Problem 60 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Addition of excess aqueous ammonia to a pink coloured aqueous solution of \(\mathrm{MCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(\mathrm{X})\) and \(\mathrm{NH}_{4} \mathrm{Cl}\) 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 \(\mathrm{HCl}\) 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 statement(s) is (are) correct? (a) Addition of silver nitrate to \(Y\) gives only two equivalents of silver chloride (b) The hybridization of the central metal ion in \(Y\) is \(d^{2} s p^{3}\) (c) \(Z\) is a tetrahedral complex (d) When \(X\) and \(Z\) are in equilibrium at \(0^{\circ} \mathrm{C}\), the colour of the solution is pink

Step-by-Step Solution

Answer

Statements (a), (c), and (d) are correct.

Step 1: Interpretation of Complex X

The compound \(\mathrm{MCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(\mathrm{X})\) appears to be a hydrated metal chloride. The magnetic moment \(3.87\) B.M. suggests it has two unpaired electrons.\footnote{The spin-only magnetic moment \(\mu\) is given by \(\mu = \sqrt{n(n+2)}\), where \(n\) is the number of unpaired electrons. \(3.87\, \mathrm{B.M.}\) roughly corresponds to two unpaired electrons.} Thus, \(X\) is probably a transition metal with +2 oxidation state.

Step 2: Analyzing Reaction to Form Y

The formation of \(Y\) involves the addition of excess ammonia. Given \(Y\) is an octahedral complex with a 1:3 electrolyte behavior, \(Y\) likely has three anionic ligands and a neutral ammonia surrounding the metal center. The lack of magnetic moment suggests all paired electrons in \(Y\). The reaction with silver nitrate yielding two equivalents of silver chloride suggests \(\mathrm{Co(NH_3)_6]^{3+}\) because Co(III) forms the Y complex which has no unpaired electrons.

Step 3: Examining Formation and Properties of Z

Formation of a blue complex \(Z\) by adding \(\mathrm{HCl}\) suggests \(Z\) has a different environment for the metal than \(Y\). With a tetrahedral geometry and a spin-only magnetic moment of \(3.87\) B.M., \(Z\) should match the electronic configuration of \(X\), indicating unpaired electrons. \(\mathrm{CoCl_4]^{2-}}\) may form as \(Z\) since tetrahedral cobalt chloride complexes result in a blue color due to split \(\mathrm{d}-\) orbitals.

Step 4: Determining Equilibrium Color

At \(0^\circ \mathrm{C}\), equilibrium between \(X\) and \(Z\) might revert \(Z\) back to the original hydration state of \(X\), which is pink. Therefore, at these conditions, a predominantly pink solution suggests \(X\) reformation.

Step 5: Evaluating Correct Statements

Each statement is assessed based on our reaction understanding: (a) is true because \(Y\) should form a 1:3 electrolyte, thus having 2 chlorides replaceable by silver; (b) is false since \(Y\) has no unpaired electrons at \(d^2sp^3\); (c) is true, \(Z\) forms a tetrahedral shape with observed properties; (d) holds true as pink is the equilibrium color.

Key Concepts

Magnetic Moment

The magnetic moment is an important concept in coordination chemistry, helping us understand the behavior of complexes regarding magnetism.
It is linked to unpaired electrons in a metal complex. The more unpaired electrons present, the higher the magnetic moment. The spin-only magnetic moment (\((\mu)\)) is calculated using the formula:

• \( \mu = \sqrt{n(n+2)} \)

where \(n\) represents the number of unpaired electrons.
For example, if the calculated magnetic moment is 3.87 Bohr Magnetons (B.M.), it suggests the presence of two unpaired electrons. This provides crucial clues about the electronic configuration and oxidation state of the metal ion in a given complex. In our exercise, the magnetic moment of complex \(X\) and \(Z\) is 3.87 B.M., indicating that both have the same number of unpaired electrons. This implies both exhibit paramagnetism, meaning they are attracted to magnetic fields due to these unpaired electrons. On the other hand, complex \(Y\) exhibits a magnetic moment of zero B.M., suggesting all electrons are paired, resulting in a diamagnetic nature, with the complete absence of magnetic attraction.

Complex Formation

Complex formation is the process by which metal ions bind with ligands to form coordination complexes.
This process significantly alters properties like color, solubility, and reactivity. The coordination number and geometry of a complex depend on the nature of the metal ion and the ligands involved. Upon adding ammonia to complex \(X\) in the exercise, an octahedral complex \(Y\) forms. This change involves ligands attaching to the central metal ion, often altering its oxidation state and coordination geometry.

• The octahedral geometry, as seen with \(Y\), typically features six ligands bonded to a central metal ion.
• An octahedral configuration accommodates both neutral ligands like ammonia and anionic ligands like chloride ions, impacting the complex's overall charge and electron configuration.
Complex \(Z\), however, shifts to a different environment upon adding excess hydrochloric acid. It adopts a tetrahedral geometry, only involving coordination with a reduced number of ligands.
Such structural variations among \(X\), \(Y\), and \(Z\) lead to different observable properties and reactivities, demonstrating how complex formation can drastically alter a compound's characteristics in coordination chemistry.

Electronic Configuration

Electronic configuration plays a vital role in understanding the properties and behaviors of coordination complexes. It involves the distribution of electrons among the orbitals of an atom or ion.
In transition metal complexes, electronic configurations are particularly important because they directly affect color, magnetism, and chemical reactivity.For complex \(X\), with a magnetic moment of 3.87 B.M., we deduce that it has two unpaired electrons. Such configurations often correspond to a partially filled \(d\) subshell.
In cobalt(II) complexes, common configurations might involve electrons occupying \(d\) orbitals such as \(d^7\) in a high-spin state. The central metal ion in complex \(Y\) assumes a different electronic configuration due to ligand influence, moving to a low-spin state where all electrons are paired within \(t_{2g}\) orbitals.
This results in no unpaired electrons and, consequently, no magnetic moment.Finally, complex \(Z\) rebuilds a similar configuration to \(X\), restoring its magnetic nature. This change might be due to shifts in field strengths offered by different ligands in tetrahedral versus octahedral arrangements. Understanding configurations helps explain complex ion behaviors during reactions and the color changes observed through electron transitions in visible spectra.