
Which of the following is a high spin complex?
A.\[{\left[ {{\text{Co}}{{\left( {{\text{N}}{{\text{H}}_3}} \right)}_6}} \right]^{3 - }}\]
B.\[{\left[ {{\text{Fe}}{{\left( {{\text{C}}{{\text{N}}_3}} \right)}_6}} \right]^{4 - }}\]
C.\[{\left[ {{\text{Ni}}{{\left( {{\text{C}}{{\text{N}}_3}} \right)}_4}} \right]^{2 - }}\]
D.\[{\left[ {{\text{Fe}}{{\text{F}}_6}} \right]^{3 - }}\]
Answer
577.5k+ views
Hint: High spin complex will be formed by ligands which have low splitting strength which is determined by the spectrochemical series. Halide ions mostly behave as weak field ligands in the presence of 3 d transition metal complexes.
Complete step by step solution:
According to valence bond theory intermixing of orbitals occurs. Intermixing or overlapping of vacant orbital of Central metal atom and orbital of ligand. Strength of the ligands that will donate electrons to metal will determine whether the pairing of electrons of Central metal atom will occur or not. The ligands such as cyanide, carbonyl, ammonia and EDTA etc. are strong field ligands and the ligands such as iodide, bromide, chloride and fluoride are weak field ligands. Spectrochemical series is a series which is found experimentally which helps in determining the strong and weak field ligands.
A strong field ligand pairs up the electron whereas weak field ligands keep the electron unpaired. This is due to the difference in the energy of the pairing and splitting energy. If pairing energy is high that means we have to provide more energy for pairing then the complex will prefer to have unpaired electron and will be called high field complex
Among the above given options fluoride is a weak field ligand and hence will not pair up the electrons and so the compound will be paramagnetic in nature due to the presence of unpaired electrons and hence high spin complex will form.
So the correct option is D.
Note: Magnetic moment of diamagnetic substances is zero and the magnetic moment of paramagnetic substances can be calculated using the formula \[\mu = \sqrt {{\text{n}}\left( {{\text{n}} + 2} \right)} \] , n is the number of unpaired electrons. Paramagnetic substances are that substance which feels attraction when placed in an external magnetic field.
Complete step by step solution:
According to valence bond theory intermixing of orbitals occurs. Intermixing or overlapping of vacant orbital of Central metal atom and orbital of ligand. Strength of the ligands that will donate electrons to metal will determine whether the pairing of electrons of Central metal atom will occur or not. The ligands such as cyanide, carbonyl, ammonia and EDTA etc. are strong field ligands and the ligands such as iodide, bromide, chloride and fluoride are weak field ligands. Spectrochemical series is a series which is found experimentally which helps in determining the strong and weak field ligands.
A strong field ligand pairs up the electron whereas weak field ligands keep the electron unpaired. This is due to the difference in the energy of the pairing and splitting energy. If pairing energy is high that means we have to provide more energy for pairing then the complex will prefer to have unpaired electron and will be called high field complex
Among the above given options fluoride is a weak field ligand and hence will not pair up the electrons and so the compound will be paramagnetic in nature due to the presence of unpaired electrons and hence high spin complex will form.
So the correct option is D.
Note: Magnetic moment of diamagnetic substances is zero and the magnetic moment of paramagnetic substances can be calculated using the formula \[\mu = \sqrt {{\text{n}}\left( {{\text{n}} + 2} \right)} \] , n is the number of unpaired electrons. Paramagnetic substances are that substance which feels attraction when placed in an external magnetic field.
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