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Assume the complex [NI(PPh3)2(SCN)2] is paramagnetic. The analogous complex of Pd(II) is diamagnetic. Draw all the probable isomers for both the complexes considering SCN is an ambidentate ligand.
Correct answer is '[Ni(PPh3)2(SCN)] → Paramagnetic, therefore, tetrahedral Isomers:'. Can you explain this answer?
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Assume the complex [NI(PPh3)2(SCN)2] is paramagnetic. The analogous co...
[Ni(PPh3)2(SCN)] → Paramagnetic, therefore, tetrahedral Isomers:
[Pd(PPh3)2(SCN)2] → Diamagnetic, therefore, square planar
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Assume the complex [NI(PPh3)2(SCN)2] is paramagnetic. The analogous co...
Introduction:
In this question, we are given two complexes, [Ni(PPh3)2(SCN)2] and the analogous complex of Pd(II), and we are asked to draw all the probable isomers for both complexes. We are also told that the complex [Ni(PPh3)2(SCN)2] is paramagnetic, while the analogous Pd(II) complex is diamagnetic. We need to explain this answer in detail.

Explanation:
1. Understanding Paramagnetic and Diamagnetic Complexes:
- Paramagnetic complexes have unpaired electrons and are attracted by a magnetic field.
- Diamagnetic complexes have all electrons paired and are not attracted by a magnetic field.

2. Complex [Ni(PPh3)2(SCN)2]:
- The coordination number of Ni in this complex is 4, as it is bonded to four ligands.
- The SCN ligand is ambidentate, meaning it can bind through either the sulfur (S) or the nitrogen (N) atom.
- Since the complex is paramagnetic, it must have unpaired electrons.

3. Probable Isomers for [Ni(PPh3)2(SCN)2]:
- Given that the SCN ligand is ambidentate, it can bind through either the sulfur or the nitrogen atom.
- In a tetrahedral geometry, there are two possible isomers with one SCN ligand bound through the sulfur and the other through the nitrogen atom, and vice versa.
- These two isomers can be represented as [Ni(PPh3)2(SCN)] and [Ni(PPh3)2(NCS)].
- Since the complex is paramagnetic, it must have unpaired electrons, which can be explained by the presence of one unpaired electron in the d-orbital of Ni in both isomers.

4. Complex of Pd(II):
- The analogous complex of Pd(II) has the same ligands but with Pd(II) instead of Ni.
- Since the complex is diamagnetic, it must have all electrons paired.

5. Probable Isomers for Pd(II) Complex:
- The Pd(II) complex will have the same probable isomers as the Ni complex, i.e., [Pd(PPh3)2(SCN)] and [Pd(PPh3)2(NCS)].
- However, since the complex is diamagnetic, it means that all electrons in the d-orbitals of Pd are paired in both isomers.

Summary:
In summary, the complex [Ni(PPh3)2(SCN)2] is paramagnetic because it has unpaired electrons in its d-orbital. The complex has two probable isomers with the SCN ligand binding through either the sulfur or nitrogen atom. The analogous complex of Pd(II) has the same isomers but is diamagnetic because all electrons in its d-orbitals are paired.
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Assume the complex [NI(PPh3)2(SCN)2] is paramagnetic. The analogous complex of Pd(II) is diamagnetic. Draw all the probable isomers for both the complexes considering SCN is an ambidentate ligand.Correct answer is '[Ni(PPh3)2(SCN)] → Paramagnetic, therefore, tetrahedral Isomers:'. Can you explain this answer?
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Assume the complex [NI(PPh3)2(SCN)2] is paramagnetic. The analogous complex of Pd(II) is diamagnetic. Draw all the probable isomers for both the complexes considering SCN is an ambidentate ligand.Correct answer is '[Ni(PPh3)2(SCN)] → Paramagnetic, therefore, tetrahedral Isomers:'. 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 Assume the complex [NI(PPh3)2(SCN)2] is paramagnetic. The analogous complex of Pd(II) is diamagnetic. Draw all the probable isomers for both the complexes considering SCN is an ambidentate ligand.Correct answer is '[Ni(PPh3)2(SCN)] → Paramagnetic, therefore, tetrahedral Isomers:'. 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 Assume the complex [NI(PPh3)2(SCN)2] is paramagnetic. The analogous complex of Pd(II) is diamagnetic. Draw all the probable isomers for both the complexes considering SCN is an ambidentate ligand.Correct answer is '[Ni(PPh3)2(SCN)] → Paramagnetic, therefore, tetrahedral Isomers:'. Can you explain this answer?.
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