Mix Examples - Periodic Classification of Elements Questions in English

(C) The atomic radius of an atom is defined as the distance from the center of the nucleus to the outermost shell containing electrons.
In the given diagrams:
- Diagram $(ii)$ shows the distance from the nucleus $(N)$ to the first shell $(K)$,which is incorrect for the total atomic radius.
- Diagram $(iii)$ correctly shows the distance from the center of the nucleus $(N)$ to the outermost shell $(M)$,which represents the atomic radius $(r)$.
- Diagram $(iv)$ shows the distance between two shells,which is incorrect.
- Diagram $(i)$ shows the distance from the nucleus to the inner shell,which is incorrect.
However,looking closely at the provided options and the standard definition,diagram $(ii)$ and $(iii)$ are often compared in such problems. Re-evaluating the diagrams: $(ii)$ shows the radius of the $K$ shell,$(iii)$ shows the radius of the $M$ shell. Both are valid representations of the radius of their respective shells. Given the options,$(ii)$ and $(iii)$ represent the distance from the nucleus to a shell,which is the definition of a radius.

(N/A) Sodium $(Na)$ is located in Group $1$ and Period $3$,or Potassium $(K)$ is located in Group $1$ and Period $4$. These are soft alkali metals that react vigorously with air and moisture,hence stored under kerosene.
$(b)$ Phosphorus $(P)$ is located in Group $15$ and Period $3$. White phosphorus exhibits variable valency and is highly reactive,catching fire in air,so it is stored under water.
$(c)$ Carbon $(C)$ is located in Group $14$ and Period $2$. It is tetravalent (valency of $4$) and its unique ability to form long chains (catenation) makes it the fundamental element of organic chemistry.

(N/A) The element with atomic number $2$ is Helium $(He)$. It belongs to Group $18$ and Period $1$ of the periodic table.
$(b)$ The element whose thin oxide layer is used for 'anodising' to prevent corrosion is Aluminium $(Al)$. It belongs to Group $13$ and Period $3$ of the periodic table.

(N/A) The element in the $2^{nd}$ Group and $3^{rd}$ Period is Magnesium $(Mg)$.
$(a)$ When magnesium oxide $(MgO)$ is dissolved in water,it forms magnesium hydroxide $(Mg(OH)_2)$:
$MgO(s) + H_2O(l) \rightarrow Mg(OH)_2(aq)$
$(b)$ The electron dot structure for the formation of magnesium oxide involves the transfer of $2$ electrons from the magnesium atom to the oxygen atom:
$Mg: + :\ddot{O}: \rightarrow [Mg]^{2+} [: \ddot{O} :]^{2-}$

(N/A) The elements corresponding to the atomic numbers $10, 20, 7,$ and $14$ are Neon $(Ne)$,Calcium $(Ca)$,Nitrogen $(N)$,and Silicon $(Si)$ respectively.
$(b)$ Group identification:
- Neon $(Z=10)$: Electronic configuration $(2, 8)$,Group $18$ (Noble gas).
- Calcium $(Z=20)$: Electronic configuration $(2, 8, 8, 2)$,Group $2$ (Alkaline earth metal).
- Nitrogen $(Z=7)$: Electronic configuration $(2, 5)$,Group $15$.
- Silicon $(Z=14)$: Electronic configuration $(2, 8, 4)$,Group $14$.
$(c)$ Period identification (based on the number of shells):
- Neon $(Z=10)$: $2$ shells,Period $2$.
- Calcium $(Z=20)$: $4$ shells,Period $4$.
- Nitrogen $(Z=7)$: $2$ shells,Period $2$.
- Silicon $(Z=14)$: $3$ shells,Period $3$.

(N/A) The answers to the crossword puzzle are as follows:
**Across:**
$(1)$ Magnesium ($Mg$,atomic number $12$)
$(3)$ Tin ($Sn$,Group $14$)
$(4)$ Iodine ($I$,halogen with $7$ valence electrons)
**Down:**
$(2)$ Sodium ($Na$,highly reactive,yellow flame,stored in kerosene)
$(5)$ Lithium ($Li$,first element of Period $2$)
$(6)$ Neon ($Ne$,used in fluorescent bulbs,second member of Group $18$)
$(7)$ Astatine ($At$,radioactive halogen)
$(8)$ Iron ($Fe$,constituent of steel,forms rust)
$(9)$ Boron ($B$,metalloid,used in bullet-proof vests)

(N/A) The elements present in the ladder are: $H, He, Li, Be, B, C, N, O, F, Ne, Na, Mg, Al, Si, P, S, Cl, Ar, K, Ca$.
Arranging them in increasing order of their atomic number $(Z)$:
$H (1), He (2), Li (3), Be (4), B (5), C (6), N (7), O (8), F (9), Ne (10), Na (11), Mg (12), Al (13), Si (14), P (15), S (16), Cl (17), Ar (18), K (19), Ca (20)$.
$(b)$ Arranging them according to their groups:
Group $1: H, Li, Na, K$
Group $2: Be, Mg, Ca$
Group $13: B, Al$
Group $14: C, Si$
Group $15: N, P$
Group $16: O, S$
Group $17: F, Cl$
Group $18: He, Ne, Ar$

(N/A)

Mendeleev's Periodic Table	Modern Periodic Table
$1$. Elements are arranged in the increasing order of their atomic masses.	$1$. Elements are arranged in the increasing order of their atomic numbers.
$2$. There are a total of $7$ groups and $6$ periods.	$2$. There are a total of $18$ groups and $7$ periods.
$3$. Elements with similar properties were placed in the same group.	$3$. Elements with the same number of valence electrons are placed in the same group.
$4$. The position of hydrogen was ambiguous.	$4$. Hydrogen is placed in group $1$ above alkali metals.
$5$. No clear distinction between metals and non-metals.	$5$. Metals are on the left and non-metals are on the right.

(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.

(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.

(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.

(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 $(He)$).

$(a)$ 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) 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.