The Crystal Field Stabilization Energy (CFSE) of $$[CoF_3(H_2O)_3]$$ $$(\Delta_0 < P)$$ is:

First, we decide the oxidation state of cobalt in the complex $$[CoF_3(H_2O)_3]$$. Each fluoride ion carries a charge of $$-1$$ while water is neutral. Because no overall charge is written on the bracket, the entire complex is electrically neutral. Hence

$$\text{Oxidation state of Co}+3(-1)=0 \;\Longrightarrow\; \text{Oxidation state of Co}=+3.$$

Neutral cobalt has atomic number $$27$$ and the ground-state configuration $$[Ar]\,3d^7\,4s^2$$. Removing three electrons to give $$\text{Co}^{3+}$$ leaves a $$3d^6$$ ion, so inside the complex the metal ion possesses a $$d^6$$ electronic configuration.

The complex is octahedral (six ligands) and both $$F^-$$ and $$H_2O$$ are weak-field ligands. The condition $$\Delta_0 < P$$ is explicitly stated, meaning the ligand field splitting energy is smaller than the pairing energy. Therefore the complex will adopt the high-spin arrangement: electrons prefer to stay unpaired as far as possible.

In an octahedral field, the five d orbitals split into lower-energy $$t_{2g}$$ (three orbitals) and higher-energy $$e_g$$ (two orbitals). For a high-spin $$d^6$$ ion the electrons are distributed as follows:

$$t_{2g}: \uparrow\,\uparrow\,\uparrow\,\downarrow \quad\Longrightarrow\quad t_{2g}^4$$ $$e_g: \uparrow\,\uparrow \qquad\;\;\;\;\;\;\;\;\;\;\;\;\Longrightarrow\quad e_g^2$$

So,

Number of electrons in $$t_{2g}=4$$,  number in $$e_g=2$$.

Crystal Field Stabilization Energy (CFSE) in an octahedral field is given by the general formula

$$\text{CFSE}=(-0.4\times n_{t_{2g}}+0.6\times n_{e_g})\Delta_0 +(N_p)\,P,$$

where:

• $$n_{t_{2g}}$$ and $$n_{e_g}$$ are the numbers of electrons in the respective sets,
• $$N_p$$ is the additional number of electron pairs created in the complex relative to the free gaseous ion,
• $$P$$ is the pairing energy.

Substituting $$n_{t_{2g}}=4$$ and $$n_{e_g}=2$$ we obtain the splitting contribution:

$$(-0.4\times4+0.6\times2)\Delta_0 =(-1.6+1.2)\Delta_0 =-0.4\Delta_0.$$

Next we compare the number of paired electrons inside the complex with that in the free ion. In a gaseous $$d^6$$ ion the first five electrons occupy separate orbitals and the sixth electron pairs with one of them, giving exactly one pair. Inside the high-spin complex we also found exactly one pair (present in the $$t_{2g}$$ set). Thus

$$N_p=0,$$

and the pairing-energy term contributes nothing:

$$N_p\,P=0.$$

Combining both parts, the total crystal field stabilization energy is simply

$$\text{CFSE}=-0.4\Delta_0.$$

Hence, the correct answer is Option B.