The total number of unpaired electrons present in [Co(NH3)6]Cl2 and [Co(NH3)6]Cl3 is ....... .

Q: Why does $NH_3$ act as a strong field ligand for $Co^{2+}$?

While $NH_3$ is typically a borderline ligand, the high charge density and small size of the $Co^{2+}$ ion cause a stronger electrostatic interaction. This leads to a larger crystal field splitting ($\Delta_o$) that overcomes the pairing energy, forcing the electrons to pair up.

Q: What is the primary difference between $Co^{2+}$ and $Co^{3+}$ in these complexes?

The primary difference is the oxidation state, which changes the number of $d$-electrons. $Co^{2+}$ has a $3d^7$ configuration, while $Co^{3+}$ has a $3d^6$ configuration. This directly impacts how electrons are distributed in the split $d$-orbitals.

Q: How do we know the coordination geometry is octahedral?

Both complexes have a coordination number of 6, as indicated by the six $NH_3$ ligands inside the square brackets. A coordination number of 6 almost always corresponds to an octahedral geometry.

Q: Why do we ignore the chloride ions when applying CFT?

The chloride ions are counter ions located outside the coordination sphere. They balance the overall charge of the complex but do not participate in the coordinate covalent bonding with the central metal ion, so they don't affect the $d$-orbital splitting.

Q: Can a complex with zero unpaired electrons be attracted to a magnet?

No, a complex with zero unpaired electrons is diamagnetic, meaning it is weakly repelled by a magnetic field. Only complexes with one or more unpaired electrons are paramagnetic and attracted to a magnetic field.

Question

The total number of unpaired electrons present in $[Co(NH_3)_6]Cl_2$ and $[Co(NH_3)_6]Cl_3$ is ....... .

Sigma Percentile · Accepted Answer

### The Enigma of Coordination Complexes

Coordination chemistry is a fascinating realm where the rules of standard covalent bonding are bent, and the behavior of electrons is dictated by the geometric arrangement of surrounding ligands. In this problem, we are presented with two seemingly similar cobalt complexes: $[Co(NH_3)_6]Cl_2$ and $[Co(NH_3)_6]Cl_3$. Our mission is to determine the total number of unpaired electrons across both complexes.

To crack this, we need a systematic game plan. First, we must determine the oxidation state of the central cobalt atom in each complex. Second, we write down the corresponding $d$-electron configuration. Finally, we apply **Crystal Field Theory (CFT)** to understand how the $d$-orbitals split and how the electrons populate these split energy levels.

### Decoding the First Complex: Cobalt(II)

Let's analyze the first complex, $[Co(NH_3)_6]Cl_2$. The coordination sphere contains six ammonia ($NH_3$) molecules, which are neutral ligands. Outside the sphere, we have two chloride ($Cl^-$) counter ions, each carrying a $-1$ charge. To maintain overall neutrality, the central cobalt atom must have an oxidation state of $+2$.

Cobalt has an atomic number of 27, giving it a ground-state electronic configuration of $[Ar] 4s^2 3d^7$. When it loses two electrons to form the $Co^{2+}$ ion, it loses them from the outermost $4s$ orbital first. Thus, the configuration for $Co^{2+}$ becomes $[Ar] 3d^7$.

### The Crystal Field Splitting of Cobalt(II)

Now comes the critical part: applying Crystal Field Theory. In an octahedral field, the five degenerate $d$-orbitals split into two sets: a lower energy $t_{2g}$ set (three orbitals) and a higher energy $e_g$ set (two orbitals). The energy gap between them is denoted by $\Delta_o$.

Here is where many students fall into a trap. Ammonia is generally considered a borderline or weak field ligand. However, due to the relatively small size and high charge density of the $Co^{2+}$ ion in this specific coordination environment, $NH_3$ acts as a **strong field ligand**. This means the crystal field splitting energy ($\Delta_o$) is greater than the pairing energy ($P$).

Because $\Delta_o > P$, it is energetically more favorable for electrons to pair up in the lower $t_{2g}$ orbitals rather than jump to the higher $e_g$ orbitals. We have 7 electrons to place. The first 6 electrons will completely fill the $t_{2g}$ level, pairing up. The 7th electron has no choice but to occupy the higher $e_g$ level.

This gives us an electronic configuration of $t_{2g}^6 e_g^1$. Counting the unpaired electrons, we find exactly **$1$ unpaired electron** in this complex.

### Decoding the Second Complex: Cobalt(III)

Moving on to the second complex, $[Co(NH_3)_6]Cl_3$. The setup is identical, except we now have three chloride counter ions outside the coordination sphere. This forces the central cobalt atom to adopt an oxidation state of $+3$.

Starting again from the ground state of cobalt ($[Ar] 4s^2 3d^7$), losing three electrons (two from $4s$ and one from $3d$) leaves us with the $Co^{3+}$ ion, which has a configuration of $[Ar] 3d^6$.

### The Crystal Field Splitting of Cobalt(III)

For the $Co^{3+}$ ion, ammonia unequivocally acts as a strong field ligand. The higher positive charge on the metal center pulls the ligands closer, increasing the electrostatic interaction and resulting in a very large crystal field splitting ($\Delta_o$).

Once again, $\Delta_o > P$. We have 6 electrons to distribute. Because the energy gap is so large, all 6 electrons will pair up in the lower energy $t_{2g}$ orbitals. The $e_g$ orbitals remain completely empty.

This results in an electronic configuration of $t_{2g}^6 e_g^0$. Since all electrons are paired, there are **$0$ unpaired electrons** in this complex.

### The Final Tally and Magnetic Implications

We have successfully analyzed both complexes. The first complex, $[Co(NH_3)_6]Cl_2$, contains $1$ unpaired electron. The second complex, $[Co(NH_3)_6]Cl_3$, contains $0$ unpaired electrons.

Adding them together, the total number of unpaired electrons is $1 + 0 = 1$.

Beyond just a number, this electron distribution dictates the physical properties of the compounds. Because $[Co(NH_3)_6]Cl_2$ possesses an unpaired electron, it will interact with external magnetic fields, making it **paramagnetic**. Conversely, $[Co(NH_3)_6]Cl_3$, with all its electrons neatly paired, will be weakly repelled by magnetic fields, making it **diamagnetic**. Always pay close attention to the oxidation state of the metal, as it can drastically alter the behavior of the surrounding ligands!

Animated Solution for Chemistry - Coordination Compounds: The total number of unpaired electrons present in $[C o (N H_{3})_{6}] C l_{2}$ and $[C o (N H_{3})_{6}] C l_{3}$ is ....... .

Enter Numerical Value:

Visualized Solution

$[C o (N H_{3})_{6}] C l_{2} & [C o (N H_{3})_{6}] C l_{3}$

$Δ_{o} & Pairing$

$[C o (N H_{3})_{6}] C l_{2} Setup$

$C o^{2 +} (3 d^{7}) Splitting$

$[C o (N H_{3})_{6}] C l_{3} Setup$

$C o^{3 +} (3 d^{6}) Splitting$

$Σ e^{-}$

$μ = n (n + 2)$

The Sigma Insight: Bonding and Crystal field

The Enigma of Coordination Complexes

Decoding the First Complex

The Crystal Field Splitting of Cobalt(II)

Decoding the Second Complex

The Crystal Field Splitting of Cobalt(III)

The Final Tally and Magnetic Implications

Similar Questions

The total number of unpaired electrons present in the complex $K_{3} [C r (o x a l a t e)_{3}]$ is ........... .

Which one of the following metal complexes is most stable?

Choose the correct statement(s) among the following :

An aqueous solution of $NiCl_{2}$ was heated with excess sodium cyanide in presence of strong oxidising agent to form $[Ni(CN)_{6}]^{2 -}$ . The total change in number of unpaired electrons on metal centre is …… .

Number of bridging CO ligands in $[M n_{2} (C O)_{10}]$ is ............ .

Match the electronic configurations in List-I with appropriate metal complex ions in List-II and choose the correct option. [Atomic Number: Fe = 26, Mn = 25, Co = 27]

List-I

List-II

The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is

The number of bridging CO ligand(s) and $Co—Co$ bond(s) in $Co_{2} (CO)_{8}$ , respectively are

The metal $d$ -orbitals that are directly facing the ligands in $K_{3} [C o (C N)_{6}]$ are

Animated Solution for Chemistry - Coordination Compounds: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The total number of unpaired electrons present in [Co(NH3​)6​]Cl2​ and [Co(NH3​)6​]Cl3​ is ....... .