IIT JAM Exam > IIT JAM Questions > (a) The spin-only magnetic moments of K3[Fe(o...
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(a) The spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 µB and 1.73 µB, respectively.
Write down their ligand field electronic configuration. Justify your answer. Atomic number of Fe and Ru are 26 and 44 respectively.
Write down their ligand field electronic configuration. Justify your answer. Atomic number of Fe and Ru are 26 and 44 respectively.
(b) Draw the structures of NO2+ , NO2 and NO 2- . Arrange them in the increasing order of O–N–O bond angles.
Correct answer is '(a) K 3[FeIII (oxalate)3 ]'. Can you explain this answer?
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(a) The spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 B and ...
(a) K 3[FeIII (oxalate)3 ]
= 5.91 BM
Ligand field electronic configuration =
Ligand field electronic configuration =
In both the complexes ligand is same but one complex is high spin and another complex is low spin. Fe and Ru belongs to same group. On moving down the group crystal field splitting energy (Δ0) increases. hence Ru complex is a low spin complex.
increasing order of ONO bond angles NO2- < NO2 < NO2+
Most Upvoted Answer
(a) The spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 B and ...
(a) Ligand field electronic configuration of K3[Fe(oxalate)3] and justification:
The ligand field electronic configuration of a coordination compound is determined by the pairing of electrons in the metal's d orbitals. The spin-only magnetic moment, μ, is related to the number of unpaired electrons, n, through the formula:
μ = √(n(n+2)) BM
Given that the spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 B and 1.73 B, we can determine the number of unpaired electrons for each case.
For the spin-only magnetic moment of 5.91 B:
5.91 = √(n(n+2))
Squaring both sides:
34.9481 = n^2 + 2n
Rearranging and solving the quadratic equation:
n^2 + 2n - 34.9481 = 0
Using the quadratic formula:
n = (-2 ± √(2^2 - 4(1)(-34.9481))) / (2(1))
n = (-2 ± √(4 + 139.7924)) / 2
n = (-2 ± √143.7924) / 2
n = (-2 ± 11.995) / 2
Taking the positive value:
n = (11.995 - 2) / 2
n = 9.995 / 2
n ≈ 4.998
For the spin-only magnetic moment of 1.73 B:
1.73 = √(n(n+2))
Squaring both sides:
2.9929 = n^2 + 2n
Rearranging and solving the quadratic equation:
n^2 + 2n - 2.9929 = 0
Using the quadratic formula:
n = (-2 ± √(2^2 - 4(1)(-2.9929))) / (2(1))
n = (-2 ± √(4 + 11.9716)) / 2
n = (-2 ± √15.9716) / 2
n = (-2 ± 3.995) / 2
Taking the positive value:
n = (3.995 - 2) / 2
n = 1.995 / 2
n ≈ 0.998
Justification:
In the case of K3[Fe(oxalate)3], the iron (Fe) ion is in the +III oxidation state. The electronic configuration of Fe(III) is [Ar] 3d^5. According to Hund's rule, the five 3d orbitals are filled with electrons singly before any pairing occurs. Therefore, in the case of the spin-only magnetic moment of 5.91 B, there are approximately 5 unpaired electrons, and in the case of the spin-only magnetic moment of 1.73 B, there is approximately 1 unpaired electron.
(b) Structures of NO2, NO2+, and NO2-
- NO2:
The structure of NO2 is linear, with a N=O double bond and a lone electron pair on nitrogen.
- NO2+:
The structure of NO
The ligand field electronic configuration of a coordination compound is determined by the pairing of electrons in the metal's d orbitals. The spin-only magnetic moment, μ, is related to the number of unpaired electrons, n, through the formula:
μ = √(n(n+2)) BM
Given that the spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 B and 1.73 B, we can determine the number of unpaired electrons for each case.
For the spin-only magnetic moment of 5.91 B:
5.91 = √(n(n+2))
Squaring both sides:
34.9481 = n^2 + 2n
Rearranging and solving the quadratic equation:
n^2 + 2n - 34.9481 = 0
Using the quadratic formula:
n = (-2 ± √(2^2 - 4(1)(-34.9481))) / (2(1))
n = (-2 ± √(4 + 139.7924)) / 2
n = (-2 ± √143.7924) / 2
n = (-2 ± 11.995) / 2
Taking the positive value:
n = (11.995 - 2) / 2
n = 9.995 / 2
n ≈ 4.998
For the spin-only magnetic moment of 1.73 B:
1.73 = √(n(n+2))
Squaring both sides:
2.9929 = n^2 + 2n
Rearranging and solving the quadratic equation:
n^2 + 2n - 2.9929 = 0
Using the quadratic formula:
n = (-2 ± √(2^2 - 4(1)(-2.9929))) / (2(1))
n = (-2 ± √(4 + 11.9716)) / 2
n = (-2 ± √15.9716) / 2
n = (-2 ± 3.995) / 2
Taking the positive value:
n = (3.995 - 2) / 2
n = 1.995 / 2
n ≈ 0.998
Justification:
In the case of K3[Fe(oxalate)3], the iron (Fe) ion is in the +III oxidation state. The electronic configuration of Fe(III) is [Ar] 3d^5. According to Hund's rule, the five 3d orbitals are filled with electrons singly before any pairing occurs. Therefore, in the case of the spin-only magnetic moment of 5.91 B, there are approximately 5 unpaired electrons, and in the case of the spin-only magnetic moment of 1.73 B, there is approximately 1 unpaired electron.
(b) Structures of NO2, NO2+, and NO2-
- NO2:
The structure of NO2 is linear, with a N=O double bond and a lone electron pair on nitrogen.
- NO2+:
The structure of NO
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(a) The spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 B and 1.73 B, respectively.Write down their ligand field electronic configuration. Justify your answer. Atomic number of Fe and Ru are 26 and 44 respectively.(b) Draw the structures of NO2+ , NO2 and NO 2- . Arrange them in the increasing order of ONO bond angles.Correct answer is '(a) K 3[FeIII (oxalate)3 ]'. Can you explain this answer?
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(a) The spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 B and 1.73 B, respectively.Write down their ligand field electronic configuration. Justify your answer. Atomic number of Fe and Ru are 26 and 44 respectively.(b) Draw the structures of NO2+ , NO2 and NO 2- . Arrange them in the increasing order of ONO bond angles.Correct answer is '(a) K 3[FeIII (oxalate)3 ]'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about (a) The spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 B and 1.73 B, respectively.Write down their ligand field electronic configuration. Justify your answer. Atomic number of Fe and Ru are 26 and 44 respectively.(b) Draw the structures of NO2+ , NO2 and NO 2- . Arrange them in the increasing order of ONO bond angles.Correct answer is '(a) K 3[FeIII (oxalate)3 ]'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for (a) The spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 B and 1.73 B, respectively.Write down their ligand field electronic configuration. Justify your answer. Atomic number of Fe and Ru are 26 and 44 respectively.(b) Draw the structures of NO2+ , NO2 and NO 2- . Arrange them in the increasing order of ONO bond angles.Correct answer is '(a) K 3[FeIII (oxalate)3 ]'. Can you explain this answer?.
(a) The spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 B and 1.73 B, respectively.Write down their ligand field electronic configuration. Justify your answer. Atomic number of Fe and Ru are 26 and 44 respectively.(b) Draw the structures of NO2+ , NO2 and NO 2- . Arrange them in the increasing order of ONO bond angles.Correct answer is '(a) K 3[FeIII (oxalate)3 ]'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about (a) The spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 B and 1.73 B, respectively.Write down their ligand field electronic configuration. Justify your answer. Atomic number of Fe and Ru are 26 and 44 respectively.(b) Draw the structures of NO2+ , NO2 and NO 2- . Arrange them in the increasing order of ONO bond angles.Correct answer is '(a) K 3[FeIII (oxalate)3 ]'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for (a) The spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 B and 1.73 B, respectively.Write down their ligand field electronic configuration. Justify your answer. Atomic number of Fe and Ru are 26 and 44 respectively.(b) Draw the structures of NO2+ , NO2 and NO 2- . Arrange them in the increasing order of ONO bond angles.Correct answer is '(a) K 3[FeIII (oxalate)3 ]'. Can you explain this answer?.
Solutions for (a) The spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 B and 1.73 B, respectively.Write down their ligand field electronic configuration. Justify your answer. Atomic number of Fe and Ru are 26 and 44 respectively.(b) Draw the structures of NO2+ , NO2 and NO 2- . Arrange them in the increasing order of ONO bond angles.Correct answer is '(a) K 3[FeIII (oxalate)3 ]'. Can you explain this answer? in English & in Hindi are available as part of our courses for IIT JAM.
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(a) The spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 B and 1.73 B, respectively.Write down their ligand field electronic configuration. Justify your answer. Atomic number of Fe and Ru are 26 and 44 respectively.(b) Draw the structures of NO2+ , NO2 and NO 2- . Arrange them in the increasing order of ONO bond angles.Correct answer is '(a) K 3[FeIII (oxalate)3 ]'. Can you explain this answer?, a detailed solution for (a) The spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 B and 1.73 B, respectively.Write down their ligand field electronic configuration. Justify your answer. Atomic number of Fe and Ru are 26 and 44 respectively.(b) Draw the structures of NO2+ , NO2 and NO 2- . Arrange them in the increasing order of ONO bond angles.Correct answer is '(a) K 3[FeIII (oxalate)3 ]'. Can you explain this answer? has been provided alongside types of (a) The spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 B and 1.73 B, respectively.Write down their ligand field electronic configuration. Justify your answer. Atomic number of Fe and Ru are 26 and 44 respectively.(b) Draw the structures of NO2+ , NO2 and NO 2- . Arrange them in the increasing order of ONO bond angles.Correct answer is '(a) K 3[FeIII (oxalate)3 ]'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice (a) The spin-only magnetic moments of K3[Fe(oxalate)3] are 5.91 B and 1.73 B, respectively.Write down their ligand field electronic configuration. Justify your answer. Atomic number of Fe and Ru are 26 and 44 respectively.(b) Draw the structures of NO2+ , NO2 and NO 2- . Arrange them in the increasing order of ONO bond angles.Correct answer is '(a) K 3[FeIII (oxalate)3 ]'. Can you explain this answer? tests, examples and also practice IIT JAM tests.
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