Which of the following molecules are polar?
CBr4, SO3,XeF4, ClF5, PBr5.
Can some explain this for me?
Thanks
CBr4, SO3,XeF4, ClF5, PBr5.
Can some explain this for me?
Thanks
You need to count up the total valence electrons. For example,,,,lets do CBr4. C =4..... Br is 7,,,,,times 4...is 28.....thus we have 32 electrons. Place the carbon in the center,,,,,and give the Brs the octet. If done correctly, you will notice that there are no lone pairs left over.....thus the molecule is nonpolar and has no dipole. Do this for each one, and look for a specie with a lone pair. XeF4 is a special case.,,,,,it has 2 lone pairs but is non polar since it is square planar. You will find that in ClF5, a lone pair exists...thus it is polar and has a dipole. Please consult a General Chemistry text and return to this problem...it is VERY IMPORTANT for the DAT exam. I would recommend the text by Raymond Chang.Which of the following molecules are polar?
CBr4, SO3,XeF4, ClF5, PBr5.
Can some explain this for me?
Thanks
Thanks a lot for answering that Dr. Romano. in order to write all the Lewis dot structures would consume time, is their an shortcut? i feel like something like this would take 3 minutes max to draw all the lewis dot.You need to count up the total valence electrons. For example,,,,lets do CBr4. C =4..... Br is 7,,,,,times 4...is 28.....thus we have 32 electrons. Place the carbon in the center,,,,,and give the Brs the octet. If done correctly, you will notice that there are no lone pairs left over.....thus the molecule is nonpolar and has no dipole. Do this for each one, and look for a specie with a lone pair. XeF4 is a special case.,,,,,it has 2 lone pairs but is non polar since it is square planar. You will find that in ClF5, a lone pair exists...thus it is polar and has a dipole. Please consult a General Chemistry text and return to this problem...it is VERY IMPORTANT for the DAT exam. I would recommend the text by Raymond Chang.
Hope this helps..
Dr. Romano
So if theres a lone pair on the central atom, that atom considered a electron rich and everything else is e- poor, and it is a polar?You need to count up the total valence electrons. For example,,,,lets do CBr4. C =4..... Br is 7,,,,,times 4...is 28.....thus we have 32 electrons. Place the carbon in the center,,,,,and give the Brs the octet. If done correctly, you will notice that there are no lone pairs left over.....thus the molecule is nonpolar and has no dipole. Do this for each one, and look for a specie with a lone pair. XeF4 is a special case.,,,,,it has 2 lone pairs but is non polar since it is square planar. You will find that in ClF5, a lone pair exists...thus it is polar and has a dipole. Please consult a General Chemistry text and return to this problem...it is VERY IMPORTANT for the DAT exam. I would recommend the text by Raymond Chang.
Hope this helps..
Dr. Romano
NO.....PRACTICE THIS !!!!! Many times the lone pairs hide themselves !!! Consider AsCl3. No way in Hell would you predict the lone electron pair unless you draw out the structure , but have an electron count. Here, we count As = 5... Cl is 7 x3 = 21...thus we have 26 valence electrons. Placing As in the middle, and the Cl on the outside, and giving the Cl 's an octet... we get......1 lone pair on As......hence this specie is sp3 and trigonal pyramidal. No easy way out here......you must do it the correct way.Thanks a lot for answering that Dr. Romano. in order to write all the Lewis dot structures would consume time, is their an shortcut? i feel like something like this would take 3 minutes max to draw all the lewis dot.
Thank,
thanks a lot! i appreciate the response and all the hard work you put in to help us all out.NO.....PRACTICE THIS !!!!! Many times the lone pairs hide themselves !!! Consider AsCl3. No way in Hell would you predict the lone electron pair unless you draw out the structure , but have an electron count. Here, we count As = 5... Cl is 7 x3 = 21...thus we have 26 valence electrons. Placing As in the middle, and the Cl on the outside, and giving the Cl 's an octet... we get......1 lone pair on As......hence this specie is sp3 and trigonal pyramidal. No easy way out here......you must do it the correct way.
Hope this helps.
Keep up the great work! We do our best to answer asap!thanks a lot! i appreciate the response and all the hard work you put in to help us all out.