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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) \(\mathrm{NiSO}_{4} .6 \mathrm{H}_{2} \mathrm{O}\)

Short Answer

Expert verified
CuSO4 路 5H2O has the lowest degree of paramagnetism per mole at 298 K.

Step by step solution

01

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.
02

Analyzing MnSO4 路 4H2O

Manganese in MnSO4 路 4H2O is in the +2 oxidation state, or Mn虏鈦. The electronic configuration of Mn is \[ \text{[Ar]} 3d^5 4s^2 \], so Mn虏鈦 has the configuration \[ \text{[Ar]} 3d^5 \], which means it has 5 unpaired electrons.
03

Analyzing CuSO4 路 5H2O

Copper in CuSO4 路 5H2O is in the +2 oxidation state, or Cu虏鈦. The electronic configuration of Cu is \[ \text{[Ar]} 3d^{10} 4s^1 \], so Cu虏鈦 has the configuration \[ \text{[Ar]} 3d^9 \], which means it has 1 unpaired electron.
04

Analyzing FeSO4 路 6H2O

Iron in FeSO4 路 6H2O is in the +2 oxidation state, or Fe虏鈦. The electronic configuration of Fe is \[ \text{[Ar]} 3d^6 4s^2 \], so Fe虏鈦 has the configuration \[ \text{[Ar]} 3d^6 \], which means it has 4 unpaired electrons.
05

Analyzing NiSO4 路 6H2O

Nickel in NiSO4 路 6H2O is in the +2 oxidation state, or Ni虏鈦. The electronic configuration of Ni is \[ \text{[Ar]} 3d^8 4s^2 \], so Ni虏鈦 has the configuration \[ \text{[Ar]} 3d^8 \], which means it has 2 unpaired electrons.
06

Comparing Unpaired Electrons

Let's summarize the unpaired electrons for each compound: Mn虏鈦 has 5, Cu虏鈦 has 1, Fe虏鈦 has 4, and Ni虏鈦 has 2 unpaired electrons. The compound with the fewest unpaired electrons will show the lowest degree of paramagnetism.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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, MnSO鈧 路 4H鈧侽, CuSO鈧 路 5H鈧侽, FeSO鈧 路 6H鈧侽, and NiSO鈧 路 6H鈧侽, we need to determine the number of unpaired electrons in the central metal ions to assess their paramagnetic nature. Copper (Cu虏鈦) in CuSO鈧 路 5H鈧侽, 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 (Mn虏鈦) has \[ \text{[Ar]} 3d^5 \], meaning all five d orbitals contain one unpaired electron each.
  • Copper (Cu虏鈦) has \[ \text{[Ar]} 3d^9 \], indicating one of the d orbitals has a single unpaired electron.
  • Iron (Fe虏鈦) has \[ \text{[Ar]} 3d^6 \], where four of the electrons remain unpaired.
  • Nickel (Ni虏鈦) 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 Cu虏鈦, 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 MnSO鈧 路 4H鈧侽 has a high degree of paramagnetism due to its five unpaired d electrons.
  • Copper's compound CuSO鈧 路 5H鈧侽 has a much lower degree of paramagnetism with only one unpaired electron.
  • Iron's FeSO鈧 路 6H鈧侽 and nickel's NiSO鈧 路 6H鈧侽 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.

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Most popular questions from this chapter

\(\mathrm{Fe}^{3+}\) can be detected by the reaction (1) \(\mathrm{Fe}^{3+}+\left[\mathrm{Fe}^{\mathrm{II}}(\mathrm{CN})_{6}\right]^{4} \longrightarrow \mathrm{Fe}^{\mathrm{III}}\left[\mathrm{Fe}^{\mathrm{ll}}(\mathrm{CN})_{6}\right]\) Deep blue (2) \(\mathrm{Fe}^{3+}+\left[\mathrm{Fe}^{\mathrm{ll}}(\mathrm{CN})_{6}\right]^{3} \longrightarrow \mathrm{Fe}^{\mathrm{Il}}\left[\mathrm{Fe}^{\mathrm{III}}(\mathrm{CN})_{6}\right]\) (3) \(\mathrm{Fe}^{3+}+\mathrm{SCN}^{-} \stackrel{\mathrm{H}_{2} \mathrm{O}}{\mathrm{C}} \begin{gathered}\text { Brown } \\\ {\left[\mathrm{Fe}(\mathrm{SCN})\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}\right]^{2^{+}}}\end{gathered}\) Blood red (a) 1 and 2 only (b) 2 and 3 only (c) 1,2 and 3 (d) 3 only

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