Problem 45
Question
Which one has the highest paramagnetism among the following ions? (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (b) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (c) \(\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (d) \(\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\)
Step-by-Step Solution
Verified Answer
(b) \\(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) has the highest paramagnetism with 4 unpaired electrons.
1Step 1: Determine the Electronic Configuration
Identify the electronic configuration for the central metal ions in each complex. - (a) \([ ext{Cr}^{3+}] : [Ar] 3d^{3}\)- (b) \([ ext{Fe}^{2+}] : [Ar] 3d^{6}\)- (c) \([ ext{Cu}^{2+}] : [Ar] 3d^{9}\)- (d) \([ ext{Zn}^{2+}] : [Ar] 3d^{10}\)
2Step 2: Count the Unpaired Electrons
Determine the number of unpaired electrons in the 3d orbitals for each metal ion. - (a) \([ ext{Cr}^{3+}] ext{ has } 3 ext{ unpaired electrons}\) - (b) \([ ext{Fe}^{2+}] ext{ has } 4 ext{ unpaired electrons}\) - (c) \([ ext{Cu}^{2+}] ext{ has } 1 ext{ unpaired electron}\) - (d) \([ ext{Zn}^{2+}] ext{ has } 0 ext{ unpaired electrons}\)
3Step 3: Analyze Paramagnetism
Paramagnetism is determined by the presence of unpaired electrons. The more unpaired electrons, the greater the paramagnetism. Therefore, compare the number of unpaired electrons in each complex.
4Step 4: Determine the Highest Paramagnetism
Since (b) \([ ext{Fe}^{2+}] ext{ has the most unpaired electrons (4), it is the most paramagnetic among the options.}\)
Key Concepts
Unpaired ElectronsElectronic ConfigurationTransition Metal Complexes
Unpaired Electrons
Unpaired electrons are electrons in an orbital that do not have a corresponding electron with opposite spin. When we discuss paramagnetism, the concept of unpaired electrons becomes pivotal. This is because unpaired electrons create a magnetic moment.
In a magnetic field, substances with unpaired electrons may become attracted due to this magnetic moment, thus exhibiting paramagnetism.
Transition metal ions often have unpaired electrons due to their partially filled d orbitals. Here's how you can assess the number of unpaired electrons in a transition metal complex:
In a magnetic field, substances with unpaired electrons may become attracted due to this magnetic moment, thus exhibiting paramagnetism.
Transition metal ions often have unpaired electrons due to their partially filled d orbitals. Here's how you can assess the number of unpaired electrons in a transition metal complex:
- Identify the electronic configuration of the metal ion.
- Count the electrons present in d orbitals.
- Determine which electrons are unpaired.
Electronic Configuration
The electronic configuration describes how an atom's or ion's electrons are arranged in its orbitals, and this arrangement dictates many of its chemical properties, including magnetism.
In coordination compounds, we often look at the transition metals' electronic configuration to understand their magnetic properties. Transition metals are unique as their d orbitals are filled in a specific way, impacting the number of unpaired electrons.
For instance, the electronic configurations for the complexes in our example are like this:
In coordination compounds, we often look at the transition metals' electronic configuration to understand their magnetic properties. Transition metals are unique as their d orbitals are filled in a specific way, impacting the number of unpaired electrons.
For instance, the electronic configurations for the complexes in our example are like this:
- For \(\text{Cr}^{3+}\) (option a), it has an electronic configuration of [Ar] 3d\(^{3}\), leaving us with 3 unpaired electrons.
- For \(\text{Fe}^{2+}\) (option b), it follows [Ar] 3d\(^{6}\), resulting in 4 unpaired electrons.
- For \(\text{Cu}^{2+}\) (option c), it has [Ar] 3d\(^{9}\), leaving only 1 unpaired electron.
- For \(\text{Zn}^{2+}\) (option d), it is [Ar] 3d\(^{10}\), with 0 unpaired electrons.
Transition Metal Complexes
Transition metal complexes are formed when transition metal ions coordinate with ligands, which are atoms, ions, or molecules that donate pairs of electrons.
These complexes have intriguing electronic properties due to the arrangement of electrons in their d orbitals, significantly influencing their magnetic properties.
These complexes have intriguing electronic properties due to the arrangement of electrons in their d orbitals, significantly influencing their magnetic properties.
- A complex is paramagnetic if it has unpaired electrons, as in the case of \(\left[\text{Fe}\left(\text{H}_2\text{O}\right)_6\right]^{2+}\) with 4 unpaired electrons.
- While, a complex with all paired electrons, such as \(\text{Zn}^{2+}\), is diamagnetic.
Other exercises in this chapter
Problem 43
The number of d electrons in \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) [atomic no. of \(\mathrm{Cr}=24]\) is (a) 2 (b) 3 (c) 4
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The compound which does not show paramagnetism is (a) \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}\) (b) \(\left[\mathrm{Ag}\left(
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According to IUPAC nomenclature sodium nitroprusside is named as (a) sodium pentacyanonitrosyl ferrate (II) (b) sodium pentacyanonitrosyl ferrate (III) (c) sodi
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