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Step-by-Step Solution
1. Identify the Complexes and Their Central Metal Ions
The complexes given are:
1. [V(CN)6]4– (central ion V2+)
2. [Fe(CN)6]4– (central ion Fe2+)
3. [Ru(NH3)6]3+ (central ion Ru3+)
4. [Cr(NH3)6]2+ (central ion Cr2+)
We need to find the spin-only magnetic moment order of these metal ions in their respective complexes.
2. Understand Why These Complexes Are Low-Spin
Cyanide (CN–) is a strong field ligand, and ammonia (NH3) is also typically considered a stronger ligand compared to water or halides. In each octahedral complex, if the ligand is strong enough (i.e., it causes a large splitting $ \Delta_{0} $), pairing of electrons occurs in the lower-energy t2g orbitals, leading to a low-spin configuration.
3. Determine the Electronic Configuration and Number of Unpaired Electrons
• [V(CN)6]4– (V2+ has d3 configuration): In a low-spin complex with three electrons, there are typically 3 unpaired electrons before any additional pairing occurs. Calculations show it remains with 3 unpaired electrons.
Number of unpaired electrons = 3
Spin-only magnetic moment formula:
$ \mu = \sqrt{n(n + 2)} \text{ B.M.} $
where $n$ is the number of unpaired electrons.
Thus,
$ \mu = \sqrt{3(3 + 2)} = \sqrt{15} \text{ B.M.} $
• [Fe(CN)6]4– (Fe2+ has d6 configuration): With CN– being a strong field ligand, all electrons pair up in the lower t2g orbitals, giving a low-spin (paired) arrangement.
Number of unpaired electrons = 0
Therefore,
$ \mu = \sqrt{0(0 + 2)} = 0 \text{ B.M.} $
• [Ru(NH3)6]3+ (Ru3+ often considered d5 or d6 depending on oxidation state; here effectively d5): Because NH3 is a relatively strong field ligand, most of the electrons pair up in the t2g set. Calculations show it ends with 1 unpaired electron in a low-spin arrangement.
Number of unpaired electrons = 1
Hence,
$ \mu = \sqrt{1(1 + 2)} = \sqrt{3} \text{ B.M.} $
• [Cr(NH3)6]2+ (Cr2+ has d4 configuration): In a low-spin scenario with NH3 as the ligand, it is possible to find partial pairing, leaving 2 unpaired electrons.
Number of unpaired electrons = 2
Thus,
$ \mu = \sqrt{2(2 + 2)} = \sqrt{8} \text{ B.M.} $
4. Arrange the Complexes Based on Their Magnetic Moments
Comparing $ \mu $ values in terms of the number of unpaired electrons:
V2+ (d3) : 3 unpaired electrons → $ \sqrt{15} \text{ B.M.} $
Cr2+ (d4) : 2 unpaired electrons → $ \sqrt{8} \text{ B.M.} $
Ru3+ (d5) : 1 unpaired electron → $ \sqrt{3} \text{ B.M.} $
Fe2+ (d6) : 0 unpaired electrons → 0 B.M.
So, the order of magnetic moments is:
$ V^{2+} > Cr^{2+} > Ru^{3+} > Fe^{2+} $
5. Final Answer
The correct order of the spin-only magnetic moment of the metal ions in these low-spin complexes is:
V2+ > Cr2+ > Ru3+ > Fe2+
Step-by-Step Solution
1. Identify the Complexes and Their Central Metal Ions
The complexes given are:
1. [V(CN)6]4– (central ion V2+)
2. [Fe(CN)6]4– (central ion Fe2+)
3. [Ru(NH3)6]3+ (central ion Ru3+)
4. [Cr(NH3)6]2+ (central ion Cr2+)
We need to find the spin-only magnetic moment order of these metal ions in their respective complexes.
2. Understand Why These Complexes Are Low-Spin
Cyanide (CN–) is a strong field ligand, and ammonia (NH3) is also typically considered a stronger ligand compared to water or halides. In each octahedral complex, if the ligand is strong enough (i.e., it causes a large splitting $ \Delta_{0} $), pairing of electrons occurs in the lower-energy t2g orbitals, leading to a low-spin configuration.
3. Determine the Electronic Configuration and Number of Unpaired Electrons
• [V(CN)6]4– (V2+ has d3 configuration): In a low-spin complex with three electrons, there are typically 3 unpaired electrons before any additional pairing occurs. Calculations show it remains with 3 unpaired electrons.
Number of unpaired electrons = 3
Spin-only magnetic moment formula:
$ \mu = \sqrt{n(n + 2)} \text{ B.M.} $
where $n$ is the number of unpaired electrons.
Thus,
$ \mu = \sqrt{3(3 + 2)} = \sqrt{15} \text{ B.M.} $
• [Fe(CN)6]4– (Fe2+ has d6 configuration): With CN– being a strong field ligand, all electrons pair up in the lower t2g orbitals, giving a low-spin (paired) arrangement.
Number of unpaired electrons = 0
Therefore,
$ \mu = \sqrt{0(0 + 2)} = 0 \text{ B.M.} $
• [Ru(NH3)6]3+ (Ru3+ often considered d5 or d6 depending on oxidation state; here effectively d5): Because NH3 is a relatively strong field ligand, most of the electrons pair up in the t2g set. Calculations show it ends with 1 unpaired electron in a low-spin arrangement.
Number of unpaired electrons = 1
Hence,
$ \mu = \sqrt{1(1 + 2)} = \sqrt{3} \text{ B.M.} $
• [Cr(NH3)6]2+ (Cr2+ has d4 configuration): In a low-spin scenario with NH3 as the ligand, it is possible to find partial pairing, leaving 2 unpaired electrons.
Number of unpaired electrons = 2
Thus,
$ \mu = \sqrt{2(2 + 2)} = \sqrt{8} \text{ B.M.} $
4. Arrange the Complexes Based on Their Magnetic Moments
Comparing $ \mu $ values in terms of the number of unpaired electrons:
V2+ (d3) : 3 unpaired electrons → $ \sqrt{15} \text{ B.M.} $
Cr2+ (d4) : 2 unpaired electrons → $ \sqrt{8} \text{ B.M.} $
Ru3+ (d5) : 1 unpaired electron → $ \sqrt{3} \text{ B.M.} $
Fe2+ (d6) : 0 unpaired electrons → 0 B.M.
So, the order of magnetic moments is:
$ V^{2+} > Cr^{2+} > Ru^{3+} > Fe^{2+} $
5. Final Answer
The correct order of the spin-only magnetic moment of the metal ions in these low-spin complexes is:
V2+ > Cr2+ > Ru3+ > Fe2+
