Animated Solution for Chemistry - Coordination Compounds: For octahedral $\text{Mn(II)}$ and tetrahedral $\text{Ni(II)}$ complexes, consider the following statements : (I) Both the complexes can be high spin. (II) $\text{Ni(II)}$ complex can very rarely be low spin. (III) With strong field ligands, $\text{Mn(II)}$ complexes can be low spin. (IV) Aqueous solution of $\text{Mn(II)}$ ions is yellow in colour. The correct statements is

Welcome to a fascinating journey into the heart of Coordination Chemistry! Today, we are going to dissect a conceptual problem that tests our understanding of Crystal Field Theory (CFT), spin states, and the optical properties of transition metal complexes. We are presented with four statements regarding Manganese(II) and Nickel(II) complexes, and our mission is to separate fact from fiction.

The Realm of High Spin

Let's begin with Statement I, which asserts that both octahedral $\text{Mn(II)}$ and tetrahedral $\text{Ni(II)}$ complexes can be high spin. To verify this, we must first look at their electronic configurations. Manganese has an atomic number of 25, making the ${\text{Mn}}^{2+}$ ion a $3d^5$ system. Nickel, with an atomic number of 28, gives us a ${\text{Ni}}^{2+}$ ion that is a $3d^8$ system.

Imagine these ions surrounded by weak field ligands, such as chloride ions (${\text{Cl}}^{-}$). For the $3d^5$ $\text{Mn(II)}$ ion, a weak field means the crystal field splitting energy (${\Delta }_o$) is relatively small—smaller than the pairing energy ($P$). Consequently, the electrons prefer to remain unpaired, occupying all five d-orbitals singly. The complex utilizes its outer $4d$ orbitals for bonding, resulting in an $s{p}^3{d}^2$ hybridized, high spin octahedral complex like $[{\text{MnCl}}_6{]}^{4-}$.

Similarly, for the $3d^8$ $\text{Ni(II)}$ ion in a tetrahedral field, the splitting energy (${\Delta }_t$) is inherently small. It forms an $s{p}^3$ hybridized complex like $[{\text{NiCl}}_4{]}^{2-}$, which is also high spin. Therefore, Statement I is absolutely correct!

The Rarity of Low Spin Tetrahedral Complexes

Statement II claims that tetrahedral $\text{Ni(II)}$ complexes can very rarely be low spin. Is there truth to this?

To answer, we must recall the fundamental relationship between tetrahedral and octahedral splitting energies: ${\Delta }_t\approx \frac{4}{9}{\Delta }_o$. Because the tetrahedral splitting energy is mathematically so small, it is almost universally less than the pairing energy ($P$). This means that in a tetrahedral geometry, electrons will almost always prefer to jump to higher energy orbitals rather than pair up in the lower ones. Consequently, low spin tetrahedral complexes are exceedingly rare across the board. Statement II is indeed correct.

The Power of Strong Field Ligands

Next, we evaluate Statement III: With strong field ligands, $\text{Mn(II)}$ complexes can be low spin.

What happens to our $\text{Mn(II)}$ complex if we introduce a strong field ligand, like cyanide (${\text{CN}}^{-}$)? A strong field ligand induces a massive crystal field splitting, meaning ${\Delta }_o$ becomes significantly greater than the pairing energy ($P$). The five d-electrons of Manganese are now forced to pair up in the lower energy $t_{2g}$ orbitals, leaving the higher energy $e_g$ orbitals completely empty.

This configuration leaves two inner $3d$ orbitals vacant, allowing for $d^{2}s{p}^3$ hybridization. The result is a low spin, inner orbital complex such as $[{\text{Mn(CN)}}_6{]}^{4-}$. Thus, Statement III is perfectly correct.

The True Color of Aqueous Mn(II)

Finally, let's examine Statement IV, which boldly claims that the aqueous solution of $\text{Mn(II)}$ ions is yellow.

When $\text{Mn(II)}$ is dissolved in water, it forms the hexaaquamanganese(II) complex, $[{\text{Mn(H}}_2{\text{O)}}_6{]}^{2+}$. Water acts as a weak field ligand, so this remains a high spin $d^5$ system with five unpaired electrons, all possessing parallel spins.

For a complex to absorb visible light via a d-d transition, an electron must jump to a higher energy d-orbital. However, because all orbitals are singly occupied with parallel spins, any jumping electron would have to flip its spin to pair up. According to quantum mechanics, such "spin-forbidden" transitions are highly improbable. Because of this spin-forbidden nature, the complex absorbs very little light and appears as a very faint, pale pink color—definitely not yellow! Therefore, Statement IV is incorrect.

Final Verdict: Statements (I), (II), and (III) are correct, making option (c) the right answer.