Atomic and Ionic radii Questions in English

(D) The atomic size or radius decreases in a period from left to right due to an increase in effective nuclear charge.
For the given elements in the second period,the order of atomic number is $Li (3) < B (5) < N (7) < F (9)$.
Therefore,the correct order of atomic size is $Li > B > N > F$.

(C) Atomic radii generally decrease in a period from left to right due to an increase in the effective nuclear charge.
In the second period,the elements are arranged as $Be (Z=4), B (Z=5), C (Z=6), N (Z=7)$.
As we move from left to right,the effective nuclear charge increases,which pulls the valence electrons closer to the nucleus,resulting in a decrease in atomic size.
Therefore,the correct order of atomic radii is $N < C < B < Be$.

(B) For isoelectronic species,the size of the ion increases as the negative charge increases and decreases as the positive charge increases.
Comparing $P^{3-}, S^{2-}$,and $Cl^-$,all have $18$ electrons.
The size order is $P^{3-} > S^{2-} > Cl^-$.
Therefore,the statement that $Cl^-$ is the largest is incorrect.
Thus,option $(B)$ is the incorrect statement.

(D) For a particular element,the order of size of anion,neutral atom,and cation is $\text{anion} > \text{neutral atom} > \text{cation}$.
Cation has the smallest size due to a greater effective nuclear charge $(Z_{eff})$.
Anion has the largest size because the addition of electrons increases electron-electron repulsion,leading to a larger ionic radius.
Thus,the correct order is $I^{-} > I > I^{+}$.

(B) The ionic radius of species depends on the effective nuclear charge and the number of shells.
For ions with the same number of electrons (isoelectronic series),the ionic radius decreases as the atomic number increases because the effective nuclear charge increases.
Comparing $S^{2-}$ $(18 \ e^-)$ and $K^+$ $(18 \ e^-)$: Since $S^{2-}$ has a lower atomic number $(Z=16)$ than $K^+$ $(Z=19)$,$S^{2-}$ has a larger ionic radius.
For the transition metal ions $Sc^{3+}$,$V^{5+}$,and $Mn^{7+}$,as the oxidation state increases,the ionic radius decreases due to the increase in effective nuclear charge and loss of electrons.
Thus,the order is $S^{2-} > K^{+} > Sc^{3+} > V^{5+} > Mn^{7+}$.

(A) For an isoelectronic series,the ionic radius decreases as the nuclear charge (atomic number) increases.
All the given ions ($Mg^{2+}$,$Na^{+}$,$Cl^{-}$,$S^{2-}$) are not isoelectronic. Let us check their electron configurations:
$Mg^{2+}$: $1s^2 2s^2 2p^6$ ($10$ electrons)
$Na^{+}$: $1s^2 2s^2 2p^6$ ($10$ electrons)
$Cl^{-}$: $1s^2 2s^2 2p^6 3s^2 3p^6$ ($18$ electrons)
$S^{2-}$: $1s^2 2s^2 2p^6 3s^2 3p^6$ ($18$ electrons)
Comparing the isoelectronic pairs:
Among $Mg^{2+}$ and $Na^{+}$,$Mg^{2+}$ has a higher nuclear charge $(Z=12)$ compared to $Na^{+}$ $(Z=11)$,so $Mg^{2+}$ is smaller.
Among $Cl^{-}$ and $S^{2-}$,$Cl^{-}$ has a higher nuclear charge $(Z=17)$ compared to $S^{2-}$ $(Z=16)$,so $Cl^{-}$ is smaller.
Comparing the two groups,the ions with $10$ electrons are significantly smaller than those with $18$ electrons due to fewer shells.
Thus,the overall order of ionic radii is $Mg^{2+} < Na^{+} < Cl^{-} < S^{2-}$.
The smallest ion is $Mg^{2+}$.

(D) As we move down the group in the periodic table,the atomic size increases due to an increase in the principal quantum number.
Oxygen $(O)$,sulfur $(S)$,selenium $(Se)$,and tellurium $(Te)$ all belong to Group $16$ of the periodic table.
Since $Te$ is at the bottom of this group,it has the largest atomic size.
Thus,the correct option is $(D)$.

(B) The correct order is $O^{2-} > F^{-} > Na^{+} > Mg^{2+} > Al^{3+}$.
All these ions are isoelectronic species,meaning they have the same number of electrons ($10$ electrons) and the same electronic configuration $(1s^2, 2s^2, 2p^6)$.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
The nuclear charges for these ions are: $O^{2-} (+8)$,$F^{-} (+9)$,$Na^{+} (+11)$,$Mg^{2+} (+12)$,and $Al^{3+} (+13)$.
Since the nuclear charge increases from $O^{2-}$ to $Al^{3+}$,the attraction between the nucleus and the electrons increases,leading to a decrease in the ionic radius.

(B) The covalent radius is defined as half the distance between the nuclei of two covalently bonded atoms. Since there is orbital overlap,the distance is the smallest.
In metallic bonding,atoms are in contact but there is no orbital overlap,making the metallic radius larger than the covalent radius.
Van der Waals radius corresponds to the distance between non-bonded atoms in adjacent molecules,which is the largest due to the absence of any chemical bond.
Therefore,the correct order is: $Covalent \ radius < Metallic \ radius < Van \ der \ Waals \ radius$.

(B) The given ions $O^{2-}, Na^{+}, F^{-}, N^{3-}, Mg^{2+}$ 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: $N^{3-} (7), O^{2-} (8), F^{-} (9), Na^{+} (11), Mg^{2+} (12)$.
Since $Mg^{2+}$ has the highest nuclear charge ($12$ protons),it has the smallest ionic radius.
Since $N^{3-}$ has the lowest nuclear charge ($7$ protons),it has the largest ionic radius.
Therefore,the ions with the smallest and largest radii are $Mg^{2+}$ and $N^{3-}$ respectively.

(D) The atomic radius decreases on moving from left to right in a period because the effective nuclear charge increases while the number of shells remains the same. Thus,the size of $O < N$ and $S < P$.
On moving down a group,the atomic radius increases due to the addition of a new shell. Thus,the size of $O < S$ and $N < P$.
Combining these trends,we compare the elements: $O$ and $N$ are in the $2^{nd}$ period,while $S$ and $P$ are in the $3^{rd}$ period.
Since $O < N$ and $S < P$,and elements in the $2^{nd}$ period are smaller than those in the $3^{rd}$ period,the overall order is $O < N < S < P$.

(C) All the given species are isoelectronic,meaning they all contain $10$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
As the positive charge on the cation increases,the effective nuclear charge increases,which pulls the electrons closer to the nucleus,resulting in a smaller size.
Among the given species,$Si^{4+}$ has the highest atomic number $(Z = 14)$ and the highest positive charge,therefore it has the smallest size.

(C) For isoelectronic species,the ionic radii decrease as the nuclear charge increases.
Since all these ions have the same number of electrons ($18$ electrons),the ionic radius depends on the effective nuclear charge $(Z_{eff})$.
As the atomic number $(Z)$ increases from $S$ $(16)$ to $Ca$ $(20)$,the nuclear charge increases,which pulls the electrons closer to the nucleus.
Therefore,the ionic radii show a significant decrease from $S^{2-}$ to $Ca^{2+}$.

(A) $\because$ $Z$ represents the number of protons in a nucleus.
For isoelectronic species,the size depends on the effective nuclear charge.
As the number of protons $(Z)$ decreases for the same number of electrons,the attraction between the nucleus and the electrons decreases,leading to an increase in ionic size.
Thus,the species with the smallest atomic number will have the largest size.
Comparing the given atomic numbers: $Z-2 < Z-1 < Z < Z+1$.
Therefore,the ion with atomic number $Z-2$ will have the largest size.
Hence,option $(A)$ is the correct answer.

(B) $Na^{+}$ and $Mg^{2+}$ are isoelectronic species having $10 \ e^{-}$ each.
However,$Mg^{2+}$ has $12$ protons while $Na^{+}$ has $11$ protons.
Due to a higher number of protons in $Mg^{2+}$,the effective nuclear charge $(Z_{eff})$ is greater for $Mg^{2+}$ than for $Na^{+}$.
Since ionic radius is inversely proportional to the effective nuclear charge,the ionic radius of $Na^{+}$ is greater than that of $Mg^{2+}$.
Therefore,both $(A)$ and $(R)$ are correct,and $(R)$ is the correct explanation of $(A)$.

(A) The atomic radius decreases across a period from left to right due to an increase in effective nuclear charge,and it increases down a group due to the addition of new shells.
Comparing the given elements:
$F$ (Fluorine,Group $17$,Period $2$)
$N$ (Nitrogen,Group $15$,Period $2$)
$P$ (Phosphorus,Group $15$,Period $3$)
$Si$ (Silicon,Group $14$,Period $3$)
$Na$ (Sodium,Group $1$,Period $3$)
Across Period $2$: $F < N$.
Across Period $3$: $P < Si < Na$.
Since Period $3$ elements have more shells than Period $2$ elements,they are larger.
Combining these,the increasing order of atomic radius is $F < N < P < Si < Na$.

(A) The elements and their atomic numbers are: $A(13)$,$B(11)$,$C(9)$,$D(7)$,$E(16)$.
To achieve the nearest inert gas configuration,these elements form the following ions:
$A^{3+}$ ($10$ electrons),$B^+$ ($10$ electrons),$C^-$ ($10$ electrons),$D^{3-}$ ($10$ electrons),$E^{2-}$ ($18$ electrons).
For isoelectronic ions $(A^{3+}, B^+, C^-, D^{3-})$,the ionic size increases as the nuclear charge $(Z)$ decreases.
$Z$ values are: $A=13, B=11, C=9, D=7$.
Since $D$ has the lowest $Z$ $(7)$,$D^{3-}$ has the largest size among the isoelectronic series.
Comparing $D^{3-}$ $(Z=7)$ and $E^{2-}$ $(Z=16)$: $D^{3-}$ has $10$ electrons in the $n=2$ shell,while $E^{2-}$ has $18$ electrons in the $n=3$ shell. Thus,$D^{3-}$ is larger than $E^{2-}$.
Therefore,$X = D$.
For the smallest size,we compare the ions. $A^{3+}$ has the highest nuclear charge $(Z=13)$ among the $10$-electron species,making it the smallest among them. $E^{2-}$ $(Z=16)$ has more shells $(n=3)$,making it larger than $A^{3+}$.
Therefore,$Y = A$.
Thus,$X$ and $Y$ are $D$ and $A$ respectively.

(B) $i$. Atomic radii: $Cl$ $(1.81 \ \mathring{A})$,$F$ $(0.72 \ \mathring{A})$,$Li$ $(1.52 \ \mathring{A})$. The correct order is $Cl > Li > F$. Thus,$i$ is incorrect.
$ii$. Atomic radii: $P$ $(1.28 \ \mathring{A})$,$C$ $(0.77 \ \mathring{A})$,$N$ $(0.75 \ \mathring{A})$. The order $P > C > N$ is correct.
$iii$. Lanthanoid contraction: $Eu$ $(2.04 \ \mathring{A})$,$Sm$ $(1.98 \ \mathring{A})$,$Tm$ $(1.74 \ \mathring{A})$. The correct order is $Eu > Sm > Tm$. Thus,$iii$ is incorrect.
$iv$. Atomic radii: $Sr$ $(2.15 \ \mathring{A})$,$Ca$ $(1.97 \ \mathring{A})$,$Mg$ $(1.60 \ \mathring{A})$. The order $Sr > Ca > Mg$ is correct.
Therefore,$ii$ and $iv$ are correct.

(D) $1.$ $C$ and $N$ are in the same period. The atomic radius of $C$ is $\approx 77 \ pm$ and $N$ is $\approx 75 \ pm$. The difference is $\approx 2 \ pm$.
$2.$ $O$ and $F$ are in the same period. The atomic radius of $O$ is $\approx 73 \ pm$ and $F$ is $\approx 71 \ pm$. The difference is $\approx 2 \ pm$.
$3.$ $P$ and $S$ are in the same period. The atomic radius of $P$ is $\approx 110 \ pm$ and $S$ is $\approx 103 \ pm$. The difference is $\approx 7 \ pm$.
$4.$ $Li$ and $Be$ are in the same period. The atomic radius of $Li$ is $\approx 152 \ pm$ and $Be$ is $\approx 111 \ pm$. The difference is $\approx 41 \ pm$.
Comparing all pairs, the difference in atomic radii is maximum for the pair $(Li, Be)$.

(B) For isoelectronic species,the size decreases as the nuclear charge (number of protons) increases.
$F^{-}$ has $9$ protons and $10$ electrons.
$Na^{+}$ has $11$ protons and $10$ electrons.
Since $Na^{+}$ has a higher nuclear charge than $F^{-}$,the electrons are pulled more strongly towards the nucleus,making $Na^{+}$ smaller than $F^{-}$.
Thus,in the pair $(F^{-}, Na^{+})$,the second ion is smaller than the first.

(D) The electronic configuration of $Na^{+}$ is $1s^2 2s^2 2p^6$ ($10$ electrons).
The electronic configuration of $F^{-}$ is $1s^2 2s^2 2p^6$ ($10$ electrons).
Since both have the same number of electrons,they are isoelectronic species,so $(R)$ is correct.
However,the ionic radius depends on the nuclear charge $(Z)$.
For $Na^{+}$,$Z = 11$,and for $F^{-}$,$Z = 9$.
As the nuclear charge increases,the attraction of the nucleus on the electrons increases,leading to a smaller ionic radius.
Therefore,the ionic radius of $Na^{+}$ $(102 \ pm)$ is smaller than that of $F^{-}$ $(133 \ pm)$.
Thus,$(A)$ is incorrect.

(C) The atomic radius of gallium ($Ga$,$Z=31$) is slightly smaller than that of aluminium ($Al$,$Z=13$).
This is because,in gallium,the $3d$-orbitals are filled before the $4p$-orbital.
These $d$-electrons have a poor shielding effect compared to $s$- and $p$-electrons.
As a result,the effective nuclear charge increases,which pulls the valence electrons closer to the nucleus,leading to a decrease in the atomic radius.

(B) The given species are $K^{+}$,$Na^{+}$,$Mg^{2+}$,and $Al^{3+}$.
$K^{+}$ has electrons in the $n=3$ shell,while $Na^{+}$,$Mg^{2+}$,and $Al^{3+}$ have electrons in the $n=2$ shell.
Therefore,$K^{+}$ has the largest ionic radius.
For the isoelectronic species $Na^{+}$,$Mg^{2+}$,and $Al^{3+}$,the ionic radius decreases as the nuclear charge (atomic number) increases.
The atomic numbers are $Na (11)$,$Mg (12)$,and $Al (13)$.
Thus,the order of ionic radii for these is $Na^{+} > Mg^{2+} > Al^{3+}$.
Combining these,the correct order is $K^{+} > Na^{+} > Mg^{2+} > Al^{3+}$.

(A) All these ions 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: $O (8), F (9), Na (11), Mg (12)$.
As the nuclear charge increases,the electrons are pulled more strongly towards the nucleus,resulting in a smaller ionic radius.
Therefore,the order of increasing ionic radii is: $Mg^{2+} < Na^{+} < F^{-} < O^{2-}$.

(A) Ionic radius increases with the number of shells. $I^{-}$ has $5$ shells,$Se^{2-}$ and $Br^{-}$ have $4$ shells,and $O^{2-}$ and $F^{-}$ have $2$ shells.
For isoelectronic species (ions with the same number of electrons),the ionic radius decreases as the atomic number $(Z)$ increases because of the greater nuclear pull.
$1.$ For $Se^{2-}$ $(Z=34)$ and $Br^{-}$ $(Z=35)$: $Se^{2-} > Br^{-}$.
$2.$ For $O^{2-}$ $(Z=8)$ and $F^{-}$ $(Z=9)$: $O^{2-} > F^{-}$.
Therefore,the correct decreasing order is: $I^{-} > Se^{2-} > Br^{-} > O^{2-} > F^{-}$.

(A) Isoelectronic species are ions or atoms that have the same number of electrons. Both $Mg^{2+}$ and $Al^{3+}$ have $10$ electrons $(1s^2 2s^2 2p^6)$.
For isoelectronic species,the ionic radius decreases as the atomic number $(Z)$ increases because the effective nuclear charge $(Z_{eff})$ increases,pulling the electrons closer to the nucleus.
$Z_{Mg} = 12$ and $Z_{Al} = 13$.
Since $Z_{Al} > Z_{Mg}$,the effective nuclear charge on the outermost electrons in $Al^{3+}$ is greater than in $Mg^{2+}$.
This higher $Z_{eff}$ results in a smaller ionic radius for $Al^{3+}$ compared to $Mg^{2+}$.
Therefore,both $A$ and $R$ are true,and $R$ is the correct explanation for $A$.

(B) The atomic radius decreases across a period from left to right due to an increase in effective nuclear charge.
Across $Period-2$,the order is $B < Be$.
Across $Period-3$,the order is $Al < Na$.
Down a group,the atomic radius increases due to the addition of a new shell.
Comparing the elements: $B$ (Period $2$,Group $13$) has the smallest radius. $Be$ (Period $2$,Group $2$) is larger than $B$. $Al$ (Period $3$,Group $13$) is larger than $B$ and $Be$. $Na$ (Period $3$,Group $1$) is the largest among these.
Therefore,the correct order of atomic radius is $B < Be < Al < Na$.

(A) The electronic configuration of $Be^{2+}$ is $[He]$,$Na^{+}$ is $[Ne]$,$Cl^{-}$ is $[Ar]$ and $Br^{-}$ is $[Kr]$.
As we move down the group,the number of shells increases,leading to an increase in ionic radius.
Since $Be^{2+}$ has $2$ shells,$Na^{+}$ has $3$ shells,$Cl^{-}$ has $4$ shells,and $Br^{-}$ has $5$ shells,the order of ionic radii is $Br^{-} > Cl^{-} > Na^{+} > Be^{2+}$.

(A) For isoelectronic species,the ionic radius decreases as the nuclear charge (number of protons) increases because the electrons are more strongly attracted to the nucleus.

$Species$	$N^{3-}, O^{2-}, Na^{+}, Mg^{2+}$
$Number$ of electrons	$10, 10, 10, 10$
$Number$ of protons	$7, 8, 11, 12$

As the number of protons increases,the effective nuclear charge increases,leading to a smaller ionic radius.
The order of ionic radius is: $Mg^{2+} < Na^{+} < O^{2-} < N^{3-}$.
Thus,$Mg^{2+}$ has the least ionic radius and $N^{3-}$ has the highest ionic radius.

(A) All the given species $(S^{2-}, P^{3-}, Cl^{-}, Ca^{2+}, Ar, K^{+})$ 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: $P (Z=15), S (Z=16), Cl (Z=17), Ar (Z=18), K (Z=19), Ca (Z=20)$.
Since the number of electrons is constant $(18)$,the species with the lowest atomic number will have the weakest nuclear attraction for the electrons,resulting in the largest radius.
Therefore,the decreasing order of radius is: $P^{3-} > S^{2-} > Cl^{-} > Ar > K^{+} > Ca^{2+}$.

(B) When an atom loses one or more electrons to form a cation,the number of protons remains the same,but the number of electrons decreases. This results in an increased effective nuclear charge,which leads to a stronger electrostatic attraction between the nucleus and the remaining electrons,pulling them closer and causing the ionic radius to be smaller than the atomic radius.

(C) The ionic radii of isoelectronic species decrease as the nuclear charge increases.
For the given ions,$K^{+}$ ($18$ electrons),$Li^{+}$ ($2$ electrons),$Mg^{2+}$ ($10$ electrons),and $Al^{3+}$ ($10$ electrons) have different electronic configurations.
Comparing them based on their positions in the periodic table:
$K^{+}$ is in the $4^{th}$ period,so it has the largest radius.
$Li^{+}$ is in the $2^{nd}$ period.
$Mg^{2+}$ and $Al^{3+}$ are isoelectronic ($10$ electrons each) and belong to the $3^{rd}$ period.
For isoelectronic species,the ionic radius decreases as the atomic number $(Z)$ increases: $Mg^{2+}$ $(Z=12)$ > $Al^{3+}$ $(Z=13)$.
Thus,the correct order is $K^{+} > Li^{+} > Mg^{2+} > Al^{3+}$.

(D) Ionic radius depends on the number of shells and the effective nuclear charge.
$Na^{+}$ $(1s^2 2s^2 2p^6)$ has $3$ shells,while $Li^{+}$,$Mg^{2+}$,and $Be^{2+}$ have $2$ shells.
Among ions with $2$ shells ($Li^{+}$,$Mg^{2+}$,$Be^{2+}$),the ionic radius decreases as the positive charge increases.
$Li^{+}$ $(Z=3)$ has the lowest charge,followed by $Mg^{2+}$ ($Z=12$,but smaller due to higher charge),and $Be^{2+}$ $(Z=4)$ is the smallest.
Thus,the correct order is $Na^{+} > Li^{+} > Mg^{2+} > Be^{2+}$.

(B) The atomic radii of the given elements are approximately: $Al = 143 \ pm$, $Si = 117 \ pm$, $N = 74 \ pm$, and $F = 64 \ pm$.
$Al$ and $Si$ belong to the $3^{rd}$ period, while $N$ and $F$ belong to the $2^{nd}$ period.
Within a period, the atomic radius decreases from left > right due to an increase in effective nuclear charge.
Thus, the order of atomic radii is $Al (143 \ pm) > Si (117 \ pm) > N (74 \ pm) > F (64 \ pm)$.

(A) In the periodic table,as we move from left to right across a period,the effective nuclear charge increases while the number of shells remains the same.
This results in a stronger attraction between the nucleus and the valence electrons,causing the atomic radius to decrease.
All the given elements $(Na, Mg, Al, Si, S)$ belong to the $3^{rd}$ period.
Following the trend of decreasing atomic radius from left to right,the order is $Na > Mg > Al > Si > S$.
Therefore,the increasing order is $S < Si < Al < Mg < Na$.

(D) The electronic configurations are:
$_{20}Ca = [Ar] 4s^2$
$_{21}Sc = [Ar] 4s^2 3d^1$
$_{22}Ti = [Ar] 4s^2 3d^2$
As we move from $Ca$ to $Ti$ across the period,the effective nuclear charge $(Z_{eff})$ increases because the $d$-orbitals have a diffused shape and provide poor shielding of the nuclear charge.
Consequently,the atomic size decreases as the atomic number increases.
The order of atomic size is $Ca > Sc > Ti$.
Therefore,the order of increasing covalent radius is $(I) < (III) < (II)$.

(C) The species $N^{3-}, O^{2-}$,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.
$N^{3-}$ has $7$ protons,$O^{2-}$ has $8$ protons,and $F^{-}$ has $9$ protons.
Since the nuclear charge increases from $N^{3-}$ to $F^{-}$,the attraction between the nucleus and the electrons increases,leading to a decrease in ionic size.
Therefore,the correct order of ionic radii is $N^{3-} (1.71 \mathring{A}) > O^{2-} (1.40 \mathring{A}) > F^{-} (1.36 \mathring{A})$.

(A) The given species $Ca^{2+}$,$Cl^{-}$,and $S^{2-}$ are isoelectronic,meaning they all contain $18$ electrons.

Species	Atomic number $(Z)$	Number of electrons
$Ca^{2+}$	$20$	$18$
$Cl^{-}$	$17$	$18$
$S^{2-}$	$16$	$18$

For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases. Since the atomic numbers are $Z(S^{2-}) = 16$,$Z(Cl^{-}) = 17$,and $Z(Ca^{2+}) = 20$,the order of increasing radius is $Ca^{2+} < Cl^{-} < S^{2-}$.

(C) In a period,moving from left to right,the atomic radius decreases due to an increase in effective nuclear charge.
In period $4$,the elements range from $K$ (Group $1$) to $Kr$ (Group $18$).
Excluding noble gases (which have larger van der Waals radii),the element with the largest atomic radius is $K$ (Potassium) and the element with the smallest atomic radius is $Br$ (Bromine).
Therefore,the correct pair is $K$ & $Br$.