The crystal field stabilization energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of metal ion

Question

The crystal field stabilization energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of metal ion $(M^{Z +})$ are $- 0.8 Δ_{o}$ and $3.87 B M$ respectively identify $(M^{Z +})$ .

Smart Achievers · Accepted Answer

(2)

$H_{2} O (W.F.L) no force pairing hence C.F splitting E = C.F stabilisation energy$

$C o^{+ 2} = 3 d^{7}$ $t_{2 g}^{5} e_{g}^{2}$

$C . F . S . E = (- 0.4 \times 5 + 2 \times 0.6) Δ_{0} = - 0.8 Δ_{0}$

$μ = \sqrt{3 (3 + 2)} = 3.87 B M$

Solution

Q.
The crystal field stabilization energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of metal ion $(M^{Z +})$ are $- 0.8 Δ_{o}$ and $3.87 B M$ respectively identify $(M^{Z +})$ .

1 $T i^{2 +}$

2 $C o^{2 +}$

3 $C r^{3 +}$

4 $M n^{2 +}$

Ans.
(2)

$H_{2} O (W.F.L) no force pairing hence C.F splitting E = C.F stabilisation energy$

$C o^{+ 2} = 3 d^{7}$ $t_{2 g}^{5} e_{g}^{2}$

$C . F . S . E = (- 0.4 \times 5 + 2 \times 0.6) Δ_{0} = - 0.8 Δ_{0}$

$μ = \sqrt{3 (3 + 2)} = 3.87 B M$

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Solution

Q. The crystal field stabilization energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of metal ion (MZ+) are -0.8Δo and 3.87 BM respectively identify (MZ+).

1 Ti2+

2 Co2+

3 Cr3+

4 Mn2+

Ans. (2) H2O (W.F.L) no force pairing hence C.F splitting E = C.F stabilisation energy Co+2=3d7 t2g5 eg2 C.F.S.E=(-0.4×5+2×0.6)Δ0=-0.8Δ0 μ=3(3+2)=3.87 BM