1. Find oxidation state and $d$-electron count of Co in each complex

• $A=[CoF_6]^{3-}$

  Let oxidation state of Co be $x$: $x+6(-1)=-3 \Rightarrow x=+3$ So Co is $Co^{3+}$. Cobalt: $Z=27$, ground state $[Ar]3d^74s^2$. $Co^{3+} = 3d^6$

• $B=[Co(H_2O)_6]^{2+}$ $x+6(0)=+2 \Rightarrow x=+2$ So Co is $Co^{2+}=3d^7$.

• $C=[Co(NH_3)_6]^{3+}$ $x=+3$ So Co is $Co^{3+}=3d^6$.

• $D=[Co(en)_3]^{3+}$ Since $en$ is neutral, $x=+3$ So Co is $Co^{3+}=3d^6$.

2. Decide high spin / low spin

For octahedral complexes:

• $A=[CoF_6]^{3-}$: $F^-$ is weak field  high spin d^6$
• $B=[Co(H_2O)_6]^{2+}$: $H_2O$ is weak field and $Co^{2+}$ generally gives high spin $d^7$
• $C=[Co(NH_3)_6]^{3+}$: $Co^{3+}$ with $NH_3$ has large splitting  low spin d^6$
• $D=[Co(en)_3]^{3+}$: $en$ is stronger than $NH_3$, and $Co^{3+}$ gives low spin $d^6$

3. Write CFSE expressions

For octahedral field:

• High spin $d^6$: $t_{2g}^4e_g^2$ $CFSE = 4(-0.4\Delta_o)+2(+0.6\Delta_o)=-0.4\Delta_o$ Magnitude $=0.4\Delta_o$

• High spin $d^7$: $t_{2g}^5e_g^2$ $CFSE = 5(-0.4\Delta_o)+2(+0.6\Delta_o)=-0.8\Delta_o$ Magnitude $=0.8\Delta_o$

• Low spin $d^6$: $t_{2g}^6e_g^0$ $CFSE = 6(-0.4\Delta_o)=-2.4\Delta_o$ Magnitude $=2.4\Delta_o$

Thus,

• $A$: high spin $d^6$  CFSE magnitude $=0.4\Delta_o$
• $B$: high spin $d^7$  CFSE magnitude $=0.8\Delta_o$
• $C$: low spin $d^6$  CFSE magnitude $=2.4\Delta_o$
• $D$: low spin $d^6$  CFSE magnitude $=2.4\Delta_o$

4. Compare $C$ and $D$ using ligand strength

Both $C$ and $D$ are low-spin $d^6$, so CFSE depends on the size of $\Delta_o$.

From spectrochemical series: $en > NH_3 > H_2O > F^-$

Hence, $\Delta_o(D) > \Delta_o(C)$ So, $CFSE(D) > CFSE(C)$

5. Overall increasing order of CFSE

Now compare all:

• $A$: $0.4\Delta_o$ with weak ligand $F^-$
• $B$: $0.8\Delta_o$ with $H_2O$
• $C$: $2.4\Delta_o$ with $NH_3$
• $D$: $2.4\Delta_o$ with stronger ligand $en$

Therefore, the increasing order is $A < B < C < D$

So the correct option is A.

6. Compare with stored correct answer

Stored correct answer: B ($B < A < C < D$)

But from octahedral CFSE calculation:

• high-spin $d^6$ has CFSE magnitude $0.4\Delta_o$
• high-spin $d^7$ has CFSE magnitude $0.8\Delta_o$

Thus $B$ must have greater CFSE than $A$, not smaller.

So the stored answer appears incorrect.