1. ************************************************* Chemistry for Engineers Homework 2 B O N D I N G Answer Key Electron arrangements &

2. Q1: Fill in every blank box regarding the atoms or ions of the following isotopes:

3. Q2: The Lewis structures for the following ions are almost complete. They show the bonding pairs and lone pairs but charges on atoms are not shown. Redraw each and add charges to atoms that need them. a) H 3 O + b) NCS - d) OH - c) HCO + e) N 3 -

4. Q3: For each molecule: a) How many valence electrons are there? b) Draw out the Lewis structure c) Deduce the oxidation state of the central atom d) Indicate bond polarity using  + and  - symbols and overall molecule polarity with ClF SF 2 PH 2 Br HCOOH

5. Count up valence electrons = 14 Count up valence electron pairs = 7 Determine which atoms are bond to which (struc. formula) = Cl  F Form single bonds between neighbouring atoms: (6 electron pairs remain unused) Use remaining electron pairs to complete octets of each atom: (three lone pairs on each halogen atom) ClF c) Cl oxidation state is +1 ++ - - ClF SF 2 PH 2 Br HCOOH

6. - - SF 2 Count up valence electrons = 20 Count up valence electron pairs = 10 Determine structural formula: Form single bonds between neighbouring atoms: (8 electron pairs remain unused) Use remaining electron pairs to complete octets of each atom: (F octet must be completed first) leaves two lone pairs on S c) S oxidation state is +2 ++ - - ClF SF 2 PH 2 Br HCOOH

7. PBrH 2 Count up valence electrons = 14 Count up valence electron pairs = 7 Determine structural formula: Form single bonds between neighbouring atoms: (4 electron pairs remain unused) Use remaining electron pairs to complete octets of each atom: (Br octet must be completed first) leaves one lone pair on P ClF SF 2 PH 2 Br HCOOH c) P oxidation state is +1 - - ++

8. HCOOH Count up valence electrons = 18 Count up valence electron pairs = 9 Determine structural formula: Place C at centre and form single bonds between neighbouring atoms: (5 electron pairs remain unused) Use remaining electron pairs to complete octets of each electronegative atom: BUT - not enough electrons to complete carbon’s octet so C=O double bond must form by donating O lone pair: ClF SF 2 PH 2 Br HCOOH - - ++ - - C has incomplete octet! c) C oxidation state is +2

9. Q4: Consider the sulphite anion (SO 3 2- ). a) How many valence electron pairs are there? Draw the Lewis resonance structures of the anion. valence electrons = 26 (13 pairs) place S as central atom and use 3 pairs – single bonds S can expand valency so double bond can be drawn: now move  bond to illustrate resonance structures: (S has 6 valence electrons so retains a lone pair)

10. b) What is the charge on each oxygen atom? - 0.667 (two thirds of a negative charge) c) Given that the strength of a typical S-O single bond is 360 kJ/mol, estimate the S-O bond strength in the sulphite ion. bond order = 1.33 bond strength expected = 480 kJ/mol.

11. Q5: For the following species: I 2 IF 4 + IF 3 PI 3 a) Deduce the oxidation state of iodine b) How many valence electron pairs are there? c) Draw the Lewis structure d) What is the geometry of the electron pairs (e.g. linear, trigonal bipyramidal etc)? What are the bond angles and molecule shape? e) What related anion has I in the +1 oxidation state? IF 2 -

12. Q5: iodine oxidation state 0 +5 +3 -1 valence electrons 14 34 28 26 valence electron pairs 7 17 14 13 F-I-F = 90º F-I-F = 90º I-P-I ~102º & 120º trigonal bipyramid (atoms arranged as sawhorse) distorted tetrahedral (molecule is pyramidal) trigonal bipyramid (molecule is T-shaped) I 2 IF 4 + IF 3 PI 3