Drawing Lewis Structures for VSPER: Problem Solutions

  • Thread starter Thread starter decibel
  • Start date Start date
  • Tags Tags
    Structures
Click For Summary
The discussion focuses on challenges in drawing Lewis structures related to VSEPR theory, specifically for the species [N(CH3)4]^+, ICl2, and [ICl4]^- . A user suggests the Clarke method for determining electron counts and bond types, which involves calculating the total number of electrons and comparing it to a formula based on the number of atoms. This method helps identify whether a molecule contains double bonds or an expanded octet. However, it is noted that this approach does not provide information on atom connectivities, which requires further analysis of formal charges. The conversation emphasizes the importance of understanding both electron distribution and structural connectivity in Lewis structures.
decibel
Messages
107
Reaction score
1
i'm having trouble drawing up some lewis structures related to VSPER theory.
i was given 22 Lewis structures to draw and i had problems with these ones...

1 - [N(CHsubscript3)subscript4]^+
2 - IClsubscript2
3 - [IClsubscript4]^-

thanks if anyone can help out with these questions.
 
Physics news on Phys.org
decibel said:
i'm having trouble drawing up some lewis structures related to VSPER theory.
i was given 22 Lewis structures to draw and i had problems with these ones...

1 - [N(CHsubscript3)subscript4]^+
2 - IClsubscript2
3 - [IClsubscript4]^-

thanks if anyone can help out with these questions.

Try visiting the Chemistry forum in Other Sciences, i remember doing this exercices, but a long time ago :smile:
 
Were you taught the 5-step method for drawing Lewis structures?
 
Clarke's Method

Hi

Here's something known as the Clarke method for drawing Lewis Structures. I read it from a General Chemistry book by Dr. Hubbard (I forget the complete name). I've found it very effective and thought I should share it with you folks.

1. Find the TOTAL number of electrons (n_{e}) for a species. For ICl_{4}^{-} this would be 7+4(7)+1 = 36.
2. Next compute the quantity (6y+2) where y is the total number of atoms minus hydrogen (for H_{2}O this would be 1).
3. Compare n_{e} and (6y+2):

(a) if n_{e} < (6y+2) then the species contains a double bond. For ICl4-, 6y+2 = 32 which is less than 36 by 4 so there is either a triple bond or two double bonds (and you can see that there are actually 2 double bonds here).

(b) if n_{e} = 6y+2, the structure contains only single bonds. As an example consider the water molecule. For it, n_{e} = 8 and 6y+2 = 8.

(c) if n_{e} > (6y+2) the central atom has an expanded octet (which means it cannot belong to the second period as the second period elements do not have d-orbitals). For example, IF_{7} has n_{e} = 7+7(7) = 56 and 6y+2 = 6(8)+2 = 50.

Note that this method will only take you a bit further in making the structure but won't tell you the crucial information: that of connectivities. You must then use formal charges and some experience to rule out incorrect structures if more than one is possible.

Cheers
Vivek
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
26
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
7
Views
995
  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 3 ·
Replies
3
Views
6K
Replies
5
Views
2K