Question

Consider that d⁶ metal ion (M²⁺) forms a complex with aqua ligands and the spin only magnetic moment of the complex is 4.90 BM. The geometry and the crystal field stabilization energy of the complex is:

Options

• Tetrahedral and −1.6 Δ_t + 1P

• Octahedral and −2.4 Δ₀ + 2P

• Tetrahedral and −0.6 Δ_t

• Octahedral and −1.6 Δ₀

Answer 1

Tetrahedral and −0.6 Δ_t

Explanation:

`sqrt("n"("n" + 2))` = 4.9

⇒ n(n + 2) = 24

⇒ n² + 2n − 24 = 0

⇒ (n + 6) (n − 4) = 0

n = 4

No. of unpaired electrons = 4

C.F.S.E = Crystal field stabilization energy = {E_2F} − {E_iso}

= `{["n"_"2g" (-0.4) + "n"_"eg" (0.6)] Δ_0 + "n"_"p" "p"} - {"n'"_"p" "P"}`

Here n_2g is the number of electrons t_2g orbitals, n_eg is number of electrons in e_g orbitals. n_P is number of electrons pair in the ligand field, and `"n'"_"p"` is the number of electrons pairs in the isotopic field if the complex is Oh complex then the configuration will be `"t"_"2g"^(2.1.1)` eg^1.1 and Here eFSE = 4 × (−0.4 Δ₀) + 20.6 Δ₀ = −0.4 Δ₀ + 1.P. when if it is Td complex process configuration eg^2.1 `"t"_"2g"^(1.11)` and CFSE = {(3 × 0.6) + 3(0.4)} ∞ + 1p} = −0.6 Δ_t + 1.P