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Inorganic Chemistry

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7 Dec 2006 11:17:15 IST

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State of hybridisation

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for the coordination compound [Mn(CN) ₆] ^-3 wat will be the state of hybridisation....?? here Mn is in +3 state.... that is 2 electrons frm 4s orbital n one frm 3d will be lost.... n then CN will pair the electrons... giving three, 3d-orbitals... so how can it give an octahedral geometry???

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arun-rashi

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8 Dec 2006 09:48:29 IST

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here magnese links with six strong ligand so complex is low spin intra orbital means d2sp3 hybridisation because steric number is 6.

any how shape is octahedral with all 90 angle.remember CN is strong ligand. other example of stong ligand is CO,etylenediamine etc.

viswanathan

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23 May 2012 10:11:36 IST

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can any one explain this

manoj rohit

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26 May 2012 11:27:22 IST

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it shd be then d3sp2 which is never possible

why is it like that

can anyone explain?????

manoj rohit

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26 May 2012 11:27:22 IST

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it shd be then d3sp2 which is never possible

why is it like that

can anyone explain?????

Janardhan K

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26 May 2012 11:39:51 IST

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as it is a stong field ligand so pairing will take place. acc. to crystal field theory first 3 electrons will occupy 3 subshells in t2g. and due to strong field ligand pairing takes place and last electron pairs with first electron. it gives two empty d subshells and 2 unpaired elctron. thus hybridization is d2sp3

Janardhan K

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26 May 2012 11:44:19 IST

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actually u all have doubt because u r not following Hund's rule. even in crystal field theory in individual t2g and eg hunds rule is followed.

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