Consider low spin complex $[C o (N H_{3})_{6}]^{2 +}$ and $[C o (N H_{3})_{4}]^{2 +}$ . If $△_{t} = \frac{4}{9} △_{o}$ and difference in crystal field stabilization energy is $\frac{x}{3} △_{o}$ between both the complexes then find the value of x

Open in App

Solution

For the low spin octahedral complex $[C o (N H_{3})_{6}]^{2 +}$ CFSE will be, $= (- 0.4 \times 6 + 0.6 \times 1) △_{o} = - 1.8 △_{o}$
For tetrahedral complex ([Co(NH_3)_4]^{2+}\) CFSE will be $= (- 0.6 \times 4 + 0.4 \times 3) △_{t} = - 1.2 △_{t}$
So, difference $= - 1.2 △_{t} + 1.8 △_{o}$
$= - 1.2 \times \frac{4}{9} △_{o} + 1.8 △_{o}$
$= \frac{(- 1.6 + 5.4) △_{o}}{3}$
$= \frac{3.8 △_{o}}{3}$

Suggest Corrections

Similar questions

[C o (N H_{3})_{4}]^{2 +}

[C o (N H_{3})_{6}]^{2 +}

[C o (N H_{3})_{6}]^{3 +}

x Δ_{o}

∣ x ∣

[C o (N H_{3})_{6}]^{3 +}

x Δ_{o}

∣ x ∣

d^{4}

d^{4}

Join BYJU'S Learning Program