Find the electronic configurations

• $\text{Mn}^{2+}$: Mn has $Z=25$. $\text{Mn}: [\text{Ar}]\,3d^5 4s^2$ $\text{Mn}^{2+}: [\text{Ar}]\,3d^5$

• $\text{Ni}^{2+}$: Ni has $Z=28$. $\text{Ni}: [\text{Ar}]\,3d^8 4s^2$ $\text{Ni}^{2+}: [\text{Ar}]\,3d^8$

Examine statement (I): both the complexes can be high spin

• Octahedral $\text{Mn}^{2+}$ is $d^5$.
• For a $d^5$ octahedral complex, with weak-field ligands it is high spin: $t_{2g}^3 e_g^2$
• Tetrahedral splitting is generally small, so tetrahedral complexes are almost always high spin.
• Thus tetrahedral $\text{Ni}^{2+}$ ($d^8$) is high spin.

Therefore, (I) is correct.

Examine statement (II): Ni(II) complex can very rarely be low spin

• For tetrahedral complexes, $\Delta_t$ is small.
• Hence electron pairing is usually not favored; tetrahedral complexes are generally high spin.
• So a tetrahedral $\text{Ni}^{2+}$ complex being low spin is extremely uncommon / very rare.

Therefore, (II) is correct.

Examine statement (III): with strong field ligands, Mn(II) complexes can be low spin

• Octahedral $\text{Mn}^{2+}$ is $d^5$.
• With strong-field ligands, pairing can occur: $t_{2g}^5 e_g^0$
• This is the low-spin configuration.

Therefore, (III) is correct.

Examine statement (IV): aqueous solution of Mn(II) ions is yellow in colour

• In aqueous solution, $\text{Mn}^{2+}$ exists mainly as $[\text{Mn}(H_2O)_6]^{2+}$.
• This ion is very pale pink, not yellow.

Therefore, (IV) is incorrect.

Select the correct option

Correct statements are: $\boxed{(I), (II), (III)}$

Hence the correct option is: $\boxed{\text{D}}$