Question

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

• A. 1
• B. 3
• C. 2
• D. 5

Answer 1

The Correct Option is B

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.