Addition of excess aqueous ammonia to a pink coloured aqueous solution of MCl2. 6H2O (X) and NH4Cl g

Q: Why does the magnetic moment of 3.87 B.M. mean there are 3 unpaired electrons?

The spin-only magnetic moment formula is $\mu = \sqrt{n(n+2)}$ B.M., where $n$ is the number of unpaired electrons. If you plug in $n=3$, you get $\sqrt{3(5)} = \sqrt{15} \approx 3.87$ B.M.

Q: Why does air oxidize Co(II) to Co(III) only when ammonia is added?

Ammonia is a strong field ligand that forms a highly stable, low-spin $d^6$ complex with Co(III). The massive crystal field stabilization energy (CFSE) released makes the oxidation of Co(II) to Co(III) thermodynamically very favorable in the presence of ammonia.

Q: How do we know complex Y is a 1:3 electrolyte?

The problem states it behaves as a 1:3 electrolyte. This means one formula unit dissociates into 4 ions total: 1 complex cation and 3 anions. For an octahedral Co(III) complex with ammonia, the formula must be $[Co(NH_3)_6]Cl_3$, which gives one $[Co(NH_3)_6]^{3+}$ and three $Cl^-$ ions.

Q: Why is complex Z tetrahedral and not square planar?

Complex Z is $[CoCl_4]^{2-}$. Cobalt is in the +2 state ($3d^7$), and chloride is a weak field ligand. Weak field ligands generally do not force electron pairing, leading to a high-spin tetrahedral geometry rather than a square planar one.

Q: How do we know the equilibrium shifts to pink at 0°C?

The formation of the blue $[CoCl_4]^{2-}$ complex from the pink $[Co(H_2O)_6]^{2+}$ complex involves breaking strong Co-O bonds and forming weaker Co-Cl bonds, making the forward reaction endothermic ($\Delta H > 0$). Lowering the temperature favors the exothermic (backward) reaction, shifting the equilibrium back to the pink complex.

Question

Addition of excess aqueous ammonia to a pink coloured aqueous solution of MCl_2. 6H_2O (X) and NH_4Cl gives an octahedral complex Y in the presence of air. In aqueous solution, complex Y behaves as 1 : 3 electrolyte. The reaction of X with excess HCl at room temperature results in the formation of a blue coloured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whereas it is zero for complex Y.
Among the following options, which statements is(are) correct ?

Options:

(A) The hybridization of the central metal ion in Y is $d^2sp^3$

(B) Z is tetrahedral complex

(C) Addition of silver nitrate to Y gives only two equivalents of silver chloride

(D) When X and Z are in equilibrium at $0^\circ	ext{C}$, the colour of the solution is pink

Sigma Percentile · Accepted Answer

The world of coordination chemistry is filled with vibrant colors and fascinating magnetic properties. In this problem, we are taken on a journey through the transformations of a cobalt complex, observing how changes in ligands and oxidation states dramatically alter its physical and chemical behavior. Let's break down this colorful mystery step by step.

### Decoding the Pink Mystery (Complex X)

We are introduced to a pink-colored aqueous solution of a metal chloride, given by the formula $MCl_2 \cdot 6H_2O$. The key to unlocking the identity of the metal lies in its magnetic moment, which is given as $3.87 	ext{ B.M.}$

The spin-only magnetic moment is calculated using the formula:

$\mu = \sqrt{n(n+2)} 	ext{ B.M.}$

Plugging in our value, we find that $n = 3$, meaning there are exactly three unpaired electrons. A pink chloride salt with a $+2$ oxidation state and three unpaired electrons is a classic signature of **Cobalt(II)**, which has a $3d^7$ electron configuration. Therefore, our starting complex X is the hexaquacobalt(II) chloride, $[Co(H_2O)_6]Cl_2$.

### The Aerial Oxidation (Complex Y)

Next, we add excess aqueous ammonia to our pink solution in the presence of air. The oxygen in the air acts as an oxidizing agent, stripping an electron from Cobalt to convert it from $Co^{2+}$ to $Co^{3+}$.

Ammonia is a strong field ligand. It aggressively replaces the water molecules in the coordination sphere, forming a new octahedral complex Y. The problem states that Y behaves as a $1:3$ electrolyte, which perfectly matches the formula $[Co(NH_3)_6]Cl_3$. This complex dissociates into one $[Co(NH_3)_6]^{3+}$ cation and three $Cl^-$ anions.

Let's evaluate the properties of Y to check the given options. Cobalt(III) has a $3d^6$ configuration. Because ammonia is a strong field ligand, it forces all six electrons to pair up in the lower energy $t_{2g}$ orbitals.

$Co^{3+} (3d^6) + 	ext{SFL} \implies t_{2g}^6 e_g^0$

This complete pairing results in a magnetic moment of zero ($\mu = 0$) and a hybridization of **$d^2sp^3$**. This confirms that **Option A is correct**. Furthermore, because there are three ionizable chloride ions outside the coordination sphere, adding silver nitrate will yield three equivalents of $AgCl$, making Option C incorrect.

### The Blue Transformation (Complex Z)

Now, let's return to our original pink complex X and react it with excess $HCl$. The high concentration of chloride ions drives out the water molecules, forming a new blue-colored complex Z.

$[Co(H_2O)_6]^{2+} + 4Cl^- ightleftharpoons [CoCl_4]^{2-} + 6H_2O$

Chloride is a weak field ligand, and the coordination number drops to four, resulting in the formation of the tetrachloridocobaltate(II) ion, $[CoCl_4]^{2-}$.

Let's analyze Z. Cobalt is still in the $+2$ state ($3d^7$). In a tetrahedral crystal field with weak ligands, it retains its three unpaired electrons.

$Co^{2+} (3d^7) + 	ext{WFL} \implies e^4 t_2^3$

This gives us back the magnetic moment of $3.87 	ext{ B.M.}$ The geometry is indeed **tetrahedral** with an $sp^3$ hybridization. This confirms that **Option B is correct**.

### The Temperature Tug-of-War (Equilibrium)

Finally, we must consider the equilibrium between the pink complex X and the blue complex Z. The forward reaction, where water is replaced by chloride ions, involves breaking strong $Co-O$ bonds and forming weaker $Co-Cl$ bonds. This process requires energy, making the forward reaction **endothermic** ($\Delta H > 0$).

According to Le Chatelier's principle, if we lower the temperature to $0^\circ	ext{C}$, the system will try to generate heat by shifting the equilibrium in the **exothermic (backward) direction**.

This means the equilibrium will shift to the left, favoring the formation of the hexaquacobalt(II) ion. As a result, the solution will revert to its original **pink color**. This confirms that **Option D is correct**.

By carefully tracking the oxidation states, the strength of the ligands, and the principles of chemical equilibrium, we have successfully navigated this complex problem. The correct statements are A, B, and D.

Select Answer:

Visualized Solution

Decoding Complex $X$

Formation of Complex $Y$

Properties of $Y$

Formation of Complex $Z$

Properties of $Z$

Equilibrium Shift

Conclusion

The Sigma Insight: Bonding and Crystal field

Decoding the Pink Mystery (Complex X)

The Aerial Oxidation (Complex Y)

The Blue Transformation (Complex Z)

The Temperature Tug-of-War (Equilibrium)

Similar Questions

Which of the following facts about the complex $[C r (N H_{3})_{6}] C l_{3}$ is wrong?

Consider that a $d^{6}$ metal ion ( $M^{2 +}$ ) forms a complex with aqua ligands and the spin only magnetic moment of the complex is $4.90 BM$ . The geometry and the crystal field stabilisation energy of the complex is

The type of hybridisation and magnetic property of the complex $[M n C l_{6}]^{3 -}$ , respectively, are

Two complexes $[C r (H_{2} O)_{6}] C l_{3}$ (A) and $[C r (N H_{3})_{6}] C l_{3}$ (B) are violet and yellow coloured, respectively. The incorrect statement regarding them is

The crystal field stabilisation energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of a metal ion ( $M^{2 +}$ ) are $- 0.8 Δ_{o}$ and $3.87$ BM, respectively. Identify ( $M^{2 +}$ ).

Choose the correct statement(s) among the following :

The pair of metal ions that can given a spin-only magnetic moment of 3.9 BM for the complex $[M (H_{2} O)_{6}] C l_{2}$ , is

Select Answer:

Visualized Solution

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Formation of Complex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y

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