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![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Dec 2006 12:15:25 IST
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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 $s 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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Manasi....
NIT-Allahabad...
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12 Dec 2006 18:22:37 IST
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In[ Mn(CN)6]-3 the electrons in Mn get paired up as the CN is strong field ligand so the hybridisation is d3sp2 which is octahedral and it is diamagnetic.
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19 Jan 2007 12:56:30 IST
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Hybridization = number of sigma bonds + number of lone pairs Here no of sigma bonds = 6 and lone pairs = 0 So total = 6 Hence sp3d2 Bon voyage
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CHINMOY RAJWANSHI
DELHI
"nobody" is perfect and i m nobody |
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