Mix Examples - Periodic Classification of Elements Questions in English

(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 $8 - 6 = 2$.

(D) The element with one electron in the outermost shell is $D$ (atomic number $19$). Its electronic configuration is $2, 8, 8, 1$.
$(b)$ Elements $A$ (atomic number $4$,configuration $2, 2$) and $E$ (atomic number $20$,configuration $2, 8, 8, 2$) belong to the same group because both have $2$ valence electrons in their outermost shell.
$(c)$ Elements $D$ (atomic number $19$,period $4$) and $E$ (atomic number $20$,period $4$) belong to the same period. Between $D$ and $E$,$D$ has a bigger atomic radius because atomic radius decreases across a period from left to right due to an increase in nuclear charge.

(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).

(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 $AD$.

(N/A) $(i)$ Element $E$ (Carbon group,valency $4$) forms only covalent compounds.
$(ii)$ Element $D$ is in Group $13$ (Period $3$),which is Aluminum $(Al)$. It is a metal with a valency of $3$.
$(iii)$ Element $B$ is in Group $15$ (Period $2$),which is Nitrogen $(N)$. It is a non-metal with a valency of $3$.
$(iv)$ $D$ is bigger in size than $E$. Across a period from left to right,the atomic size decreases due to an increase in effective nuclear charge.
$(v)$ Elements $C$ and $F$ belong to Group $18$,which are known as Noble gases.

(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.

(A-D) $Cl$ and $O$ are non-metals with high electronegativity and will gain electrons to form anions ($Cl^-$ and $O^{2-}$).
$Na$ and $Al$ are metals with low ionization energy and will lose electrons to form cations ($Na^+$ and $Al^{3+}$).
$(b)$ Inert elements are those that have a stable valence shell configuration (octet or duplet) and do not react under normal conditions. Examples include $He$ (Helium) and $Ne$ (Neon).

(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.

(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) $Ca$ 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).

(B) Elements $B$ and $C$ belong to the same group (Group $2$) because they both have $2$ valence electrons in their outermost shell. $B$ is Beryllium $(Be)$ and $C$ is Magnesium $(Mg)$.
$(b)$ Elements $A$ and $B$ belong to the same period (Period $2$) because they both have $2$ shells. $A$ is Lithium $(Li)$ and $B$ is Beryllium $(Be)$.
$(c)$ Element $D$ is inert because its outermost shell is completely filled (it is Helium $(He)$).

(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.

(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.

(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) 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) Classification of elements helps in:
$(i)$ Systematic study of the properties of elements.
$(ii)$ Making it easier to understand and remember the properties of elements.
$(b)$ The two criteria used by Mendeleev were:
$(i)$ Increasing order of atomic mass.
$(ii)$ Similarity in the chemical properties of the elements.
$(c)$ Mendeleev left gaps in his periodic table for elements that were yet to be discovered,such as gallium and germanium.
$(d)$ There was no mention of noble gases like helium,neon,and argon because they were not known at that time and were discovered much later.
$(e)$ According to Mendeleev's periodic table,$Cl-35$ and $Cl-37$ should be placed in different positions because the basis of classification was the increasing order of atomic masses. However,since their chemical properties are identical,they are placed in the same position in the modern periodic table,which is based on atomic number.

(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.

(II, V) Oxygen $(O)$ and chlorine $(Cl)$ would form anions.
$O + 2e^- \rightarrow O^{2-}$ (Oxide ion)
$Cl + e^- \rightarrow Cl^-$ (Chloride ion)
Reason: Anions are negatively charged ions formed by the gain of electrons. Non-metals like oxygen and chlorine gain electrons to achieve a stable noble gas electronic configuration (octet rule).

(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) 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)

< strong>Characteristic	< strong>Metallic character in a group	< strong>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) 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).

$(a)$ Inert elements (noble gases) are placed in group $18$.
$(b)$ Elements in group $16$ ($B$ and $E$) gain $2$ electrons to form $B^{2-}$ and $E^{2-}$ ions. Elements in group $17$ ($C$ and $F$) gain $1$ electron to form $C^{-}$ and $F^{-}$ ions.
$(c)$ $F$ would have chemical properties similar to $C$ because they belong to the same group $(17)$.
$(d)$ $A$ is Helium $(He)$, which is in period $1$, so it has only $1$ shell ($K$-shell).
$(e)$ Both $A$ and $D$ are inert noble gases belonging to group $18$, meaning they have stable electronic configurations (complete valence shells).
$(f)$ $B$ (Oxygen) is the most abundant element in the earth's crust.

(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) 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.

(C) Electronic configuration:

Element	$K$	$L$	$M$
$X_{11}$	$2$	$8$	$1$
$Y_{16}$	$2$	$8$	$6$

$(b)$ They will form an ionic bond.
Reason: $X$ has $1$ valence electron and will lose it to form $X^+$ ion. $Y$ has $6$ valence electrons and will gain $2$ electrons to form $Y^{2-}$ ion. The bond formed by the transfer of valence electrons from a metal $(X)$ to a non-metal $(Y)$ is called an ionic or electrovalent bond.
$(c)$ Formula of the compound:

Symbol	Valency
$X$	$+1$
$Y$	$-2$

By criss-cross method,the formula is $X_2Y$.

(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 $8 - (\text{valence electrons}) = 8 - 7 = 1$.

(N/A) The electronic configuration of $X$ is $2, 8, 6$.
Number of valence electrons $= 6$.
Valency $= 8 - 6 = 2$.
$(b)$ When $X$ reacts with hydrogen,it forms a compound with the molecular formula $H_2X$. Since $X$ is Sulphur,the formula is $H_2S$. The electron dot structure is shown in the image provided.
$(c)$ The element $X$ is Sulphur $(S)$. It is a non-metal.

(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 $Na > Mg > Al > Si$.

(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.

(C) When Dmitri Mendeleev started his work on the classification of elements,there were $63$ elements known to science.
He used these $63$ elements to formulate his Periodic Law,which states that the properties of elements are a periodic function of their atomic masses.

(A) John Newlands,an English chemist,proposed the Law of Octaves in $1864$.
He arranged the elements known at that time in increasing order of their atomic masses.
Newlands assumed that only $56$ elements existed in nature and that no more elements would be discovered in the future.
Therefore,the correct option is $A$.

(D) Johann Wolfgang $D$öbereiner attempted to arrange elements with similar properties into groups of three,known as triads.
In his system,the atomic mass of the middle element was approximately the arithmetic mean of the atomic masses of the other two elements.
However,this system was not widely applicable because $D$öbereiner could identify only $5$ such triads from the elements known at that time.
Therefore,the correct option is $D$.

(C) John Newlands proposed the Law of Octaves in $1866$.
He arranged the elements in increasing order of their atomic masses and observed that every eighth element had properties similar to the first element.
This law was found to be applicable only up to $Calcium$ $(Ca)$, which is the $20^{th}$ element in the periodic table.
Beyond $Calcium$, the properties of the eighth element did not correspond to the first element.

(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 $Li < Be < C < O < F$.

(C) Eka-aluminium was the name given by Mendeleev to the element Gallium $(Ga)$.
Gallium belongs to Group $13$ of the periodic table,similar to Aluminium $(Al)$.
Elements in Group $13$ have a valency of $3$.
When an element $E$ with a valency of $3$ reacts with Oxygen (which has a valency of $2$),it forms an oxide with the formula $E_2O_3$ according to the criss-cross method of chemical formula writing.
Therefore,Eka-aluminium forms an oxide of the type $E_2O_3$.

(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.

(A) Non-metals generally form acidic oxides,while metals form basic or amphoteric oxides.
$1$. Atomic number $7$ corresponds to Nitrogen $(N)$,which is a non-metal. It forms acidic oxides like $NO_2$ and $N_2O_5$.
$2$. Atomic number $3$ corresponds to Lithium $(Li)$,which is an alkali metal and forms a basic oxide $(Li_2O)$.
$3$. Atomic number $12$ corresponds to Magnesium $(Mg)$,which is an alkaline earth metal and forms a basic oxide $(MgO)$.
$4$. Atomic number $19$ corresponds to Potassium $(K)$,which is an alkali metal and forms a basic oxide $(K_2O)$.
Therefore,the element with atomic number $7$ will form an acidic oxide.

(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.