Ch8 chemical bonding

Question 1

ionic bond

Answer 1

electrostatic forces that exist between ions of opposite charge. Result from interaction of metals with nonmetals, usually by the transfer of one or more electrons from one atom to the other.

Question 2

covalent bond

Answer 2

results from sharing of electron between two atoms, usually when non metals interact with nonmetals

Question 3

metallic bond

Answer 3

found in metals, such as copper, iron, and aluminum. Each atom in a metal is bonded to several neighboring atoms. Bonding electrons are free to move throughout the 3D structure of the metal.

Question 4

octet rule

Answer 4

Atoms tend to gain/lost/share electrons until they are surrounded by eight valence electrons (unless they are trying to be like He, which only has 2)

Question 5

heat of ionic bonding

Answer 5

When a nonmetal gains an electron, it is endothermic. When removing an electron from a metal, it is endothermic. However, Ionic bonding is exothermic because the lattice energy is endothermic

Question 6

lattice energy3

Answer 6

-energy required to completely separate a mole of a solid ionic compound into its gaseous ions.
-For a given arrangement of ions, the lattice energy increases as the CHARGES INCREASE and as their radii decrease (but mostly on charges) because Eel=(kQ1Q2)/d
k=8.99*10^9J–m/C^2
-highly endothermic

Question 7

transition metals in bonding

Answer 7

generally don’t form ions that have a noble-gas configuration. In forming ions, transition metals lose the valence-shell s electrons first, then as many d electrons as are required to reach the charge of the ion. Fe: loses two 4s electrons and one 3d electron to become Fe3+

Question 8

born-haber cycle

Answer 8

• direct path to form compound: heat of formation of compound
• indirect path: 5 stpes: 1)vaporize elements 2)break bonds (no subscrpits) 3)remove e from metal 4)add e to non metal 5)add ions together
• sum of indirect steps should equal direct path

Question 9

bond length

Answer 9

The distance between bonded atoms (bond length) decreases as the number of shared electron pairs increases(single-double-triple). Going down a column (HF→ HI), bond length increases bc ionic size increases

Question 10

nonpolar covalent bond

Answer 10

electrons are shared equally between two atoms, as in Cl2, electronegs are equal

Question 11

polar covalent bond

Answer 11

one of the atoms exerts greater attraction for the bonding electrons than the other, as in HF

Question 12

electronegativity

Answer 12

• ability of an atom in a molecule to attract electrons to itself.
• Greater electronegativity =greater ability to attract electrons to itself.
• Increase left to right on PT and decreases top to bottom, with some exceptions in transition elements
• Highly negative electron affinity=high electronegativity

Question 13

pauling scale2

Answer 13

• electronegativity scale

- values are unitless. P, most electronegative is 4.0 and Cs, the least electronegative, is 0.7

Question 14

electro negativity difference4

Answer 14

• 0: nonpolar bond (Cl2)
• <1.9: polar covalent (HF). More electronegative has partial negative charge and less electroneg has partial positive charge
• > 3.0: ionic (LiF). electrons are completely transferred and elements have ionic charges
• greater difference in electronegativity=more polar their bond

Question 15

dipole moment3

Answer 15

• dipole: Whenever two electrical charges of equal magnitude but opposite sign are separated by a distance
• The magnitude of a dipole is called its dipole moment, denoted. mew=Qr where Q+ and Q- are two equal and opposite charges and r is the distance between them. Units: mew: debyes (D) which equals 3.3410^-30 (C-m), Q: e, which equals 1.6010^-19 C, r: angstroms
• increase as the magnitude of charges increase and as the distance increases. Going down a column (HF-HI), bond length increases however electronegativity difference decreases so dipole moment decreases

Question 16

Formal charge4

Answer 16

• charge the atom would have if all the atoms in the molecule had the same electronegativity. Bookkeeping convention to help choose between Lewis structures.
• To calculate:
  1) All nonbonded electrons are assigned to the atom on which they are found. 2)For any bond, half of the bonding electrons are assigned to each atom in the bond. 3) Subtract the number of electrons assigned to the atom from the number of valence electrons in the isolated atom
• The sum of the formal charges equals the overall charge on the ion. Formal charges on a neutral molecule add to zero, is not same as charge though
• to decide between lewis structures: 1) choose Lewis structure in which the atoms bear formal charges closest to zero. 2) choose the Lewis structure in which any negative charges reside on the more electronegative atoms

Question 17

resonance structure

Answer 17

different placement of electrons in the lewis structures of the same compound. This is because for some compounds (O3) have bonds that are in between double and single bonds, but since there are no rules for that, just write two resonance structures to average what it actually is. All resonance structures TOGETHER can accurately describe the compound in which all the bond lengths are the same

Question 18

exceptions to octet rule3

Answer 18

• Molecules and polyatomic ions containing an odd number of electrons (NO, ClO2)
• Molecules and polyatomic ions in which an atom has fewer than an octet of valence electrons: BF3. If following the regular rules, BF3 would have 3 resonance structures with 2 single bonds and 1 double bond. However, it is unlike F to have a double bond, so BF3 only has 1 structure with all single bonds, even though B will not have an octet
• Molecules of polyatomic ions in which an atom has more than an octet of valence electrons: PCl5. These only exist for atoms in the third row and lower because third period elements have ns, np, and unfilled nd orbitals that can be used in bonding. The number of molecules/ions with expanded valence shells increases as the central atom size increases, and when the surrounding atoms are the smallest and most electronegative, F, Cl, and O

Question 19

bond enthalpy4

Answer 19

• enthalpy change for the breaking of a particular bond in one mole of a gaseous substance (Cl–Cl(g)→ 2Cl(g)). It is represented as D
• always a positive quantity, energy is always required to break chemical bonds (endothermic) and released when a bond forms (exothermic) . Greater bond enthalpy=stronger bond
• derived for GASEOUS molecules and they are only averaged values
• As the number of bonds between two atoms increases, the bond grows shorter and stronger

Question 20

estimate enthalpy of reaction using bond enthalpy

Answer 20

1) Bonds are broken in the reactants that are not present in the products (endothermic)
2) Form bonds in the products that were not present in the reactants(exothermic)
- The enthalpy of reaction is the sum of the bond enthalpies of the bonds broken minus the sum of the bond enthalpies of the bonds formed: Hrxn=sum(bond H of bonds broken)-sum(bond H of bonds formed)