Question

Low spin complex of $\sf d^6$ - cation in an octahedral field will have the following energy :
1. $- \dfrac{12}{5} \Delta{}_o + P$
2. $- \dfrac{12}{5} \Delta{}_o + 3P$
3. $- \dfrac{2}{5} \Delta{}_o + 2P$
4. $- \dfrac{2}{5} \Delta{}_o + P$

Here,
$\sf \Delta{}_o \longrightarrow Crystal \ Field \ Splitting \ Energy \\ \sf P \longrightarrow Electron \ Pairing \ Energy$ ​

Answer 1

Answer:

Low spin complex of d6 cation having Δ0> P.E

Configuration is t2g6eg0 and 3 electron are paired in t2g orbital

=5−2Δ0×6+3P

=5−12Δ0+3P

Hence, the correct option is A.

Answer 2

Answer:

$\bullet\;\boxed{\sf CFSE= -\dfrac{12}{5}\;\Delta_{o}+3P}$

Explanation:

$\rule{300}{1.5}$

As it is given that it is a low spin complex so, here the ligands are strong field ligands hence the Crystal field splitting energy will be greater than the Electron pairing energy i.e. Δ₀ > P .

Now, the d⁶ splits into t₂g and eg orbitals, and the energy difference is  - 2/5 Δ₀ and 3/5 Δ₀ respectively.

Now see the diagram, as it is strong field ligands pairing will occur in t₂g orbitals, and electrons enters eg orbitals only when t₂g are completely filled.

Before, we start finding the energy we must know that the electrons are paired so, there will be addition of the P (Electron Pairing Energy).

So, lets find the CFSE,

$\longrightarrow\sf CFSE =\Bigg(-\dfrac{2}{5}t_{2g}\;e^{-}+\dfrac{3}{5}e_{g}\;e^{-}\Bigg)\Delta_{0}+ NP$

Here,

• N Denotes No. of paired electrons.

Simplifying,

$\longrightarrow\sf CFSE =\Bigg(-\dfrac{2}{5}\times 6+\dfrac{3}{5}\times 0\Bigg)\Delta_{0}+ 3\times P\\\\\\\\\longrightarrow\sf CFSE =\Bigg(-\dfrac{12}{5}+0\Bigg)\Delta_{0}+ 3P\\\\\\\\\longrightarrow\sf CFSE =\Bigg(-\dfrac{12}{5}\Bigg)\Delta_{0}+ 3P\\\\\\\\\longrightarrow\large{\underline{\boxed{\red{\sf CFSE =-\dfrac{12}{5}\;\Delta_{0}+ 3P}}}}$

Hence, Option - 2 is correct!

$\rule{300}{1.5}$