Trends in the Modern Periodic Table Questions in English

(N/A) $(i)$ Electropositive character increases down a group because the atomic size increases and the effective nuclear charge decreases,making it easier to lose electrons.
$(ii)$ Electropositive character decreases across a period because the atomic size decreases and the effective nuclear charge increases,making it harder to lose electrons.
$(b)$ Atomic number and atomic mass increase both across a period and down a group.

(N/A) $Ca$ has a smaller atomic size. As we move from left to right across a period,the atomic size decreases because the effective nuclear charge increases,pulling the electrons closer to the nucleus.
$(b)$ $K$ is more electropositive. Electropositive character (metallic character) decreases across a period from left to right because the effective nuclear charge increases,making it harder to lose electrons.
$(c)$ $K$ belongs to group $1$ (Alkali Metals). $Ca$ belongs to group $2$ (Alkaline Earth Metals).

(A) $(i)$ $(a)$ The correct order of elements in the second period is: $Li(3), Be(4), B(5), C(6), N(7), O(8), F(9)$.
$(b)$ The largest atom is Lithium $(Li)$ and the smallest atom is Fluorine $(F)$.
$(ii)$ As we move from left to right in a period,the atomic radius decreases. This happens because the nuclear charge increases with the addition of protons in the nucleus,which exerts a stronger pull on the valence electrons,drawing them closer to the nucleus and thereby reducing the atomic size.

(A) $(i)$ The atomic number of $Q$ is $13$. Its electronic configuration is $2, 8, 3$. Since it has $3$ valence electrons,it loses $3$ electrons to achieve a stable octet. Thus,the valency of $Q$ is $3$.
$(ii)$ Elements with atomic numbers $12$ ($P$: Magnesium) and $13$ ($Q$: Aluminium) are metals. Element $14$ ($R$: Silicon) is a metalloid. Element $15$ ($S$: Phosphorus) is a non-metal.
$(iii)$ $P$ (Magnesium) will form the most basic oxide. In a period,metallic character decreases from left to right as the nuclear charge increases,making it harder to lose electrons. Since $P$ is the leftmost element among these,it is the most metallic and forms the most basic oxide.

(A) The electronic configuration of the element with atomic number $17$ is $2, 8, 7$. Since it has $7$ valence electrons,it needs $1$ electron to complete its octet. Therefore,its valency is $1$.
$(b)$ It is a non-metal. Elements with $5, 6,$ or $7$ valence electrons have a tendency to gain electrons to achieve a stable configuration,which is a characteristic property of non-metals.
$(c)$ It belongs to Group $17$ (Halogens). In a group,the atomic size increases down the group due to the addition of new shells. Since this element (Chlorine) is in the $3^{rd}$ period,it is smaller than the elements below it (like Bromine and Iodine) but larger than the element above it (Fluorine).

(N/A) $(i)$ The decreasing order of atomic radii is: $Li (152) > Be (111) > B (88) > C (77) > N (74) > O (66) > F (64)$.
$(ii)$ Yes, the elements are now arranged in the order of a period in the periodic table (from left to right).
$(iii)$ $(a)$ The element with the smallest atomic number is $Li$ (atomic number $3$).
$(b)$ The element with the largest atomic number is $F$ (atomic number $9$).
$(iv)$ From the data, we can infer that the atomic radius decreases as we move from left to right in a period.
$(v)$ Metal: $Li$ (Lithium), Non-metal: $F$ (Fluorine), Metalloid: $B$ (Boron).
$(vi)$ As we move from left to right in a period, the nuclear charge increases due to the addition of protons in the nucleus. This increased nuclear charge pulls the valence electrons closer to the nucleus, resulting in a decrease in atomic radius.

(N/A) Due to an increase in nuclear charge,the tendency to lose electrons decreases. This reduces the metallic character and,consequently,the basic character decreases.
$(b)$ Due to the increased nuclear charge,the atomic size decreases,which leads to a decrease in the electropositive character.

(N/A)

Characteristic	Metallic character in a group	Metallic character in a period
$1.$ Effective nuclear charge	The effective nuclear charge experienced by valence electrons decreases because the outermost electrons are farther from the nucleus due to the addition of new shells.	The effective nuclear charge experienced by valence electrons increases,holding the outermost electrons more tightly. This occurs because the number of shells remains constant across a period.
$2.$ Tendency to lose or gain electrons	The tendency to lose electrons increases as we move down a group. Consequently,metallic character and basic character increase.	The tendency to gain electrons increases due to the increased nuclear charge. Consequently,non-metallic character and acidic character increase.

(N/A) Electronic configuration of elements:
$X (Z=7): 2, 5$. It has $2$ shells,so it belongs to Period $2$. Since it has $5$ valence electrons,its Group number is $5 + 10 = 15$.
$Y (Z=8): 2, 6$. It has $2$ shells,so it belongs to Period $2$. Since it has $6$ valence electrons,its Group number is $6 + 10 = 16$.
$Z (Z=9): 2, 7$. It has $2$ shells,so it belongs to Period $2$. Since it has $7$ valence electrons,its Group number is $7 + 10 = 17$.
$(b)$ Atomic radius decreases across a period from left to right due to an increase in effective nuclear charge. Since all three elements belong to Period $2$ and their atomic numbers increase from $X$ to $Y$ to $Z$,the decreasing order of atomic radii is $X > Y > Z$.
$(c)$ $X$ has a valency of $3$ (needs $3$ electrons to complete its octet) and $Z$ has a valency of $1$ (needs $1$ electron to complete its octet). By cross-multiplying their valencies,the formula of the compound formed is $XZ_3$.

(N/A) The atomic number of an element is calculated as: Atomic number = Mass number - Number of neutrons.
Atomic number of $X = 35 - 18 = 17$.
The element with atomic number $17$ is Chlorine $(Cl)$.
The electronic configuration is determined by distributing $17$ electrons in shells: $2, 8, 7$.
Since the valence shell has $7$ electrons, the group number is $10 + 7 = 17$.
The number of occupied shells is $3$, so the period number is $3$.
The valency is calculated as $8 - (\text{valence electrons}) = 8 - 7 = 1$.

(A) The atomic size decreases across a period from left to right. Therefore,the increasing order of atomic size is: $C < B < Be < Li$.
$(b)$ These elements belong to the $2nd$ period of the Modern Periodic Table. Therefore,the last electron enters the $L$ shell (the $2nd$ shell).
$(c)$ The number of valence electrons corresponds to the group number for these elements: $Li = 1, Be = 2, B = 3, C = 4$.
$(d)$ Electropositivity decreases across a period from left to right. Thus,$Li$ is the most electropositive element among them.

(A) The increasing order of atomic radii is: $Li (52\, pm) < Na (61\, pm) < K (231\, pm) < Rb (244\, pm) < Cs (262\, pm)$.
$(i)$ The element with the smallest atom is Lithium $(Li)$ with an atomic radius of $52\, pm$, and the element with the largest atom is Caesium $(Cs)$ with an atomic radius of $262\, pm$.
$(ii)$ As we move down a group in the periodic table, the atomic size increases. This happens because a new electron shell is added to the atom at each step, which increases the distance between the outermost electrons and the nucleus, thereby increasing the atomic radius.

(A) $(i)$ The elements of the second period in the decreasing order of their atomic radii are:

Element	$Li$	$Be$	$B$	$C$	$N$	$O$
Radius $(pm)$	$152$	$111$	$88$	$77$	$74$	$66$

$(ii)$ Yes,the elements are now arranged in the pattern of a period in the Periodic Table.
$(iii)$ $Li$ has the largest atomic radius $(152\,pm)$ and $O$ has the smallest atomic radius $(66\,pm)$.
$(iv)$ The atomic radius decreases as we move from left to right in a period because the nuclear charge increases,which pulls the electrons closer to the nucleus.

(C) Moving from left to right across a period in the Periodic Table:
$1$. The metallic character of elements decreases because the nuclear charge increases,making it harder to lose electrons.
$2$. The number of valence electrons increases from $1$ to $8$.
$3$. Since the nuclear charge increases and the atomic size decreases,the effective nuclear attraction on valence electrons increases,making it harder for atoms to lose electrons. Thus,the tendency to lose electrons decreases.
$4$. The oxides of elements on the left are basic,while those on the right are acidic.
Therefore,the statement that 'atoms lose their electrons more easily' is incorrect.

(A) Atomic size generally increases down a group in the periodic table because the number of electron shells increases.
Fluorine $(F)$,Chlorine $(Cl)$,Bromine $(Br)$,and Iodine $(I)$ all belong to Group $17$ (Halogens).
The order of these elements down the group is $F, Cl, Br, I$.
Therefore,the atomic size increases in the order: $F < Cl < Br < I$.
Since the question asks for the order of increasing atomic size,the correct sequence is $F < Cl < Br < I$.

(B) In the periodic table,as we move from left to right across a period,the atomic size decreases.
This is because the nuclear charge increases with the addition of protons,which pulls the electrons closer to the nucleus.
$Na$,$Mg$,$Al$,and $Si$ all belong to the $3^{rd}$ period.
Their atomic numbers are $Na (11)$,$Mg (12)$,$Al (13)$,and $Si (14)$.
Therefore,the order of decreasing atomic size is $Na > Mg > Al > Si$.

(D) The electronic configuration of element $X$ is $2, 8, 8, 2$.
$1$. The number of shells occupied by electrons determines the period number. Here,there are $4$ shells $(K, L, M, N)$,so the element belongs to the $4^{th}$ period.
$2$. The number of valence electrons (electrons in the outermost shell) determines the group number. Here,there are $2$ valence electrons,so the element belongs to the $2^{nd}$ group.
Therefore,element $X$ belongs to the $4^{th}$ period and $2^{nd}$ group.

(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 positions from left to right are: $Na$ (Group $1$),$Mg$ (Group $2$),$Al$ (Group $13$),$Si$ (Group $14$),and $Cl$ (Group $17$).
Therefore,the order of decreasing metallic character is $Na > Mg > Al > Si > Cl$.

(B) Non-metallic character is the tendency of an element to gain electrons.
In the periodic table,non-metallic character increases as we move from left to right across a period.
All the given elements $(Li, Be, C, O, F)$ belong to the second period of the periodic table.
Their positions from left to right are: $Li$ (Group $1$),$Be$ (Group $2$),$C$ (Group $14$),$O$ (Group $16$),$F$ (Group $17$).
Therefore,the order of increasing non-metallic character is $Li < Be < C < O < F$.

(A) 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 $A$: $Be, Mg, Ca$ are all in Group $2$. As we move down the group from $Be$ to $Mg$ to $Ca$,the metallic character increases.
For option $B$: $Li$ is in Period $2$,while $Na$ and $K$ are in Period $3$ and $4$ respectively. The order $Na, Li, K$ is not increasing.
For option $C$: $Mg, Al, Si$ are in the same period. Metallic character decreases from $Mg$ to $Si$.
For option $D$: $C, N, O$ are in the same period. Metallic character decreases from $C$ to $O$.
Therefore,the correct set is $Be, Mg, Ca$.

(B) The statement is False.
Atomic size (or atomic radius) is defined as the distance between the centre of the nucleus and the outermost shell (valence shell) of an isolated atom,not the penultimate shell.

(A) The statement is True.
$1$. Metallic character is the tendency of an element to lose electrons.
$2$. Across a period,the effective nuclear charge increases,which pulls the valence electrons closer to the nucleus,making it harder to lose them; thus,metallic character decreases.
$3$. Down a group,the number of shells increases,which decreases the effective nuclear charge experienced by valence electrons due to increased shielding,making it easier to lose electrons; thus,metallic character increases.

(A) The statement is $True$.
Non-metals have a high tendency to attract electrons due to their high electronegativity.
They typically gain electrons to complete their valence shells,forming anions or covalent bonds.
In the periodic table,electronegativity increases from left to right across a period and decreases down a group.
Consequently,non-metals are located on the right-hand side of the periodic table,particularly towards the top,where electronegativity values are highest.