Periodic Table & Periodicity — Page 12 | JEE Chemistry

Q111

Given below are two statements : Statement I : The metallic radius of

\mathrm{Na}

is

1.86 \mathrm{~A}^{\circ}

and the ionic radius of

\mathrm{Na}^{+}

is lesser than

1.86 \mathrm{~A}^{\circ}

Statement II : Ions are always smaller in size than the corresponding elements. In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are true

B Statement I is incorrect but Statement II is true

C Both Statement I and Statement II are false

D Statement I is correct but Statement II is false

Correct Answer

Option D

Solution

To determine the correctness of the given statements, let's analyze each one individually based on atomic and ionic radii concepts.

Statement I: The metallic radius of $\mathrm{Na}$ is $1.86\, \text{Å}$ and the ionic radius of $\mathrm{Na}^+$ is lesser than $1.86\, \text{Å}$ .

Analysis: Metallic Radius of Sodium ( $\mathrm{Na}$ ): Sodium is a metal, and its metallic radius is indeed approximately $1.86\, \text{Å}$ .

The metallic radius refers to half the distance between the nuclei of two adjacent atoms in a metallic lattice.

Ionic Radius of Sodium Ion ( $\mathrm{Na}^+$ ): When sodium loses an electron to form $\mathrm{Na}^+$ , it loses its outermost electron shell (the 3s orbital).

This results in a significant decrease in size due to: Decrease in Electron-Electron Repulsion: Fewer electrons mean less repulsion among them.

Unchanged Nuclear Charge: The number of protons remains the same, so the effective nuclear charge per electron increases, pulling the remaining electrons closer to the nucleus.

The ionic radius of $\mathrm{Na}^+$ is approximately $0.95\, \text{Å}$ , which is significantly smaller than $1.86\, \text{Å}$ .

Conclusion: Statement I is correct.

Statement II: Ions are always smaller in size than the corresponding elements.

Analysis: General Trends: Cations ( $\text{Positive Ions}$ ): Formed by the loss of one or more electrons.

Result: Cations are smaller than their parent atoms due to loss of electron(s) and decreased electron-electron repulsion.

Example: $\mathrm{Na} \rightarrow \mathrm{Na}^+$ (size decreases).

Anions ( $\text{Negative Ions}$ ): Formed by the gain of one or more electrons.

Result: Anions are larger than their parent atoms due to added electron(s) and increased electron-electron repulsion.

Example: $\mathrm{Cl} \rightarrow \mathrm{Cl}^-$ (size increases).

Exceptions to Statement II: Since anions are larger than their corresponding neutral atoms, the statement that "ions are always smaller in size than the corresponding elements" is incorrect.

Conclusion: Statement II is false.

Final Answer: Statement I is correct but Statement II is false.

Q112

The electron affinity value are negative for A.

\mathrm{Be} \rightarrow \mathrm{Be}^{-}

B.

\mathrm{N} \rightarrow \mathrm{N}^{-}

C.

\mathrm{O} \rightarrow \mathrm{O}^{2-}

D.

\mathrm{Na} \rightarrow \mathrm{Na}^{-}

E.

\mathrm{Al} \rightarrow \mathrm{Al}^{-}

Choose the most appropriate answer from the options given below :

A A, B, D and E only

B D and E only

C A and D only

D A, B and C only

Correct Answer

Option D

Solution

To determine which elements have negative electron affinity values, we'll analyze each option individually.

Understanding Electron Affinity Electron Affinity (EA): The energy change when an electron is added to a neutral atom in the gas phase to form a negative ion.

Convention: If energy is released (exothermic process), the electron affinity is considered positive.

If energy is absorbed (endothermic process), the electron affinity is considered negative.

Option A: $\mathrm{Be} \rightarrow \mathrm{Be}^{-}$ Electronic Configuration of Be: $1s^2\,2s^2$ Analysis: Adding an electron to beryllium means placing it into the higher energy $2p$ orbital.

Beryllium has a filled $2s$ subshell, so the added electron experiences higher energy and electron-electron repulsion.

Result: Energy is absorbed; the process is endothermic.

Conclusion: Electron affinity of Be is negative.

Option B: $\mathrm{N} \rightarrow \mathrm{N}^{-}$ Electronic Configuration of N: $1s^2\,2s^2\,2p^3$ Analysis: Nitrogen has a half-filled $2p$ subshell.

Adding an electron introduces repulsion in the half-filled orbital.

Result: Energy is absorbed; the process is endothermic.

Conclusion: Electron affinity of N is negative.

Option C: $\mathrm{O} \rightarrow \mathrm{O}^{2-}$ Electronic Configuration of O: $1s^2\,2s^2\,2p^4$ Analysis: First Electron Affinity (O to O⁻): Exothermic (energy released).

Second Electron Affinity (O⁻ to O²⁻): Endothermic because adding an electron to a negative ion requires energy due to electron-electron repulsion.

Overall Process (O to O²⁻): The second step dominates, making the overall process endothermic.

Conclusion: Electron affinity for forming $\mathrm{O}^{2-}$ is negative.

Option D: $\mathrm{Na} \rightarrow \mathrm{Na}^{-}$ Electronic Configuration of Na: $1s^2\,2s^2\,2p^6\,3s^1$ Analysis: Adding an electron fills the $3s$ orbital.

Sodium tends to lose an electron to form $\mathrm{Na}^+$ , not gain one.

However, adding an electron is still an exothermic process due to the low energy of the $3s$ orbital.

Result: Energy is released; the process is exothermic.

Conclusion: Electron affinity of Na is positive.

Option E: $\mathrm{Al} \rightarrow \mathrm{Al}^{-}$ Electronic Configuration of Al: $1s^2\,2s^2\,2p^6\,3s^2\,3p^1$ Analysis: Adding an electron to the $3p$ orbital.

The process releases energy due to the addition of an electron to a partially filled orbital.

Result: Energy is released; the process is exothermic.

Conclusion: Electron affinity of Al is positive.

Final Conclusion Negative Electron Affinity Values (Endothermic Processes): Option A: Beryllium ( $\mathrm{Be}$ ) Option B: Nitrogen ( $\mathrm{N}$ ) Option C: Oxygen forming $\mathrm{O}^{2-}$ Therefore, the correct answer is: Option D: A, B, and C only Answer: Option D

Q113

The type of oxide formed by the element among Li, Na, Be, Mg, B and Al that has the least atomic radius is :

A AO

B AO2

C A2O

D A2O3

Correct Answer

Option D

Solution

The elements given are Li, Na, Be, Mg, B and Al 2 $^{nd}$ period elements $\to$ Li, Be, B Order of atomic radii $\to$ Li > Be > B 3 $^{rd}$ period elements $\to$ Na, Mg, Al Order of atomic radii $\to$ Na > Mg > Al In a period, atomic radii decreases on moving from left to right.

So, the element with least atomic radii in 2 $_{nd}$ period (Li, Be, B) is B.

The element with least atomic radii 3 $_{rd}$ period (Na, Mg, Al) is Al.

In B and Al $\Rightarrow$ B and Al are same group elements.

Down the group, radii increases.

So, B has least radii.

B is the element with least atomic radius to form the oxide $A_2O_3$ .

Oxidation states of the A in oxides :

{A_0} \to + 2\,\,{A_{{0_2}}} \to + 2\,\,{A_{{2^0}}} \to + 1\,\,{A_{2_3^0}} \to + 3

B outer configuration is s $^2$ P $^1$ . 3 valence electrons and form A $_2$ O $_3$ type oxide.

Q114

First ionisation enthalpy values of first four group 15 elements are given below. Choose the correct value for the element that is a main component of apatite family :

A 834 kJ mol−1

B 947 kJ mol−1

C 1402 kJ mol−1

D 1012 kJ mol−1

Correct Answer

Option D

Solution

Apalite family - A group of phosphate minerals that includes fluorapatite, chlorapatite, and hychoxyapatite.

Chemical formula

C{a_{10}}{(P{O_4})_6}{(OH,F,Cl)_2}

\left. \begin{array}{ll}Fluorapatite \to C{a_{10}}{(P{O_4})_6}{F_2} \\ Chlorapatite \to C{a_{10}}{(P{O_4})_6}C{l_2} \\ Hydroxyapatite \to C{a_{10}}{(P{O_4})_6}{(OH)_2} \end{array} \right\}\begin{array}{ll}{Main\,components:} \\ {Calcium\,and\,phosphate} \end{array}

So, the element considered is phosphorus.

The first four group 15 elements are Nitrogen (N), Phosphorus (P), Arsenic (As) and Antimony (Sb).

The outer configuration of phosphorus is half filled and hence stable

({s^2}{p^3})

.

For a stable configuration, ionization enthalpy is high.

Ionization enthalpy - Energy required to remove an electron from the gaseous atom.

Nitrogen has the highest first ionization enthalpy in group 15.

Down the group, ionization enthalpy decreases.

Down the group, atomic size increases, valence electrons are less attracted to the nucleus.

And here, electron removal will become easier.

So, ionization enthalpy is also decreases.

The highest value of ionization enthalpy 1402 KJ mol $^{-1}$ corresponds to nitrogen element.

So, the first ionization enthalpy of the element that is a main component of apatite family is option D) 1012 kJ mol $^{-1}$ .

Q115

An element ‘E’ has the ionisation enthalpy value of 374 kJ mol⁻¹. ‘E’ reacts with elements A, B, C and D with electron gain enthalpy values of −328, −349, −325 and −295 kJ mol⁻¹, respectively. The correct order of the products EA, EB, EC and ED in terms of ionic character is :

A EB > EA > EC > ED

B EA > EB > EC > ED

C ED > EC > EA > EB

D ED > EC > EB > EA

Correct Answer

Option A

Solution

Element $E \rightarrow$ Ionisation enthalpy value $\rightarrow 374 \mathrm{~kJ} \mathrm{~mol}^{-1}$ Element $A \rightarrow$ Electron gain enthalpy value $\rightarrow-328 \mathrm{~kJ} \mathrm{~mol}^{-1}$ Element $B \rightarrow$ Electron gain enthalpy value $\rightarrow-349 \mathrm{~kJ} \mathrm{~mol}^{-1}$ Element $C \rightarrow$ Electron gain enthalpy value $\rightarrow-325 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}$ Element $D \rightarrow$ Electron gain enthalpy value $\rightarrow-295 \mathrm{~kJ} \mathrm{~mol}^{-1}$ Ionization enthalpy is the energy required to remove an election from an atom or ion in the gas phase.

An element with lower ionization enthalpy indicate easier cation formation and this lead to higher ionic character.

Electron gain enthalpy is the energy released when an election is added to an atom or ion in the gas phase.

A highly negative election gain enthalpy Indicate that the atom readily gains electrons, favouring the formation of anions.

This leads to stronger ionic character in compounds.

High ionic character generally occurs when the difference in ionisation enthalpy and electron gain enthalpy between the two atoms is large.

For $E A$ , Difference between ionisation enthalpy and electron gain enthalpy

\begin{aligned} & =374 \mathrm{kJmol}^{-1}-\left(-328 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) \\ & =702 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}

For $E B$ ,

\begin{aligned} &\text{Difference between ionisation enthalpy and elation gain enthalpy } \\ & =374 \mathrm{~kJ} \mathrm{~mol}^{-1}-\left(-349 \mathrm{~kJ~mol}^1\right) \\ & =723 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}

For $E C$ , $\begin{aligned} & \text{ Difference between ionisation enthalpy and election gain enthalpy } \\ & =374 \mathrm{~kJ} \mathrm{~mol}^{-1}-\left(-325 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) \\ & =699 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$ For $ED$ , Difference between ionisation enthalpy and election gain enthalpy

\begin{aligned} & =374 \mathrm{~kJ} \mathrm{~mol}^{-1}-\left(-295 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) \\ & =669 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}

Higher difference between ionisation enthalpy and election gain enthalpy is for compound EB.

So, EB has higher ionic character.

The Lower value of difference between ionisation enthalpy and election gain enthalpy is for compound ED.

So, ED has lower ionic character.

The increasing order of compounds for the difference between ionisation enthalpy and electron gain enthalpy $\to$

E B So, increasing. order of compounds for ionic character:

E D In the decreasing order,

E B>E A>E C>E D

Correct option (1)

Q116

Given below are two statements :Statement (I) : The radii of isoelectronic species increases in the order. Mg2+ < Na+ < F

-

< O2-Statement (II) : The magnitude of electron gain enthalpy of halogen decreases in the order. Cl > F > Br > IIn the light of the above statements, choose the most appropriate answer from the options given below :

A Statement I is incorrect but Statement II is correct

B Both Statement I and Statement II are correct

C Statement I is correct but Statement II is incorrect

D Both Statement I and Statement II are incorrect

Correct Answer

Option B

Solution

Statement I : Correct The radii of isoelectronic species increases in the order $\mathrm{Mg^{2+} In general, radius of cation is smaller than anion$ \mathrm{A^+ Reason : In cations, number of electrons decreases, but number of protons remains the same.

This causes a stronger pull on the remaining electrons by the nucleus, resulting in a higher effective nuclear charge.

With increase in effective nuclear charge, radii decreases so, increase in positive charge indicate more number of electron removal and hence increase in effective nuclear charge.

As a result, radii decreases.

So, Mg $^{2+}$ has lower radii than Na $^+$ .

For anion, number of electrons are increased.

The electron-electron repulsion weakens the attraction between the nucleus and the electrons, resulting in a decrease in effective nuclear charge.

So, effective nuclear charge of O^{2-} $is lower than F^-$ .

So, O $^{2-}$ with lower effective nuclear charge causes longer radii of it.

So,

{F^ - } So,

M{g^{2 + }} Ionic radii of isoelectronic species decreases as nuclear charge increases.

The cation with greater +ve charge will have a smaller radius and the anion with the greater $-$ ve charge will have a bigger radius for isoelectronic species.

Statement II : Correct The magnitude of electron gain enthalpy of halogen decreases in the order Cl > F > Br > I.

Electron gain enthalpy is the amount of energy released when an atom accepts the electron from any neutral isolated gaseous atom to form an anion.

Generally, electron gain enthalpy decreases down the group.

The exception is for F and cl.

Electron gain enthalpy of F is less than Cl.

It is due to the comparatively small size of F atom.

For Cl, Br and I, the electron gain enthalpy decreases as go from Cl to I.

So, both the statements are correct.

Q117

The ionic mobility of alkali metal ions in aqueous solution is maximum for :

A K+

B Rb+

C Li+

D Na+

Correct Answer

Option B

Solution

Since the mobility of of ions is inversely proportional to the size of their hydrated ions, hence the increasing order of ionic mobility is

L{i^ + } < N{a^ + } < {K^ + } < R{b^ + }

Q118

Which of the following electronegativity order is incorrect?

A

\mathrm{S}<\mathrm{Cl}<\mathrm{O}<\mathrm{F}

B

\mathrm{Al}<\mathrm{Si}<\mathrm{C}<\mathrm{N}

C

\mathrm{Al}<\mathrm{Mg}<\mathrm{B}<\mathrm{N}

D

\mathrm{Mg}<\mathrm{Be}<\mathrm{B}<\mathrm{N}

Correct Answer

Option C

Solution

Let’s compare the known electronegativities (Pauling scale) of the elements in each option: F ~ 3.98 O ~ 3.44 N ~ 3.04 Cl ~ 3.16 C ~ 2.55 B ~ 2.04 S ~ 2.58 (sometimes listed as 2.5–2.58) Si ~ 1.90 Al ~ 1.61 Be ~ 1.57 Mg ~ 1.31 Now, check each statement: Option A: $\mathrm{S} S (2.58) < Cl (3.16) < O (3.44) < F (3.98) This order is correct. Option B:$ \mathrm{Al} Al (1.61) < Si (1.90) < C (2.55) < N (3.04) This order is correct.

Option C: $\mathrm{Al} Actual electronegativities are Al (1.61) and Mg (1.31). The statement says “Al < Mg,” which would mean 1.61 < 1.31, which is wrong. Therefore, Option C is incorrect. Option D:$ \mathrm{Mg} Mg (1.31) < Be (1.57) < B (2.04) < N (3.04) This order is correct.

Answer: Option C is the incorrect order.

Q119

Given below are two statements :Statement I : According to the Law of Octaves, the elements were arranged in the increasing order of their atomic number.Statement II : Meyer observed a periodically repeated pattern upon plotting physical properties of certain elements against their respective atomic numbers.In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are false

B Statement I is false but Statement II is true

C Statement I is true but Statement II is false

D Both Statement I and Statement II are true

Correct Answer

Option A

Solution

Law of octaves was arranged in the increasing order of their atomic weight.

Lothar Meyer plotted the physical properties such as atomic volume, melting point and boiling point against atomic weight.

Q120

Given below are two statements : Statement (I) : An element in the extreme left of the periodic table forms acidic oxides. Statement (II) : Acid is formed during the reaction between water and oxide of a reactive element present in the extreme right of the periodic table. In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are false

B Both Statement I and Statement II are true

C Statement I is true but Statement II is false

D Statement I is false but Statement II is true

Correct Answer

Option D

Solution

First, let us re-state the two statements clearly: Statement (I): An element in the extreme left of the periodic table forms acidic oxides.

Statement (II): Acid is formed during the reaction between water and oxide of a reactive element present in the extreme right of the periodic table.

Analyzing Statement (I) The extreme left of the periodic table corresponds to the alkali metals (Group 1) and alkaline earth metals (Group 2).

These metals typically form basic (or occasionally amphoteric, in the case of some Group 2 elements) oxides, not acidic oxides.

For example, Na₂O, K₂O, MgO, CaO, etc., all form basic solutions (e.g., Na₂O + H₂O → 2 NaOH).

Hence, Statement (I)—that an element in the extreme left forms acidic oxides—is false.

Analyzing Statement (II) The extreme right of the periodic table corresponds to the nonmetals in Groups 15, 16, 17 (and noble gases in Group 18).

Nonmetal oxides (such as those of sulfur, phosphorus, chlorine) are generally acidic.

When these oxides dissolve in water, they typically form acids.

Example: SO₃ + H₂O → H₂SO₄ (sulfuric acid) Example: P₂O₅ + 3 H₂O → 2 H₃PO₄ (phosphoric acid) Example: Cl₂O₇ + H₂O → 2 HClO₄ (perchloric acid) Thus, Statement (II)—that acid is formed when water reacts with an oxide of a reactive element in the extreme right—is true.

Conclusion Statement (I) is false.

Statement (II) is true.

Therefore, the correct choice is: (D) Statement (I) is false but Statement (II) is true.