Mix Examples - Periodic Classification of Elements Questions in English

(A) The element that replaced Eka-silicon is Germanium $(Ge)$ and the element that replaced Eka-aluminium is Gallium $(Ga)$.
$(b)$ In the Modern Periodic Table,Germanium $(Ge)$ belongs to Group $14$ and Period $4$. Gallium $(Ga)$ belongs to Group $13$ and Period $4$.

(A) Mendeleev's Eka-silicon is known as Germanium $(Ge)$,which is a metalloid. Eka-aluminium is known as Gallium $(Ga)$,which is a metal.
$(b)$ Gallium $(Ga)$ belongs to Group $13$ of the periodic table,so it has $3$ valence electrons. Germanium $(Ge)$ belongs to Group $14$ of the periodic table,so it has $4$ valence electrons.

(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.

(N/A) The element $X$ is sulphur $(S)$, which is a yellow solid at room temperature and exhibits catenation and allotropy.
$(b)$ The atomic number of sulphur is $16$. Its electronic configuration is $2, 8, 6$ or $1s^2 2s^2 2p^6 3s^2 3p^4$.
$(c)$ The thermal decomposition of ferrous sulphate crystals $(FeSO_4 \cdot 7H_2O)$ is represented by the following balanced chemical equation:
$2FeSO_4(s) \xrightarrow{\Delta} Fe_2O_3(s) + SO_2(g) + SO_3(g)$

(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 $16$.

(A) The element $X$ is Nitrogen $(N)$. Its atomic number is $7$,and its electronic configuration is $2, 5$. Therefore,it has $5$ valence electrons.
$(b)$ The diatomic molecule of nitrogen is $N_2$. The electron dot structure shows that each nitrogen atom shares $3$ electrons with the other to complete its octet,resulting in a triple covalent bond.
$(c)$ In ammonia $(NH_3)$,the nitrogen atom shares one electron with each of the $3$ hydrogen atoms. This results in the formation of $3$ single covalent bonds.

(A) The group of elements that could be placed in Mendeleev's Periodic Table without disturbing the original order is the Noble gases.
Reason:
$1$. Mendeleev's periodic law states that the properties of elements are a periodic function of their atomic masses.
$2$. Noble gases are chemically inert and possess very low reactivity due to their stable electronic configuration.
$3$. Because they do not react with other elements,they could be placed in a separate group (Group $0$ or $VIII$) without requiring any changes to the positions of the existing elements in the table.

(N/A) Mendeleev examined the relationship between the atomic masses of elements and their physical and chemical properties.
Among chemical properties,he concentrated on the compounds formed by elements with oxygen and hydrogen (oxides and hydrides).
He selected these two elements because they are very reactive and formed compounds with most elements.
He wrote down the properties of each element on separate cards and grouped together elements with similar properties.
He observed that most of the elements got arranged in the periodic table in the order of increasing atomic masses.
Based on this,he formulated the Periodic Law,which states that the properties of elements are a periodic function of their atomic masses.

$(A)$ 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.

(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 $15$.

(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) $(i)$ 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.

(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.

(N/A) The atomic number of an element is equal to the total number of electrons present in its neutral atom.
For element $X$,the electronic configuration is $2, 8, 2$. The total number of electrons is $2 + 8 + 2 = 12$. Therefore,the atomic number of $X$ is $12$.
For element $Y$,the electronic configuration is $2, 8, 6$. The total number of electrons is $2 + 8 + 6 = 16$. Therefore,the atomic number of $Y$ is $16$.

(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.

(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) According to Newlands' Law of Octaves,certain elements were placed in the same slot to fit the periodic arrangement.
The two pairs of elements placed in the same slot are:
$(i)$ Cobalt $(Co)$ and Nickel $(Ni)$
$(ii)$ Cerium $(Ce)$ and Lanthanum $(La)$

(N/A) Newlands' law of Octaves states that when elements are arranged in the order of increasing atomic masses,the properties of every eighth element are a repetition of the properties of the first element. This pattern is analogous to the musical notes in an octave (sa,re,ga,ma,pa,dha,ni,sa),where the eighth note is the same as the first.

(N/A) Mendeleev's Periodic Law states that the physical and chemical properties of elements are a periodic function of their atomic masses.

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

(N/A) Mendeleev predicted the existence of several elements that had not been discovered at the time he proposed his periodic table. He named them $Eka-boron$,$Eka-aluminium$,and $Eka-silicon$. Later,these elements were discovered and named as follows:
$1$. $Eka-boron$ was discovered as $Scandium$ $(Sc)$.
$2$. $Eka-aluminium$ was discovered as $Gallium$ $(Ga)$.
$3$. $Eka-silicon$ was discovered as $Germanium$ $(Ge)$.
Therefore,any two of these,such as $Gallium$ and $Germanium$,are correct answers.

(N/A) Elements that exhibit properties intermediate between those of metals and non-metals are known as metalloids.
Examples: $Silicon$ $(Si)$ and $Boron$ $(B)$.

(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 $Ba > Ca > Mg > Be$.

(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.

(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.

(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-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.

(N/A) The anomalies of Mendeleev's periodic table that were resolved by the modern periodic law are as follows:
$(i)$ Position of Hydrogen: Mendeleev could not assign a fixed position to hydrogen because it shows properties of both alkali metals and halogens. The modern periodic law places it based on atomic number.
$(ii)$ Position of Isotopes: Isotopes have the same atomic number but different atomic masses. Mendeleev's table was based on atomic mass,leading to confusion. The modern periodic law,based on atomic number,places all isotopes of an element in the same position.
$(iii)$ Anomalous Pairs: In Mendeleev's table,some elements with higher atomic masses were placed before elements with lower atomic masses (e.g.,Cobalt and Nickel). The modern periodic law,based on atomic number,correctly places these elements in increasing order of their atomic numbers.

(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 $X$.

(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 $2, 8, 7$.

(A) Elements that form anions are those that gain electrons to complete their valence shell. Among the given elements,$O$ (Oxygen),$F$ (Fluorine),and $Cl$ (Chlorine) are non-metals that gain electrons to form anions.
$(b)$ The electronic configurations of the identified anions are:
$O^{2-}$ (Oxide ion): Atomic number $8$,configuration $2, 6$. By gaining $2$ electrons,it becomes $2, 8$.
$F^{-}$ (Fluoride ion): Atomic number $9$,configuration $2, 7$. By gaining $1$ electron,it becomes $2, 8$.
$Cl^{-}$ (Chloride ion): Atomic number $17$,configuration $2, 8, 7$. By gaining $1$ electron,it becomes $2, 8, 8$.

(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 $MgO$.

(A-D) Elements $B$ $(Z=11)$ and $C$ $(Z=3)$ belong to the same group (Group $1$,Alkali metals) because their electronic configurations are $2, 8, 1$ and $2, 1$ respectively,both having $1$ valence electron.
$(b)$ Atomic size decreases across a period and increases down a group. The elements are: $C$ ($Z=3$,Period $2$),$B$ ($Z=11$,Period $3$),$D$ ($Z=14$,Period $3$),$A$ ($Z=16$,Period $3$). Comparing sizes: $B$ $(Z=11)$ > $D$ $(Z=14)$ > $A$ $(Z=16)$ and $C$ is in Period $2$,so it is smaller than $B$. The decreasing order is $B > D > A > C$.
$(c)$ Element $B$ has atomic number $11$,so its valency is $1$. Oxygen has a valency of $2$. Thus,the formula of the oxide is $B_2O$.
$(d)$ Element $D$ (Silicon,$Z=14$) is a metalloid.

(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.

(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.

(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 $0$: $H$

(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 $16$.

(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.

(N/A) Sodium $(Na)$ or Potassium $(K)$ are soft metals that can be easily cut with a knife.
$(b)$ Mercury $(Hg)$ is the only metal that exists in a liquid state at room temperature.
$(c)$ Aluminum $(Al)$ has an atomic number of $13$. Its electronic configuration is $(2, 8, 3)$,meaning it has $3$ valence electrons. Boron $(B)$ also has $3$ valence electrons with configuration $(2, 3)$.

(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 $QCl_2$.