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![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 Aug 2007 12:40:01 IST
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CH3-CH-=C=CH-CH3
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11 Aug 2007 12:42:08 IST
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although the C is double bonded to two C's, it's structure is linear. so it is sp hybridised
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11 Aug 2007 12:43:57 IST
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is that sp first tell the answer then i will give xplanation pls
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" I've shed many tears. But after, I realize, why am I crying? No matter how bad it got, it could be worse. No matter how much you think it's not going to be ok, eventually it will be. You just have to let it be."
EXPECT MORE FROM URSELF THAN FRM OTHERS AS EXPECTATION FRM OTHERS HURTS U WHILE FROM UR SELF INSPIRES U A lot |
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11 Aug 2007 12:44:35 IST
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hi..
the unserlined carbon is forming 2 pie-bonds, one on each side. Hence it must have two unhybridized orbitlas. Also its forming 2 sigma bonda with two carbon atoms, one on each side, so they must be formed by hybridized orbitals..
hence two hybridized orbitals and two unhybridized orbitals.. what u obtain in the case of sp-hybridisation...
hope its v.clear...
Also, these class of compounds are called as allenes or "cumulated dienes".
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A diamond is merely a lump of coal that did well under pressure. So imagine how brilliant a human being can be by sustaining pressures of life. Be hopeful.. and fight till end.
Prashant. |
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11 Aug 2007 12:50:52 IST
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sp hybridized
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* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
* Your friendly neighborhood spideyunlimited |
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11 Aug 2007 12:54:30 IST
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1. appliable for C.N.O and neutral molecule pi bond hybridisation 0 sp3 1 sp2 2 sp 2. and for all molecule N = 1/2[ no. of valence electron in d central atom + no. of electron of monovalent atom(H,F,Cl,Br,I) - charge of cation + charge of anion ] N= 2 - sp 3 - sp2 4 - sp3 5 - sp3d 6 - sp3d2 7 - sp3d3 this is applicable for molecules having one central atom 3. no. of siga bond + no. of lone pairs on central atom hope itz helpfull n rate me if so
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11 Aug 2007 12:56:19 IST
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n answer for this is sp
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11 Aug 2007 12:56:42 IST
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2 sigma bonds + 0 lone pairs = sp
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---------------------------------------------------------------
* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
* Your friendly neighborhood spideyunlimited |
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11 Aug 2007 12:57:30 IST
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yes the 3rd method of aankur verma's is the easiest.. thts wht i use too
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---------------------------------------------------------------
* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
* Your friendly neighborhood spideyunlimited |
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11 Aug 2007 13:16:17 IST
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Hello my dear... see.. i strongly suggest NOT TO USE ANY FORMULA FOR HYBTIDISATION... go by the basic funda.. as i explained in my previous reply... okk.. with ur logic (@ ankur and spidy...) .. Oxygen in furan shud b sp3 hybridised.. as it does not have any pie-bonds in da structure... and you already know its SP2 not sp3 . See, any novice with ur logic will end up in trouble.. yaar unko funda sikhaao... formula nahin !!!!!!!! and again how do you manage to calculate the hybridisation of Mn and Cr in KMnO4 and K2Cr2O7 by using formula no. 2 ?? ... and u sill dunno whether its sp3 or dsp2 ? (sorry... no hard feelings for anyone )
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A diamond is merely a lump of coal that did well under pressure. So imagine how brilliant a human being can be by sustaining pressures of life. Be hopeful.. and fight till end.
Prashant. |
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11 Aug 2007 18:36:27 IST
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The underlined carbon has double bond on either side, i.e., one sigma and one pi bond. Hence the hybid state of the carbon will be sp, because in this hybridisation carbon can form one sigma bond on either side. The pi bonds are formed using the unhybridised p orbitals. Refer to structure of allene.
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