Mix Examples - Classification of Elements and Periodicity in Properties Questions in English

(A) In the periodic table,the elements located at the far left (Groups $1$ and $2$) are metals,not non$-$metals. Therefore,statement $A$ is incorrect.
Statement $B$ is correct because elements in the same group share the same valence shell electron configuration.
Statement $C$ is correct as the majority of elements in the periodic table are metals.
Statement $D$ is correct because metalloids (semi$-$metals) act as a boundary between metals and non$-$metals.

(B) $Li$ and $Mg$ exhibit a diagonal relationship due to their similar ionic sizes and electronegativities.
This relationship occurs between certain elements of the second and third periods of the periodic table.

(C) diagonal relationship is observed between certain elements of the second and third periods.
The pairs that exhibit a diagonal relationship are:
$1. Li \text{ and } Mg$
$2. Be \text{ and } Al$
$3. B \text{ and } Si$
Therefore,there are $3$ such pairs in the given list.
The pair $Al \text{ and } S$ does not exhibit a diagonal relationship.

(A) The electronic configuration of the ion with $-2$ oxidation state is $1s^2 2s^2 2p^6 3s^2 3p^6$,which corresponds to $18$ electrons.
In the neutral state,the number of electrons is $18 - 2 = 16$.
The element with atomic number $Z = 16$ is Sulphur $(S)$.
The electronic configuration of neutral Sulphur is $1s^2 2s^2 2p^6 3s^2 3p^4$.
Since the valence shell is $n = 3$,it belongs to period $3$.
Since it has $6$ valence electrons $(3s^2 3p^4)$,it belongs to group $16$.

(C) Statement $(i)$ is correct: In the periodic table,about $78 \%$ of elements are metals.
Statement $(ii)$ is incorrect: In a group,metallic character increases from top to bottom,and in a period,non-metallic character increases from left to right.
Statement $(iii)$ is correct: The element $Ho$ (Holmium) has atomic number $67$ and belongs to the lanthanide series,which is part of the $f-$block.

(A) The correct matches are as follows:
$(A)$ Rubidium $(Rb)$ is an alkali metal,which is an $s$-block element. So,$A-3$.
$(B)$ Platinum $(Pt)$ has the atomic number $78$. So,$B-4$.
$(C)$ Eka-silicon was the name given by Mendeleev to Germanium $(Ge)$. So,$C-1$.
$(D)$ Polonium $(Po)$ is a radioactive element belonging to group $16$ (chalcogens). So,$D-2$.
Thus,the correct matching is $A-3, B-4, C-1, D-2$.

(B) Let us analyze each option:
$1$. First ionisation enthalpy: The order is $N > O > P > S$. Nitrogen $(2p^3)$ has a stable half-filled configuration,making its $IE_1$ higher than Oxygen $(2p^4)$. Similarly,Phosphorus $(3p^3)$ is higher than Sulfur $(3p^4)$. Across the period,$N > O$ and $P > S$,and down the group $N > P$ and $O > S$. Thus,$N > O > P > S$ is correct.
$2$. Negative electron gain enthalpy: The correct order is $Cl > F > S > O$. Chlorine has a higher electron gain enthalpy than Fluorine due to inter-electronic repulsion in the small $2p$ orbital of Fluorine. Similarly,Sulfur is higher than Oxygen. The given order $F > Cl > O > S$ is incorrect.
$3$. Size: For the same element in different oxidation states,the size decreases as the positive charge increases. Thus,$Fe^{3+} < Fe^{2+} < Fe$ is correct.
$4$. Non-metallic character: Non-metallic character increases across a period and decreases down a group. $O$ and $N$ are more non-metallic than $S$ and $P$. Among $O$ and $N$,$O$ is more electronegative/non-metallic. Thus,$O > N > S > P$ is correct.
Therefore,the incorrect order is option $B$.

(B) Let us analyze each option:
$1$. $Li < Na < K$ (Metallic radius): Metallic radius increases down a group as the number of shells increases. This order is correct.
$2$. $Br < F < Cl$ (Electron gain enthalpy): The electron gain enthalpy of $F$ is less negative than $Cl$ due to small size and inter-electronic repulsion. The correct order is $F < Br < Cl$. Thus,$Br < F < Cl$ is incorrect.
$3$. $C < N < O$ (First ionization enthalpy): Ionization enthalpy generally increases across a period. $N$ has a stable half-filled $p$-orbital configuration,so its ionization enthalpy is higher than $O$. The correct order is $C < O < N$. However,in many contexts,$C < N < O$ is considered incorrect due to the $N > O$ anomaly. Given the options,$B$ is the most explicitly incorrect statement regarding periodic trends.
$4$. $Mg^{2+} < Na^{+} < F^{-}$ (Ionic radius): These are isoelectronic species ($10$ electrons). For isoelectronic species,the ionic radius decreases as the nuclear charge $(Z)$ increases. $Z$ values are $Mg=12, Na=11, F=9$. Thus,$Mg^{2+} < Na^{+} < F^{-}$ is correct.

(B) Ionization enthalpy: Across a period,it increases. For $B, Be, C, O, N$,the order is $B < Be < C < O < N$. Thus,$A \rightarrow III$.
$(B)$ Metallic character: It decreases across a period and increases down a group. For $Na, Mg, Si, P$,the order of metallic character is $P < Si < Mg < Na$. Thus,$B \rightarrow I$.
$(C)$ Electron gain enthalpy: Generally becomes more negative across a period. For halogens,the order of electron gain enthalpy (magnitude) is $I < Br < F < Cl$. Thus,$C \rightarrow IV$.
$(D)$ Electronegativity: Increases across a period. For $I, N, O, F$,the order is $I < N < O < F$. Thus,$D \rightarrow II$.
Therefore,the correct match is $A-III, B-I, C-IV, D-II$.

(D) Electron gain enthalpy: In a group,electron gain enthalpy generally decreases down the group,but for $p$-block elements,the $EA_1$ of the second period element is less than that of the third period element due to smaller size and inter-electronic repulsion. Thus,the order is $S > Se > O$ (Correct).
Electronegativity: Across a period,it increases,and down a group,it decreases. The order $F > O > Cl$ is correct.
Metallic radius: It decreases from left to right and increases from top to bottom. The order $Na > Li > Al$ is correct.
Metallic character: It increases down the group. Therefore,the order should be $Ba > K > Na$. The given set $Na > K > Ba$ is incorrect.

(A) Statement $I$: Nitrogen $(Z=7)$ has a half-filled $2p^3$ configuration,which provides extra stability,leading to a higher ionization enthalpy compared to Beryllium $(Z=4)$. Also,electronegativity increases across a period,so Nitrogen $(3.04)$ is more electronegative than Beryllium $(1.57)$. Thus,statement $I$ is correct.
Statement $II$: Non-metal oxides and high oxidation state metal oxides are generally acidic. $B_2O_3$ is a non-metal oxide (acidic),and $CrO_3$ (where $Cr$ is in $+6$ oxidation state) is an acidic oxide. Thus,statement $II$ is correct.

(C) The electronic configuration of nitrogen $(N)$ is $1s^2 2s^2 2p^3$ and that of oxygen $(O)$ is $1s^2 2s^2 2p^4$.
Statement-$1$: Nitrogen has a half-filled $2p$ subshell,which makes it very stable. Adding an electron to nitrogen requires energy (endothermic),whereas adding an electron to oxygen is exothermic. Thus,nitrogen has a less negative (lesser) electron gain enthalpy than oxygen. Statement-$1$ is correct.
Statement-$2$: Nitrogen has a stable half-filled $2p^3$ configuration,making it harder to remove an electron compared to oxygen,which has a $2p^4$ configuration. Therefore,the ionization enthalpy of nitrogen is higher than that of oxygen,meaning oxygen has a lesser ionization enthalpy than nitrogen. Statement-$2$ is correct.

(B) The atomic number of $K$ is $19$ with the electronic configuration $[Ar] 4s^1$.
For $K \longrightarrow K^{+} + e^{-}$,the first ionization enthalpy $\Delta_i H_1 = 419 \ kJ \ mol^{-1}$ is low because $K$ loses its valence electron easily to achieve a stable noble gas configuration.
For $K^{+} \longrightarrow K^{2+} + e^{-}$,the second ionization enthalpy $\Delta_i H_2 = 3051 \ kJ \ mol^{-1}$ is very high because the electron is being removed from a stable,fully-filled noble gas core (argon configuration).
Potassium is an alkali metal with low electron affinity,resulting in a low negative electron gain enthalpy $\Delta_{eg} H = -48 \ kJ \ mol^{-1}$.
Thus,the given values correspond to $K$.

(D) Let us analyze the given options for the incorrect order:
$A$. $Al_2O_3$ (amphoteric) < $MgO$ (basic) < $Na_2O$ (strongly basic) < $K_2O$ (more strongly basic). This order is correct.
$B$. In a group,acidity increases down the group as the bond dissociation energy decreases. Thus,$NH_3 < PH_3 < AsH_3$ is the correct order of acidity.
$C$. Ionic radius increases down a group due to the addition of new shells. Thus,$Li^{+} < Na^{+} < K^{+} < Cs^{+}$ is correct.
$D$. The first ionisation enthalpy generally increases from left to right in a period. However,$Be$ $(2s^2)$ has a higher ionisation enthalpy than $B$ $(2s^2 2p^1)$ due to the stability of the fully filled $s$-orbital. The correct order is $Li < B < Be < C$. Therefore,the order $Li < Be < B < C$ is incorrect.

(C) Electron gain enthalpy becomes less negative on moving down a group due to an increase in atomic size.
However,the electron gain enthalpy of $F$ is less negative than $Cl$ due to inter-electronic repulsion in the small-sized $F$ atom.
Thus,the correct order of electron gain enthalpy is $I < Br < F < Cl$.
Since the given order $I < Br < Cl < F$ does not match the actual trend,option $(C)$ is the correct answer.

(B) $I$ and $II$ are correct: The properties of elements are periodic functions of their atomic numbers,and non-metals are fewer in number compared to metallic elements.
$III$ is correct: In the periodic table,the first ionisation energy of elements along a period does not vary in a regular manner due to variations in effective nuclear charge and shielding effects.
$IV$ is incorrect: The ground state electronic configuration of $Pd (Z=46)$ is $[Kr] 4d^{10} 5s^0$.
Therefore,statements $I, II,$ and $III$ are correct.

(B) The correct answer is $B$.
$(A)$ Nitrogen $(2s^2 2p^3)$ has a stable half-filled $p$-orbital,so its first ionization enthalpy is higher than carbon $(2s^2 2p^2)$. This statement is correct.
$(B)$ Electron gain enthalpy of sulphur is more negative than that of oxygen due to the small size of oxygen,which leads to inter-electronic repulsions. Thus,the electron gain enthalpy of oxygen is less than that of sulphur. This statement is incorrect.
$(C)$ $Mg^{2+}$ and $Al^{3+}$ are isoelectronic species ($10$ electrons). For isoelectronic species,ionic radius decreases as the nuclear charge increases. Since $Z$ of $Al$ $(13)$ is greater than $Mg$ $(12)$,the radius of $Mg^{2+}$ is greater than $Al^{3+}$. This statement is correct.
$(D)$ Electronegativity increases across a period. Fluorine is more electronegative than oxygen. This statement is correct.

(A) The most reactive non-metal is $B$ because it has the highest negative electron gain enthalpy $(-328 \ kJ/mol)$.
The most reactive metal is $A$ because it has the lowest first ionization enthalpy $(419 \ kJ/mol)$.
The least reactive element is $D$ because it has a very high first ionization enthalpy and a positive electron gain enthalpy (noble gas characteristic).
The metal forming a binary halide is $C$ (e.g.,alkaline earth metal like $Mg$),which has comparable successive ionization energies.

(C) Statement $(I)$ is correct: The first ionisation enthalpy order for $Na$ $(1s^2 2s^2 2p^6 3s^1)$,$Mg$ $(1s^2 2s^2 2p^6 3s^2)$,and $Al$ $(1s^2 2s^2 2p^6 3s^2 3p^1)$ is $Mg > Al > Na$ due to the stable fully-filled $3s$ orbital in $Mg$.
Statement $(II)$ is incorrect: The element with an electronegativity of $3.5$ is oxygen,whereas chlorine has an electronegativity of $3.0$.
Statement $(III)$ is correct: These are isoelectronic species with $10$ electrons. For isoelectronic ions,the ionic radius decreases as the nuclear charge $(Z)$ increases. The order of $Z$ is $O^{2-} (8) < F^{-} (9) < Na^{+} (11) < Mg^{2+} (12)$,so the size order is $Mg^{2+} < Na^{+} < F^{-} < O^{2-}$.
Statement $(IV)$ is incorrect: The element with atomic number $106$ is seaborgium $(Sg)$,while Bohrium $(Bh)$ has atomic number $107$.

(C) 'He' is the second most abundant element in the universe. The most abundant element is hydrogen,and helium makes up most of the remaining $25 \%$.
$(b)$ Darmstadtium is a chemical element with the symbol $Ds$ and atomic number $110$.
$(c)$ Under ordinary conditions,osmium is the element with the highest density $(22.59 \ g/cm^3)$.
$(d)$ Francium is the most electropositive element in the periodic table.
All four statements are correct.

(B) On moving from left to right in a period,the effective nuclear charge increases,which causes the atomic radius to decrease.
Due to the increase in effective nuclear charge and decrease in atomic size,the ability to lose electrons decreases,which leads to a decrease in metallic character.

(C) Apart from radioactivity,all are periodic properties.
Radioactivity is a process in which nuclei of certain elements undergo spontaneous disintegration and it does not depend on the electronic configuration of the atom,so it is not a periodic property.
Atomic size,electron affinity,and ionisation potential are properties that vary or show a regular trend according to the electronic configuration of elements.

(D) The explanation of the given statements is as follows:
$I$. Both beryllium $(Be)$ and barium $(Ba)$ belong to the same group (group-$2$). As we move down the group,the size of the atom/ion increases,and the covalent character decreases. Thus,$Ba$ compounds are generally less covalent than $Be$ compounds. Therefore,statement $I$ is not correct.
$II$. The electron gain enthalpy of $He$ is positive due to its very small size and fully-filled electronic configuration $(1s^2)$. Hence,statement $II$ is correct.
$III$. The oxidation state of $O$ in $OF_2$ is $+2$,whereas the oxidation state of $O$ in $Na_2O$ is $-2$. Hence,statement $III$ is not correct.
$IV$. Both $Na^{+}$ and $F^{-}$ are isoelectronic species (both have $10$ electrons). However,$Na^{+}$ has $11$ protons while $F^{-}$ has $9$ protons. Due to the higher nuclear charge in $Na^{+}$,it exerts a greater attractive force on the electrons,making its radius smaller than that of $F^{-}$. Hence,statement $IV$ is correct.
Since statements $I$ and $III$ are not correct,the correct option is $(D)$.

(D) $(i)$ Acidic property increases with the increase in oxidation number of the central atom and for the same oxidation number,acidic nature increases along the period. Thus,$Na_2O$ (basic) $< MgO$ (basic) $< ZnO$ (amphoteric) $< P_4O_6$ (acidic). This statement is True $(T)$.
$(ii)$ The electron gain enthalpy of $Cl$ is more negative than that of $F$ due to the small size and inter-electronic repulsion in $F$. The order is $Cl > F > Br$. Thus,this statement is False $(F)$.
$(iii)$ Ionic size increases with an increase in the number of electrons (anions) and decreases with a loss of electrons (cations). The order $M^{2-} > M^{-} > M^{+} > M^{2+}$ is correct. This statement is True $(T)$.
$(iv)$ The second ionisation enthalpy involves removing an electron from $M^{+}$ ion. $K^{+}$ has a stable noble gas configuration $([Ar])$,while $Cu^{+}$ has a $d^{10}$ configuration. Removing an electron from the stable noble gas configuration of $K^{+}$ requires much more energy than from $Cu^{+}$. Thus,the second ionisation enthalpy of $K$ is greater than that of $Cu$. This statement is False $(F)$.
Therefore,the correct sequence is $(i)-T, (ii)-F, (iii)-T, (iv)-F$. The correct option is $(d)$.

(B) $(i)$ First ionisation enthalpy of $He$ $(2372 \ kJ/mol)$ is less than the second ionisation enthalpy of $Li$ $(7298 \ kJ/mol)$. This is correct.
$(ii)$ $Li$ has the highest second ionisation enthalpy because removing the second electron from $Li^+$ requires breaking the stable $1s^2$ noble gas configuration. This is correct.
$(iii)$ All transition elements are $d$-block elements,but all $d$-block elements are not transition elements (e.g.,$Zn, Cd, Hg$). This is incorrect.
$(iv)$ The letter '$J$' is the only alphabet not found in the periodic table. This is correct.
$(v)$ Francium is extremely rare,and its concentration in the Earth's crust is estimated to be $\sim 10^{-18} \ ppm$. This is correct.
Therefore,statements $(i), (ii), (iv),$ and $(v)$ are correct.

(C) Given that the atomic number of $B$ is $Z$. Since $B$ is a noble gas,it belongs to group $18$.
Therefore,the element $A$ with atomic number $Z-1$ is a halogen (group $17$),which has a very high electron affinity.
The element $C$ with atomic number $Z+1$ is an alkali metal (group $1$),which exists in a $+1$ oxidation state.
The element $D$ with atomic number $Z+2$ is an alkaline earth metal (group $2$),which exists in a $+2$ oxidation state.
Thus,statement $(1)$ is correct ($A$ is a halogen) and statement $(3)$ is correct ($D$ is an alkaline earth metal). Statement $(2)$ is incorrect because $C$ exists in a $+1$ oxidation state.

(B) Statement $I$ is correct because the modern periodic law states that physical and chemical properties of elements are periodic functions of their atomic numbers,which is equivalent to their electronic configuration.
Statement $II$ is incorrect because fluorine $(F)$ has the highest electronegativity $(4.0)$ in the periodic table,which is greater than that of chlorine ($Cl$,$3.0$).
Statement $III$ is incorrect because electropositive nature (metallic character) increases from top to bottom in a group as the ionization energy decreases.
Therefore,only statement $I$ is correct.

(A) diagonal relationship exists between certain elements of the second and third periods of the periodic table. Specifically,$Li$ (Group $1$,Period $2$) shows a diagonal relationship with $Mg$ (Group $2$,Period $3$). This occurs because they have similar ionic sizes and charge-to-size ratios.

(D) The enthalpy change for the reaction $Mg + 2F \rightarrow MgF_2$ can be calculated by summing the energy changes involved in the ionization of $Mg$ and the electron gain of $F$.
$Mg_{(g)} \rightarrow Mg_{(g)}^{+2} + 2e^{-}$,$\Delta H_1 = IE_1 + IE_2 = 737 + 1451 = 2188\ kJ\ mol^{-1}$.
$2F_{(g)} + 2e^{-} \rightarrow 2F_{(g)}^{-}$,$\Delta H_2 = 2 \times (-EA) = 2 \times (-328) = -656\ kJ\ mol^{-1}$.
Adding these steps,the total enthalpy change is $\Delta H = \Delta H_1 + \Delta H_2 = 2188 - 656 = 1532\ kJ\ mol^{-1}$.

(A) Statement $I$: The electronic configuration of $Na^{+}$ is $1s^{2} 2s^{2} 2p^{6}$,which is a stable noble gas configuration. Removing the second electron from $Na^{+}$ requires a very high amount of energy. In contrast,$Mg^{+}$ has the configuration $1s^{2} 2s^{2} 2p^{6} 3s^{1}$,and removing the second electron results in a stable $Mg^{2+}$ ion $(1s^{2} 2s^{2} 2p^{6})$. Thus,the second ionisation enthalpy of $Na$ is much higher than that of $Mg$. Statement $I$ is true.
Statement $II$: Both $O^{2-}$ and $F^{-}$ are isoelectronic species with $10$ electrons. The ionic radius of isoelectronic species decreases as the nuclear charge $(Z)$ increases. Since the atomic number of $O$ $(Z=8)$ is less than that of $F$ $(Z=9)$,the ionic radius of $O^{2-}$ is larger than that of $F^{-}$. Statement $II$ is true.

(D) Statement $I$ is true: Metallic character increases down a group and decreases across a period from left to right. $K$ ($Group$ $1$) is the most metallic,followed by $Mg$ ($Group$ $2$),$Al$ ($Group$ $13$),and $B$ ($Group$ $13$,metalloid). Thus,$K > Mg > Al > B$ is correct.
Statement $II$ is false: While atomic radius is greater than cationic radius,it is smaller than anionic radius. Therefore,it is not always greater than the ionic radius for any element.

(B) Statement-$I$: The atomic radius of $Mg$ $(160 \ pm)$ is greater than $Al$ $(143 \ pm)$. The ionic radius of $Mg^{2+}$ $(72 \ pm)$ is greater than $Al^{3+}$ $(54 \ pm)$. Thus, the correct order is $Mg > Al > Mg^{2+} > Al^{3+}$. Therefore, Statement-$I$ is false.
Statement-$II$: The magnitude of electron gain enthalpy generally increases across a period and decreases down a group. Due to small size and inter-electronic repulsion, oxygen $(O)$ has a lower electron gain enthalpy than sulfur $(S)$. The correct order of magnitude of electron gain enthalpy is $Cl > Br > S > O$. Therefore, Statement-$II$ is true.

(B) Statement-$I$: The first ionization enthalpy $(IE_1)$ generally increases across a period due to increasing $Z_{eff}$. However,$N$ $(2p^3)$ has a stable half-filled configuration,making its $IE_1$ higher than $O$ $(2p^4)$. The correct order is $C < O < N < F$. Thus,Statement-$I$ is true.
Statement-$II$: The magnitude of electron gain enthalpy $(|\Delta_{eg}H|)$ generally decreases down a group. However,oxygen $(O)$ has a very small size,leading to strong inter-electronic repulsions,which makes its electron gain enthalpy value lower than the rest of the group members. The correct order is $S > Se > Te > Po > O$. Thus,Statement-$II$ is true.

(C) In a period,the first ionization enthalpy increases from left to right due to an increase in the effective nuclear charge and a decrease in atomic size.
Therefore,the element at the extreme left of a period has the lowest first ionization enthalpy.
Conversely,the electron gain enthalpy becomes more negative (higher magnitude) as we move from left to right across a period (excluding noble gases,which have positive values due to stable configurations).
Thus,the element at the extreme right (halogens) has the highest negative electron gain enthalpy.

(A) $1$. Comparing the sizes: $Mg$ $(160 \text{ pm})$ > $Al$ $(143 \text{ pm})$ > $Mg^{2+}$ $(72 \text{ pm})$ > $Al^{3+}$ $(54 \text{ pm})$. Thus,$Mg$ is the largest and $Al^{3+}$ is the smallest. Statement $A$ is incorrect.
$2$. The element with atomic number $107$ is Bohrium $(Bh)$,but its systematic $IUPAC$ name is Unnilseptium. Statement $B$ is correct.
$3$. $Li$ and $Mg$ show diagonal relationship due to similar charge-to-size ratios. Statement $C$ is correct.
$4$. In $[AlCl(H_2O)_5]^{2+}$,$Al$ is in $+3$ oxidation state and forms $6$ coordinate bonds (covalency $6$). Statement $D$ is correct.