Trends in the Modern Periodic Table Questions in English

(N/A) Calcium $(Ca)$ and strontium $(Sr)$ are expected to show chemical reactions similar to magnesium $(Mg)$.
This is because the number of valence electrons $(2)$ is the same in all these three elements.
Since chemical properties of elements are determined by their valence electrons,elements with the same number of valence electrons exhibit similar chemical reactions.

(N/A) Lithium $(Li)$,sodium $(Na)$,and potassium $(K)$ have a single electron in their outermost shells.
$(b)$ Magnesium $(Mg)$ and calcium $(Ca)$ have two electrons in their outermost shells.
$(c)$ Neon $(Ne)$,argon $(Ar)$,and xenon $(Xe)$ have filled outermost shells.

(D) Metallic character increases as we move from right to left across a period and decreases as we move from top to bottom down a group.
Among the given elements,$Be$ (Beryllium) is located in Group $2$ and Period $2$.
$Ga$,$Ge$,$As$,and $Se$ are located in Period $4$.
Since $Be$ is in the second period and is an alkaline earth metal located at the far left compared to the other elements listed,it exhibits the highest metallic character.

(C) The correct answer is $C$.
When moving from left to right across the periods of the periodic table,the effective nuclear charge increases and the atomic size decreases.
This makes it harder for atoms to lose electrons,meaning their metallic character decreases and non-metallic character increases.
Therefore,the statement that 'atoms lose their electrons more easily' is incorrect.

(A) All elements in the same group (column) of the Periodic Table exhibit similar chemical properties because they have the same number of valence electrons.
Boron $(B)$ belongs to Group $13$ and has $3$ valence electrons.
Therefore,all elements in this column share the same number of valence electrons $(3)$ and consequently possess a valency of $3$.

(B) All elements in the same column (group) as fluorine belong to Group $17$ (the halogens).
Elements in the same group have the same number of valence electrons.
Fluorine has $7$ valence electrons,so all elements in this group also have $7$ valence electrons.
Consequently,they all exhibit similar chemical properties and have a valency of $1$.

(B) The atomic number is the sum of the electrons in all shells: $2 + 8 + 7 = 17$. Thus,the atomic number is $17$.
$(b)$ Elements with the same number of valence electrons exhibit similar chemical properties. The given element has $7$ valence electrons. Among the choices,$F(9)$ has an electronic configuration of $2, 7$,meaning it also has $7$ valence electrons. Therefore,it is chemically similar to $F(9)$.

$A$ is a non-metal because it belongs to Group $17$, which consists of halogens, and all halogens are non-metals.
$(b)$ $C$ is less reactive than $A$. In Group $17$ (halogens), the reactivity of elements decreases as we move down the group because the atomic size increases and the tendency to gain an electron decreases.

(N/A) will be smaller in size than $B$. As we move from left to right across a period,the nuclear charge increases,which pulls the electrons closer to the nucleus,resulting in a decrease in atomic size.
$(b)$ $A$ will form an anion. Since $A$ belongs to Group $17$,it has $7$ valence electrons and needs one more electron to complete its octet,thus it gains an electron to form a negatively charged ion (anion).

(C) To determine the number of valence electrons,we look at the group number or the electronic configuration of the elements:
$1$. $Na$ (Sodium) has atomic number $11$. Its electronic configuration is $2, 8, 1$. It has $1$ valence electron.
$2$. $Al$ (Aluminium) has atomic number $13$. Its electronic configuration is $2, 8, 3$. It has $3$ valence electrons.
$3$. $Si$ (Silicon) has atomic number $14$. Its electronic configuration is $2, 8, 4$. It has $4$ valence electrons.
$4$. $P$ (Phosphorus) has atomic number $15$. Its electronic configuration is $2, 8, 5$. It has $5$ valence electrons.
Comparing these,$P$ has the maximum number of valence electrons $(5)$.

(D) In the periodic table,atomic radius generally decreases as we move from left to right across a period due to an increase in effective nuclear charge.
$N$,$O$,and $F$ all belong to the $2^{nd}$ period of the periodic table.
Their atomic numbers are $N (7)$,$O (8)$,and $F (9)$.
As the atomic number increases from left to right,the number of protons increases,which pulls the electrons closer to the nucleus,resulting in a smaller atomic radius.
Therefore,the order of atomic radii is $F < O < N$.
Thus,the correct increasing order is $F < O < N$.

(A) Atomic radius increases down a group due to the addition of new electron shells.
Atomic radius decreases across a period from left to right due to an increase in effective nuclear charge.
Comparing the given elements:
$Na$ (Group $1$, Period $3$)
$Mg$ (Group $2$, Period $3$)
$K$ (Group $1$, Period $4$)
$Ca$ (Group $2$, Period $4$)
Elements in Period $4$ have larger radii than those in Period $3$.
Between $K$ and $Ca$, $K$ is on the left side of the periodic table, meaning it has a lower effective nuclear charge than $Ca$, resulting in a larger atomic radius.
Therefore, $K$ has the largest atomic radius among the given options.

(B) The ease with which an element loses an electron is determined by its ionization energy.
Lower ionization energy means the element loses an electron more easily.
Ionization energy generally decreases down a group and increases across a period from left to right.
Comparing the given elements: $Na$ (Group $1$,Period $3$),$Mg$ (Group $2$,Period $3$),$K$ (Group $1$,Period $4$),and $Ca$ (Group $2$,Period $4$).
Among these,$K$ is in the lowest period (Period $4$) and Group $1$,meaning it has the largest atomic size and the weakest nuclear hold on its valence electron.
Therefore,$K$ has the lowest ionization energy and loses an electron most easily.

(A) Metallic character refers to the tendency of an element to lose electrons.
In the periodic table,metallic character decreases as we move from left to right across a period because the effective nuclear charge increases,making it harder to lose electrons.
All the given elements $(Na, Mg, Al, Si, Cl)$ belong to the $3^{rd}$ period of the periodic table.
Their order from left to right in the $3^{rd}$ period is: $Na (Group 1), Mg (Group 2), Al (Group 13), Si (Group 14), Cl (Group 17)$.
Therefore,the decreasing order of metallic character is: $Na > Mg > Al > Si > Cl$.

(B) The nonmetallic character of elements increases as we move from left to right across a period in the periodic table.
These elements belong to the $2^{nd}$ period of the periodic table.
The order of these elements in the $2^{nd}$ period from left to right is: $Li, Be, C, O, F$.
Since nonmetallic character increases from left to right,the order of increasing nonmetallic character is $Li < Be < C < O < F$.
Therefore,the correct option is $B$.

(D) $1$. Atomic radius increases down the group because the number of shells increases.
$2$. Metallic character increases down the group because the tendency to lose electrons increases due to the increase in atomic size.
$3$. The number of shells increases by $1$ for each successive element down a group.
$4$. Valence (valency) remains the same for all elements in the same group because they have the same number of valence electrons. Therefore,it does not increase.

(A) In the periodic table,as we move from left to right across a period,the atomic number increases,which means the number of protons in the nucleus increases.
This leads to an increase in the nuclear charge,which pulls the electrons more strongly towards the nucleus.
Although the number of electrons also increases,they are added to the same shell.
Consequently,the effective nuclear charge increases,causing the atomic radius to decrease across a period.

(B) Metallic character refers to the tendency of an element to lose electrons.
In a group,metallic character increases as we move down because the atomic size increases and the valence electrons are held less tightly by the nucleus.
In a period,metallic character decreases from left to right because the atomic size decreases and the effective nuclear charge increases,making it harder to lose electrons.
For option $B$: $Be$ (Group $2$,Period $2$),$Mg$ (Group $2$,Period $3$),and $Ca$ (Group $2$,Period $4$).
As we move down Group $2$ from $Be$ to $Mg$ to $Ca$,the metallic character increases.
Therefore,the order $Be < Mg < Ca$ represents increasing metallic character.

(N/A) The group number of an element in the Modern Periodic Table can be determined by its valence electrons. For elements with valence electrons greater than $2$, the group number is calculated as $(10 + \text{number of valence electrons})$.
$1$. For element $A$ (valence electrons = $3$): Group number = $10 + 3 = 13$. Valency = $3$.
$2$. For element $B$ (valence electrons = $4$): Group number = $10 + 4 = 14$. Valency = $4$.
$3$. For element $C$ (valence electrons = $2$): Group number = $2$. Valency = $2$.

Element	Group Number	Valency
$A$	$13$	$3$
$B$	$14$	$4$
$C$	$2$	$2$

(B) The radius of species $Y$ is smaller than that of species $X$.
Reasoning:
$1$. Species $X$ is a neutral magnesium atom $(Mg)$ with $12$ protons and $12$ electrons.
$2$. Species $Y$ is a magnesium cation $(Mg^{2+})$ formed by the loss of $2$ electrons,resulting in $12$ protons and $10$ electrons.
$3$. In $Y$,the nuclear charge ($12$ protons) acts on fewer electrons ($10$ electrons) compared to $X$. This increases the effective nuclear charge per electron,pulling the remaining electron shells closer to the nucleus.
$4$. Additionally,the loss of electrons may result in the loss of an entire outermost shell,further decreasing the ionic radius compared to the atomic radius.

(D) Atomic radius decreases across a period from left to right due to an increase in effective nuclear charge.
$(a)$ $Li, Be, N, F$ belong to the second period. Arranging them in increasing order of atomic radii: $F < N < Be < Li$.
$(b)$ $Cl, Br, I, At$ belong to the halogen group ($17^{th}$ group). Atomic radius increases down a group due to the addition of new shells. Arranging them in increasing order: $Cl < Br < I < At$.
Therefore,the correct sequence is $(a) F < N < Be < Li$ and $(b) Cl < Br < I < At$.

(C) Metallic character increases as we move down a group and decreases as we move from left to right across a period.
$1$. $K$ (Potassium) is in Group $1$,Period $4$.
$2$. $Ca$ (Calcium) is in Group $2$,Period $4$.
$3$. $Ga$ (Gallium) is in Group $13$,Period $4$.
$4$. $Ge$ (Germanium) is in Group $14$,Period $4$.
$5$. $Mg$ (Magnesium) is in Group $2$,Period $3$.
Comparing these,$K$ is the most metallic (Group $1$),followed by $Ca$ (Group $2$,Period $4$),then $Mg$ (Group $2$,Period $3$,which is less metallic than $Ca$ due to being in a higher period),then $Ga$ (Group $13$),and finally $Ge$ (Group $14$).
Thus,the increasing order of metallic character is $Ge < Ga < Mg < Ca < K$.

(A) The electronic configurations of Nitrogen $(N)$ and Phosphorus $(P)$ are as follows:

Element	Electronic Configuration $(K, L, M)$
Nitrogen $(Z=7)$	$2, 5$
Phosphorus $(Z=15)$	$2, 8, 5$

Nitrogen is more electronegative than phosphorus.
Reasoning:
$1$. Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself.
$2$. As we move down a group in the Periodic Table,the number of electron shells increases.
$3$. In Nitrogen,the valence electrons are in the $L$ shell (closer to the nucleus),whereas in Phosphorus,they are in the $M$ shell (further from the nucleus).
$4$. Due to the increased distance from the nucleus and increased shielding effect,the effective nuclear charge experienced by the valence electrons decreases as we move down the group.
$5$. Consequently,the ability to attract electrons decreases,making Nitrogen more electronegative than Phosphorus.

(A) The elements with atomic numbers $3$ to $9$ are: Lithium ($Li$,$Z=3$),Beryllium ($Be$,$Z=4$),Boron ($B$,$Z=5$),Carbon ($C$,$Z=6$),Nitrogen ($N$,$Z=7$),Oxygen ($O$,$Z=8$),and Fluorine ($F$,$Z=9$). These elements belong to the second period.
$(a)$ Electropositive nature decreases across a period. Thus,the first element,Lithium $(Li)$,is the most electropositive.
$(b)$ Electronegativity increases across a period. Thus,the last element,Fluorine $(F)$,is the most electronegative.
$(c)$ Atomic size decreases across a period from left to right. Thus,Fluorine $(F)$ has the smallest atomic size.
$(d)$ Boron $(B)$ is a metalloid located between metals and non-metals.
$(e)$ Valency increases from $1$ to $4$ and then decreases to $1$. Carbon $(C)$ has a valency of $4$,which is the maximum in this series.

$(A)$ The atomic number of element $A$ is $17$.
The electronic configuration of the element is determined by filling the shells $K, L, M, N$ in order:
$K = 2$
$L = 8$
$M = 7$
$1$. Period: The number of shells occupied by electrons determines the period. Since there are $3$ shells $(K, L, M)$, the element belongs to the $3rd$ period.
$2$. Group: For elements with more than $2$ valence electrons, the group number is calculated as $10 + \text{valence electrons}$. Here, the valence electrons in the outermost shell $(M)$ are $7$. Therefore, the group number is $10 + 7 = 17$.
Thus, element $A$ belongs to Group $17$ and Period $3$.

(B) Potassium $(K)$ has a stronger metallic character than lithium $(Li)$.
This is because metallic character depends on the ease with which an atom can lose its valence electrons.
As we move down a group in the periodic table,the atomic size increases due to the addition of new shells.
Consequently,the valence electrons are further away from the nucleus and are held less tightly by the nuclear charge.
Therefore,$K$ can lose its valence electron more easily than $Li$,making it more metallic.

(A) The atomic number of element $A$ is $16$.
To determine its position in the periodic table,we write its electronic configuration:
$A(16) = 2, 8, 6$.
Since there are $3$ shells $(K, L, M)$,the element belongs to the $3rd$ period.
Since there are $6$ valence electrons in the outermost shell,the group number is $10 + 6 = 16$.
Therefore,the element belongs to group $16$ and period $3$.

(N/A) $(i)$ They have a valency of $1$ (monovalent).
$(ii)$ They are highly reactive metals.
$(iii)$ They are electropositive in nature.
$(iv)$ They form basic oxides when they react with oxygen.

(C) The electronic configuration of the elements is as follows:
Element $A$ (atomic number $12$): $2, 8, 2$. It has $2$ valence electrons.
Element $B$ (atomic number $18$): $2, 8, 8$. It has $8$ valence electrons (stable noble gas configuration).
Element $C$ (atomic number $20$): $2, 8, 8, 2$. It has $2$ valence electrons.
Elements with the same number of valence electrons exhibit similar chemical properties.
Therefore,elements $A$ and $C$ show similar properties because both have $2$ valence electrons and belong to Group $2$ of the periodic table.

(B) The elements on the right-hand side of the periodic table are primarily non-metals.
Non-metals tend to form acidic oxides when they react with oxygen.
For example,carbon forms carbon dioxide $(CO_2)$ and sulfur forms sulfur dioxide $(SO_2)$,both of which are acidic in nature.

(A) Atomic size increases down a group in the periodic table because the number of electron shells increases.
All the given elements $(Be, Mg, Ca, Ba)$ belong to Group $2$ (Alkaline Earth Metals).
The order of these elements in Group $2$ from top to bottom is: $Be < Mg < Ca < Ba$.
Therefore,the decreasing order of their atomic size is $Ba > Ca > Mg > Be$.

(B) As we move from left to right across a period in the periodic table,the atomic number increases,meaning the number of protons in the nucleus increases.
These additional protons increase the effective nuclear charge,which exerts a stronger pull on the electrons in the same valence shell.
Because the electrons are pulled closer to the nucleus,the atomic radius (or atomic size) decreases.

(A-D) $(i)$ Calcium $(Ca)$ has an electronic configuration of $2, 8, 8, 2$. Since it has $2$ valence electrons,it tends to lose them to achieve a stable octet,making it a metal.
$(ii)$ The valency of an element is determined by the number of electrons it needs to lose or gain to complete its octet. Since Calcium loses $2$ electrons,its valency is $2$.
$(iii)$ The valency of Calcium $(Ca)$ is $+2$ and the valency of Chlorine $(Cl)$ is $-1$. By criss-crossing the valencies,the formula of its chloride is $CaCl_{2}$.
$(iv)$ Calcium $(Ca)$ is smaller than Potassium $(K)$. As we move from left to right across a period,the atomic radius decreases because the nuclear charge increases,pulling the electrons closer to the nucleus.

(A) The electronic configurations are:
${ }_{11}X = 2, 8, 1$ (Metal)
${ }_{7}Y = 2, 5$ (Non-metal)
${ }_{6}Z = 2, 4$ (Non-metal)
$(a)$ The increasing order of basic nature is: $Y_2O_5 < ZO_2 < X_2O$.
$(b)$ $X$ (Sodium) is a metal,so its oxide $(Na_2O)$ is strongly basic. $Y$ (Nitrogen) and $Z$ (Carbon) are non-metals,so their oxides ($N_2O_5$ and $CO_2$) are acidic. Among non-metals,the acidity decreases as we move from right to left in a period. Therefore,the basic character increases from $Y$ to $Z$ to $X$.

(N/A) The element belongs to group $2$,so it has $2$ valence electrons.
$(b)$ Elements in group $2$ are alkaline earth metals,so it is a metal.
$(c)$ The element in the third period and second group is Magnesium $(Mg)$.
$(d)$ Magnesium has a valency of $2$ and oxygen has a valency of $2$. The formula of its oxide is $MgO$.

(N/A) The electronic configuration of $Li$ $(Z=3)$ is $2, 1$,while that of $Na$ $(Z=11)$ is $2, 8, 1$. Lithium has $2$ shells,whereas Sodium has $3$ shells. As the number of shells increases down the group,the atomic size increases. Therefore,the Lithium atom is smaller than the Sodium atom.
$(b)$ The electronic configuration of $Cl$ $(Z=17)$ is $2, 8, 7$ and that of $S$ $(Z=16)$ is $2, 8, 6$. Electronegativity depends on the effective nuclear charge and atomic size. Chlorine has a smaller atomic size and a higher effective nuclear charge compared to Sulphur. Consequently,Chlorine has a greater tendency to attract shared electrons,making it more electronegative than Sulphur.

(N/A) Elements belonging to the same group have the same number of valence electrons in their outermost shell,which determines their chemical reactivity and bonding behavior.
$(b)$ Elements in the same period have a different number of valence electrons as we move from left to right,leading to a gradual change in their chemical properties.

(A) The atomic size decreases from left to right across a period. Therefore,the atom of $M$ is larger than the atom of $N$.
$(b)$ Metallic character decreases from left to right across a period. Thus,$M$ is more metallic than $N$.
$(c)$ Elements in group $I$ have a valency of $1$,while elements in group $II$ have a valency of $2$.
$(d)$ Since $M$ has a valency of $1$ and $N$ has a valency of $2$,the formulae of their chlorides are $MCl$ and $NCl_{2}$ respectively.

(A) $(i)$ Atomic size decreases across a period from left to right. Therefore,the element on the extreme left $(A)$ has the biggest atomic size,and the element on the extreme right (excluding noble gases,but here $G$ is in group $17$) has the smallest atomic size.
$(a)$ Biggest: $A$
$(b)$ Smallest: $G$
$(ii)$ Valency is determined by the number of valence electrons. Elements in group $13$ have $3$ valence electrons (valency $3$). Elements in group $18$ are noble gases with a stable octet (valency $0$).
$(a)$ Valency $3$: $C$
$(b)$ Valency $0$: $H$

(N/A) Trend of atomic size:
Along a group: From top to bottom,the atomic size increases because a new shell is added in each subsequent element.
Along a period: From left to right,the atomic size decreases because the nuclear charge increases,which pulls the electrons closer to the nucleus.
Trend of metallic character:
Along a group: Metallic character increases from top to bottom because the tendency to lose electrons increases due to the increase in atomic size.
Along a period: Metallic character decreases from left to right because the tendency to lose electrons decreases as the effective nuclear charge increases.

(N/A) When moving from left to right across a period in the periodic table:
$(a)$ Atomic radius decreases: This occurs because the nuclear charge increases,which pulls the electrons closer to the nucleus.
$(b)$ Metallic character decreases: As we move across a period,the tendency of elements to lose electrons decreases,making them less metallic and more non-metallic in nature.

(N/A) $(i)$ Atomic size.
$(ii)$ Valency or combining capacity.
$(iii)$ Metallic character.
$(iv)$ Non-metallic character.

(A-D) Magnesium $(Mg)$ has a smaller atomic size than Sodium $(Na)$. As we move from left to right across a period,the nuclear charge increases,which pulls the electrons closer to the nucleus,resulting in a decrease in atomic radius.
$(b)$ Sodium $(Na)$ is more electropositive than Magnesium $(Mg)$. Electropositivity is the tendency of an atom to lose electrons. Since $Na$ has a lower nuclear charge and a larger atomic size,it can lose its valence electron more easily than $Mg$.
$(c)$ Sodium $(Na)$ belongs to Group $1$ (Alkali metals),and Magnesium $(Mg)$ belongs to Group $2$ (Alkaline earth metals).

(N/A) The basic nature of oxides decreases across a period from left to right. It changes from basic to acidic.
The basic nature of oxides increases down a group from top to bottom.
These variations in the acidic and basic nature of oxides are related to the electronegativity of the element. As the electronegativity of the element increases along a period,the acidic character of the oxide increases. Conversely,as the electronegativity of the element decreases down a group,the basic character of the oxide increases.

(N/A) As we move from left to right across the third period,the basic character of the oxides decreases.
$1$. $Na_2O$ and $MgO$ are basic in nature.
$2$. $Al_2O_3$ and $SiO_2$ are amphoteric in nature (they show both acidic and basic properties).
$3$. $P_4O_{10}$,$SO_3$,and $Cl_2O_7$ are acidic in nature.
This trend occurs because the electronegativity of the elements increases from left to right,leading to a decrease in the ionic character and an increase in the covalent character of the oxides.

(N/A)

Property	Element $P$ (Group $1$)	Element $Q$ (Group $2$)
$(a)$ No. of electrons in the atom	$1$ more than the previous noble gas	$2$ more than the previous noble gas
$(b)$ Size of the atom	Larger	Smaller
$(c)$ Metallic character	More metallic	Less metallic
$(d)$ Tendency to lose electrons	Higher	Lower
$(e)$ Formula of oxides	$P_2O$	$QO$
$(f)$ Formula of chlorides	$PCl$	$QCl_2$

Explanation:
$1$. As we move from left to right in a period,the atomic size decreases due to an increase in effective nuclear charge.
$2$. Metallic character decreases from left to right because the tendency to lose electrons decreases.
$3$. Group $1$ elements have a valency of $1$,forming oxides $P_2O$ and chlorides $PCl$. Group $2$ elements have a valency of $2$,forming oxides $QO$ and chlorides $QCl_2$.

(N/A) The electronic configuration of an element with atomic number $16$ is $2, 8, 6$.
$1$. Period Number: The number of shells occupied by electrons determines the period number. Since this element has $3$ shells $(K, L, M)$, it belongs to the $3^{rd}$ period.
$2$. Group Number: For elements with more than $2$ valence electrons, the group number is calculated as $10 + \text{valence electrons}$. Here, the number of valence electrons is $6$, so the group number is $10 + 6 = 16$.
$3$. Valency: Valency is determined by the number of electrons required to complete the outermost shell (octet rule). For elements with $5, 6,$ or $7$ valence electrons, the valency is $8 - \text{valence electrons}$. Thus, the valency is $8 - 6 = 2$.

(A-D) $(i)$ These elements belong to Group $2$ because they all have the same number of valence electrons ($2$ valence electrons).
$(ii)$ $Be$ (Beryllium) is the least reactive element. In a group,the chemical reactivity of metals increases from top to bottom as the tendency to lose electrons increases due to the increase in atomic size.
$(iii)$ $Ca$ (Calcium) has the largest atomic radius. Atomic radius increases down a group because the number of electron shells increases ($Be$ has $2$ shells,$Mg$ has $3$ shells,and $Ca$ has $4$ shells).

(N/A) group is defined as the vertical column of elements in the modern periodic table.
$(i)$ The number of valence electrons remains the same for all elements in a group.
$(ii)$ The number of occupied shells increases by one as we move down a group.
$(iii)$ The size of atoms increases because the number of shells increases,leading to a greater distance between the nucleus and the outermost electrons.
$(iv)$ The metallic character of elements increases as we move down a group because the tendency to lose electrons increases due to the increased atomic size.
$(v)$ The effective nuclear charge experienced by valence electrons decreases as we move down a group because the distance between the nucleus and the valence electrons increases,and the inner shells provide more shielding.

(N/A) Cations are formed by the loss of valence electrons. This results in a higher effective nuclear charge per electron,which pulls the remaining electrons closer to the nucleus,thereby decreasing the size of the cation compared to the parent atom.
$(b)$ As we move down a group,a new electron shell is added to each subsequent element. This increases the distance between the valence electrons and the nucleus,leading to an increase in atomic size.
$(c)$ As we move across a period,the number of protons in the nucleus increases,which increases the nuclear charge. Since the electrons are added to the same shell,the increased nuclear attraction pulls the electron cloud closer to the nucleus,resulting in a decrease in atomic size.