Question:

Number of complexes with even number of electrons in \(t_{2g}\) orbitals is: \[[\text{Fe(H}_2\text{O)}_6]^{2+}, \, [\text{Co(H}_2\text{O)}_6]^{2+}, \, [\text{Co(H}_2\text{O)}_6]^{3+}, \, [\text{Cu(H}_2\text{O)}_6]^{2+}, \, [\text{Cr(H}_2\text{O)}_6]^{2+}\]

Updated On: Nov 3, 2025
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The Correct Option is B

Approach Solution - 1

This problem requires identifying complexes with an even number of electrons in their \(t_{2g}\) orbitals. Let's evaluate each given complex:

  1. \([\text{Fe(H}_2\text{O)}_6]^{2+}:\)
    • Iron in this complex is Fe(II), meaning it has a 3d6 electron configuration.
    • In an octahedral field, the electronic configuration for Fe2+ is \( (t_{2g})^4 (e_g)^2 \).
    • The number of \(t_{2g}\) electrons is 4, which is even.
  2. \([\text{Co(H}_2\text{O)}_6]^{2+}:\)
    • Cobalt in this complex is Co(II), meaning it has a 3d7 electron configuration.
    • In an octahedral field, the configuration is \( (t_{2g})^5 (e_g)^2 \).
    • The number of \(t_{2g}\) electrons is 5, which is odd.
  3. \([\text{Co(H}_2\text{O)}_6]^{3+}\):
    • Cobalt in this complex is Co(III), which has a 3d6 configuration.
    • In an octahedral field, the configuration is \( (t_{2g})^6 (e_g)^0 \).
    • The number of \(t_{2g}\) electrons is 6, which is even.
  4. \([\text{Cu(H}_2\text{O)}_6]^{2+}:\)
    • Copper in this complex is Cu(II), which has a 3d9 configuration.
    • In an octahedral field, the configuration is \( (t_{2g})^6 (e_g)^3 \).
    • The number of \(t_{2g}\) electrons is 6, which is even.
  5. \([\text{Cr(H}_2\text{O)}_6]^{2+}:\)
    • Chromium in this complex is Cr(II), which has a 3d4 configuration.
    • In an octahedral field, the configuration is \( (t_{2g})^3 (e_g)^1 \).
    • The number of \(t_{2g}\) electrons is 3, which is odd.

From this analysis, the complexes with an even number of electrons in \(t_{2g}\) orbitals are \([\text{Fe(H}_2\text{O)}_6]^{2+}, [\text{Co(H}_2\text{O)}_6]^{3+}, and [\text{Cu(H}_2\text{O)}_6]^{2+}\). Therefore, there are 3 such complexes.

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Approach Solution -2

To determine the number of complexes with an even number of electrons in $t_{2g}$ orbitals, we calculate the electronic configuration of the central metal ion in each complex:
$[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$: Fe$^{2+}$ has $(3d^6)$ configuration. In an octahedral field:
\[t_{2g}^4e_g^2 \quad (4 \, \text{electrons in } t_{2g})\]
Even number of electrons in $t_{2g}$.
$[\text{Co}(\text{H}_2\text{O})_6]^{2+}$: Co$^{2+}$ has $(3d^7)$ configuration. In an octahedral field:
\[t_{2g}^5e_g^2 \quad (5 \, \text{electrons in } t_{2g})\]
Odd number of electrons in $t_{2g}$.
$[\text{Co}(\text{H}_2\text{O})_6]^{3+}$: Co$^{3+}$ has $(3d^6)$ configuration. In an octahedral field:
\[t_{2g}^6e_g^0 \quad (6 \, \text{electrons in } t_{2g})\]
Even number of electrons in $t_{2g}$.
$[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$: Cu$^{2+}$ has $(3d^9)$ configuration. In an octahedral field:
\[t_{2g}^6e_g^3 \quad (6 \, \text{electrons in } t_{2g})\]
Even number of electrons in $t_{2g}$.
$[\text{Cr}(\text{H}_2\text{O})_6]^{2+}$: Cr$^{2+}$ has $(3d^4)$ configuration. In an octahedral field:
\[t_{2g}^3e_g^1 \quad (3 \, \text{electrons in } t_{2g})\]
Odd number of electrons in $t_{2g}$.
Complexes with even number of electrons in $t_{2g}$ orbitals are:
\[[\text{Fe}(\text{H}_2\text{O})_6]^{2+}, \, [\text{Co}(\text{H}_2\text{O})_6]^{3+}, \, [\text{Cu}(\text{H}_2\text{O})_6]^{2+}\]
Final Answer: 3 complexes.

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