End of chapter exercises

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Griffin Dennis
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1 End of chapter exercises Problem 1: Give one word/term for each of the following descriptions. 1. The distance between two adjacent atoms in a molecule. 2. A type of chemical bond that involves the sharing of electrons between two atoms. 3. A measure of an atom's ability to attract electrons to itself in a chemical bond. Answer 1: 1. Bond length 2. Covalent bond 3. Electronegativity Problem 2: Which ONE of the following best describes the bond formed between an H+ ion and the NH3 molecule? 1. Covalent bond 2. Dative covalent (co-ordinate covalent) bond 3. Ionic Bond 4. Hydrogen Bond Answer 2: Dative covalent (co-ordinate covalent) bond Problem 3: Explain the meaning of each of the following terms:

2 1. valence electrons 2. bond energy 3. covalent bond Answer 3: 1. The number of electrons in the outermost shell of an atom that are available for use in bonding either by sharing, donating or accepting. 2. The amount of energy needed for a bond to break. 3. A type of bond that occurs between two atoms with a difference in electronegativity between 0 and 2,1. Problem 4: Which of the following reactions will not take place? Explain your answer. 1. H+H H2 2. Ne+Ne Ne2 3. Cl+Cl Cl2 Answer 4: Ne+Ne Ne2 will not take place as neon does not have electrons available for bonding. Neon is a noble gas and has a full outer shell of electrons. Problem 5: Draw the Lewis diagrams for each of the following: 1. An atom of strontium (Sr). (Hint: Which group is it in? It will have an identical Lewis diagram to other elements in that group). 2. An atom of iodine. 3. A molecule of hydrogen bromide (HBr).

3 4. A molecule of nitrogen dioxide (NO2). (Hint: There will be a single unpaired electron). Answer 5: Problem 6: Given the following Lewis diagram, where X and Y each represent a different element: 1. How many valence electrons does X have? 2. How many valence electrons does Y have? 3. Which elements could X and Y represent? Answer 6: X could be hydrogen and Y could be nitrogen.

4 Problem 7: Determine the shape of the following molecules: 1. O2 2. MgI2 3. BCl3 4. CS2 5. CCl4 6. CH3Cl 7. Br2 8. SCl5F Answer 7: 1. This is a diatomic molecule and so the molecular shape is linear. 2. The central atom is magnesium (draw the molecules Lewis structure to see this). There are two electron pairs around magnesium and no lone pairs. There are two bonding electron pairs and no lone pairs. The molecule has the general formula AX2. Using this information and Table (Reference) we find that the molecular shape is linear. 3. The central atom is boron (draw the molecules Lewis structure to see this). There are three electron pairs around boron and no lone pairs. There are three bonding electron pairs and no lone pairs. The molecule has the general formula AX3. Using this information and Table (Reference) we find that the molecular shape is trigonal planar. 4. The central atom is carbon (draw the molecules Lewis structure to see this). There are four electron pairs around carbon forming two double bonds. There are no lone pairs.

5 The molecule has the general formula AX2. Using this information and Table (Reference) we find that the molecular shape is linear. 5. The central atom is carbon (draw the molecules Lewis structure to see this). There are four bonding electron pairs and no lone pairs. The molecule has the general formula AX4. Using this information and Table (Reference) we find that the molecular shape is tetrahedral. 6. The central atom is carbon (draw the molecules Lewis structure to see this). There are four bonding electron pairs and no lone pairs. The molecule has the general formula AX4. Using this information and Table (Reference) we find that the molecular shape is tetrahedral. 7. This is a diatomic molecule and so the molecular shape is linear. 8. The central atom is sulfur There are six electron pairs around beryllium and no lone pairs. The molecule has the general formula AX6. Using this information and Table (Reference) we find that the molecular shape is octahedral. Problem 8: Complete the following table. Element pair Hydrogen and lithium Hydrogen and boron Hydrogen and oxygen Hydrogen and sulfur Magnesium and nitrogen Electronegativity difference Type of bond that could form

6 Magnesium and chlorine Boron and fluorine Sodium and fluorine Oxygen and nitrogen Oxygen and carbon Table 1 Answer 8: Element pair Electronegativity difference Type of bond that could form Hydrogen and lithium 1,22 Strong polar covalent bond Hydrogen and boron 0,16 Weak polar covalent bond Hydrogen and oxygen 1,24 Strong polar covalent bond Hydrogen and sulfur 0,38 Weak polar covalent bond Magnesium and nitrogen Magnesium and chlorine 1,73 Strong polar covalent bond 1,73 Strong polar covalent bond Boron and fluorine 1,94 Strong polar covalent bond Sodium and fluorine 3,05 Ionic bond Oxygen and nitrogen 0,40 Weak polar covalent bond Oxygen and carbon 0,89 Weak polar covalent bond Table 2 Problem 9: Are the following molecules polar or non-polar?

7 1. O2 2. MgBr2 3. BF3 4. CH2O Answer 9: 1. The molecule is linear. There are two bonding pairs forming a double bond and two lone pairs on each oxygen atom. There is one bond. The electronegativity difference between oxygen and oxygen is 0. The bond is non-polar. The molecule is symmetrical and is non-polar. 2. The molecule is linear. There are two bonding pairs forming two single bonds and three lone pairs on each bromine atom. There are two bonds, both of which are between magnesium and bromine. The electronegativity difference between magnesium and bromine is 1,6. The bonds are polar. The molecule is symmetrical and is non-polar. 3. The molecule is trigonal planar. There are three bonding pairs forming three single bonds and three lone pairs on each fluorine atom. There are three bonds, all of which are between boron and fluorine. The electronegativity difference between boron and fluorine is 2,0. The bonds are polar. The molecule is symmetrical and is non-polar. 4. The molecule is trigonal planar. There are four bonding pairs forming two single bonds and one double bond. There are two lone pairs on the oxygen atom. There are three bonds, two of which are between carbon and hydrogen. The electronegativity difference between carbon and hydrogen is 0,4. The other bond is

8 between carbon and oxygen. The electronegativity difference between carbon and oxygen is 1,0. All the bonds are polar. The molecule is not symmetrical and is polar. Problem 10: Given the following graph for hydrogen: 1. The bond length for hydrogen is 74 pm. Indicate this value on the graph. (Remember that pm is a picometer and means m). The bond energy for hydrogen is 436 kj mol 1. Indicate this value on the graph. 2. What is important about point X? Answer 10:

9 1. 2. At point X the attractive and repulsive forces acting on the two hydrogen atoms are balanced. The energy is at a minimum. Problem 11: Hydrogen chloride has a bond length of 127 pm and a bond energy of 432 kj mol 1. Draw a graph of energy versus distance and indicate these values on your graph. The graph does not have to be accurate, a rough sketch graph will do. Answer 11:

Fill in the chart below to determine the valence electrons of elements 3-10

Fill in the chart below to determine the valence electrons of elements 3-10 Chemistry 11 Atomic Theory IV Name: Date: Block: 1. Lewis Diagrams 2. VSEPR Lewis Diagrams Lewis diagrams show the bonding between atoms of a molecule. Only the outermost electrons of an atom (called electrons)