NCERT Class 12 Chemistry: d and f Block Elements

Electronic Configuration and General Trends of Transition Elements

The d-block elements, also known as transition elements, have valence electrons in their $(n-1)d$ and $ns$ subshells. The general electronic configuration is $(n-1)d^{1-10}ns^{1-2}$. Transition metals exhibit unique characteristics because their $d$ orbitals are partially filled. An exception is Zn, Cd, and Hg, which have completely filled $d^{10}$ configurations in their ground and common oxidation states, and are therefore often not considered true transition metals. Key periodic trends include a progressive decrease in atomic and ionic radii across a period due to poor shielding by $d$-electrons, which increases the effective nuclear charge. This poor shielding effect culminates in the 'Lanthanoid Contraction' in the 5d series, causing their atomic radii to be nearly identical to those of the corresponding 4d series elements (e.g., Zr and Hf).

Key Terms and Chemical Phenomena

Transition Elements

Elements that have partly filled d-orbitals in their ground state or in any of their common oxidation states.

Lanthanoid Contraction

The steady decrease in the atomic and ionic radii of lanthanoids with an increase in atomic number, caused by the poor shielding effect of 4f electrons.

Interstitial Compounds

Compounds formed when small atoms like H, C, or N are trapped inside the crystal lattices of transition metals, making them harder and less malleable.

Disproportionation Reaction

A reaction in which an element in one oxidation state is simultaneously oxidized and reduced, such as the reaction of Mn(VI) in acidic medium.

Calculating Spin-Only Magnetic Moment

1. Determine the oxidation state of the metal ion — Identify the charge on the transition metal ion by analyzing the chemical formula of the coordination complex or salt.
2. Write the electronic configuration — Write the configuration of the neutral atom and then remove electrons (starting from the outermost s orbital, then the d orbital) to find the ion's configuration.
3. Count the number of unpaired electrons (n) — Draw the five d-orbitals using Hund's rule to determine the number of unpaired electrons ($n$).
4. Apply the spin-only formula — Use the formula $\mu = \sqrt{n(n+2)}$ Bohr Magnetons (BM) to calculate the magnetic moment.

Avoid These Common Board Exam Mistakes

• Do not write the electronic configuration of $Cr$ as $[Ar] 3d^4 4s^2$; the correct stable half-filled configuration is $[Ar] 3d^5 4s^1$. Similarly, $Cu$ is $[Ar] 3d^{10} 4s^1$.
• When removing electrons to form cations (e.g., $Fe^{2+}$), always remove electrons from the $4s$ orbital before the $3d$ orbital.
• Lanthanoid contraction explanation must highlight the 'imperfect shielding' of $4f$ electrons, not just 'poor shielding'.
• Do not confuse $d$-block elements with transition elements. Remember that $Zn$, $Cd$, and $Hg$ are $d$-block elements but NOT transition elements because they do not have partially filled d-orbitals in ground or common ionic states.

Practice Questions with Solutions

• Q: Why do transition elements show variable oxidation states? A: Step 1: Analyze the energy difference between $(n-1)d$ and $ns$ orbitals. In transition elements, the energy difference between the $(n-1)d$ subshell and the $ns$ subshell is extremely small. Step 2: Understand electron participation. Because of this tiny energy gap, electrons from both the $ns$ and $(n-1)d$ subshells can participate in chemical bond formation. Step 3: Relate to oxidation states. As a result, as more d-electrons are shared or lost along with s-electrons, the elements exhibit multiple stable oxidation states. Final answer: Transition elements show variable oxidation states because the energy difference between the $(n-1)d$ and $ns$ orbitals is very small, allowing electrons from both subshells to participate in bonding.
• Q: Calculate the spin-only magnetic moment of a $M^{2+}$ ion in aqueous solution if its atomic number is 25. A: Step 1: Write the electronic configuration of the neutral atom. The atomic number is 25, which corresponds to Manganese (Mn). Its ground-state configuration is $[Ar] 3d^5 4s^2$. Step 2: Write the configuration of the $M^{2+}$ ion. To form $Mn^{2+}$, remove 2 electrons from the outermost $4s$ orbital. The resulting configuration is $[Ar] 3d^5$. Step 3: Determine the number of unpaired electrons ($n$). A $3d^5$ subshell has 5 d-orbitals, each occupied by a single electron according to Hund's rule. Thus, $n = 5$. Step 4: Calculate the magnetic moment using the spin-only formula. $\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM. Final answer: The spin-only magnetic moment of the $M^{2+}$ ion is 5.92 BM.
• Q: What is Lanthanoid Contraction and what are its consequences? A: Step 1: Define Lanthanoid Contraction. It is the progressive, steady decrease in the atomic and ionic radii of the lanthanoid elements (from La to Lu) as the atomic number increases. Step 2: Explain the cause. This contraction is due to the poor shielding effect of $4f$ electrons, which cannot effectively screen the increasing nuclear charge, pulling the outer electrons closer. Step 3: State the consequences. 1. The atomic and ionic sizes of the 4d (second transition series) and 5d (third transition series) elements become almost identical (e.g., Zr and Hf). 2. Separating lanthanoids in their pure states becomes difficult due to very similar chemical properties. 3. The basic strength of hydroxides decreases from $La(OH)_3$ to $Lu(OH)_3$. Final answer: Lanthanoid contraction is the steady size decrease along the lanthanoid series due to poor $4f$ shielding. Its primary consequence is that 4d and 5d transition elements share nearly identical atomic radii and chemical behaviors.
• Q: Why is $Cr^{2+}$ reducing while $Mn^{3+}$ is oxidizing, even though both have $d^4$ configurations? A: Step 1: Analyze the reduction of $Cr^{2+}$. When $Cr^{2+}$ acts as a reducing agent, it loses an electron to become $Cr^{3+}$ ($d^3$ configuration). In water, this corresponds to a stable half-filled $t_{2g}^3$ level of the $d$-orbitals, which is highly stable. Step 2: Analyze the oxidation of $Mn^{3+}$. When $Mn^{3+}$ acts as an oxidizing agent, it gains an electron to become $Mn^{2+}$ ($d^5$ configuration). A $d^5$ configuration represents a highly stable half-filled d-subshell. Step 3: Draw the conclusion. Therefore, $Cr^{2+}$ readily oxidizes to $Cr^{3+}$ (reducing agent), whereas $Mn^{3+}$ readily reduces to $Mn^{2+}$ (oxidizing agent). Final answer: $Cr^{2+}$ is reducing because oxidizing to $Cr^{3+}$ yields a stable $t_{2g}^3$ configuration, while $Mn^{3+}$ is oxidizing because reducing to $Mn^{2+}$ yields a stable $3d^5$ configuration.

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