Inorganic Chemistry Practice Test 1-2 PDF

INORGANIC CHEMISTRY b 8 5 0 CHEMICAL BOND 90f d b 9 7 A. Ty...

INORGANIC CHEMISTRY b 8 5 0 CHEMICAL BOND 90f d b 9 7 A. Types of Chemical Bonds 74 f 9 B. Properties of a Covalent b 63 Bond 7 C. Octet Rule 368 66 D. Lewis Structures o m i l.c E. Molecular m a Geometry g F. Polarity np @ e G.ygwIMFA iv INORGANIC CHEMISTRY PRACTICE TEST 1-2  Electronegativity Electronegativity is a relative concept, meaning that an element’s electronegativity b 8 0 can be measured only in relation to the electronegativity of other elements. 5 0 f b 9 Linus Pauling devised a method for calculating relative electronegativities of most elements (Figure 9.5). 7 d f 9 74 39 b 6 8 7 36 66 o m i l.c m a @g p wen y g iv Chang, 2010  Electronegativity 8 There is no sharp distinction between a polar covalent and an0ionic b bond, f5 them. but the following general rule is helpful in distinguishing between 90 d b 9 7 An ionic bond forms when the ΔEN between 4 f the two bonding atoms is 2.0 or more. 9 7 This rule applies to most but not all ionic 63 7b compounds. Sometimes chemists use the 6 8 quantity percent ionic character to 6 3describe 6 the nature of a bond (Chang, 2010). o m i l.c a m of the bond ΔEN g Character @ < 0.4 np covalent bond e 0.4 – 1.7ygwpolar covalent bond i v > 1.7 ionic bond Chang, 2010  PROPERTIES OF A BOND b 8 5 0 90f d b 9 7 4f 9 7 63 7b 68 3 66 o m ↑ Bond Energy ↑ BOND ORDER i l.c m a ↓ Bond Length @g np w e y g iv  OCTET RULE b 8 5 0 90f d b 9 7 4f 9 7 63 7b 68 3 66 o m i l.c ma @g p wen y g iv Chang, 2010  EXPANDED OCTET b 8 5 0 90f d b 9 7 4f 9 7 63 7b 68 3 66 o m i l.c a gm elements cannot have an expanded octet. Yellow: second-period @ p n elements and beyond can have an expanded octet. Blue: third- eperiod g w v Green: y i the noble gases usually only have an expanded octet. Chang, 2010  LEWIS STRUCTURES The Lewis structure of a molecule represents the arrangement of8valence 0bdrawn as electrons among the atoms in a molecule. Valence electrons 5are dots around the symbol of the element. 0 f 9 b 7 d f 9 74 39 b 6 8 7 36 66 o m i l.c m a @g p wen y g iv Gilbert Newton Lewis was best known for his discovery of the covalent bond and his concept of electron pairs, and his Lewis dot structures.  Ionic Compounds b 8 5 0 90f d b 9 7 4f 9 7 63 7b 68 3 66 o m i l.c ma @g p wen y g iv  Resonance Structures b 8 There are cases wherein there is more than one Lewis structure to represent a 0 vary in molecule or polyatomic ion. These are called resonance structures,5which 0 f their contribution to the true structure of the molecule or ion. b 9 Some resonance structures are equivalent and contribute equally to the truedstructure. 9 7 4f 9 7 63 7b 68 3 66 o m i l.c m a The importance of a resonance structure may be determined from the formal g @that comprise the molecule or ion. charges on the atoms np we y g 𝑏𝑜𝑛𝑑𝑖𝑛𝑔 𝑒 ! i v 𝒇𝒐𝒓𝒎𝒂𝒍 𝒄𝒉𝒂𝒓𝒈𝒆 = 𝑣𝑎𝑙𝑒𝑛𝑐𝑒 𝑒 ! − 𝑛𝑜𝑛𝑏𝑜𝑛𝑑𝑖𝑛𝑔 𝑒 ! − 2  Resonance Structures A resonance structure is more important when: b 8 5 0 Adjacent atoms do not have similar formal charges. 0 f Electronegative atoms bear negative formal charges. b9 7 d 9 molecules. Formal charges on atoms are close to zero for neutral f 7 4 the magnetic property Number of unpaired electrons is consistent with of the species. 3 9 6 b 8 7 36 66 o m i l.c m a @g p wen y g 𝑏𝑜𝑛𝑑𝑖𝑛𝑔 𝑒 ! i v 𝒇𝒐𝒓𝒎𝒂𝒍 𝒄𝒉𝒂𝒓𝒈𝒆 = 𝑣𝑎𝑙𝑒𝑛𝑐𝑒 𝑒 ! − 𝑛𝑜𝑛𝑏𝑜𝑛𝑑𝑖𝑛𝑔 𝑒 ! − 2  Sample Problem 8 (atomic Draw three resonance structures for the molecule nitrous oxide, Nb2O 5 0 arrangement is NNO). Indicate formal charges. f 0 With respect to the properties of the molecule, which is thedmost 9 b important one? 9 7 4 f 9 7 63 7b 6 8 6 3 6 o m i l.c m a Structure (b) is the @ g most important because the negative charge is on the more np atom. electronegativeeoxygen g w iv y Structure (c) is the least important because it has a larger separation of formal charges. Also, the positive charge is on the more electronegative oxygen atom. Chang, 2010  Sample Problem 8 b O atoms 0 Draw a Lewis structure for the sulfate ion (SO42–) in which all four 5 are bonded to the central S atom. 0 f b 9 7d f 9 74 39 b 6 8 7 36 66 o m i l.c a This structure involves m an expanded octet on S but may be considered more plausible because it@ g fewer formal charges. The general rule for elements in bears np beyond is: the third period and e wg structure that obeys the octet rule is preferred over one that A resonancey involvesivan expanded octet but bears fewer formal charges. Chang, 2010  Sample Problem b 8 Rank the resonance structures in each group in order of increasing5 0 contribution to the resonance hybrid. 90 f d b 9 7 4 f 9 7 63 7b 6 8 6 3 6 o m i l.c m a @ g np w e y g i v Smith, 2011  MOLECULAR GEOMETRY b 8 of the The geometry or shape of a molecule is the three-dimensional arrangement 5 0 geometry f atoms in a molecule. The model used for the approximation of molecular 0 is VSEPR, or valence shell electron pair repulsion 9 b 7 d f 9 74 3 9 b 6 8 7 36 66 o m i l.c To predict the molecular m a geometry using the VSEPR Model: @ Draw the most plausible g Lewis structure of the molecule. Get the sum e p ofnthe electron pairs around the central atom. Determine g wthe arrangement of all the electron pairs in a way that minimizes i v y repulsions. QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER  b 8 5 0 90f d b 9 7 4f 9 7 63 7b 68 3 66 o m i l.c ma @g p wen y g iv  b 8 5 0 90f d b 9 7 4f 9 7 63 7b 68 3 66 o m i l.c ma @g p wen y g iv  b 8 5 0 90f d b 9 7 4f 9 7 63 7b 68 3 66 o m i l.c ma @g p wen y g iv Chang, 2010  b 8 5 0 90f d b 9 7 4f 9 7 63 7b 68 3 66 o m i l.c ma @g p wen y g iv Chang, 2010  Sample Problem Draw the Lewis structures of CH4, NH3, H2O. b8 5 0 f Determine their molecular geometry and bond0angles. b 9 7d f 9 74 39 b 6 8 7 36 66 o m Methane i l.c Ammonia Water ma CH4 @g NH3 H2O 109.5° np 107.3° 104.5° w e y g i v Repulsion bonding pair-bonding pair < lone pair-bonding pair < lone pair-lone pair  Sample Problem b8 Draw the Lewis structure of NF3. Determine its molecular geometry. 0 5 1. Determine the number of valence electrons. 0f 9 d b 2. Central atom has the least electronegativity. 7 9 O, and F) of terminal 3. Complete the octet/duet of strict followers (C, H,4fN, atoms first then central atom. 9 7 6 4. Adjust the number of bonds in order to complete3 the octet if necessary. 7b a sum of formal charges equal to 5. Adjust the number of bonds in order to8have 36a “desirable” set of formal charges. the charge of the molecule and to have 66 o m i l.c m a @g p wen y g iv  POLARITY 8 effect A molecule is polar when it has a net dipole, resulting from thebnet 5 0 0 f of the polar bonds. If the dipoles of polar bonds do not completely cancel out each other, then the molecule is polar. b9 d 9 7 4 f 3 97 b 6 8 7 36 66 o m i l.c m a @g p w en y g iv QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER  Sample Problem b Draw the Lewis structures of the following compounds 8 5 0 Classify as polar or nonpolar. 0f 9 d b 9 7 4 f BF3 CH4 PCl7 95 SF6 63 b 687 3 66 o m i l.c m a @g p w en y g iv QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER  Sample Problem Draw the Lewis structure of SF4 b 8 5 0 Classify as polar or nonpolar. 90f d b 9 7 4 f 9 7 63 7b 68 3 66 o m i l.c m a @g p w en y g iv QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER  Sample Problem Draw the Lewis structure of ClF3 b 8 5 0 Classify as polar or nonpolar. 90f d b 9 7 4 f 9 7 63 7b 68 3 66 o m i l.c m a @g p w en y g iv QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER  Sample Problem Draw the Lewis structure of BrF5 b 8 5 0 Classify as polar or nonpolar. 90f d b 9 7 4 f 9 7 63 7b 68 3 66 o m i l.c m a @g p w en y g iv QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER  Sample Problem Draw the Lewis structure of XeF4 b 8 5 0 Classify as polar or nonpolar. 90f d b 9 7 4 f 9 7 63 7b 68 3 66 o m i l.c m a @g p w en y g iv QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER  DIPOLE MOMENT b8which is A quantitative measure of bond polarity is its dipole moment,0𝝁, the product of the charge 𝑄 and the distance 𝑟 between the f5 charges: 0 9 𝝁 = 𝑸×𝒓 d b 9 7 4f 𝑸 refers only to the magnitude of the charge 7 9and not to its sign, so 𝝁 is 3 6 expressed in debye units, 𝐷. always positive. Dipole moments are usually 7b 8 6 !"# 3 1𝐷 = 3.336×10 66 𝐶-𝑚 o m i l.c Peter Debye made m a significant contributions in the study of many molecular structure, @g polymer chemistry, X-ray analysis, and electrolyte solution. He was npawarded the Nobel Prize in Chemistry in 1936. w e y g i v Chang, 2010 Diatomic molecules made up of different elements (HCl, CO, NO) have dipole moments and are polar, whereas those with the same elements (H2, O2, F2) are nonpolar because they do not have dipole moments. b 8 0 For a molecule made up of three or more atoms both the polarity of the bonds and the 5 0f molecular geometry determine whether there is a dipole moment. Even if polar bonds are 9 present, the molecule will not necessarily have a dipole moment. d b 9 7 4 f 9 7 63 7b 68 3 66 o m i l.c m a @g p w en y g In NH3 and NF3, the shift of electron density in NH3 is toward N and so contributes a larger iv dipole moment, whereas the NF bond moments are directed away from the N atom. So together they offset the contribution of the lone pair to the dipole moment. Thus, the resultant dipole moment in NH3 is larger than that in NF3. Chang, 2010  b 8 5 0 90f d b 9 7 4 f 9 7 63 7b 68 3 Dipole moments can be used to distinguish between molecules that have the same formula 66 but different structures. For example, the following molecules both exist; they have the o m same molecular formula (C2H2Cl2), the same number and type of bonds, but different molecular structures: i l.c m a @g p wen y g iv Chang, 2010  INTERMOLECULAR FORCES b London forces usually increase with molar mass because molecules8 with 5 0 forces f larger molar mass tend to have more electrons, and dispersion 0 increase in strength with the number of electrons. b9d 9 7 Furthermore, larger molar mass often 4 f 9 7 means a bigger atom whose electron 63 distribution is more easily disturbed b 8 7 because the outer electrons are less6 tightly held by the nuclei. 6 3 6 o m As expected, melting point i l.c increases m a as the number of gelectrons in the molecule increases. p @ e n g w ↑ MM i v y ↑ dispersion forces ↑ melting point Chang, 2010  ↑ MM ↑ dispersion forces ↑ boiling point for compounds with elements in the same period. Hydrogen compounds of Group 14 follow this trend. The lightest compound, CH4, has the lowest boiling point. The heaviest compound, SnH4, has the highest boiling point. b 8 5 0 90f d b 9 7 4 f 9 7 63 7b 68 3 66 o m i l.c m a @g p wen Hydrogen compounds of the elements in Groups 15, 16, and 17 do not follow this trend. y g The lightest compound has the highest boiling point and this observation must mean that iv there are stronger intermolecular attractions in NH3, H2O, and HF, compared to other molecules in the same groups. Chang, 2010  Additional Problem b 8 Which has a higher boiling point? CH3F or CCl4 0 f5 9 0 d b CH3F CCl4 f97 32 g/mol 154 g/mol 74 3 9 -78.4°C b 6 76.7°C polar 8 7 nonpolar 36 In many cases, dispersion forces are 6 6 comparable to or even greater than the dipole- o m dipole forces between polar molecules. Although CH3F is polar, it boils at a much c lower temperature than CCl4i,la. nonpolar molecule. a gm because Cl atoms being larger than F contains more CCl4 boils at a higher@temp n p the dispersion forces among CCl4 molecules are stronger than electrons. As a result, the dispersionw e plus the dipole-dipole forces among CH3F molecules. forces y g iv ↑ MM ↑ van der Waals forces ↑ boiling point QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER  INTERMOLECULAR FORCES b 8 Table 1.18. Summary of the types of intermolecular forces (Smith, 2011). 5 0 0 f b 9 Type of force Relative strength Exhibited by 7 d Example f 9 74 CH3CH2CH2CH2CH3 van der Waals Weak 9 All molecules 3 CH3CH2CH2CHO b 6 CH3CH2CH2CH2OH 8 7 with a Dipole-dipole Moderate 6 Molecules 3 dipole CH3CH2CH2CHO 66net CH3CH2CH2CH2OH o m Molecules with Strong il.c Hydrogen CH3CH2CH2CH2OH bonding m a O–H, N–H, or H–F g @ strong Ion-ion n p Very Ionic compounds NaCl, LiF w e y g v i ↑ IMFA strength ↑ melting or boiling point QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER  INTERMOLECULAR FORCES b 8 Hydrogen bonding is the electrostatic interaction between 0 the X–H 5 pair of (X = N, O, F) bond of one molecule and the unshared electron 0 f 9 b increases as another molecule, Y. The strength of a hydrogen dbond 9 7 the X–H bond dipole increases. 4 f 9 7 63 ↑ X–H bond dipole ↑ hydrogen 7b bond strength 6 8 6 3 6 o m i l.c m a @g np w e y g i v Chang, 2010  INTERMOLECULAR FORCES b 8 5 0 90f d b 9 7 4f 9 7 63 7b 68 3 66 o m i l.c a m g @ in water, ammonia, and hydrogen fluoride Hydrogen bonding np e gw Solid linesyrepresent covalent bonds, and dotted lines represent hydrogen bonds. iv Chang, 2010  Sample Problem b 8 Arrange HF, H2O, NH3, and CH4 in order of increasing boiling 5 0 point. 0 f 9 bH O CH4 NH3 HF 7 d 2 f 9 -161°C -33°C -19°C74 100°C 3 9 b 6 In H2O, two hydrogens and two lone 8pairs 7 of oxygen allow formation of hydrogen bond interactions in a63lattice 6 of water molecules. Water is considered an ideal hydrogen bonded 6 system. o m In NH3, the amount of hydrogen i l.c bonding is limited since N only has one a menough lone pairs to go around to satisfy all the H. lone pair. There are not @ g np In HF, there is shortage of H. w e y g i v ↑ IMFA strength ↑ boiling point QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER QUANTUM REVIEW CENTER