ and ${Na^+}$ are isoelectronic,as each contains $10$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
The atomic numbers are $N (7), O (8), F (9),$ and $Na (11)$.
Therefore,the order of ionic radii is ${N^{3-}} > {O^{2-}} > {F^-} > {Na^+}$.

(B) Isoelectronic ions have the same number of electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
As we move from $O^{2-}$ to $Al^{3+}$,the nuclear charge increases while the number of electrons remains constant ($10$ electrons).
Therefore,the ionic radius decreases from $O^{2-}$ to $Al^{3+}$.

(B) The ionic radius $(r)$ is inversely proportional to the effective nuclear charge $(Z_{eff})$,which can be expressed as $r \propto \frac{1}{Z_{eff}}$.
As the effective nuclear charge increases,the electrons are pulled more strongly towards the nucleus,resulting in a decrease in the ionic radius.

(B) In a period,the covalent radius decreases from left to right due to an increase in effective nuclear charge.
The atomic numbers of the given elements are: $Ca (Z=20)$,$Sc (Z=21)$,and $Ti (Z=22)$.
Since they belong to the same period ($4^{th}$ period),the order of their covalent radii is $Ti < Sc < Ca$.
Therefore,the increasing order is $(I) < (III) < (II)$.

(D) The species $O^{2-}$ and $F^-$ are isoelectronic,having $10$ electrons each. For isoelectronic species,the ionic radius decreases as the nuclear charge $(Z)$ increases. Since $Z$ for $O$ is $8$ and $F$ is $9$,the order is $O^{2-} > F^-$.
Comparing neutral atoms,the atomic radius decreases across a period from left to right. Thus,$O > F$.
Furthermore,an anion is always larger than its parent neutral atom because of increased electron-electron repulsion and decreased effective nuclear charge per electron. Therefore,$O^{2-} > O$ and $F^- > F$.
Combining these,the overall order of radii is $O^{2-} > F^- > O > F$.

(D) Atomic size increases down the group in the periodic table.
In the given options,$Li, Na, K$ belong to Group $1$ and are arranged in increasing order of their atomic size as we move down the group.

(D) The ionic radius increases as we move down a group in the periodic table.
Since $Be^{2+}$,$Mg^{2+}$,$Ca^{2+}$,and $Sr^{2+}$ belong to Group $2$,the ionic radius increases in the order: $Be^{2+} < Mg^{2+} < Ca^{2+} < Sr^{2+}$.
Therefore,$Sr^{2+}$ has the largest ionic radius.

(C) As we move down a group,the atomic radius increases because a new principal energy level is added for each subsequent element,which increases the distance between the nucleus and the outermost electrons.

(A) The size of an anion is always larger than its parent atom because the addition of an electron increases the electron-electron repulsion,while the nuclear charge remains the same,leading to a decrease in the effective nuclear charge per electron.

(C) The ions $Li^{+}$,$Be^{2+}$,$O^{2-}$,and $F^{-}$ are compared based on their electronic configuration and nuclear charge.
$O^{2-}$ and $F^{-}$ are isoelectronic with $10$ electrons $(1s^2 2s^2 2p^6)$.
For isoelectronic species,the ionic radius increases as the nuclear charge $(Z)$ decreases.
The atomic number of $O$ is $8$ and $F$ is $9$.
Since $Z_{O} < Z_{F}$,the ionic radius of $O^{2-}$ is greater than that of $F^{-}$.
$Li^{+}$ and $Be^{2+}$ have fewer shells $(n=1)$ compared to $O^{2-}$ and $F^{-}$ $(n=2)$,so they have smaller radii.
Therefore,$O^{2-}$ has the largest ionic radius.

(D) 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.
$N^{3-}$ $(Z=7)$,$O^{2-}$ $(Z=8)$,$F^{-}$ $(Z=9)$,$Na^{+}$ $(Z=11)$.
Since $Na^{+}$ has the highest atomic number $(Z=11)$,it has the strongest nuclear attraction for the electrons,making it the smallest in size.

(A) For isoelectronic species,the ionic radius decreases as the nuclear charge $(Z)$ increases.
The species $Mg^{2+}$ $(Z=12)$,$Na^{+}$ $(Z=11)$,and $F^{-}$ $(Z=9)$ are isoelectronic,each having $10$ electrons.
Since $F^{-}$ has the lowest nuclear charge $(Z=9)$,it has the largest size.
$Mg^{2+}$ has the highest nuclear charge $(Z=12)$,so it has the smallest size.
$Al$ is a neutral atom with $13$ electrons,which is significantly larger than the ions.
Thus,the correct order of increasing size is $Al < Mg^{2+} < Na^{+} < F^{-}$.

(C) In the periodic table,atomic size decreases from left to right across a period and increases from top to bottom down a group. $Li$ and $Be$ belong to period $2$,while $Na$ and $Mg$ belong to period $3$. Among $Li$ and $Be$,$Be$ is to the right,so it is smaller. Thus,$Be$ has the smallest atomic size.

(C) The size of an anion is always larger than its parent neutral atom because of the increased electron-electron repulsion and decreased effective nuclear charge per electron.
For the species $Na^-$,$Na$,and $Na^+$,the order of size is $Na^- > Na > Na^+$.
Therefore,$Na^-$ has the largest size.

(D) The ionic radius increases down a group and decreases across a period from left to right.
For the given elements,$Mg$ ($2^{nd}$ group,$3^{rd}$ period) has the smallest size.
$Na$ ($1^{st}$ group,$3^{rd}$ period) is larger than $Mg$.
$K$ ($1^{st}$ group,$4^{th}$ period) is larger than $Na$.
$Rb$ ($1^{st}$ group,$5^{th}$ period) is the largest.
Therefore,the increasing order of ionic radii is $Mg < Na < K < Rb$.

(D) All the given ions ($Na^{+}$,$Mg^{2+}$,$Al^{3+}$,and $Si^{4+}$) are isoelectronic species,meaning they all contain $10$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
Since $Si^{4+}$ has the highest atomic number $(Z = 14)$ among the given ions,it exerts the strongest attraction on the electrons,resulting in the lowest ionic radius.
Therefore,the correct option is $(D)$.

(A) The species $Ar$,$K^{+}$,and $Ca^{2+}$ are isoelectronic,meaning they all contain $18$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge $(Z)$ increases.
The atomic numbers are: $Ar$ $(Z=18)$,$K^{+}$ $(Z=19)$,and $Ca^{2+}$ $(Z=20)$.
Since the nuclear charge increases in the order $Ar < K^{+} < Ca^{2+}$,the ionic radii decrease in the same order.
Therefore,the order of increasing radii is $Ca^{2+} < K^{+} < Ar$.

(D) Cations lose electrons and are smaller in size than the parent atom,whereas anions gain electrons and are larger in size than the parent atom.
For the hydrogen species,the order is $H^{-} > H > H^{+}$.
For isoelectronic species,the ionic radius decreases as the atomic number (nuclear charge) increases.
Comparing $O^{2-}$,$F^{-}$,and $Na^{+}$,all have $10$ electrons. Their atomic numbers are $8$,$9$,and $11$ respectively.
Therefore,the correct order of ionic radii is $O^{2-} > F^{-} > Na^{+}$.
Since none of the provided options $A$,$B$,or $C$ correctly represent this order,the correct answer is $D$.

(A) Atomic radius of the elements decreases across a period from left to right due to an increase in effective nuclear charge.
On moving down a group,the number of shells increases,so the atomic radius increases.
Amongst isoelectronic species,the ionic radius increases with an increase in negative charge or a decrease in positive charge.
Therefore,the statement that 'smaller the positive charge on the cation,smaller is the ionic radius' is incorrect,because a smaller positive charge means a larger ionic radius.

(C) For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
All the given species $(S^{2-}, Cl^{-}, K^{+}, Ca^{2+})$ have $18$ electrons.
The atomic numbers are: $S (16), Cl (17), K (19), Ca (20)$.
As the nuclear charge increases,the force of attraction between the nucleus and the electrons increases,which pulls the electron cloud closer to the nucleus,resulting in a smaller ionic radius.
Therefore,the order of decreasing ionic radii is $S^{2-} > Cl^{-} > K^{+} > Ca^{2+}$.

(B) Atomic radius decreases across a period from left to right due to an increase in effective nuclear charge,while it increases down a group due to the addition of new shells.
The electronic configurations are:
$Mg (Z=12): [Ne] 3s^2$
$P (Z=15): [Ne] 3s^2 3p^3$
$Cl (Z=17): [Ne] 3s^2 3p^5$
$Ca (Z=20): [Ar] 4s^2$
$Mg, P,$ and $Cl$ belong to the $3^{rd}$ period. Within the same period,the atomic radius decreases as $Z$ increases: $Mg > P > Cl$.
$Ca$ belongs to the $4^{th}$ period,so it has a larger atomic radius than the $3^{rd}$ period elements.
Therefore,the increasing order of atomic radii is $Cl < P < Mg < Ca$.

(A) All the given species ($Ca^{2+}$,$K^{+}$,$Ar$,$Cl^{-}$,$S^{2-}$) 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: $Ca$ $(Z=20)$,$K$ $(Z=19)$,$Ar$ $(Z=18)$,$Cl$ $(Z=17)$,$S$ $(Z=16)$.
Since the nuclear charge follows the order $Ca^{2+} > K^{+} > Ar > Cl^{-} > S^{2-}$,the ionic size follows the inverse order:
$Ca^{2+} < K^{+} < Ar < Cl^{-} < S^{2-}$.

(D) All the given ions $(O^{2-}, F^{-}, Na^{+}, Mg^{2+}, Al^{3+})$ are isoelectronic,meaning they all have $10$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
$O^{2-}$ $(Z=8)$,$F^{-}$ $(Z=9)$,$Na^{+}$ $(Z=11)$,$Mg^{2+}$ $(Z=12)$,$Al^{3+}$ $(Z=13)$.
Since the nuclear charge increases from $O^{2-}$ to $Al^{3+}$,the ionic radius decreases in the order: $O^{2-} > F^{-} > Na^{+} > Mg^{2+} > Al^{3+}$.

(C) For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number,$Z$) increases.
All the given species,$S^{2-}, Cl^{-}, K^{+},$ and $Ca^{2+}$,contain $18$ electrons.
Their respective atomic numbers $(Z)$ are: $S = 16, Cl = 17, K = 19, Ca = 20$.
Since the ionic radius is inversely proportional to the nuclear charge for isoelectronic species,the increasing order of ionic radii is: $Ca^{2+} < K^{+} < Cl^{-} < S^{2-}$.

(A) $N^{3-}$,$O^{2-}$,and $F^{-}$ are isoelectronic species,as each contains $10$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
The atomic numbers are $N (7)$,$O (8)$,and $F (9)$.
Therefore,the ionic radii decrease in the order $N^{3-} > O^{2-} > F^{-}$.
The values are $1.71 \ \mathring{A}$,$1.40 \ \mathring{A}$,and $1.36 \ \mathring{A}$ respectively.

(C) The number of electrons in the given ions are:
$O^{2-} = 8 + 2 = 10$ electrons
$F^{-} = 9 + 1 = 10$ electrons
$B^{3+} = 5 - 3 = 2$ electrons
$Li^{+} = 3 - 1 = 2$ electrons
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
Comparing $O^{2-}$ (atomic number $Z = 8$) and $F^{-}$ (atomic number $Z = 9$):
Since $O^{2-}$ has a lower nuclear charge than $F^{-}$,it has a larger ionic radius.
Therefore,$O^{2-}$ has the highest ionic radius among the given options.

(B) For isoelectronic species,the ionic size decreases as the nuclear charge (atomic number) increases.
All these ions have $10$ electrons ($1s^2 2s^2 2p^6$ configuration).
The atomic numbers are: $Na (Z=11)$,$Mg (Z=12)$,$Al (Z=13)$,and $Si (Z=14)$.
As the nuclear charge increases,the electrostatic attraction between the nucleus and the electrons increases,which pulls the electron cloud closer to the nucleus,resulting in a smaller ionic radius.
Therefore,the order of ionic size is $Na^{+} > Mg^{2+} > Al^{3+} > Si^{4+}$.

(D) $1$. For option $A$: The atomic radius increases down a group. However,due to the lanthanide contraction,the size of $Ga$ is slightly smaller than expected,but $B < Al < Ga$ is not the standard trend as $Al$ and $Ga$ are very similar in size.
$2$. For option $B$: These are isoelectronic species. For isoelectronic species,the size decreases as the nuclear charge $(Z)$ increases. The order should be $O^{2-} > S^{2-} > Cl^{-}$. Thus,$Cl^{-} < S^{2-} < O^{2-}$ is incorrect.
$3$. For option $C$: Atomic size increases down a group. The correct order is $Sc < Y < La$. Thus,$La < Y < Sc$ is incorrect.
$4$. For option $D$: For the same element,the size of the ion depends on the charge. The order of size is $I^{+} < I < I^{-}$. As the number of electrons increases (negative charge),the inter-electronic repulsion increases,leading to a larger size. As the number of electrons decreases (positive charge),the effective nuclear charge per electron increases,leading to a smaller size. This is the correct order.

(D) For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
Since the number of electrons is the same ($10$ electrons in each),the effective nuclear charge increases from $O^{2-}$ to $Al^{3+}$.
Therefore,the ionic radii follow the order: $O^{2-} > F^{-} > Na^{+} > Mg^{2+} > Al^{3+}$.
This represents a significant decrease from $O^{2-}$ to $Al^{3+}$.

(B) $C^{4-}$,$N^{3-}$,and $O^{2-}$ are isoelectronic species,all having $10$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number,$Z$) increases.
The atomic numbers are: $C (Z=6)$,$N (Z=7)$,and $O (Z=8)$.
Since the atomic number of $N$ $(Z=7)$ lies between $C$ $(Z=6)$ and $O$ $(Z=8)$,its ionic radius will lie between the ionic radii of $C^{4-}$ $(2.60 \ \mathring{A})$ and $O^{2-}$ $(1.40 \ \mathring{A})$.
Among the given options,$1.71 \ \mathring{A}$ is the only value that falls between $1.40$ and $2.60$.

(A) The size of an atom or ion depends on the effective nuclear charge and the number of shells.
For isoelectronic species $(O^{2-}, F^{-})$,the species with the lower nuclear charge has a larger radius. Since $O$ $(Z=8)$ has a lower nuclear charge than $F$ $(Z=9)$,$O^{2-}$ is larger than $F^{-}$.
Comparing neutral atoms,$O$ $(Z=8)$ is larger than $F$ $(Z=9)$ due to the decrease in atomic radius across a period.
Furthermore,an anion is always larger than its parent neutral atom due to increased inter-electronic repulsion and decreased effective nuclear charge per electron.
Therefore,the correct order of radii is $O^{2-} > F^{-} > O > F$.
Hence,option $A$ is correct.

(B) The given ions $O^{2-}$,$F^{-}$,$Na^{+}$,and $Al^{3+}$ are isoelectronic species,as each contains $10$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number,$Z$) increases.
The atomic numbers are: $O (Z=8)$,$F (Z=9)$,$Na (Z=11)$,and $Al (Z=13)$.
Since the nuclear charge increases in the order $O < F < Na < Al$,the ionic size decreases in the order $O^{2-} > F^{-} > Na^{+} > Al^{3+}$.
Therefore,the increasing order of ionic size is $Al^{3+} < Na^{+} < F^{-} < O^{2-}$.