Grade 12Inorganic Chemistry
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Arun
Approved Tutor Answer8 Years agoDear student
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Arun (askIITians forum expert)
Valence shell configuration of central metals in the following compounds :
1)Fe(CO)5 : Electronic configuration of Fe = 3d?6 4s2?
CO is a strong field ligand therefore, electronic config. of Fe in Fe(CO)5 = 3d84s0 4p0
Therefore, 5 CO will donate their lone pair in 5 orbitals namely 1 d, 1 s and 3 p and so its hybridisation is dsp3.
2) IF7 : Electronic configuration of I = 5s2 5p55d0
Electronic configuration of I in +7 oxidation state = 5s0 5p0 5d0
7 F donate their electron in 1 1 s , 3 p and 3 d orbitals and so hybridisation is d3sp3
3) Ni(CO)4 : Electronic configuration of Ni : 3d84s2
CO is a strong field ligand therefore, electronic config. of Ni in Ni(CO)4 = 3d10 4s0 4p0
Therefore, 4 CO will donate their electron in 1 s and three p orbitals giving sp3 hybridisation.
4)XeO4 : Electronic configuration of Xe = 5s25p6
Electronic configuration of Xe in +8 oxidation state = 5s0 5p0.
4 O atoms donate their electrons and this results in sp3 hybridisation.
As it can be seen from the above, only Fe in Fe(CO)5 uses orbital with different quantum number i.e. 3 d and 4s and 4p i.e. the principal quantum number is different, therefore, the answer is 1) Fe(CO)5
1)Fe(CO)5 : Electronic configuration of Fe = 3d?6 4s2?
CO is a strong field ligand therefore, electronic config. of Fe in Fe(CO)5 = 3d84s0 4p0
Therefore, 5 CO will donate their lone pair in 5 orbitals namely 1 d, 1 s and 3 p and so its hybridisation is dsp3.
2) IF7 : Electronic configuration of I = 5s2 5p55d0
Electronic configuration of I in +7 oxidation state = 5s0 5p0 5d0
7 F donate their electron in 1 1 s , 3 p and 3 d orbitals and so hybridisation is d3sp3
3) Ni(CO)4 : Electronic configuration of Ni : 3d84s2
CO is a strong field ligand therefore, electronic config. of Ni in Ni(CO)4 = 3d10 4s0 4p0
Therefore, 4 CO will donate their electron in 1 s and three p orbitals giving sp3 hybridisation.
4)XeO4 : Electronic configuration of Xe = 5s25p6
Electronic configuration of Xe in +8 oxidation state = 5s0 5p0.
4 O atoms donate their electrons and this results in sp3 hybridisation.
As it can be seen from the above, only Fe in Fe(CO)5 uses orbital with different quantum number i.e. 3 d and 4s and 4p i.e. the principal quantum number is different, therefore, the answer is 1) Fe(CO)5
Arun (askIITians forum expert)
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