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Nickel(Z = 28) combines with a uninegative monodentate ligand to form a dia magnetic complex [NiL4]2–. The hybridisation involved and the number of unpaired electrons present in the complex are respectively :
- a)sp3, zero
- b)sp3. two
- c)dsp2 one
- d)dsp2, zero
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
Nickel(Z = 28) combines with a uninegative monodentate ligand to form ...
Monodentate ligand is uninegative and hence, the oxidation number of Ni in [NiX
4
]
2−
is +2.
The electronic configuration of Ni
2+
is [Ar]3d
8
.
If the compound is diamagnetic it means all the electrons are paired. Hence, there will be one 3d orbital empty and one 4s orbital empty.
The hybridisation involved in the formation of complex [NiL
4
]
2−
is dsp
2
.
solution
4
]
2−
is +2.
The electronic configuration of Ni
2+
is [Ar]3d
8
.
If the compound is diamagnetic it means all the electrons are paired. Hence, there will be one 3d orbital empty and one 4s orbital empty.
The hybridisation involved in the formation of complex [NiL
4
]
2−
is dsp
2
.
solution
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Nickel(Z = 28) combines with a uninegative monodentate ligand to form ...
Hybridisation and Unpaired Electrons in [NiL4]2− Complex
Hybridisation:
- The coordination number of nickel in the complex is 4, indicating sp3 hybridization.
- Each ligand is monodentate, meaning it can donate one electron pair to the central nickel atom.
- Since there are four ligands, the nickel atom will use four of its d orbitals to form bonds with the ligands, leading to sp3 hybridization.
Number of Unpaired Electrons:
- The complex is diamagnetic, which means it has no unpaired electrons.
- In a diamagnetic complex, all the electrons are paired up, resulting in a net magnetic moment of zero.
- Therefore, the [NiL4]2− complex has zero unpaired electrons.
Therefore, the correct hybridization involved in the [NiL4]2− complex is sp3, and the number of unpaired electrons present in the complex is zero.
Hybridisation:
- The coordination number of nickel in the complex is 4, indicating sp3 hybridization.
- Each ligand is monodentate, meaning it can donate one electron pair to the central nickel atom.
- Since there are four ligands, the nickel atom will use four of its d orbitals to form bonds with the ligands, leading to sp3 hybridization.
Number of Unpaired Electrons:
- The complex is diamagnetic, which means it has no unpaired electrons.
- In a diamagnetic complex, all the electrons are paired up, resulting in a net magnetic moment of zero.
- Therefore, the [NiL4]2− complex has zero unpaired electrons.
Therefore, the correct hybridization involved in the [NiL4]2− complex is sp3, and the number of unpaired electrons present in the complex is zero.
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Nickel(Z = 28) combines with a uninegative monodentate ligand to form a dia magnetic complex [NiL4]2–. The hybridisation involved and the number of unpaired electrons present in the complex are respectively :a)sp3, zerob)sp3. twoc)dsp2 oned)dsp2, zeroCorrect answer is option 'C'. Can you explain this answer?
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Nickel(Z = 28) combines with a uninegative monodentate ligand to form a dia magnetic complex [NiL4]2–. The hybridisation involved and the number of unpaired electrons present in the complex are respectively :a)sp3, zerob)sp3. twoc)dsp2 oned)dsp2, zeroCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Nickel(Z = 28) combines with a uninegative monodentate ligand to form a dia magnetic complex [NiL4]2–. The hybridisation involved and the number of unpaired electrons present in the complex are respectively :a)sp3, zerob)sp3. twoc)dsp2 oned)dsp2, zeroCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Nickel(Z = 28) combines with a uninegative monodentate ligand to form a dia magnetic complex [NiL4]2–. The hybridisation involved and the number of unpaired electrons present in the complex are respectively :a)sp3, zerob)sp3. twoc)dsp2 oned)dsp2, zeroCorrect answer is option 'C'. Can you explain this answer?.
Nickel(Z = 28) combines with a uninegative monodentate ligand to form a dia magnetic complex [NiL4]2–. The hybridisation involved and the number of unpaired electrons present in the complex are respectively :a)sp3, zerob)sp3. twoc)dsp2 oned)dsp2, zeroCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Nickel(Z = 28) combines with a uninegative monodentate ligand to form a dia magnetic complex [NiL4]2–. The hybridisation involved and the number of unpaired electrons present in the complex are respectively :a)sp3, zerob)sp3. twoc)dsp2 oned)dsp2, zeroCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Nickel(Z = 28) combines with a uninegative monodentate ligand to form a dia magnetic complex [NiL4]2–. The hybridisation involved and the number of unpaired electrons present in the complex are respectively :a)sp3, zerob)sp3. twoc)dsp2 oned)dsp2, zeroCorrect answer is option 'C'. Can you explain this answer?.
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Nickel(Z = 28) combines with a uninegative monodentate ligand to form a dia magnetic complex [NiL4]2–. The hybridisation involved and the number of unpaired electrons present in the complex are respectively :a)sp3, zerob)sp3. twoc)dsp2 oned)dsp2, zeroCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for Nickel(Z = 28) combines with a uninegative monodentate ligand to form a dia magnetic complex [NiL4]2–. The hybridisation involved and the number of unpaired electrons present in the complex are respectively :a)sp3, zerob)sp3. twoc)dsp2 oned)dsp2, zeroCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of Nickel(Z = 28) combines with a uninegative monodentate ligand to form a dia magnetic complex [NiL4]2–. The hybridisation involved and the number of unpaired electrons present in the complex are respectively :a)sp3, zerob)sp3. twoc)dsp2 oned)dsp2, zeroCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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