The Modern Periodic Table Questions in English

(N/A) Noble gases are chemically inert elements because they have a stable electronic configuration with a complete octet (or duplet in the case of Helium).
Their chemical properties are significantly different from all other elements in the periodic table.
Due to their unique stability and lack of reactivity,they are placed in a separate group,known as Group $18$.

(N/A) Mendeleev was unable to assign a fixed position to hydrogen and isotopes in his periodic table.
In Mendeleev's periodic table,the increase in atomic mass of the elements was not always regular from one element to the next.
It was believed that a more fundamental property than atomic mass could explain periodic properties in a better manner.
Henry Moseley demonstrated that the atomic number of an element is a more fundamental property than its atomic mass for explaining periodic properties.
By arranging elements in the increasing order of their atomic numbers,the Modern Periodic Table successfully resolved the anomalies found in Mendeleev's Periodic Table.

(N/A) Yes. The atoms of all three elements—$lithium$,$sodium$,and $potassium$—have one electron in their outermost shells. This similarity in their electronic configuration (all belong to Group $1$) is the reason they exhibit similar chemical properties,such as reacting with water to release hydrogen gas.

(C) Both helium $(He)$ and neon $(Ne)$ are noble gases.
Their atoms have completely filled outermost shells.
Helium has a stable duplet in its $K$ shell ($2$ electrons),while neon has a stable octet in its $L$ shell ($8$ electrons).
This stable electronic configuration makes them chemically unreactive or inert.

(A) The first ten elements of the Modern Periodic Table are Hydrogen $(H)$,Helium $(He)$,Lithium $(Li)$,Beryllium $(Be)$,Boron $(B)$,Carbon $(C)$,Nitrogen $(N)$,Oxygen $(O)$,Fluorine $(F)$,and Neon $(Ne)$.
Among these,Lithium $(Li)$ and Beryllium $(Be)$ are classified as metals,while the others are non-metals or metalloids.

(N/A) Neon $(Ne)$ has two shells,$K$ and $L$,both of which are completely filled with electrons ($2$ electrons in $K$ shell and $8$ electrons in $L$ shell,total $10$ electrons).
$(b)$ Magnesium $(Mg)$ has the electronic configuration $2, 8, 2$ (total $12$ electrons).

(N/A) Silicon $(Si)$ has a total of three shells,with four electrons in its valence shell ($2$ electrons in $K$ shell,$8$ electrons in $L$ shell,and $4$ electrons in $M$ shell).
$(b)$ Boron $(B)$ has a total of two shells,with three electrons in its valence shell ($2$ electrons in $K$ shell and $3$ electrons in $L$ shell).
$(c)$ Carbon $(C)$ has twice as many electrons in its second shell as in its first shell ($2$ electrons in $K$ shell and $4$ electrons in $L$ shell).

(D) The Modern Periodic Law states that the properties of elements are a periodic function of their atomic number.
Statement $(i)$ is incorrect because elements are arranged in increasing,not decreasing,order of atomic number.
Statement $(ii)$ is incorrect because the Modern Periodic Table is based on atomic number,not atomic mass (which was the basis for Mendeleev's Periodic Table).
Statement $(iii)$ is incorrect because isotopes have the same atomic number and are placed in the same group,not in adjoining groups.
Statement $(iv)$ is correct as it accurately describes the arrangement of elements in the Modern Periodic Table.
Therefore,statements $(i), (ii),$ and $(iii)$ are incorrect.

(A) The Modern Periodic Table is organized based on the atomic number of elements.
It consists of $18$ vertical columns,which are referred to as groups.
It also consists of $7$ horizontal rows,which are referred to as periods.
Therefore,the statement that it has $18$ vertical columns known as groups is correct.

(B) To determine the period of an element,we look at its electronic configuration.
$1$. Element $A$ (Atomic number $2$): Configuration is $2$ (Period $1$).
$2$. Element $B$ (Atomic number $3$): Configuration is $2, 1$ (Period $2$).
$3$. Element $C$ (Atomic number $7$): Configuration is $2, 5$ (Period $2$).
$4$. Element $D$ (Atomic number $10$): Configuration is $2, 8$ (Period $2$).
$5$. Element $E$ (Atomic number $30$): Configuration is $2, 8, 18, 2$ (Period $4$).
Elements $B$,$C$,and $D$ all have electrons in $2$ shells,meaning they belong to the $2$nd period.

(C) To determine the group of an element,we write its electronic configuration:
$A$ (atomic number $9$): $2, 7$ (Group $17$,Halogens)
$B$ (atomic number $11$): $2, 8, 1$ (Group $1$,Alkali metals)
$C$ (atomic number $17$): $2, 8, 7$ (Group $17$,Halogens)
$D$ (atomic number $12$): $2, 8, 2$ (Group $2$,Alkaline earth metals)
$E$ (atomic number $13$): $2, 8, 3$ (Group $13$)
Elements belonging to the same group have the same number of valence electrons.
$A$ and $C$ both have $7$ valence electrons,so they belong to the same group (Group $17$).

(D) The electronic configuration of the element is $2, 8$.
This means the element has a total of $10$ electrons,which corresponds to the atomic number $10$.
The element with atomic number $10$ is Neon $(Ne)$.
In the Modern Periodic Table,elements with a completely filled outermost shell (octet) are placed in Group $18$,which consists of noble gases.
Since the valence shell of this element is full ($8$ electrons),it belongs to Group $18$.

(B) The period number of an element in the periodic table corresponds to the number of shells occupied by electrons in its atom.
For elements in period $1$,the outermost shell is the $K$ shell $(n=1)$.
For elements in period $2$,the outermost shell is the $L$ shell $(n=2)$.
For elements in period $3$,the outermost shell is the $M$ shell $(n=3)$,and so on.
Therefore,for elements of period $2$,the outermost shell is the $L$ shell.

(D) Isotopes are atoms of the same element that have the same atomic number but different mass numbers.
$(i)$ Incorrect: Isotopes have different atomic masses due to a different number of neutrons.
$(ii)$ Correct: Isotopes of the same element must have the same atomic number (number of protons).
$(iii)$ Incorrect: Isotopes often show different physical properties (like density or boiling point) because these properties depend on the mass of the atom.
$(iv)$ Correct: Isotopes show the same chemical properties because chemical properties are determined by the electronic configuration,which remains the same for isotopes of the same element.
Therefore,statements $(ii)$ and $(iv)$ are correct.

(D) The elements $B$ (Boron),$Si$ (Silicon),and $Ge$ (Germanium) are all classified as metalloids.
Metalloids are elements that exhibit properties intermediate between those of metals and non-metals.
Boron is a metalloid,Silicon is a metalloid,and Germanium is also a metalloid.
Therefore,the correct option is $D$.

(B) The element with atomic number $14$ is Silicon $(Si)$.
Silicon is located in group $14$ and period $3$ of the periodic table.
It exhibits properties intermediate between metals and non-metals,making it a metalloid.
It forms an acidic oxide $(SiO_2)$ and a covalent halide $(SiCl_4)$.
Therefore,it belongs to the category of metalloids.

(N/A) Hydrogen is placed in Group $1$ at the top of the alkali metals because it shares properties with both alkali metals and halogens.
$1$. Resemblance with alkali metals: Like alkali metals ($Li, Na, K$, etc.), hydrogen has $1$ valence electron in its outermost shell and forms positive ions $(H^+)$ by losing this electron.
$2$. Resemblance with halogens: Like halogens ($F, Cl, Br$, etc.), hydrogen is a non-metal and requires $1$ electron to complete its valence shell, often forming covalent bonds and negative ions $(H^-)$ in specific conditions.
Due to this dual nature, its position remains unique and debated in the periodic table.

(A) The element is Magnesium $(Mg)$,as it belongs to Group $2$ and Period $3$.
$(b)$ The electronic configuration of Magnesium $(Z=12)$ is $2, 8, 2$ in the $K, L, M$ shells respectively.
$(c)$ The balanced chemical equation for the combustion of Magnesium in air is: $2Mg(s) + O_2(g) \rightarrow 2MgO(s)$.

(N/A) Electronic configuration:
- For atomic number $10$ (Neon): $(2, 8)$
- For atomic number $20$ (Calcium): $(2, 8, 8, 2)$
- For atomic number $7$ (Nitrogen): $(2, 5)$
- For atomic number $14$ (Silicon): $(2, 8, 4)$
$(b)$ Valency:
- For $10$ (Neon): Valence electrons = $8$,so valency = $8 - 8 = 0$ (Inert gas).
- For $20$ (Calcium): Valence electrons = $2$,so valency = $2$.
- For $7$ (Nitrogen): Valence electrons = $5$,so valency = $8 - 5 = 3$.
- For $14$ (Silicon): Valence electrons = $4$,so valency = $4$.

(N/A) The position of an element in the Modern Periodic Table is determined by its electronic configuration:
$1$. The number of valence electrons determines the group number. Elements with the same number of valence electrons are placed in the same group (column).
$2$. The number of shells occupied by electrons determines the period number. Elements in the same period have the same number of electron shells.
$3$. As we move across a period,the number of valence electrons increases by one,while the number of shells remains constant.

(A) Calcium $(Ca)$ has an atomic number of $20$ and its electronic configuration is $2, 8, 8, 2$. It belongs to Group $2$ of the periodic table.
Elements in the same group have the same number of valence electrons and exhibit similar chemical properties.
- Atomic number $12$ (Magnesium,$Mg$): Electronic configuration $2, 8, 2$. It has $2$ valence electrons and belongs to Group $2$.
- Atomic number $38$ (Strontium,$Sr$): Electronic configuration $2, 8, 18, 8, 2$. It has $2$ valence electrons and belongs to Group $2$.
Therefore,elements with atomic numbers $12$ and $38$ have physical and chemical properties resembling calcium.

(A) The element $X$ is Sulphur $(S)$. When Sulphur burns in air,it forms Sulphur dioxide $(SO_2)$ and Sulphur trioxide $(SO_3)$. These oxides are non-metallic oxides,and non-metallic oxides are acidic in nature.
$(b)$ Sulphur has an atomic number of $16$. Its electronic configuration is $2, 8, 6$. Since it has $3$ shells,it belongs to the $3^{rd}$ period. Since it has $6$ valence electrons,it belongs to group $16$.

(B) The atomic number of an element is equal to the total number of electrons present in its energy shells (orbits).
Given electronic configuration is $2, 8, 5$.
Atomic number $= 2 + 8 + 5 = 15$.
Therefore,the element with this electronic configuration is Phosphorus,which has an atomic number of $15$.

(N/A) $(i)$ Noble gases are chemically inert due to their stable electronic configuration.
$(ii)$ They are present in extremely low concentrations in the atmosphere,making them difficult to detect.

(B) The modern periodic table consists of $18$ vertical columns.
These vertical columns are known as groups.

(B) The first period of the periodic table consists of only two elements: Hydrogen $(H)$ and Helium $(He)$.
Hydrogen has an atomic number of $1$,with an electronic configuration of $1s^1$,meaning it has $1$ valence electron.
Helium has an atomic number of $2$,with an electronic configuration of $1s^2$,meaning it has $2$ valence electrons.
Since the first shell ($K$-shell) can hold a maximum of $2$ electrons,the maximum number of valence electrons for any atom in the first period is $2$.

(N/A) The Modern Periodic Law states that the physical and chemical properties of elements are a periodic function of their atomic numbers.

(C) In the modern periodic table,metalloids (such as boron,silicon,germanium,arsenic,antimony,tellurium,and polonium) are located along a zigzag line that separates metals from non-metals. Metals are found on the left side of this line,while non-metals are found on the right side.

(N/A) Lithium $(Li)$ and potassium $(K)$ are placed in Group $1$ of the periodic table because both of them have one valence electron in their outermost shell.

Name of the element	Atomic number	Electronic configuration $(K, L, M, N)$
Lithium	$3$	$2, 1$
Potassium	$19$	$2, 8, 8, 1$

$(b)$ The first two elements in Group $2$ are Beryllium $(Be)$ and Magnesium $(Mg)$.

Name of the element	Atomic number
Beryllium $(Be)$	$4$
Magnesium $(Mg)$	$12$

(A) The element with atomic number $17$ is Chlorine $(Cl)$.
$(b)$ It belongs to the $3rd$ period because it has $3$ electron shells $(K, L, M)$.
$(c)$ It belongs to group $17$ (Halogens) because it has $7$ valence electrons.
$(d)$ The electronic configuration is $2, 8, 7$.

(N/A) The modern periodic table is based on the atomic number of elements and the electronic configuration (number of valence electrons).
$(b)$ Hydrogen shows dual behavior:
$1.$ Like alkali metals,it has one valence electron and forms positive ions $(H^+)$ by losing an electron.
$2.$ Like halogens,it is a non-metal,exists as a diatomic molecule $(H_2)$,and forms covalent compounds by sharing electrons.
Due to these properties resembling both groups,it cannot be assigned a fixed position.

(B) The Modern Periodic Law states that the properties of elements are a periodic function of their atomic numbers.
$(b)$ The odd element is $B$ (atomic number $8$).
Reason: The electronic configuration of $A$ $(Z=1)$ is $1$,$C$ $(Z=11)$ is $2, 8, 1$,and $D$ $(Z=19)$ is $2, 8, 8, 1$. All these elements $(A, C, D)$ have $1$ electron in their valence shell and belong to Group $1$ (alkali metals). However,$B$ $(Z=8)$ has an electronic configuration of $2, 6$,meaning it has $6$ electrons in its valence shell and belongs to Group $16$.

(A) $(i)$ Since all four elements have electrons distributed in three shells,they all belong to the $3^{rd}$ period of the modern periodic table.
$(ii)$ The electronic configuration is determined by filling the shells $(K=2, L=8, M=n)$:
For element $A$ (outermost shell has $1$ electron): $2, 8, 1$.
For element $D$ (outermost shell has $7$ electrons): $2, 8, 7$.
$(iii)$ Element $A$ has $1$ valence electron,so it forms a cation $A^+$. Element $D$ has $7$ valence electrons,so it gains $1$ electron to form an anion $D^-$. The compound formed by the combination of $A$ and $D$ is $AD$.

(D) An element with two completely filled shells follows the electronic configuration $(2, 8)$. The total number of electrons is $2 + 8 = 10$,which corresponds to Neon $(Ne)$. The number of valence electrons in this element is $8$.
$(b)$ The element with $8$ valence electrons is inert (noble gas) because its outermost shell is completely filled,making it chemically stable and non-reactive. Thus,the element with $8$ valence electrons is inert.

(N/A) $(i)$ Modern Periodic Law states that the properties of elements are a periodic function of their atomic numbers.
$(ii)$ $(a)$ There are $18$ groups in the modern periodic table.
$(b)$ There are $7$ periods in the modern periodic table.
$(iii)$ $(a)$ Hydrogen is placed in the $1st$ group because its electronic configuration resembles that of alkali metals.
$(b)$ Isotopes of an element have the same atomic number,so they occupy the same position in the modern periodic table,thus resolving the anomaly found in Mendeleev's periodic table.

(N/A) The element $X$ belongs to the $4th$ period because the number of occupied electron shells is $4$.
$(b)$ The element $X$ belongs to the $2nd$ group because it has $2$ valence electrons in its outermost shell.
The valency of element $X$ is $2$ (or $+2$) because it needs to lose $2$ electrons from its outermost shell to achieve a stable noble gas configuration.

(N/A) Sodium $(Na)$ has the electronic configuration $2, 8, 1$.
$(b)$ Carbon $(C)$ has atomic number $6$. Its electronic configuration is $2, 4$. It has a total of two shells ($K$ and $L$),with $4$ electrons in the valence shell.
$(c)$ Aluminium $(Al)$ has atomic number $13$. Its electronic configuration is $2, 8, 3$. It has a total of three shells ($K, L,$ and $M$),with $3$ electrons in the valence shell.
$(d)$ Helium $(He)$ has atomic number $2$. Its electronic configuration is $2$. It has one shell $(K)$ which is completely filled with $2$ electrons.
$(e)$ Carbon $(C)$ has atomic number $6$. Its electronic configuration is $2, 4$. The first shell $(K)$ has $2$ electrons and the second shell $(L)$ has $4$ electrons. Thus,it has twice as many electrons in the second shell as in the first shell.

(N/A) The basis of arranging elements in the Modern Periodic Table is the increasing order of their atomic number $(Z)$. This approach allows for more precise predictions of elemental properties.
$(b)$ Isotopes are atoms of the same element that have the same atomic number $(Z)$ but different atomic masses. Since the Modern Periodic Table is organized based on increasing atomic numbers,all isotopes of a given element share the same position. Therefore,they are placed in the same group and the same period,effectively resolving the problem of their positioning.

(N/A) Since $X$ is in group $13$,it has $3$ valence electrons. Its valency is $3$.
$(b)$ Element $Y$ has an atomic number of $8$,so its electronic configuration is $2, 6$. It needs $2$ electrons to complete its octet,so its valency is $2$. Element $X$ has a valency of $3$. By criss-crossing the valencies,the molecular formula is $X_2Y_3$. Since $X$ is Aluminum $(Al)$ and $Y$ is Oxygen $(O)$,the compound is $Al_2O_3$ (Aluminum oxide).
$(c)$ Chlorine $(Cl)$ has a valency of $1$. Element $X$ has a valency of $3$. By criss-crossing the valencies,the formula is $XCl_3$. Since $X$ is Aluminum,the compound is $AlCl_3$ (Aluminum chloride).

(N/A) Argon $(Ar)$ is an inert gas because its outermost shell is completely filled with $8$ electrons,making it stable.
$(b)$ Silicon $(Si)$ has a valency of $4$. It forms covalent bonds because it shares its $4$ valence electrons to complete its octet.
$(c)$ Sodium $(Na)$ and Magnesium $(Mg)$ are electropositive elements because they have a tendency to lose electrons to attain a stable noble gas configuration.

(B) The electronic configuration of the element with atomic number $18$ is $2, 8, 8$.
Since the outermost shell contains $8$ electrons,it has a stable octet configuration.
Elements with a stable octet configuration belong to Group $18$ of the periodic table,which are known as Noble gases (or Inert gases).

(C) Henry Moseley,in $1913$,conducted experiments using $X$-ray spectra of elements.
He discovered that the frequency of the $X$-rays emitted by an element is related to its atomic number $(Z)$ rather than its atomic mass.
This led to the formulation of the Modern Periodic Law,which states that the properties of elements are a periodic function of their atomic numbers.
Therefore,he demonstrated that atomic number is a more fundamental property of an element than atomic mass.

(D) The elements $B$ (Boron),$Si$ (Silicon),and $Ge$ (Germanium) are all classified as metalloids.
Metalloids are elements that exhibit properties intermediate between those of metals and non-metals.
In the periodic table,these elements are located along the 'zig-zag' line that separates metals from non-metals.
Therefore,all three elements $B, Si$,and $Ge$ are metalloids.

(B) The element with atomic number $14$ is Silicon $(Si)$.
Silicon is a member of Group $14$ in the periodic table.
It exhibits properties intermediate between metals and non-metals,making it a metalloid.
Silicon forms an acidic oxide $(SiO_2)$ and covalent halides (e.g.,$SiCl_4$).
Therefore,the correct category is metalloid.

(C) The Modern Periodic Table is organized based on the atomic number of elements.
It consists of $7$ horizontal rows,which are called Periods.
It consists of $18$ vertical columns,which are called Groups.
Therefore,the statement that it has $18$ vertical columns known as Groups is correct.

(D) To determine the period of an element,we look at its electronic configuration.
$A$ (Atomic number $2$): Configuration is $2$ ($K$-shell). It belongs to Period $1$.
$B$ (Atomic number $3$): Configuration is $2, 1$. It belongs to Period $2$.
$C$ (Atomic number $7$): Configuration is $2, 5$. It belongs to Period $2$.
$D$ (Atomic number $10$): Configuration is $2, 8$. It belongs to Period $2$.
$E$ (Atomic number $30$): Configuration is $2, 8, 18, 2$. It belongs to Period $4$.
Elements with the same number of shells belong to the same period.
Elements $B, C,$ and $D$ all have $2$ shells,so they belong to Period $2$.

(A) To determine if elements belong to the same group,we must write their electronic configurations:
$A$ (atomic number $9$): $2, 7$ (Group $17$)
$B$ (atomic number $11$): $2, 8, 1$ (Group $1$)
$C$ (atomic number $17$): $2, 8, 7$ (Group $17$)
$D$ (atomic number $12$): $2, 8, 2$ (Group $2$)
$E$ (atomic number $13$): $2, 8, 3$ (Group $13$)
Elements with the same number of valence electrons belong to the same group.
Here,$A$ and $C$ both have $7$ valence electrons,so they belong to the same group (Group $17$).

(B) The electronic configuration of the element is $2, 8$.
This indicates that the element has $2$ shells ($K$ and $L$ shells).
The valence shell ($L$ shell) is completely filled with $8$ electrons,which represents a stable octet configuration.
Elements with a completely filled valence shell are classified as noble gases.
In the Modern Periodic Table,all noble gases are placed in Group $18$.

(D) The period number of an element corresponds to the number of shells present in its atom.
For elements in period $2$,the electrons are filled in the first two shells.
The first shell is the $K$ shell $(n=1)$,and the second shell is the $L$ shell $(n=2)$.
Since the period is $2$,the $L$ shell is the outermost shell for these elements.

(FALSE) The statement is False.
Hydrogen has an electronic configuration of $1s^1$,which means it has one valence electron in its outermost shell.
Alkaline earth metals (Group $2$ elements) have a general valence shell electronic configuration of $ns^2$,meaning they have two valence electrons.
Hydrogen's configuration actually resembles that of alkali metals (Group $1$),which have a general valence shell configuration of $ns^1$.