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Q4.36:
Write down the number of 3d electrons in each of the following ions: Ti^{2+}, V^{2+}, Cr^{3+}, Mn^{2+}, Fe^{2+}, Fe^{3+}, Co^{2+}, Ni^{2+} and Cu^{2+}. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).

Solution :

Step 1: Electronic Configuration of Neutral Atoms

To determine the number of $3d$ electrons in the specified ions, we first identify the atomic number ($Z$) and the ground-state electronic configuration of the neutral transition metals. [Per the Aufbau Principle and Hund’s Rule].

Element Symbol $Z$ Ground State Configuration ($[Ar] 3d^n 4s^x$)
Titanium Ti 22 $3d^2 4s^2$
Vanadium V 23 $3d^3 4s^2$
Chromium Cr 24 $3d^5 4s^1$
Manganese Mn 25 $3d^5 4s^2$
Iron Fe 26 $3d^6 4s^2$
Cobalt Co 27 $3d^7 4s^2$
Nickel Ni 28 $3d^8 4s^2$
Copper Cu 29 $3d^{10} 4s^1$

Step 2: Calculating $3d$ Electrons in Ions

When ions are formed, electrons are removed first from the $4s$ orbital followed by the $3d$ orbital [Per the principle of energetic stability in transition metal ion formation].

  • $Ti^{2+}$: $[Ar] 3d^2 4s^2 \rightarrow [Ar] 3d^2 \implies$ 2 electrons
  • $V^{2+}$: $[Ar] 3d^3 4s^2 \rightarrow [Ar] 3d^3 \implies$ 3 electrons
  • $Cr^{3+}$: $[Ar] 3d^5 4s^1 \rightarrow [Ar] 3d^3 \implies$ 3 electrons
  • $Mn^{2+}$: $[Ar] 3d^5 4s^2 \rightarrow [Ar] 3d^5 \implies$ 5 electrons
  • $Fe^{2+}$: $[Ar] 3d^6 4s^2 \rightarrow [Ar] 3d^6 \implies$ 6 electrons
  • $Fe^{3+}$: $[Ar] 3d^6 4s^2 \rightarrow [Ar] 3d^5 \implies$ 5 electrons
  • $Co^{2+}$: $[Ar] 3d^7 4s^2 \rightarrow [Ar] 3d^7 \implies$ 7 electrons
  • $Ni^{2+}$: $[Ar] 3d^8 4s^2 \rightarrow [Ar] 3d^8 \implies$ 8 electrons
  • $Cu^{2+}$: $[Ar] 3d^{10} 4s^1 \rightarrow [Ar] 3d^9 \implies$ 9 electrons

Step 3: Crystal Field Splitting in Octahedral Hydrated Ions

In an octahedral field ($[H_2O]$ is a weak field ligand), the five $3d$ orbitals split into lower energy $t_{2g}$ ($d_{xy}, d_{yz}, d_{zx}$) and higher energy $e_g$ ($d_{z^2}, d_{x^2-y^2}$) sets. Occupancy follows Hund’s Rule and the Aufbau Principle.

Crystal Field Splitting ($O_h$ Symmetry) $t_{2g}$ $e_g$
Ion $d^n$ $t_{2g}$ Configuration $e_g$ Configuration
$Ti^{2+}$ $d^2$ $(t_{2g})^2$ $(e_g)^0$
$V^{2+}$ $d^3$ $(t_{2g})^3$ $(e_g)^0$
$Cr^{3+}$ $d^3$ $(t_{2g})^3$ $(e_g)^0$
$Mn^{2+}$ $d^5$ $(t_{2g})^3$ $(e_g)^2$
$Fe^{2+}$ $d^6$ $(t_{2g})^4$ $(e_g)^2$
$Fe^{3+}$ $d^5$ $(t_{2g})^3$ $(e_g)^2$
$Co^{2+}$ $d^7$ $(t_{2g})^5$ $(e_g)^2$
$Ni^{2+}$ $d^8$ $(t_{2g})^6$ $(e_g)^2$
$Cu^{2+}$ $d^9$ $(t_{2g})^6$ $(e_g)^3$

Final Solution: The number of 3d electrons are: $Ti^{2+}(2)$, $V^{2+}(3)$, $Cr^{3+}(3)$, $Mn^{2+}(5)$, $Fe^{2+}(6)$, $Fe^{3+}(5)$, $Co^{2+}(7)$, $Ni^{2+}(8)$, and $Cu^{2+}(9)$. Occupancy in hydrated octahedral complexes is defined by the high-spin $t_{2g}^a e_g^b$ configuration as tabulated above.


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