Q.(a) Elements of group 16 generally show lower value of first ionisation enthalpy compared to the corresponding periods of group 15. Why? [2]

The key to this question lies not in memorizing numbers but in understanding what the third ionization enthalpy actually measures. It is the energy required to remove the third electron from a gaseous atom — that is, to go from the +2 ion to the +3 ion. So we are really comparing the stability of the M²⁺ ions of these elements. The more stable the M²⁺ ion, the harder it is to pull off another electron, and the higher the third ionization enthalpy.

Now, stability in transition metal ions is heavily influenced by the d-electron configuration. A half-filled d⁵ or fully filled d¹⁰ subshell confers exceptional stability. Let us examine each element’s +2 ion.

1. Vanadium (Z = 23)

   Electronic configuration of V: $[\text{Ar}]\,3d^3\,4s^2$

   V²⁺: remove two 4s electrons → $[\text{Ar}]\,3d^3$

   This is neither half-filled nor fully filled. It is a relatively ordinary configuration.

2. Chromium (Z = 24)

   Cr has a special ground state: $[\text{Ar}]\,3d^5\,4s^1$ (half-filled d gives extra stability).

   Cr²⁺: remove the 4s electron and one 3d electron → $[\text{Ar}]\,3d^4$

   This is not half-filled. The half-filled stability of Cr atom is lost in Cr²⁺.

3. Manganese (Z = 25)

   Mn: $[\text{Ar}]\,3d^5\,4s^2$

   Mn²⁺: remove two 4s electrons → $[\text{Ar}]\,3d^5$

   This is exactly half-filled! The d⁵ configuration is exceptionally stable. Removing a third electron would break this stable half-filled shell, requiring a large amount of energy.

4. Iron (Z = 26)

   Fe: $[\text{Ar}]\,3d^6\,4s^2$

   Fe²⁺: remove two 4s electrons → $[\text{Ar}]\,3d^6$ …

The question tests your understanding of how the lanthanoid contraction affects the chemistry of post-lanthanoid elements, particularly the 4d and 5d transition metals.

Why Zr and Hf are Chemical Twins

Zirconium (Zr, atomic number 40) sits in the second transition series (4d block), while hafnium (Hf, atomic number 72) belongs to the third transition series (5d block). Normally, when you move down a group in the periodic table, atomic and ionic radii increase because you're adding entire electron shells. This is why sodium is larger than lithium, and potassium larger than sodium.

But something unusual happens between the 4d and 5d series. Between Zr and Hf, the periodic table inserts the entire lanthanoid series (elements 57–71) — fourteen f-block elements. As electrons fill the poorly shielding 4f orbitals across the lanthanoids, the effective nuclear charge experienced by outer electrons increases steadily. This pulls all the electron shells inward, causing a cumulative contraction in atomic size. By the time we reach Hf, this lanthanoid contraction has almost exactly compensated for the addition of an extra shell.

The result? The atomic radius of Zr is approximately 160 pm, while Hf is about 159 pm — virtually identical. Their ionic radii ($\text{Zr}^{4+}$ ≈ 72 pm, $\text{Hf}^{4+}$ ≈ 71 pm) are equally similar.

Why Similar Radii Make Separation Difficult

Chemical behavior depends heavily on ionic size and charge. When two elements have:

• The same oxidation states (both commonly +4)
• Nearly identical ionic radii
• The same coordination preferences

...they form compounds with almost indistinguishable properties. Their oxides, halides, and complexes have similar solubilities, crystal structures, and stabilities. Traditional separation methods like fractional crystallization or precipitation rely on differences in these properties, so when the properties are nearly identical, separation becomes extraordinarily challenging.

Historically, chemists struggled for decades to separate Zr and Hf. Even today, industrial separation requires sophisticated techniques like solvent extraction or ion exchange with carefully chosen ligands that can exploit the tiny remaining differences. …

The key to this question lies in understanding how transition metals exhibit variable oxidation states. Unlike main group elements where the outermost s and p electrons are the only ones involved, transition metals can use both the ns and (n−1)d electrons for bonding. The highest possible oxidation state for a given configuration is simply the total number of electrons in the outermost s and d orbitals — because in principle, all of them can be lost.

Let’s examine each option carefully.

1. Option A: $3d^3 4s^2$

   Total electrons in the valence shell = $3 + 2 = 5$. So the maximum oxidation state possible is +5. This is seen in elements like vanadium (V), which indeed shows +5 in compounds like $\text{V}_2\text{O}_5$.

2. Option B: $3d^5 4s^1$

   Total = $5 + 1 = 6$. Maximum oxidation state = +6. Chromium (Cr) has this configuration and shows +6 in $\text{CrO}_3$ and dichromates. Notice that chromium’s actual ground state is $3d^5 4s^1$, not $3d^4 4s^2$, due to the extra stability of a half-filled d-subshell.

3. Option C: $3d^5 4s^2$

   Total = $5 + 2 = 7$. Maximum oxidation state = +7. Manganese (Mn) has this configuration and exhibits +7 in permanganate ion ($\text{MnO}_4^-$). This is the highest among the given options.

4. Option D: $3d^6 4s^2$ …