Skeletal structure for urea [CO(NH2)2] molecule is shown to the right. Complete the rest of the Lewis structure, to answer questions 26 and 27.
NH2
26) This molecule, when ‘cleaned-up’ (octet; formal charge), will have bonds and lone ‘pairs’ while the formal charge of oxygen is single bonds,
multiple
A) 6, 1.8,0
B) 7,1,8,-2
C) 6, 1, 4,0
D) 7,0,8,0
E) 7, 1.8,-2
27) The electron pair geometry (epg) around each nitrogen is (a). arrange to have a (b) while it will re-molecular shape (ms) around the each nitrogen center.
A) (a) linear; (b) trigonal planar
B) (a) tetrahedral; (b) bent
C) (a) trigonal planar, (b) tetrahedral
D) (a) tetrahedral; (b) trigonal pyramidal
E) (a) tetrahedral; (b) trigonal planar
The Correct Answer and Explanation is :
Let’s break down both questions based on the urea molecule (CO(NH2)2).
Lewis Structure:
The structure of urea consists of a carbonyl group (C=O) attached to two amine groups (NH2). The carbon is double-bonded to oxygen, and each nitrogen is single-bonded to carbon and bonded to two hydrogens.
Question 26: Formal Charges and Bonds
You are asked to determine the bond and lone pair arrangement in the urea molecule to satisfy the octet rule and minimize formal charges.
- Carbon in urea is double-bonded to oxygen and single-bonded to each nitrogen.
- Nitrogen atoms are each bonded to two hydrogens and one carbon.
- Oxygen will have two lone pairs, while nitrogen atoms each have one lone pair and are involved in single bonds with carbon and hydrogen.
Now let’s determine the formal charge:
- Carbon: 4 valence electrons (from the periodic table), and in the Lewis structure, it is bonded to two atoms (oxygen and nitrogen) and no lone pairs. The formal charge on carbon is 0.
- Oxygen: 6 valence electrons, 2 lone pairs, and it forms one double bond with carbon. Formal charge = 6 – (4 from bonding) = 0.
- Nitrogen: 5 valence electrons, one lone pair, and 3 bonds (2 single bonds to hydrogen and 1 single bond to carbon). Formal charge = 5 – (2 bonding electrons + 2 from the lone pair) = 0.
So, the formal charges are 0 for all atoms, meaning no charges on the structure. Thus, C is correct: 7, 0, 8, 0.
Question 27: Electron Pair Geometry (EPG) and Molecular Shape (MS)
For the nitrogen atoms in urea:
- The nitrogen atoms are each bonded to three atoms (two hydrogens and one carbon) and have one lone pair. This makes the electron pair geometry (EPG) around each nitrogen tetrahedral, as there are four regions of electron density around each nitrogen (three bonds and one lone pair).
- The molecular shape (MS) around each nitrogen will be trigonal pyramidal because there is a lone pair that pushes the bonding electrons into a pyramid shape.
Thus, D is correct: (a) tetrahedral; (b) trigonal pyramidal.
Summary of Answers:
- Question 26: D) 7, 0, 8, 0
- Question 27: D) (a) tetrahedral; (b) trigonal pyramidal
Let me generate the Lewis structure for urea as you requested.
Here is the detailed Lewis structure for the urea molecule, showing the bonding and lone pairs as discussed. Let me know if you need further clarification!
