Page 1 of 1

Why is (CH3)2Be is sp?

Posted: Wed Sep 14, 2011 9:36 am
by Chem_Mod
Question: Can you please explain to me why (CH3)2Be has a hybridization of sp rather than sp3? Is it because Be has an empty p orbital and lone pairs do not effect it?

Re: Why is (CH3)2Be is sp?

Posted: Wed Sep 14, 2011 9:36 am
by Chem_Mod
Answer: Be is in group 2 and forms two bonds, one to each CH3. Placing Be in the center forms a linear molecule with two single bonds.Therefore Be is sp hybridization.

Re: Why is (CH3)2Be is sp?

Posted: Sat Nov 18, 2017 7:49 pm
by Dang Lam
Can you please explain why we can put Be in the middle? thanks

Re: Why is (CH3)2Be is sp?

Posted: Sat Nov 18, 2017 9:04 pm
by Cynthia Tsang
The central atom is always the one with the lowest electronegativity. Be is lower than C so it is in between the two carbons

Re: Why is (CH3)2Be is sp?

Posted: Wed Nov 22, 2017 12:33 pm
by Evelyn L 1H
On number 4.19, it says that (CH3)2BE is tetrahedral - how is that right if it's VSEPR is sp?

Re: Why is (CH3)2Be is sp?  [ENDORSED]

Posted: Wed Nov 22, 2017 12:55 pm
by Chem_Mod
Please read things carefully.

The solution manual says the shape is a tetrahedral about the carbon atoms.

However, as a whole, the molecule is linear. Meaning with respect to the central Be atom in (CH3)2Be.

Re: Why is (CH3)2Be is sp?

Posted: Tue Nov 20, 2018 7:31 pm
by Destiny Diaz 4D
Chem_Mod wrote:Please read things carefully.

The solution manual says the shape is a tetrahedral about the carbon atoms.

However, as a whole, the molecule is linear. Meaning with respect to the central Be atom in (CH3)2Be.


So to better understand, if the question were to ask of the shape in respect to the carbon atoms then the shape would be tetrahedral and if in respect to Be then the shape would be linear? Since the question doesnt specify would you just assume Be and just specify in the answer what your bond angle is refering to ?

Re: Why is (CH3)2Be is sp?

Posted: Sat Nov 24, 2018 11:19 pm
by Chem_Mod
Your statement is correct. You can also ask for clarification if it appears on the exam