Problem 120 - A Complete Resource Book in Chemistry for JEE Main

Question

Amongst the following, the lowest degree of paramagnetism per mole of the compound at \(298 \mathrm{~K}\) will be shown by (a) \(\mathrm{MnSO}_{4} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{CuSO}_{4} .5 \mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{FeSO}_{4} .6 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{NiSO}_{4} .6 \mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

CuSO4 · 5H2O has the lowest degree of paramagnetism per mole at 298 K.

Step 1: Understanding Paramagnetism

Paramagnetism in compounds is due to the presence of unpaired electrons. More unpaired electrons result in stronger paramagnetism. To compare paramagnetism, we need to find the number of unpaired electrons in each compound.

Step 2: Analyzing MnSO4 · 4H2O

Manganese in MnSO4 · 4H2O is in the +2 oxidation state, or Mn²⁺. The electronic configuration of Mn is \[ \text{[Ar]} 3d^5 4s^2 \], so Mn²⁺ has the configuration \[ \text{[Ar]} 3d^5 \], which means it has 5 unpaired electrons.

Step 3: Analyzing CuSO4 · 5H2O

Copper in CuSO4 · 5H2O is in the +2 oxidation state, or Cu²⁺. The electronic configuration of Cu is \[ \text{[Ar]} 3d^{10} 4s^1 \], so Cu²⁺ has the configuration \[ \text{[Ar]} 3d^9 \], which means it has 1 unpaired electron.

Step 4: Analyzing FeSO4 · 6H2O

Iron in FeSO4 · 6H2O is in the +2 oxidation state, or Fe²⁺. The electronic configuration of Fe is \[ \text{[Ar]} 3d^6 4s^2 \], so Fe²⁺ has the configuration \[ \text{[Ar]} 3d^6 \], which means it has 4 unpaired electrons.

Step 5: Analyzing NiSO4 · 6H2O

Nickel in NiSO4 · 6H2O is in the +2 oxidation state, or Ni²⁺. The electronic configuration of Ni is \[ \text{[Ar]} 3d^8 4s^2 \], so Ni²⁺ has the configuration \[ \text{[Ar]} 3d^8 \], which means it has 2 unpaired electrons.

Step 6: Comparing Unpaired Electrons

Let's summarize the unpaired electrons for each compound: Mn²⁺ has 5, Cu²⁺ has 1, Fe²⁺ has 4, and Ni²⁺ has 2 unpaired electrons. The compound with the fewest unpaired electrons will show the lowest degree of paramagnetism.

Key Concepts

Unpaired Electrons

In the world of chemistry, unpaired electrons play a pivotal role in dictating the magnetic properties of compounds. When electrons in an atom's orbitals are unpaired, they induce a magnetic moment, making the substance paramagnetic. Essentially, the more unpaired electrons are present, the stronger the paramagnetic property.
This phenomenon is crucial for the study of transition metal compounds. In the compounds mentioned in the exercise, MnSO₄ · 4H₂O, CuSO₄ · 5H₂O, FeSO₄ · 6H₂O, and NiSO₄ · 6H₂O, we need to determine the number of unpaired electrons in the central metal ions to assess their paramagnetic nature. Copper (Cu²⁺) in CuSO₄ · 5H₂O, with just one unpaired electron, exhibits the weakest paramagnetic property among them.

Electronic Configuration

Understanding the electronic configuration of an atom helps us predict its chemical behavior, including its magnetic properties.
For transition metals, this configuration can be a little tricky due to the filling of the 3d orbitals. Typically, these metals have valence electrons in their d and sometimes s orbitals.
Taking the examples from the step-by-step solution:

• Manganese (Mn²⁺) has \[ \text{[Ar]} 3d^5 \], meaning all five d orbitals contain one unpaired electron each.
• Copper (Cu²⁺) has \[ \text{[Ar]} 3d^9 \], indicating one of the d orbitals has a single unpaired electron.
• Iron (Fe²⁺) has \[ \text{[Ar]} 3d^6 \], where four of the electrons remain unpaired.
• Nickel (Ni²⁺) comes with \[ \text{[Ar]} 3d^8 \], with two unpaired electrons.

The degree of paramagnetism can thus be judged by counting these unpaired electrons. Fewer unpaired electrons, as seen in Cu²⁺, lead to a lower degree of paramagnetism.

Transition Metal Compounds

Transition metal compounds are fascinating because of their variable oxidation states and complex electronic configurations. Such properties allow these metals to form diverse compounds with distinct magnetic behaviors.
Transition metals are characterized by the incomplete filling of their d orbitals, which grants unique reactivity and bonding capabilities, distinct from other elements.
In the given exercise, the compounds are sulfates of some common transition metals like manganese, copper, iron, and nickel.

• Manganese's compound MnSO₄ · 4H₂O has a high degree of paramagnetism due to its five unpaired d electrons.
• Copper's compound CuSO₄ · 5H₂O has a much lower degree of paramagnetism with only one unpaired electron.
• Iron's FeSO₄ · 6H₂O and nickel's NiSO₄ · 6H₂O fall somewhere in between.

Through analyzing their electronic configurations, one can predict their magnetic properties. Transition metal chemistry is a testament to how electron configuration shapes the whole structure and properties of compounds.