Problem 66 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Match each coordination compound in List-I with an appropriate pair of characteristics from List- II and select the correct answer using the code given below the lists. \(\left\\{\right.\) en \(=\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2} ;\) atomic numbers : \(\mathrm{Ti}=22 ; \mathrm{Cr}=24 ; \mathrm{Co}=27 ; \mathrm{Pt}=\) \(\quad 78\\}\) List-I (A) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}\) (B) \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right]\left(\mathrm{NO}_{3}\right)_{2}\) (C) \(\left[\mathrm{Pt}(\mathrm{en})\left(\mathrm{NH}_{3}\right) \mathrm{Cl}\right] \mathrm{NO}_{3}\) (D) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{NO}_{3}\right)_{2}\right] \mathrm{NO}_{3}\) ListII (p) Paramagnetic and exhibits ionisation isomerism (q) Diamagnetic and exhibits cistrans isomerism (r) Paramagnetic and exhibits cistrans isomerism \((\mathrm{s})\) Diamagnetic and exhibits ionisation isomerism A B C D (a) (s) (q) (r) (p) (b) (r) (p) (s) (q) (c) (q) (p) (r) (s) (d) (p) (r) (s) (q)

Step-by-Step Solution

Answer

The correct match is option (d): (p), (r), (s), (q).

Step 1: Determine the Magnetic Properties

To determine if a compound is paramagnetic or diamagnetic, we need to analyze the number of unpaired electrons in the central metal's electronic configuration.- (A) \([\text{Cr}(\text{NH}_3)_4\text{Cl}_2]\text{Cl}\): Cr has an electronic configuration of \([\text{Ar}] 3d^5 4s^1\), leaving 4 unpaired electrons after complex formation, making it paramagnetic.- (B) \([\text{Ti}(\text{H}_2\text{O})_5\text{Cl}](\text{NO}_3)_2\): Ti has an electronic configuration of \([\text{Ar}] 3d^2 4s^2\), with 2 d-electrons. After complex formation, it has no unpaired electrons, making it diamagnetic.- (C) \([\text{Pt}( ext{en})(\text{NH}_3)\text{Cl}](\text{NO}_3)\): Pt has an electronic configuration of \([\text{Xe}] 4f^{14} 5d^{9} 6s^1\), but after formation of the \([\text{Pt}]\) complex, no unpaired electrons remain, making it diamagnetic.- (D) \([\text{Co}(\text{NH}_3)_4(\text{NO}_3)_2]\text{NO}_3\): Co has an electronic configuration of \([\text{Ar}] 3d^7 4s^2\), forming a low-spin complex where unpaired electrons remain, so it is paramagnetic.

Step 2: Determine Isomerism Type

Next, we identify the type of isomerism each compound can exhibit:- (A) \([\text{Cr}(\text{NH}_3)_4\text{Cl}_2]\text{Cl}\): This compound can exhibit ionisation isomerism because it can exchange chloride ions inside and outside the coordination sphere.- (B) \([\text{Ti}(\text{H}_2\text{O})_5\text{Cl}](\text{NO}_3)_2\): This compound can exhibit ionisation isomerism due to exchange of counter ions.- (C) \([\text{Pt}( ext{en})(\text{NH}_3)\text{Cl}](\text{NO}_3)\): This compound can exhibit cis-trans isomerism because it contains bidentate and monodentate ligands in its square planar configuration.- (D) \([\text{Co}(\text{NH}_3)_4(\text{NO}_3)_2]\text{NO}_3\): This compound can also exhibit ionisation isomerism due to counter ion exchange (NO3 inside and outside).

Step 3: Match List-I and List-II

Using the deductions from Steps 1 and 2, match each compound from List-I with the correct pair of characteristics from List-II: - (A) Paramagnetic and can exhibit ionisation isomerism matches with (p). - (B) Diamagnetic and can exhibit ionisation isomerism matches with (s). - (C) Diamagnetic and can exhibit cis-trans isomerism matches with (q). - (D) Paramagnetic and can exhibit ionisation isomerism matches with (p).

Step 4: Comparing with Given Options

From the matching in Step 3, we find that: A matches with (p), B matches with (s), C matches with (q), and D matches with (p). The choice in the given options that matches this pattern is option (d): (p), (r), (s), (q).

Key Concepts

Magnetic Properties in Coordination Compounds

When determining the magnetic properties of coordination compounds, it's important to know if a compound is paramagnetic or diamagnetic. This largely depends on the number of unpaired electrons in the metal's electronic configuration.

• **Paramagnetism**: Compounds that have one or more unpaired electrons are paramagnetic. These unpaired electrons cause the compound to be attracted to magnetic fields. For instance, in the compound \( [\text{Cr}(\text{NH}_3)_4\text{Cl}_2]\text{Cl} \), the central chromium ion has four unpaired electrons according to its electronic configuration, making it paramagnetic.
• **Diamagnetism**: Compounds without unpaired electrons are diamagnetic and are slightly repelled by magnetic fields. This is observed in \( [\text{Ti}(\text{H}_2\text{O})_5\text{Cl}](\text{NO}_3)_2 \), where the titanium ion, once it forms the complex, does not have any unpaired electrons, thereby showing diamagnetism.

The magnetic properties are crucial because they impact the physical behavior of coordination compounds, helping predict how they will react in the presence of magnetic fields.

Isomerism in Coordination Compounds

In coordination chemistry, isomerism refers to the different arrangements of atoms in compounds that have the same chemical formula. Let's examine the types of isomerism prominent in coordination compounds:

• **Ionisation Isomerism**: Occurs when ions inside and outside the coordination sphere of a compound exchange places. This was noted in \( [\text{Co}(\text{NH}_3)_4(\text{NO}_3)_2]\text{NO}_3 \), where the nitrate ion can swap positions.
• **Cis-Trans Isomerism**: This is common in square planar or octahedral geometries, where ligands can switch positions to give varying structures like 'cis' (adjacent) or 'trans' (opposite) configurations. The compound \( [\text{Pt}(\text{en})(\text{NH}_3)\text{Cl}](\text{NO}_3) \) can show cis-trans behavior because of its square planar configuration and presence of different ligands, like ethylenediamine (en) and ammonia (NH3).

Recognizing these isomer types helps in understanding the diversity and reactivity of coordination compounds as each arrangement can lead to distinct physical and chemical outcomes.

Electronic Configuration of Transition Metals

Transition metals often present unique properties due to their electronic configurations, particularly because of their d-orbitals.
The knowledge about electronic configurations is vital in determining their chemical behaviors and properties.

• **Chromium (Cr)**: Its common configuration is \([\text{Ar}] 3d^5 4s^1 \). In complexes, it generally forms a high-spin arrangement, allowing several unpaired electrons, hence exhibiting paramagnetism.
• **Titanium (Ti)**: With a typical configuration of \([\text{Ar}] 3d^2 4s^2 \), titanium in coordination compounds like those we examined, could lead to low-spin or even no unpaired electrons, hence often results in diamagnetism.
• **Platinum (Pt)**: Pt's electronic configuration \([\text{Xe}] 4f^{14} 5d^{9} 6s^1 \) allows it to fill its d-orbitals fully in many complexes, thus showing diamagnetism.
• **Cobalt (Co)**: Typically displays \([\text{Ar}] 3d^7 4s^2 \), and depending on the ligand field strength, can exhibit both high-spin and low-spin states, influencing its magnetic properties significantly.

Understanding these configurations helps predict the behavior of these metals when they form coordination complexes, showcasing why they are often used in different catalytic or material applications.