Question

Low spin complex of $$d^6$$-cation in an octahedral field will have the following energy:
($${\mathrm{\Delta }}_0$$= Crystal Field Splitting Energy in an octahedral field,
P = Electron pairing energy)
(a) -12/5 $${\mathrm{\Delta }}_0$$ + P
(b) -12/5 $${\mathrm{\Delta }}_0$$+ 3P
(c) -2/5 $${\mathrm{\Delta }}_0$$ + 2P
(d) -2/5 $${\mathrm{\Delta }}_0$$ + P

Answer 1

Hint: In an octahedral complex, when Δ is large (strong field ligand), the electrons will first fill the lower energy d orbitals before any electrons are placed on the higher energy d orbitals. It is then classified as low spin because there is a minimal number of unpaired electrons.
Complete answer:
Let us first look into the Crystal Field Theory (CFT) and what low spin and high spin complexes really mean in terms of orbital splitting before moving onto the calculation of Crystal Field Stabilisation Energy of the given complex of d⁶-cation.
The CFT categorizes, qualitatively, how the metal d orbitals are filled in crystal field theory after they are split by what the theory proposes are the ligand-induced electron repulsions with the usual Hund's rule and Aufbau Principle being applied.
Some basic facets of the CFT are as follows:
Ligands come in, and their important orbitals interact with the metal d orbitals.
Electrons repel electrons to destabilize certain metal d orbitals. In an octahedral field, these are known as the $$e{\ast }_g$$ orbitals.
Electrons are attracted to the electropositive metal centre to stabilize certain metal d orbitals. In an octahedral field, these are known as the $$t_{2g}$$ orbitals.
Let us now look into complex spin and the specifics of orbital splitting before trying to solve this question
The crystal field splitting energy is called $${\mathrm{\Delta }}_0$$ in an octahedral field for simplicity.
High spin complexes half-fill the lower energy d orbitals first, and then move up to the higher energy d orbitals to half-fill those next, before pairing starts occurring, as those orbitals are so similar in energy to the lower energy orbitals.
Low spin complexes fill the lower energy orbitals completely first, before moving on to the higher energy orbitals, as those orbitals are so much higher in energy.
Now, with all this information, let us try and answer this question:
In a low spin complex with $$d^6$$- configuration, all the 6 electrons will fill up the 3 lower $$t_{2g}$$ orbitals. For,
$$\begin{array}{rl}& Octahedralcomplexenergy={\mathrm{\Delta }}_0(-0.4\times n{t}_{2g}+0.6\times ne{\ast }_g),\\ & \text{where n is the number of electrons in those particular orbitals}\text{.}\end{array}$$
Here, n $$e{\ast }_g$$ = 0 and n $$t_{2g}$$ = 6 so energy is:
CFSE = ( -0.4×6 $${\mathrm{\Delta }}_0$$)
CFSE = -2.4 $${\mathrm{\Delta }}_0$$ = -12$${\mathrm{\Delta }}_0$$/5.
Now, since all the 3 orbitals have paired electrons, therefore pairing energy = 3P.
Thus, the net energy of the given low spin complex is:
-12$${\mathrm{\Delta }}_0$$/5 + 3P
 Therefore, the answer to this question is (b).
Note:
According to crystal field theory, the interaction between a transition metal and ligands arises from the attraction between the positively charged metal cation and the negative charge on the non-bonding electrons of the ligand. The theory is developed by considering energy changes of the five-degenerate d-orbitals upon being surrounded by an array of point charges consisting of the ligands.