Question

Low spin complean of d6- cation in an octahedral
field will have the following energy
(a) - 12/5 delta +p(b) -12/5delta +3p (c) - 2/5delta +2p(d) - 2/5delta +p​

Answer 1

→ Low spin complex of d6 cation having

→ Δ0 >

→ P.E

→ Configuration is t2g6eg0 and 3 electron are paired in t2g

→ Orbital

==> -2 / 5 Δ0 x 6 + 3P

==> -12 / 5 + 3p

So , Option [B] Is Correct_!!

Answer 2

Answer:

option b is correct

Explanation:

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

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.

$\begin{gathered}\\\end{gathered}$

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,

$\begin{gathered}\\\end{gathered}$

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

Here,

N Denotes No. of paired electrons.

Simplifying,

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

$\begin{gathered}\\\end{gathered}$

Hence, Option - 2 is correct.