JEE advanced problems on Classification of Elements

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Masterclass: 25 Ultra-Challenging JEE Advanced Problems on Classification of Elements

From the mathematical rigor of Slater's Rules to the thermodynamic paradoxes of Electron Gain Enthalpy. Master the periodic trends of Inorganic Chemistry.

The Blueprint of the Universe!
Inorganic chemistry is not merely memorization. At the JEE Advanced level, periodicity requires deep quantum mechanical intuition. You must understand how shielding ($Z_{eff}$) alters ionic radii, why half-filled stability creates kinks in ionization energy curves, and how Pauling derived his electronegativity scale from bond dissociation enthalpies. Prepare to decode the Periodic Table.

Problem 1: Effective Nuclear Charge ($Z_{eff}$) via Slater's Rules

Using Slater's rules, calculate the shielding constant ($\sigma$) and the effective nuclear charge ($Z_{eff}$) experienced by a $4s$ electron and a $3d$ electron in a Zinc atom ($Zn$, $Z = 30$). Why are $4s$ electrons lost before $3d$ electrons during ionization?

View Solution

Strategy: Write the electronic configuration grouped according to Slater's rules: $(1s)(2s, 2p)(3s, 3p)(3d)(4s, 4p)$. Apply the specific shielding contributions for $ns/np$ versus $nd/nf$ electrons.

Step 1: Shielding for a 4s electron
Configuration: $(1s^2) (2s^2, 2p^6) (3s^2, 3p^6) (3d^{10}) (4s^2)$.
For an $ns$ electron, the rules are:
- Electrons in the same group ($n=4$): $0.35$ each.
- Electrons in the $(n-1)$ group ($n=3$): $0.85$ each.
- Electrons in $(n-2)$ or lower groups ($n \le 2$): $1.00$ each.
$\sigma_{4s} = (1 \times 0.35) + (18 \times 0.85) + (10 \times 1.00) = 0.35 + 15.3 + 10 = 25.65$.
$Z_{eff}(4s) = Z - \sigma = 30 - 25.65 = \mathbf{4.35}$.

Step 2: Shielding for a 3d electron
For an $nd$ electron, the rules are different:
- Electrons in the same group ($3d$): $0.35$ each.
- ALL electrons in lower groups (including $3s, 3p$): $1.00$ each.
$\sigma_{3d} = (9 \times 0.35) + (18 \times 1.00) = 3.15 + 18 = 21.15$.
$Z_{eff}(3d) = 30 - 21.15 = \mathbf{8.85}$.

Step 3: Ionization Rationale
The $3d$ electrons experience a vastly higher effective nuclear charge ($8.85$) compared to the $4s$ electrons ($4.35$). Therefore, the $3d$ electrons are pulled much closer to the nucleus and bound more tightly. The $4s$ electrons are loosely held at the periphery and are always removed first during ionization.

Final Answer: For $4s$: $Z_{eff} = 4.35$. For $3d$: $Z_{eff} = 8.85$. The $4s$ electrons experience a much weaker nuclear pull, leading to their preferential removal.

Problem 2: Expanding the Periodic Table (The 8th Period)

Assuming the Aufbau principle continues to hold perfectly, what would be the total number of elements in the hypothetical 8th period of the periodic table? Which subshells would be filled in this period?

View Solution

Strategy: Use the $(n+l)$ rule to determine the sequence of subshell filling for $n=8$. The number of elements equals the total number of electrons that can be accommodated in these subshells.

Step 1: Identify the Subshells for $n=8$
A new period always starts with the filling of the $ns$ subshell and ends with the filling of the $np$ subshell.
For the 8th period, the filling starts with $8s$ and must end with $8p$.

Step 2: Determine Intervening Subshells using $(n+l)$ Rule
The subshells filled between $8s$ and $8p$ must have an $(n+l)$ value equal to $8$ or $9$, strictly following the sequence:
- $8s$: $n+l = 8+0 = 8$
- $5g$: $n+l = 5+4 = 9$ (Since $l=4$ is $g$, $l=3$ is $f$)
- $6f$: $n+l = 6+3 = 9$
- $7d$: $n+l = 7+2 = 9$
- $8p$: $n+l = 8+1 = 9$

Step 3: Calculate Total Electron Capacity
Total elements = sum of electrons in these subshells:
- $8s$ can hold $2$ electrons.
- $5g$ can hold $2(2l+1) = 2(2 \times 4 + 1) = 18$ electrons.
- $6f$ can hold $14$ electrons.
- $7d$ can hold $10$ electrons.
- $8p$ can hold $6$ electrons.
Total = $2 + 18 + 14 + 10 + 6 = 50$ electrons.

Final Answer: The 8th period will contain exactly $50$ elements. The filling order will be $8s \rightarrow 5g \rightarrow 6f \rightarrow 7d \rightarrow 8p$.

Problem 3: Anomalous Ionization Enthalpy (Period 2 vs 3)

The first ionization enthalpy ($IE_1$) of Nitrogen is greater than that of Oxygen, but the first ionization enthalpy of Phosphorus is almost identical to (or only slightly greater than) Sulfur. Explain why the anomaly is highly pronounced in Period 2 but dampened in Period 3.

View Solution

Strategy: Evaluate the stability of exactly half-filled subshells ($p^3$) versus the interelectronic repulsion in $p^4$ configurations. Then factor in atomic size.

Step 1: The Base Anomaly (Nitrogen vs Oxygen)
Nitrogen has a $2p^3$ configuration (exactly half-filled, highly stable). Oxygen has a $2p^4$ configuration. The 4th electron in oxygen is forced to pair up in a small $2p$ orbital. The intense interelectronic repulsion in this compact $2p$ orbital makes it much easier to remove the paired electron from Oxygen than an unpaired electron from Nitrogen. Thus, $IE_1(N) > IE_1(O)$.

Step 2: The Period 3 Scenario (Phosphorus vs Sulfur)
Phosphorus is $3p^3$ and Sulfur is $3p^4$. The same half-filled logic applies, but the $3p$ orbitals are significantly larger and more diffuse than the $2p$ orbitals.

Step 3: The Dampening Effect
Because the $3p$ orbital is large, placing two electrons in the same orbital in Sulfur causes far less interelectronic repulsion than it did in Oxygen. Therefore, the "penalty" for being $p^4$ is drastically reduced. While Phosphorus still benefits from half-filled exchange energy, the overriding increase in nuclear charge ($Z$) in Sulfur almost perfectly compensates for it. Thus, the $IE$ values of P and S are very close.

Final Answer: The anomaly relies on interelectronic repulsion of paired electrons. In Period 2, the $2p$ orbital is tiny, causing massive repulsion in Oxygen ($p^4$), sharply lowering its $IE$. In Period 3, the larger $3p$ orbital disperses the paired electrons in Sulfur, minimizing repulsion and dampening the half-filled anomaly.

Problem 4: Successive Electron Gain Enthalpies ($\Delta_{eg}H$)

The first electron gain enthalpy ($\Delta_{eg1}H$) of Oxygen is $-141 \text{ kJ mol}^{-1}$ (exothermic). However, the second electron gain enthalpy ($\Delta_{eg2}H$) to form the $O^{2-}$ ion is $+780 \text{ kJ mol}^{-1}$ (highly endothermic). If forming $O^{2-}$ requires so much energy, why do stable metal oxides containing $O^{2-}$ exist abundantly in nature?

View Solution

Strategy: Analyze the electrostatic forces during electron addition, and then apply the Born-Haber cycle to evaluate the overall thermodynamic stability of the solid lattice.

Step 1: The Endothermic $\Delta_{eg2}H$
Adding the first electron to neutral Oxygen is favorable because the nucleus attracts the incoming electron (exothermic). However, adding a second electron to the $O^-$ anion involves pushing a negatively charged electron into a region that is already negatively charged. The massive electrostatic repulsion between the incoming electron and the $O^-$ ion must be overcome by supplying energy, making $\Delta_{eg2}H$ highly endothermic.

Step 2: The Stabilizing Factor in Nature
In nature, $O^{2-}$ does not exist as an isolated gas. It exists in solid ionic lattices (like $MgO, Al_2O_3$).

Step 3: The Role of Lattice Energy
The $O^{2-}$ ion is small and highly charged. When it packs into a crystal lattice with metal cations, the resulting electrostatic attraction releases a colossal amount of energy, known as Lattice Enthalpy. This highly exothermic Lattice Enthalpy vastly overcompensates for the endothermic $\Delta_{eg2}H$, making the overall formation of the solid oxide thermodynamically highly favorable ($\Delta G < 0$).

Final Answer: $\Delta_{eg2}H$ is endothermic due to severe inter-anionic electrostatic repulsion. However, $O^{2-}$ forms stable oxides because its high charge density leads to a massively exothermic Lattice Enthalpy that overrides the energy cost of adding the second electron.

Problem 5: Isoelectronic Series Radii

Arrange the following isoelectronic species in strictly increasing order of their ionic radii: $N^{3-}, O^{2-}, F^-, Na^+, Mg^{2+}, Al^{3+}$. Provide the mathematical ratio that governs this trend.

View Solution

Strategy: For isoelectronic species, the number of electrons is constant. The only variable is the number of protons in the nucleus (Nuclear Charge, $Z$). Determine the ratio of protons to electrons ($Z/e$).

Step 1: Verify Isoelectronic Nature
All listed species possess exactly $10$ electrons ($1s^2 2s^2 2p^6$). The shielding effect ($\sigma$) is identical for all of them.

Step 2: Analyze the $Z/e$ Ratio
The ionic radius is inversely proportional to the effective nuclear charge ($Z_{eff}$). Since shielding is constant, radius is inversely proportional to the actual nuclear charge ($Z$). The higher the $Z/e$ ratio, the stronger the nucleus pulls the electron cloud inward, shrinking the ion.
- $N^{3-}$: $Z=7, e=10 \rightarrow Z/e = 0.70$
- $O^{2-}$: $Z=8, e=10 \rightarrow Z/e = 0.80$
- $F^-$: $Z=9, e=10 \rightarrow Z/e = 0.90$
- $Na^+$: $Z=11, e=10 \rightarrow Z/e = 1.10$
- $Mg^{2+}$: $Z=12, e=10 \rightarrow Z/e = 1.20$
- $Al^{3+}$: $Z=13, e=10 \rightarrow Z/e = 1.30$

Step 3: Arrange the Radii
The highest $Z/e$ ratio corresponds to the smallest radius. Therefore, highly charged cations are smallest, and highly charged anions are largest.

Final Answer: Increasing order of radii: $Al^{3+} < Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}$. The size is inversely proportional to the $Z/e$ ratio (Nuclear charge / Number of electrons).

Problem 6: Pauling Electronegativity Calculation

Given the bond dissociation enthalpies: $E_{A-A} = 104 \text{ kcal/mol}$, $E_{B-B} = 38 \text{ kcal/mol}$, and $E_{A-B} = 111 \text{ kcal/mol}$. If the electronegativity of element $A$ ($\chi_A$) is $3.0$, calculate the exact electronegativity of element $B$ ($\chi_B$) using Pauling's formula. (Assume $\chi_A > \chi_B$).

View Solution

Strategy: Pauling defined electronegativity difference based on the extra resonance energy of a polar covalent bond compared to a purely covalent theoretical bond.

Step 1: Write Pauling's Equation
The difference in electronegativity is given by:
$|\chi_A - \chi_B| = 0.208 \sqrt{\Delta}$ (If energies are in $\text{kcal/mol}$).
Where $\Delta$ is the resonance energy of the $A-B$ bond.

Step 2: Calculate $\Delta$
$\Delta = E_{\text{experimental}(A-B)} - E_{\text{theoretical}(A-B)}$
Pauling used the geometric mean (sometimes arithmetic, but geometric is rigorous) for the theoretical pure covalent bond: $E_{\text{theoretical}} = \sqrt{E_{A-A} \times E_{B-B}}$.
$E_{\text{theoretical}} = \sqrt{104 \times 38} = \sqrt{3952} \approx 62.86 \text{ kcal/mol}$.
$\Delta = 111 - 62.86 = 48.14 \text{ kcal/mol}$.

Step 3: Calculate $\chi_B$
$\chi_A - \chi_B = 0.208 \times \sqrt{48.14}$
$3.0 - \chi_B = 0.208 \times 6.938$
$3.0 - \chi_B = 1.443$
$\chi_B = 3.0 - 1.443 = 1.557$.

Final Answer: The electronegativity of element $B$ is approximately $1.56$.

Problem 7: Mulliken-Jaffe vs Pauling Scale

For an element $X$, the Ionization Enthalpy ($IE$) is $12.0 \text{ eV}$ and the Electron Affinity ($EA$) is $3.6 \text{ eV}$. Calculate its electronegativity on the Mulliken scale ($\chi_M$) and approximate its value on the Pauling scale ($\chi_P$).

View Solution

Strategy: Mulliken defined electronegativity as the arithmetic average of $IE$ and $EA$ (when expressed in electron-volts). Convert this to the Pauling scale using the empirical relation.

Step 1: Calculate Mulliken Electronegativity ($\chi_M$)
$\chi_M = \frac{IE + EA}{2}$ (where values are in $\text{eV/atom}$).
$\chi_M = \frac{12.0 + 3.6}{2} = \frac{15.6}{2} = 7.8$.

Step 2: Convert to Pauling Scale ($\chi_P$)
Pauling's values are empirically derived to fit around $4.0$ for Fluorine. The relationship between the two scales is roughly:
$\chi_P = \frac{\chi_M}{2.8}$
$\chi_P = \frac{7.8}{2.8} \approx 2.78$.

Final Answer: Mulliken Electronegativity $\chi_M = 7.8$. Pauling Electronegativity $\chi_P \approx 2.78$.

Problem 8: Decoding Successive Ionization Enthalpies

The successive ionization enthalpies of a third-period element $M$ (in $\text{kJ mol}^{-1}$) are: $IE_1 = 577$, $IE_2 = 1816$, $IE_3 = 2744$, $IE_4 = 11577$, $IE_5 = 14842$. Identify the element $M$ and explain the rationale for the massive leap between $IE_3$ and $IE_4$.

View Solution

Strategy: Look for the largest sudden "jump" or ratio between successive ionization energies. This jump indicates that an inner, noble-gas core shell is being broken into.

Step 1: Analyze the Energy Gaps
- Gap $1 \rightarrow 2$: $1816 - 577 = 1239$
- Gap $2 \rightarrow 3$: $2744 - 1816 = 928$
- Gap $3 \rightarrow 4$: $11577 - 2744 = \mathbf{8833}$ (Massive Leap!)
- Gap $4 \rightarrow 5$: $14842 - 11577 = 3265$

Step 2: Relate Jump to Valence Electrons
A sudden massive leap between $IE_3$ and $IE_4$ means the first three electrons were relatively easy to remove (they were valence electrons in the outermost shell). Removing the fourth electron required an immense amount of energy, meaning it had to be extracted from a highly stable, completely filled inner noble-gas core.

Step 3: Identify the Element
The element has exactly 3 valence electrons. It belongs to Group 13. Since it is explicitly stated to be in the third period, the element is Aluminum ($Al$). Its configuration is $[Ne] 3s^2 3p^1$. Removing the 4th electron requires breaking into the ultra-stable $[Ne]$ core ($2p^6$).

Final Answer: The element is Aluminum ($Al$). The massive leap between $IE_3$ and $IE_4$ occurs because the 4th electron is removed from the deeply buried, highly stable noble gas core ($[Ne] 2p^6$).

Problem 9: The Diagonal Relationship (Ionic Potential)

Lithium differs from other alkali metals but exhibits striking chemical similarities to Magnesium (Group 2). This is known as a diagonal relationship. Calculate the Ionic Potential ($\Phi$) for $Li^+$ ($r = 76 \text{ pm}$) and $Mg^{2+}$ ($r = 72 \text{ pm}$) to physically justify why they behave similarly.

View Solution

Strategy: The diagonal relationship is fundamentally driven by the similarity in charge-to-size ratio, known as Ionic Potential or Polarizing Power ($\Phi$).

Step 1: Define Ionic Potential ($\Phi$)
$\Phi = \frac{\text{Charge of Ion (z)}}{\text{Ionic Radius (r)}}$

Step 2: Calculate for $Li^+$
Charge $z = 1$. Radius $r = 76 \text{ pm}$.
$\Phi_{Li^+} = \frac{1}{76} \approx 0.0131 \text{ pm}^{-1}$.

Step 3: Calculate for $Mg^{2+}$
Charge $z = 2$. Radius $r = 72 \text{ pm}$.
$\Phi_{Mg^{2+}} = \frac{2}{72} \approx 0.0277 \text{ pm}^{-1}$.
Correction Note: Wait, traditional values use $A^{\circ}$ or specific constants, but the pure ratio shows $Mg^{2+}$ has roughly double the polarizing power? Actually, in classical texts, the diagonal relationship is supported because moving right increases $Z/r$ and moving down decreases it, somewhat canceling out. Let's look at the Charge Density (Charge/Volume).
Volume $\propto r^3$.
Density for $Li^+ \propto 1 / (76)^3 = 2.2 \times 10^{-6}$.
Density for $Mg^{2+} \propto 2 / (72)^3 = 5.3 \times 10^{-6}$.
While not perfectly identical, their small size and high polarizing power compared to rest of Group 1 force high covalent character (Fajans' Rules) in their salts, causing similarities like carbonate decomposition and hydration.

Final Answer: The similarity arises because their Charge Density (Polarizing Power, $\Phi = z/r$) is comparably high. Both are exceptionally small, heavily polarizing the electron clouds of anions, leading to significant covalent character in their compounds, unlike standard alkali metals.

Problem 10: Amphoteric Oxides Identification

Classify the following oxides as Acidic, Basic, Neutral, or Amphoteric: $CO, Al_2O_3, Cl_2O_7, BaO, N_2O, ZnO$. State the general periodic trend that predicts this nature.

View Solution

Strategy: The acid-base nature of an oxide correlates with the electronegativity (and metallic character) of the central atom. High EN = Acidic, Low EN = Basic, Intermediate EN = Amphoteric. Some specific low-oxidation-state non-metal oxides are neutral.

Step 1: Analyze Each Oxide
- $CO$ (Carbon Monoxide): A classic Neutral oxide. It does not form acids or bases with water.
- $Al_2O_3$ (Aluminum Oxide): Lies on the metalloid boundary. It reacts with both acids and bases. Amphoteric.
- $Cl_2O_7$ (Dichlorine Heptoxide): Formed by a highly electronegative halogen in its highest oxidation state ($+7$). Dissolves in water to form Perchloric acid ($HClO_4$). Highly Acidic.
- $BaO$ (Barium Oxide): Formed by a highly electropositive s-block metal. Dissolves to form $Ba(OH)_2$. Highly Basic.
- $N_2O$ (Nitrous Oxide): Like $CO$, it is a low-oxidation-state non-metal oxide. Neutral.
- $ZnO$ (Zinc Oxide): A border-line transition metal oxide that dissolves in both acids and bases (forming zincates). Amphoteric.

Step 2: State the Trend
As you move from left to right across a period, metallic character decreases and non-metallic character increases. Therefore, the nature of oxides transitions from strongly Basic $\rightarrow$ Amphoteric $\rightarrow$ Acidic.

Final Answer: Basic ($BaO$), Acidic ($Cl_2O_7$), Amphoteric ($Al_2O_3, ZnO$), Neutral ($CO, N_2O$). The trend moves from basic to acidic across a period as electronegativity increases.

Problem 11: Electron Gain Enthalpy Paradox (Halogens)

Fluorine is the most electronegative element in the periodic table. However, Chlorine has a significantly higher (more negative) Electron Gain Enthalpy ($\Delta_{eg}H$) than Fluorine. Explain the quantum mechanical reason behind this paradox.

View Solution

Strategy: Distinguish between Electronegativity (pulling shared electrons in a bond) and Electron Gain Enthalpy (adding a new electron to an isolated gaseous atom). Consider the physical size of the orbitals.

Step 1: Define the Contradiction
Fluorine pulls electrons better than Chlorine in a covalent bond ($\chi_F = 4.0, \chi_{Cl} = 3.0$). However, when adding a free electron, Chlorine releases more energy ($\Delta_{eg}H = -349 \text{ kJ/mol}$) than Fluorine ($\Delta_{eg}H = -328 \text{ kJ/mol}$).

Step 2: Analyze Orbital Size and Sterics
In Fluorine, the incoming electron must enter the $2p$ subshell. The $2p$ orbital is extremely compact and small. It already contains 5 electrons tightly packed together. Adding a 6th electron into this tiny volume creates massive interelectronic repulsion. This repulsion counteracts the strong pull of the nucleus, reducing the net energy released.

Step 3: The Chlorine Advantage
In Chlorine, the incoming electron enters the $3p$ subshell. The $3p$ orbital is significantly larger and more diffuse. The 5 existing electrons are spread over a much larger volume, so the incoming 6th electron experiences very little interelectronic repulsion. The nucleus can pull the electron in efficiently, releasing maximum energy.

Final Answer: Due to the extremely small size of the $2p$ orbital in Fluorine, massive interelectronic repulsion weakens the energy released upon gaining an electron. Chlorine's larger $3p$ orbital minimizes this repulsion, allowing it to have the highest $\Delta_{eg}H$ in the periodic table.

Problem 12: Lanthanoid Contraction Consequences

Normally, atomic radii increase significantly as you move down a group. However, Zirconium (Zr, Period 5) and Hafnium (Hf, Period 6) have almost identical atomic radii ($160 \text{ pm}$ vs $159 \text{ pm}$). What is the name of this phenomenon, and what is its specific quantum mechanical cause?

View Solution

Strategy: Identify what gets filled between Period 5 and Period 6 that wasn't there before. Look at the $4f$ subshell shielding capabilities.

Step 1: The Standard Trend vs The Reality
Moving from Zr ($4d$ series) to Hf ($5d$ series), an entire principal quantum shell is added ($n=5 \rightarrow n=6$). The radius should logically increase significantly.

Step 2: The Intervening Elements
Between Barium ($Z=56$) and Hafnium ($Z=72$) lie the 14 Lanthanoid elements. In these elements, the $4f$ subshell is being progressively filled with 14 electrons.

Step 3: The Shielding Effect of f-orbitals
The shape of $f$-orbitals is highly diffuse and complex. Electrons in $f$-orbitals provide exceptionally poor shielding (screening) for the outer valence electrons against the pull of the nucleus.
As 14 protons are added to the nucleus across the lanthanide series, the 14 new $4f$ electrons fail to adequately shield the outer $6s$ and $5d$ electrons. The Effective Nuclear Charge ($Z_{eff}$) spikes dramatically, pulling the outer electron cloud inward.

Step 4: The Cancellation
This steady contraction across the f-block (the Lanthanoid Contraction) perfectly cancels out the expected size increase of adding the 6th principal shell. Thus, Post-Lanthanide elements (like Hf, Ta, W) are nearly identical in size to their Group counterparts in the previous period (Zr, Nb, Mo).

Final Answer: This is the Lanthanoid Contraction. It is caused by the exceptionally poor shielding effect of $4f$ electrons, which allows a massive increase in effective nuclear charge ($Z_{eff}$) that shrinks the atom, perfectly canceling the expected increase from a new principal shell.

Problem 13: Inert Pair Effect (Oxidation State Stability)

In Group 14, Aluminum is exclusively stable in the $+3$ oxidation state. However, for Thallium (Tl) at the bottom of the group, the $+1$ oxidation state is far more stable than the $+3$ state (and $Tl^{3+}$ acts as a strong oxidizing agent). Explain the thermodynamic and quantum causes of this "Inert Pair Effect."

View Solution

Strategy: Evaluate the penetration effect of $s$-orbitals and the poor shielding of internal $d$ and $f$ orbitals for heavy elements. Relate this to bond energy.

Step 1: Define the Inert Pair Effect
The reluctance of the valence $ns^2$ electrons to participate in bonding, leading to a stable oxidation state that is 2 less than the group valency, is called the Inert Pair Effect. It is highly pronounced in the 6th period ($Tl, Pb, Bi$).

Step 2: Quantum Cause (Poor Shielding & Relativistic Effects)
Thallium ($Z=81$) has fully filled $4f^{14}$ and $5d^{10}$ inner subshells. As established, $d$ and $f$ electrons shield poorly. The Effective Nuclear Charge ($Z_{eff}$) experienced by the valence shell is exceptionally high.
Furthermore, $s$-orbitals are highly penetrating (they spend time very close to the nucleus). Because of the high $Z_{eff}$, the $6s^2$ electrons are pulled tightly to the nucleus, rendering them chemically "inert" and unwilling to ionize or hybridize. Only the single $6p^1$ electron is easily lost, leading to the $+1$ state.

Step 3: Thermodynamic Cause (Bond Energy)
To exhibit the $+3$ state, Tl must unpair its $6s^2$ electrons and form three bonds. Because Tl is a massive atom, the bonds it forms with other elements are very long and weak. The energy released by forming these weak bonds is insufficient to "pay back" the massive promotion/ionization energy required to unpair the tightly held $6s^2$ electrons.

Final Answer: The Inert Pair Effect. It is caused by the poor shielding of inner $d$ and $f$ orbitals, leading to a high $Z_{eff}$ that tightly binds the highly penetrating $ns^2$ electrons, preventing them from participating in bonding.

Problem 14: Hydration Energy vs Ionic Mobility

Arrange the alkali metal ions ($Li^+, Na^+, K^+, Rb^+, Cs^+$) in increasing order of their ionic mobility in an aqueous solution. Explain why the smallest isolated ion does not move the fastest when an electric field is applied.

View Solution

Strategy: Distinguish between crystal ionic radius and hydrated ionic radius. Small, highly charged ions attract more water molecules.

Step 1: Analyze Charge Density
In the gas phase, $Li^+$ is the smallest ion and $Cs^+$ is the largest. Because $Li^+$ is so small, its positive charge is concentrated over a tiny volume. It has the highest charge density.

Step 2: The Hydration Shell
When placed in water, the highly polar water molecules are intensely attracted to the high charge density of $Li^+$. A massive, thick shell of water molecules tightly binds to the $Li^+$ ion. Conversely, the large $Cs^+$ ion has low charge density and attracts very few water molecules.

Step 3: Hydrated Radius and Mobility
The effective size of the ion in solution is the ion plus its entire attached water shell (Hydrated Radius).
Because $Li^+$ attracts the most water, its Hydrated Radius is the largest of the group! $Cs^+_{(aq)}$ has the smallest hydrated radius.
Under an electric field, a bulky, heavily hydrated ion moves sluggishly due to high viscous drag. The smaller hydrated ion moves the fastest.

Final Answer: Increasing order of ionic mobility: $Li^+_{(aq)} < Na^+_{(aq)} < K^+_{(aq)} < Rb^+_{(aq)} < Cs^+_{(aq)}$. $Li^+$ has the highest charge density, resulting in maximum hydration. Its massive hydrated radius causes maximum drag, making it the slowest moving ion.

Problem 15: Nomenclature of Superheavy Elements

Using IUPAC systematic nomenclature rules, deduce the name and the three-letter symbol for the hypothetical element with atomic number $Z = 119$. Predict the block and group it would belong to if discovered.

View Solution

Strategy: Use the IUPAC numerical roots (0=nil, 1=un, 2=bi, 3=tri, 4=quad, etc.). Predict its position based on the noble gas preceding it.

Step 1: Apply IUPAC Roots for $119$
- $1 = \text{un}$
- $1 = \text{un}$
- $9 = \text{enn}$
Combine the roots and add the suffix "-ium": $\text{un} + \text{un} + \text{enn} + \text{ium} = $ Ununennium.
The three-letter symbol takes the first letter of each root: Uue.

Step 2: Predict Position in Periodic Table
The heaviest known noble gas is Oganesson ($Og$) at $Z = 118$, which completes the 7th period ($7p^6$).
Therefore, element 119 must begin a brand new period. The next available subshell according to Aufbau is $8s$.
Its configuration would end in $[Og] 8s^1$.

Step 3: Determine Block and Group
Since the differentiating electron enters an $s$-orbital, it is in the s-block. Since it has 1 valence electron, it belongs to Group 1 (Alkali Metals).

Final Answer: Name: Ununennium (Uue). It would belong to Group 1, s-block (and Period 8).

Problem 16: Electronegativity of Hybridized Carbons

Explain why a terminal alkyne hydrogen ($sp$-C) is acidic enough to react with $NaNH_2$, whereas an alkane hydrogen ($sp^3$-C) is completely inert. Frame your answer strictly around the concept of orbital hybridization and effective electronegativity.

View Solution

Strategy: Relate the percentage of s-character in a hybrid orbital to the nucleus's ability to hold onto electron pairs.

Step 1: Analyze s-character
- Alkane Carbon ($sp^3$): $25\%$ s-character, $75\%$ p-character.
- Alkene Carbon ($sp^2$): $33.3\%$ s-character.
- Alkyne Carbon ($sp$): $50\%$ s-character.

Step 2: Physical meaning of s-character
$s$-orbitals are spherical and have high probability density extremely close to the nucleus (high penetration effect). $p$-orbitals are elongated and extend far from the nucleus. An orbital with higher s-character sits physically closer to the positively charged nucleus.

Step 3: Effect on Electronegativity
Because an $sp$-hybridized orbital is closer to the nucleus, the carbon atom pulls the shared electrons of the $C-H$ bond much more strongly toward itself. This dramatically increases the effective electronegativity of the $sp$-carbon (making it roughly as electronegative as Nitrogen!).

Step 4: Acidity Conclusion
Because the $sp$-carbon hogs the bond electrons, the attached Hydrogen becomes highly electron-deficient ($\delta+$), allowing strong bases like the amide ion (${NH_2^-}$) to easily abstract it as a proton ($H^+$).

Final Answer: The $sp$-carbon has $50\%$ s-character, placing its orbital closer to the nucleus. This massively increases its effective electronegativity, polarizing the $C-H$ bond and rendering the terminal proton highly acidic compared to the $25\%$ s-character of an inert $sp^3$-carbon.

Problem 17: Anomalous Density Trend of Group 1

Moving down the Alkali Metals (Group 1), density generally increases ($Li < Na < K < Rb < Cs$). However, there is one distinct anomaly in this trend. Identify the anomaly and explain the structural reason causing it.

View Solution

Strategy: Density is Mass/Volume. Both atomic mass and atomic volume increase down a group. If density drops, volume must have increased proportionally more than mass.

Step 1: The General Trend
Down the group, atomic mass increases rapidly due to added protons and neutrons. Atomic volume also increases as new shells are added. Because the increase in mass usually outpaces the increase in volume, overall density increases.

Step 2: Identify the Anomaly
The actual density order is: $Li \ (0.53) < K \ (0.86) < Na \ (0.97) < Rb \ (1.53) < Cs \ (1.90)$.
Potassium ($K$) is anomalously lighter (less dense) than Sodium ($Na$).

Step 3: Explain the Volume Spike
When moving from Sodium (Period 3) to Potassium (Period 4), the $3d$ subshell becomes available. Even though the $3d$ subshell is empty in Potassium (configuration $[Ar] 4s^1$), the sheer presence of these empty $d$-orbitals causes a sudden, massive expansion in the atomic volume. This anomalous spike in volume completely overrides the expected increase in atomic mass, causing the density to dip unexpectedly before rising again at Rubidium.

Final Answer: Potassium is less dense than Sodium. This anomaly occurs because the sudden availability of empty $3d$ orbitals in Potassium causes a massive, disproportionate increase in atomic volume that overrides the increase in atomic mass.

Problem 18: Penetration Effect on Ionization Enthalpy

Even though Boron ($Z=5$) has a higher nuclear charge than Beryllium ($Z=4$), the first ionization enthalpy ($IE_1$) of Beryllium is higher than that of Boron. Explain this deviation based on the concept of orbital penetration.

View Solution

Strategy: Evaluate the electronic configurations of both elements and look at which specific subshell the outermost electron resides in.

Step 1: Write Configurations
- Beryllium ($Be$): $1s^2 2s^2$
- Boron ($B$): $1s^2 2s^2 2p^1$

Step 2: The Penetration Effect
For a given principal quantum shell ($n=2$), the radial probability distribution shows that $s$-orbitals penetrate much closer to the nucleus than $p$-orbitals. The order of penetration is $s > p > d > f$. Because the $2s$ electrons penetrate deeply, they are held much more tightly by the nucleus than a $2p$ electron.

Step 3: Applying to the Elements
In Beryllium, the electron must be removed from a highly penetrating, fully paired $2s$ orbital. In Boron, the electron is removed from a $2p$ orbital. The $2p$ electron is further away from the nucleus and heavily shielded by the completely filled $2s^2$ subshell underneath it. The ease of removing the shielded $2p$ electron vastly overrides the single extra proton Boron possesses.

Final Answer: Beryllium loses an electron from the highly penetrating, stable $2s$ orbital. Boron loses an electron from a $2p$ orbital, which is further from the nucleus and effectively shielded by the inner $2s^2$ electrons, requiring less energy to remove.

Problem 19: Born-Haber / Lattice vs Hydration Dilemma

The solubility of Alkaline Earth Metal hydroxides increases down the group ($Mg(OH)_2 \rightarrow Ba(OH)_2$), but the solubility of their sulfates strictly decreases down the group ($MgSO_4 \rightarrow BaSO_4$). Reconcile these opposing solubility trends using Lattice Enthalpy and Hydration Enthalpy.

View Solution

Strategy: Solubility depends on whether the energy gained from hydration ($\Delta H_{hyd}$) is greater than the energy required to break the lattice ($\Delta H_{lat}$). Both decrease down the group, but at different rates depending on anion size.

Step 1: Size Mismatch Rule
Lattice energy is highest when the cation and anion are of similar size. A large mismatch in size leads to poor packing and lower lattice energy.

Step 2: Analyze Hydroxides (Small Anion)
The $OH^-$ ion is small. Down the group, the cation gets larger ($Mg^{2+} \rightarrow Ba^{2+}$). The mismatch increases, causing Lattice Energy to drop rapidly. Hydration energy also drops as cation size increases, but not as fast as the lattice energy plummets. Since breaking the lattice becomes increasingly easy, solubility increases down the group.

Step 3: Analyze Sulfates (Large Anion)
The $SO_4^{2-}$ ion is massive. $Mg^{2+}$ is small, so there is a massive mismatch, meaning $MgSO_4$ has weak lattice energy and dissolves easily (high hydration energy of $Mg^{2+}$ helps). As we move down to $Ba^{2+}$, the massive cation perfectly matches the massive sulfate anion. The Lattice Energy remains exceptionally strong and high. However, the hydration energy drops rapidly because $Ba^{2+}$ is too large to attract water effectively. Because Hydration energy drops much faster than Lattice energy, solubility decreases heavily down the group.

Final Answer: For small anions ($OH^-$), Lattice Energy decreases faster than Hydration Energy down the group, increasing solubility. For massive anions ($SO_4^{2-}$), Lattice Energy remains constant/strong due to size matching, while Hydration Energy plummets, rendering heavy sulfates insoluble.

Problem 20: Master Deduction (The Unknown Elements)

Element X has $IE_1 = 496 \text{ kJ/mol}$ and forms a strongly basic oxide. Element Y has $IE_1 = 1256 \text{ kJ/mol}$, $EA = -349 \text{ kJ/mol}$, and forms a highly acidic oxide. Element Z has an exceptionally high $IE_1 = 2080 \text{ kJ/mol}$ and a positive $EA$. Identify the group each element belongs to, and predict the formula of the compound formed by X and Y.

View Solution

Strategy: Match the thermodynamic properties to broad periodic table trends (metals, non-metals, halogens, noble gases).

Step 1: Identify Element X
$IE_1$ is very low. It forms a strongly basic oxide. These are the hallmark characteristics of highly electropositive metals. Specifically, a very low $IE_1$ points to Group 1 (Alkali Metals). (The value $496$ exactly matches Sodium).

Step 2: Identify Element Y
$IE_1$ is very high (non-metal). The $EA$ is exceptionally high and negative (highly exothermic). It forms a highly acidic oxide. These traits define the halogens. Specifically, $-349 \text{ kJ/mol}$ is the highest electron affinity in the table, belonging to Chlorine. Y is in Group 17 (Halogens).

Step 3: Identify Element Z
$IE_1$ is massive (hardest to remove an electron). The $EA$ is positive (endothermic), meaning the atom violently resists accepting an extra electron. This completely full, perfectly stable state defines Group 18 (Noble Gases). (The value $2080$ matches Neon).

Step 4: Formula of X and Y
X is an alkali metal ($X^+$). Y is a halogen ($Y^-$). They combine in a 1:1 ratio to form an ionic salt.
Formula: $XY$ (e.g., $NaCl$).

Final Answer: X is Group 1 (Alkali). Y is Group 17 (Halogen). Z is Group 18 (Noble Gas). The compound formed between X and Y is strictly an ionic salt with the formula XY.

Mastering the Map of Matter

Congratulations on completing these ultra-challenging problems on Periodicity. The Periodic Table is not merely a chart; it is a visual spreadsheet of quantum mechanics. Anomalies in ionization energies, electron affinities, and atomic radii are never random—they are rigorous consequences of shielding ($Z_{eff}$), orbital penetration, and interelectronic repulsion. Master these concepts, and predicting chemical reactivity becomes second nature. Visit Chemca.in for more elite masterclasses!

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