Q.The second and third rows of transition elements resemble each other much more than they resemble the first row. Explain why?

The key concept here is periodic trends in atomic (metallic) radius down Group 13. Normally, as you go down a group, each element adds a new electron shell, so the atomic radius increases. However, Group 13 has a famous exception: gallium is actually smaller than aluminum. This happens because gallium has a filled d-subshell (3d¹⁰) that doesn’t shield the nuclear charge well, pulling the outer electrons in tighter. Additionally, gallium’s crystal structure is less metallic and more molecular, leading to a smaller metallic radius than expected.

Let’s work through the order step by step.

1. General trend down Group 13 From aluminum (Al) to indium (In) to thallium (Tl), each element is in a higher period, so the atomic radius should increase:

This is the baseline expectation.

2. The gallium anomaly Gallium (Ga) sits directly below Al in the periodic table. However, its metallic radius is actually smaller than Al’s. Why?
   • Ga has the electron configuration [Ar] 3d¹⁰ 4s² 4p¹. The 3d electrons are poor at shielding the 4p electron from the nuclear charge, so the effective nuclear charge felt by the outer electron is higher.
   • Also, gallium’s crystal structure is orthorhombic (not close-packed like Al), leading to shorter interatomic distances. The result:

3. Comparing Ga with In and Tl …

Concept and Intuition

Beryllium (Be, group 2) and aluminium (Al, group 13) are diagonal neighbours in the periodic table. This diagonal relationship arises because the increase in nuclear charge down a group is offset by the increase in atomic size across a period, leading to similar charge-to-size ratios and thus similar chemical behaviour. However, covalency limits depend on the number of available empty orbitals: Be has only 2s and 2p orbitals (max 4), while Al has 3s, 3p, and 3d orbitals (max 6). This distinction is crucial.

Step-by-step reasoning

1. Statement I: Chlorides with $\ce{Cl^-}$ bridged structure in vapour phase

   • Beryllium chloride ($\ce{BeCl2}$) in the vapour phase exists as a dimer $\ce{Be2Cl4}$, where each Be is surrounded by four Cl atoms, with two Cl atoms bridging the two Be atoms.
   • Aluminium chloride ($\ce{AlCl3}$) also dimerises in the vapour phase to $\ce{Al2Cl6}$, with chlorine bridges.
   • Both exhibit electron-deficient character and form bridged structures to complete their octet.
   • Conclusion: Statement I is correct.

2. Statement II: Strong tendency to form complexes

   • Be²⁺ has a high charge density (small size, +2 charge), making it a strong Lewis acid that readily accepts electron pairs from ligands (e.g., $\ce{[Be(H2O)4]^{2+}}$, $\ce{[BeF4]^{2-}}$).
   • Al³⁺ has an even higher charge density and forms many complexes (e.g., $\ce{[AlF6]^{3-}}$, $\ce{[Al(H2O)6]^{3+}}$).
   • Both show a strong tendency to form coordination compounds.
   • Conclusion: Statement II is correct. …

The key concept here is periodic trends in atomic radii, but with a twist. Normally, as you go down a group, atomic radius increases because each new element has an additional electron shell. However, group 13 (B, Al, Ga, In, Tl) shows an anomaly: gallium (Ga) is actually smaller than aluminium (Al), and thallium (Tl) is only slightly larger than indium (In). This happens because of the d-block contraction (for Ga and In) and the f-block contraction (for Tl). These contractions occur when electrons fill inner d or f orbitals, which are poor at shielding the nuclear charge, so the outer electrons are pulled in more tightly.

Let’s work through the reasoning step by step.

1. Start with the general trend.

   Down a group, atomic radius should increase: B < Al < Ga < In < Tl. But experimental data show this is not true. The first break occurs at Ga.

2. Why is Ga smaller than Al?

   Gallium comes after the first row of transition metals (Sc to Zn). In those elements, the 3d orbitals are filled. The 3d electrons are poor at shielding the 4s and 4p electrons from the nuclear charge. This d-block contraction (also called scandide contraction) makes the effective nuclear charge on the outer electrons higher, shrinking the atom. So Ga (atomic radius ~135 pm) is actually smaller than Al (~143 pm).

3. What about indium?

   Indium is below Ga, so it has an extra shell (n=5). Despite the d-block contraction from the 4d electrons, the addition of a shell wins out, so In (~167 pm) is larger than both Al and Ga.

4. Now thallium — the second twist.

   Thallium comes after the lanthanides (Ce to Lu). The 4f orbitals are filled in the lanthanide series, and these f-electrons are even worse at shielding than d-electrons. This f-block contraction (lanthanide contraction) is so strong that Tl (~170 pm) is only slightly larger than In (~167 pm), despite having an extra shell. In fact, Tl is smaller than you’d expect from a simple periodic trend. …

The key idea is that each set of elements belongs to the same vertical column (group) in the modern periodic table. The group number tells you the number of valence electrons (for main‑group elements) or the typical oxidation states (for transition metals). Instead of memorizing, you can reason from the element’s position: for example, Mn is directly above Tc and Re, so they share group 7. Similarly, Zn, Cd, Hg are the “volatile” group 12 metals; Ti, Zr, Hf are group 4; Ga, In, Tl are group 13 (the boron family). The Roman numerals I–V are just the group numbers 12, 4, 17, 7, 13 respectively.

1. Identify the group of Mn, Tc, Re (Set A).

   Manganese (Mn) is in the first transition series, atomic number 25. Its electron configuration ends in $3d^5 4s^2$, placing it in group 7. Technetium (Tc) and rhenium (Re) are directly below Mn in the same column. So Set A belongs to group 7, which is listed as IV (since IV = 7).

   Thus A → IV.

2. Identify the group of Zn, Cd, Hg (Set B).

   Zinc (Zn) ends in $3d^{10}4s^2$, a full d‑subshell, placing it in group 12. Cadmium (Cd) and mercury (Hg) are below Zn. Group 12 is listed as I (since I = 12).

   Thus B → I.

3. Identify the group of Ti, Zr, Hf (Set C).

   Titanium (Ti) has configuration $3d^2 4s^2$, making it group 4. Zirconium (Zr) and hafnium (Hf) are directly below. Group 4 is listed as II (since II = 4).

   Thus C → II.

4. Identify the group of Ga, In, Tl (Set D). …

The question asks for the correct order of atomic radii among the group 13 elements: Al, Ga, In, and Tl. This is a classic case where the simple periodic trend — radius increases down a group — hits a snag. The reason lies in the electronic configurations and the effect of inner electron shells.

Let’s recall the trend: as you go down a group, new electron shells are added, so atomic radius should increase. For group 13, that would predict Al < Ga < In < Tl. But experiment shows otherwise. Why?

The key is the d-block contraction. Gallium sits right after the first transition series (elements Sc to Zn). The 3d electrons in Ga are poor at shielding the nuclear charge, so the effective nuclear charge felt by the outermost electrons is higher than expected. This pulls the electron cloud inward, making Ga’s atomic radius smaller than Al’s, despite Ga being one period below.

Indium and thallium follow after the 4d and 5d series, but the effect is less dramatic for In (4d) and Tl (5d) because the 4f and 5f contractions also play a role, but the overall trend of increasing size resumes after Ga.

Let’s break it down step by step.

1. Identify the group and period positions.

   Al is in period 3, Ga in period 4, In in period 5, Tl in period 6. Normally, period 4 elements are larger than period 3 elements. But Ga is an exception.

2. Understand the d-block contraction.

   Ga has the configuration $[\text{Ar}] 3d^{10} 4s^2 4p^1$. The 3d electrons are in a full d-subshell, which is diffuse and does not shield the 4p electron effectively from the nuclear charge. This increases the effective nuclear charge ($Z_{\text{eff}}$) on the outermost electron, shrinking the atomic radius. Al has $[\text{Ne}] 3s^2 3p^1$ — no d-electrons, so its $Z_{\text{eff}}$ is lower, and its radius is larger than Ga’s.

3. Compare Al and Ga.

   Experimental atomic radii (in picometers, typical values):

   • Al: 143 pm
   • Ga: 135 pm So Ga < Al.

4. Compare In and Tl. …

The key concept here is shielding (or screening) effect and how different types of orbitals (s, p, d) contribute to it. In the periodic table, atomic radius generally increases down a group because new electron shells are added. However, gallium (Ga) is an exception: it is smaller than aluminium (Al), even though it lies directly below Al in Group 13. This anomaly is due to the presence of a filled set of 3d orbitals in gallium, which are absent in aluminium.

Let’s break down why this happens.

1. Recall the trend and the anomaly

   Aluminium (atomic number 13) has the electron configuration $[Ne] 3s^2 3p^1$. Gallium (atomic number 31) has $[Ar] 3d^{10} 4s^2 4p^1$. Normally, adding a shell (from n=3 to n=4) should increase the radius. But gallium’s radius is actually about 135 pm vs. aluminium’s 143 pm — a clear contraction.

2. Understand shielding and effective nuclear charge

   Electrons in inner shells “shield” outer electrons from the full positive charge of the nucleus. The effective nuclear charge ($Z_{\text{eff}}$) felt by an outer electron is the actual nuclear charge minus the shielding constant. If shielding is poor, $Z_{\text{eff}}$ is higher, pulling electrons closer and shrinking the atom.

3. Identify the role of d-electrons

   In gallium, the 3d subshell is completely filled. d-orbitals are diffuse and penetrate poorly toward the nucleus compared to s- or p-orbitals. This means d-electrons are not very effective at shielding the outer 4s and 4p electrons from the nuclear charge. As a result, the 4p electron in gallium experiences a higher $Z_{\text{eff}}$ than expected, contracting the atomic radius.

4. Compare with aluminium

   Aluminium has no d-electrons; its inner electrons are 1s, 2s, and 2p — all of which are better shielders. So aluminium’s outer electron feels a lower $Z_{\text{eff}}$ relative to its nuclear charge, giving a larger radius.

5. Evaluate the options …

The concept here is stability of oxidation states in transition metals, governed by electronic configuration and the extra stability of half-filled and fully-filled d-subshells. While melting points generally increase with more unpaired electrons (due to stronger metallic bonding), and Cr, Mo, W indeed have very high melting points, the stability of Mn²⁺ over Mn³⁺ is a classic exception. Let’s check each statement.

1. Statement (A): Cr, Mo and W have high melting points

   These elements belong to Group 6. They have a high number of unpaired d-electrons (Cr: d⁵, Mo: d⁵, W: d⁴ in their metallic states), leading to strong metallic bonding. This results in exceptionally high melting points (e.g., W melts at ~3422°C). This statement is true.

2. Statement (B): With increase in number of unpaired electrons melting point increases

   In transition metals, metallic bonding strength depends on the number of delocalized electrons contributed by d-orbitals. More unpaired electrons mean more electrons available for bonding, raising the melting point. This trend generally holds across a period (e.g., from Sc to Cr). This statement is true.

3. Statement (C): Mn³⁺ is more stable than Mn²⁺ …

The concept first

In a metallic lattice, positive kernels sit in a "sea" of delocalised electrons. The strength of the metallic bond — and hence the melting point, boiling point and enthalpy of atomisation — grows with the number of electrons each atom contributes to that sea, particularly the number of unpaired electrons available for bonding.

For $s$-block metals only the $ns$ electrons (one or two) are available. For a transition metal, the $(n-1)d$ and $ns$ orbitals lie very close in energy, so the $d$ electrons can also take part. That gives a much larger pool of bonding electrons.

Step-by-step

1. Is the Assertion true? Yes. Compare (approximate melting points):

Transition metals are, in general, hard, dense and high-melting. (A) is true.

2. Is the Reason true? Yes — the near-degeneracy of $(n-1)d$ and $ns$ means both sets of electrons enter the delocalised bonding, so more electrons per atom participate in interatomic metallic bonding. (R) is true.

3. Does (R) explain (A)? Yes, and there are two beautiful pieces of supporting evidence within the block itself: …

Understanding Electronic Configuration Anomalies

Molybdenum (atomic number 42) presents a classic case of an anomalous electron configuration. We'd naively expect it to follow the Aufbau principle strictly and have $4d^4\ 5s^2$, but nature has other plans.

The key concept here is exchange energy — the quantum mechanical stabilization that occurs when electrons with parallel spins can exchange positions. The more unpaired electrons with the same spin, the greater the exchange energy stabilization. This is a manifestation of the Pauli exclusion principle and electron correlation effects.

Evaluating Assertion (A)

Let's verify Mo's ground state configuration:

1. Expected configuration: Following the simple filling order, we'd predict $[\text{Kr}]\ 4d^4\ 5s^2$

2. Actual configuration: Molybdenum actually has $[\text{Kr}]\ 4d^5\ 5s^1$

3. Why the promotion? Moving one electron from the $5s$ to the $4d$ orbital creates:

   • Five unpaired electrons in the $4d$ subshell (half-filled)
   • One unpaired electron in the $5s$ orbital
   • Total: six unpaired electrons

4. Energy balance: The exchange energy gained from having six unpaired electrons (instead of four) more than compensates for the small energy cost of promoting an electron from $5s$ to $4d$

Assertion (A) is TRUE.

Evaluating Reason (R)

Now let's examine the exchange energy claim across the second transition series (Y through Cd):

The concept: periodic trends in ionic radii and nuclear charge.

Ionic radius depends on two competing factors: the number of protons in the nucleus (nuclear charge) and the number of electrons. In the same period, as you move from left to right, the nuclear charge increases while electrons are added to the same principal shell. A higher nuclear charge pulls the electron cloud inward more strongly, shrinking the ionic radius. So the key idea is: greater nuclear charge → smaller ionic radius, for ions with the same number of electron shells.

Now let’s apply this to the given statements.

1. Check the Assertion (A): “The ionic radii of the alkaline earth metals are smaller than those of alkali metals in the same period.”

   In the same period (e.g., Period 3), the alkali metal is sodium (Na) and the alkaline earth metal is magnesium (Mg). The common ions are Na⁺ and Mg²⁺. Both have the same electron configuration as neon (2,8), but Mg²⁺ has 12 protons pulling on 10 electrons, while Na⁺ has only 11 protons. The stronger pull in Mg²⁺ makes its ionic radius smaller. This trend holds across all periods: Li⁺ > Be²⁺, Na⁺ > Mg²⁺, K⁺ > Ca²⁺, etc. So Assertion (A) is true.

2. Check the Reason (R): “Alkali metals have higher nuclear charge than that of the alkaline earth metals.” …

The question tests your understanding of the actinoid series (elements 89–103). Unlike lanthanoids, where 4f electrons are deeply buried, actinoid 5f orbitals are more extended and can participate in covalent bonding, especially early in the series. This leads to several distinctive trends.

1. Statement A: "The separation of actinoid elements from each other is difficult"

   This is correct. Actinoids have very similar chemical properties (e.g., all form +3 ions, and early ones also show +4, +5, +6 states). Their ionic radii decrease slowly across the series (actinoid contraction), making separation by ion exchange or solvent extraction challenging — much like lanthanoids, but with added complexity from multiple oxidation states.

2. Statement B: "The covalency of the compounds of actinoid metals decrease from left to right along actinoid series"

   This is incorrect. Covalency increases from left to right. Early actinoids (Th, Pa, U) have more diffuse 5f orbitals that overlap poorly with ligands, so bonding is more ionic. As you move right, the 5f orbitals contract and become more like 4f orbitals, but they also become more available for covalent interaction due to better energy matching with ligand orbitals. Actually, the trend is: covalency is highest in the middle of the series (around Pu, Am) and then decreases again? Let’s be precise: For actinoids, the 5f orbitals are more extended than 4f, so early members show some covalency, but the maximum covalency occurs around the middle (e.g., in Np, Pu, Am) because the 5f orbitals are still relatively diffuse and the energy gap to ligand orbitals is smallest. However, the statement says "decrease from left to right" — that is false; it generally increases then decreases, but certainly not a monotonic decrease. The correct trend is that covalency increases initially, then decreases. So B is wrong.

3. Statement C: "The compounds of Lawrencium are most covalent" …