Q.Using the Periodic Table, predict the formulas of compounds which might be formed by the following pairs of elements;

The question tests a subtle trend in group 13 (boron family). Most students memorise that melting and boiling points decrease down a group, but group 13 is famous for an exception: gallium melts in your hand at about 30 °C, far lower than its neighbours. Boiling points, however, follow a different logic — they depend on the strength of metallic bonding in the liquid state, which weakens as atomic size increases and the inert pair effect stabilises lower oxidation states.

Let’s walk through the reasoning step by step.

1. Melting point trend in group 13

   Melting points generally decrease from B to Al, then drop sharply at Ga, rise slightly at In, and drop again at Tl. The anomaly at gallium arises because its crystal structure (orthorhombic) is held together by weak van der Waals forces between Ga₂ dimers, not by strong metallic bonding. This makes Ga’s melting point the lowest in the group — about 30 °C. So X, the element with the lowest melting point, is gallium (Ga).

2. Boiling point trend in group 13

   Boiling points reflect the energy needed to separate atoms in the liquid state. Down the group, atomic size increases and metallic bonding weakens, so boiling points generally decrease. Thallium, being the heaviest, has the largest atomic radius and the weakest metallic bonding. Additionally, the inert pair effect makes Tl’s +1 oxidation state more stable, further reducing its tendency to form strong metallic bonds. Thus Tl has the lowest boiling point in the group — about 1457 °C, compared to Ga’s 2400 °C. So Y, the element with the lowest boiling point, is thallium (Tl).

3. Matching to the options
   • (A) B, Tl — Boron has a very high melting point (diamond-like covalent network), not the lowest.
   • (B) Ga, Tl — Correct: Ga has lowest melting point, Tl has lowest boiling point.
   • (C) Tl, Tl — Tl’s melting point is low but not the lowest (Ga is lower).
   • (D) In, Ga — Indium’s melting point is higher than Ga’s; Ga’s boiling point is higher than Tl’s.

Concept and Intuition

The size of an ion, known as its ionic radius, depends on several factors:

1. Number of electron shells: As we move down a group in the periodic table, new electron shells are added, leading to an increase in atomic and ionic size.
2. Nuclear charge (number of protons): For ions with the same number of electron shells (or isoelectronic species), a higher nuclear charge pulls the electrons more strongly towards the nucleus, resulting in a smaller ionic radius.
3. Number of electrons:
   • Cations are formed by losing electrons. The removal of electrons reduces electron-electron repulsion and often leads to the loss of an entire electron shell, making cations significantly smaller than their parent atoms. A higher positive charge (e.g., $Mg^{2+}$ vs $Na^+$) means more electrons have been removed or the remaining electrons are held more tightly by the same nucleus, leading to a smaller size.
   • Anions are formed by gaining electrons. The addition of electrons increases electron-electron repulsion, causing the electron cloud to expand, making anions larger than their parent atoms. A higher negative charge (e.g., $O^{2-}$ vs $F^-$) means more electrons have been added, leading to greater repulsion and a larger size.

When comparing ions, especially those that are isoelectronic (have the same number of electrons), the key factor is the nuclear charge. The species with the lowest nuclear charge (fewest protons) will have the largest radius because the electrons are less strongly attracted to the nucleus.

Step-by-Step Solution

1. Categorize the given ions into cations and anions.

   • Cations (positively charged ions): $Ca^{2+}$, $K^+$, $Ba^{2+}$, $Mg^{2+}$, $Cs^+$
   • Anions (negatively charged ions): $S^{2-}$, $P^{3-}$, $Cl^-$

2. Identify the largest cation.

   Let's list the cations with their atomic numbers ($Z$) and electron configurations (number of electrons):

   • $Mg^{2+}$: $Z=12$, 10 electrons (like Neon)

   • $K^+$: $Z=19$, 18 electrons (like Argon)

   • $Ca^{2+}$: $Z=20$, 18 electrons (like Argon)

   • $Cs^+$: $Z=55$, 54 electrons (like Xenon)

   • $Ba^{2+}$: $Z=56$, 54 electrons (like Xenon)

   • Compare $K^+$ and $Ca^{2+}$: Both have 18 electrons. $K^+$ has $Z=19$ and $Ca^{2+}$ has $Z=20$. Since $K^+$ has a smaller nuclear charge, it will exert less pull on its 18 electrons, making it larger than $Ca^{2+}$. So, $K^+ > Ca^{2+}$.

   • Compare $Cs^+$ and $Ba^{2+}$: Both have 54 electrons. $Cs^+$ has $Z=55$ and $Ba^{2+}$ has $Z=56$. Similarly, $Cs^+$ has a smaller nuclear charge, making it larger than $Ba^{2+}$. So, $Cs^+ > Ba^{2+}$.

   • Compare $Mg^{2+}$, $K^+$, and $Cs^+$: These ions belong to different periods. $Mg^{2+}$ is from Period 3, $K^+$ is from Period 4, and $Cs^+$ is from Period 6. Ionic size increases down a group due to the addition of new electron shells. Therefore, $Cs^+$ (Period 6) will be significantly larger than $K^+$ (Period 4), which in turn is larger than $Mg^{2+}$ (Period 3).

   Combining these comparisons, $Cs^+$ is the largest cation among the given options.

3. Identify the largest anion.

   Let's list the anions with their atomic numbers ($Z$) and electron configurations (number of electrons):

   • $P^{3-}$: $Z=15$, 18 electrons (like Argon)
   • $S^{2-}$: $Z=16$, 18 electrons (like Argon)
   • $Cl^-$: $Z=17$, 18 electrons (like Argon)

   All three anions are isoelectronic with Argon (18 electrons). For isoelectronic species, the ionic radius decreases as the nuclear charge increases.

   • $P^{3-}$ has $Z=15$.
   • $S^{2-}$ has $Z=16$.
   • $Cl^-$ has $Z=17$.

   Since $P^{3-}$ has the smallest nuclear charge ($Z=15$) among these isoelectronic anions, it will experience the weakest pull from the nucleus on its 18 electrons, resulting in the largest ionic radius.

   Therefore, $P^{3-} > S^{2-} > Cl^-$.

   $P^{3-}$ is the largest anion among the given options.

4. Combine the results.

   The largest cation is $Cs^+$ and the largest anion is $P^{3-}$.

Ions formed (nearest inert-gas configuration)

Comparing sizes

• $\text{S}^{2-}$ has 18 electrons occupying three shells ($n=3$), so it is larger than every 10-electron (two-shell) ion. Hence the largest ion is $\text{S}^{2-}$ → element E.
• Among the isoelectronic 10-electron ions, radius decreases as nuclear charge rises. The highest charge is $Z=13$, so $\text{Al}^{3+}$ is the smallest ion → element A.

Therefore $X$ (largest) $= \text{E}$ and $Y$ (smallest) $= \text{A}$.

Electron gain enthalpy ($\Delta_{eg}H$) is the energy change when an electron is added to a neutral gaseous atom; a more negative value means more energy released (greater affinity). Approximate values:

Oxygen is a second-period element whose valence 2p orbitals are small and compact. The added electron experiences strong repulsion from the electrons already crowded in that small region, so comparatively little energy is released. Its $\Delta_{eg}H$ is therefore the least negative (smallest magnitude) among the four, i.e. the least electron gain enthalpy.

Concept and Intuition

This question tests your understanding of periodic trends in the carbon group (Group 14). Melting point in these elements is governed by the strength of metallic bonding and the structure of the solid. Carbon (diamond) has a giant covalent network with very strong bonds, so it melts at an extremely high temperature. As we go down the group, the atoms become larger, and the bonding changes from purely covalent to more metallic. The key insight: melting point does not simply decrease monotonically down the group. Instead, it drops sharply from carbon to silicon, continues to fall through germanium to tin, but then rises slightly for lead. This is because tin and lead are true metals, but lead’s heavier nucleus and relativistic effects strengthen its metallic bonds a bit. So the lowest melting point among the four given elements is tin.

Step-by-step reasoning

1. Recall the trend in Group 14

   The melting points (°C) are roughly:

   • Carbon (diamond): ~3550
   • Silicon (Si): 1414
   • Germanium (Ge): 938
   • Tin (Sn): 232
   • Lead (Pb): 327 So the pattern is: C >> Si > Ge > Sn < Pb. The minimum occurs at tin.

2. Why does tin have such a low melting point?

   Tin exists in two common allotropes: white tin (β-Sn, metallic) and gray tin (α-Sn, nonmetallic). The metallic form has relatively weak bonding compared to the covalent networks of Si and Ge. The large atomic size of tin means its valence electrons are far from the nucleus and less effective at holding the lattice together. This makes tin’s metallic bonds quite weak, hence a low melting point.

3. Why does lead’s melting point rise above tin’s?

   Lead is heavier and has a much higher nuclear charge. Relativistic effects contract the 6s orbitals, making the electrons more tightly bound. This strengthens the metallic bonding in lead compared to tin, raising its melting point slightly. So lead melts at 327°C, about 95°C higher than tin.

4. Compare the given options

   • Si: 1414°C
   • Ge: 938°C
   • Sn: 232°C
   • Pb: 327°C Clearly, tin has the lowest value.

Concept & Intuition

In solid-state chemistry, the formula of a compound in a cubic unit cell is found by counting how many atoms of each element actually belong to one unit cell. Atoms at corners are shared by 8 cells, so each contributes 1/8. Atoms on edges are shared by 4 cells, so each contributes 1/4. An atom at the center belongs entirely to that cell (contribution = 1). This avoids double-counting and gives the simplest whole-number ratio.

Step-by-step reasoning

1. Identify positions and contributions

   • W atoms: at cube corners. There are 8 corners. Each corner atom is shared by 8 adjacent unit cells, so contribution per corner = $\frac{1}{8}$.

     Total W atoms per cell = $8 \times \frac{1}{8} = 1$.

   • O atoms: at cube edges. A cube has 12 edges. Each edge atom is shared by 4 unit cells, so contribution per edge = $\frac{1}{4}$.

     Total O atoms per cell = $12 \times \frac{1}{4} = 3$.

   • Na atoms: at cube centers. There is 1 center per cell, not shared.

     Total Na atoms per cell = $1 \times 1 = 1$.

2. Write the ratio

   From the counts: Na : W : O = 1 : 1 : 3.

   Therefore the empirical formula is NaWO₃.

3. Match to options

   Option (A) is NaWO₃, which matches exactly. The others have different atom ratios.

Concept & Intuition

The key is knowing the geometry of a cubic close-packed (CCP) structure (also called face-centered cubic, FCC). In CCP, atoms are arranged in layers with the pattern ABCABC. The number of octahedral voids in a CCP unit cell equals the number of atoms in that cell. So if Y atoms form the CCP lattice, the number of Y atoms per unit cell is 4, and the number of octahedral voids is also 4. If X fills all these octahedral voids, then there are 4 X atoms per unit cell as well. The simplest whole-number ratio of X to Y is therefore 1:1, giving the formula XY.

Step-by-step reasoning

1. Determine the number of Y atoms in a CCP unit cell.

   In a CCP (FCC) arrangement, atoms are at the corners and face centers.

   • 8 corners × $\frac{1}{8}$ per corner = 1 atom
   • 6 face centers × $\frac{1}{2}$ per face = 3 atoms Total Y atoms per unit cell = $1 + 3 = 4$.

2. Count the octahedral voids in a CCP unit cell.

   In a CCP structure, octahedral voids are located at the body center and at the midpoints of each edge.

   • 1 body center (fully inside the cell) = 1 void
   • 12 edges × $\frac{1}{4}$ per edge (each edge void is shared by 4 cells) = 3 voids Total octahedral voids = $1 + 3 = 4$. This matches the general rule: number of octahedral voids = number of atoms in the close-packed lattice.

3. Place X atoms into the octahedral voids.

   The problem states that X occupies the octahedral voids. It does not say “partially” or “half” — it implies all such voids are filled. Therefore, number of X atoms per unit cell = number of octahedral voids = 4.

4. Write the formula from the ratio.

   Per unit cell: X = 4, Y = 4.

   Simplest ratio X : Y = 4 : 4 = 1 : 1.

   Hence the formula is XY.

Concept & Intuition

The key is knowing the geometry of a cubic close-packed (CCP) structure. In CCP (also called face-centered cubic, FCC), atoms are arranged in layers with the pattern ABCABC. The number of octahedral voids in any close-packed structure equals the number of atoms in the packing. So if Y atoms form the CCP lattice, the number of octahedral voids is exactly the same as the number of Y atoms. If X fills all those voids, the ratio of X to Y is 1:1.

Step-by-step reasoning

1. Determine the number of Y atoms per unit cell in CCP.

   In a CCP (FCC) arrangement, atoms are at the corners and face centers.

   • 8 corners × $\frac{1}{8}$ = 1 atom
   • 6 faces × $\frac{1}{2}$ = 3 atoms Total = 4 Y atoms per unit cell.

2. Find the number of octahedral voids in a CCP unit cell.

   In any close-packed structure (CCP or HCP), the number of octahedral voids equals the number of atoms in the packing.

   • For CCP: 4 atoms → 4 octahedral voids. (Alternatively, octahedral voids are located at the body center and at the midpoints of each edge: 1 body center + 12 edges × $\frac{1}{4}$ = 1 + 3 = 4.)

3. Place X cations in all octahedral voids.

   The problem states X occupies octahedral voids. If it occupies all of them, then there are 4 X atoms per unit cell.

4. Write the simplest formula.

   Ratio X : Y = 4 : 4 = 1 : 1 → formula is XY.