Chapters 7, 8 & 9

- For each of the compounds below, draw the Lewis structure.

- PBr
_{3}, tripodal, sp^{3}

- CCl
_{4}, tetrahedral, sp^{3}

- NH
_{3, }tripodal, sp^{3}

- BF
_{3}, trigonal planar, sp^{2 }

- PF
_{5, }trigonal bipyramid, sp^{3}d

- SF
_{4}, wedge, sp^{3}d

- SeCl
_{6}, octahedral, sp^{3}d^{2}

- I
_{3}^{-}, linear, sp^{3}d

- XeF
_{4}, square planar, sp^{3}d^{2}

- ICl
_{3}, t-shaped, sp^{3}d

- ClO
_{3}^{-}, trigonal planar, sp^{2}plus d orbital interactions in p bonds. Only one of three possible resonance structures is shown.

- PBr
- Properly name the following compounds.

- MnSO
_{4}manganese (II)sulfate

- AlPO
_{4}alluminum phosphate

- P
_{2}O_{3}diphosphorus trioxide

- K
_{3}PO_{4}potassium phosphate

- KClO
_{3}potassium chlorate

- Sc
_{2}S_{3}Scandium (III) sulfide

- N
_{2}O_{5}dinitrogen pentoxide

- Ca(NO
_{3})_{2}calcium nitrate

- CoNO
_{3}cobalt (I) nitrate

- FeSO
_{4}iron (II) sulfate

- MnSO
- Draw Lewis structures for these more complex molecules:

- hydrazine

- ethylene

- propylene

- hydrazine
- Define:

- valence electrons - reactive electrons above the previous noble gas, likely to participate in bonding

- bonding pair of electrons - a pair of electyrons shared between two atoms to form a bond

- lone pair or non-bonding pair (electrons)- a pari of electrons resident on an atom that does not participate in bonding

- electronegativity - the attraction an element has for the electrons in a bond with a different element

- first ionization energy - the energy needed to remove the outermost electron of an atom

- quantum number - a number designating an allowed energy

- n - prinicpal quantum number, n = 1, 2, 3,...

- l - angular momentum (azimuthal) quantum number, l = 0 to n -1

- m
_{l}- magnetic quantum number, m_{l}= -l to +l

- m
_{s}- spin quantum number, m_{s}= +1/2 or -/12

- Pauli exculsion priciple - each electyron in an atom, ion or molecule has a unique set of quantum numbers

- Hund's rule - electrons are placed in orbitals with the same (degenerate) energies with parallel spins prior to pairing electrons in orbitals

- periodicity - the property that similar chemical properties recur within the periodic table. Such properties lead to groups or families (vertical columns) on the periodic table.

- period - a horizontal row on the periodic table.

- group or family - a vertical column on the periodic table.

- isoelectronic - having the same electron configuration.

- paramagnetic - containing at least one unpaired electron.

- diamagnetic - containing no unpaired electrons.

- valence electrons - reactive electrons above the previous noble gas, likely to participate in bonding
- Predict the products for the reaction between the paired elements below.

- O and Li

Li_{2}O

- P and S

P_{2}S_{3}or P_{2}S_{5}

- S and F

SF_{2}, SF_{4}or SF_{6}

- Al and F

AlF_{3}

- Mg and N

Mg_{3}N_{2}

- O and Li
- What are the electron configurations for the elments below? Follow the Aufbau (building-up) principle.

- Mn = [Ar]4s
^{2}3d^{5}

- Se = [Ar]4
^{2}3d^{10}4s^{2}

- Pr = [Xe]6s
^{2}4f^{3}

- Pb = [Xe]6s
^{2}4f^{14}5d^{10}6p^{2}

- As = [Ar]4s
^{2}3d^{10}4p^{3}

- Sr = [Kr]5s
^{2}

- In = [Kr]5s
^{2}4d^{10}5p^{1}

- He = 1s
^{2}

- Mg = [Ne]3s
^{2}

- Sc = [Ar]4s
^{1}3d^{1}

- Mn = [Ar]4s
- Provide the quantum numbers for the last electron placed in each of the elements in problem above.

- n = 3, l = 2, m
_{l}= 2, m_{s}= +1/2

- n = 4, l = 1, m
_{l}= -1, m_{s}= -1/2

- n = 4, l = 3, m
_{l}= -1, m_{s}= +1/2

- n = 6, l = 1, m
_{l}= 0, m_{s}= +1/2

- n = 4, l = 1, m
_{l}= 1, m_{s}= +1/2

- n = 5, l = 0, m
_{l}= 0, m_{s}= -1/2

- n = 5, l = 1, m
_{l}= -1, m_{s}= +1/2

- n = 1, l = 0, m
_{l}= 0, m_{s}= -1/2

- n = 3, l = 0, m
_{l}= 0, m_{s}= -1/2

- n = 3, l = 0, m
_{l}= 3, m_{s}= +1/2

- n = 3, l = 2, m
- Tell if each element in the problem two problems above is diamagnetic or paramagnetic.

- paramagnetic

- paramagnetic

- paramagnetic

- paramagnetic

- paramagnetic

- diamagnetic

- paramagnetic

- diamagnetic

- diamagnetic

- paramagnetic>

- paramagnetic
- Rationalize your way to the proper electron configurations for these exceptions to the Aufbau (building-up) principle

- Cr = [He]4s
^{1}3d^{5}

- Gd = [He]6s
^{2}4f^{7}5d^{1}

- Au = [He]6s
^{1}4f^{14}5d^{10}

- Pd = [Kr]5s
^{0}4d^{10}

- Mo = [Kr]5s
^{1}4d^{5}

- Cr = [He]4s
- What is the trend for electronegativity?

Increases traveling up a columns and increasing the farther right in a row, for the s and p blocks. For the d and f blocks a chart needs to be consulted.

. - What is the trend for first ionization energy?

Increases traveling up a columns and increasing the farther right in a row.

- What is the trend for electron affinity?

Increases traveling up a columns and increasing the farther right in a row for the absolute value of the electron affinity.

- Given the formula for the compound below in each example, predict the formula for the compound asked for

- If the formula for the oxide of aluminum is Al
_{2}O_{3}, what would the formula for the oxide of indium, In, be?

In_{2}O_{3}

- If the formula for the sulfide of beryllium is BeS, what would the formula for the sulfide of strontium, Sr, be?

SrS

- If the formula for the nitride of sodium is Na
_{3}N, what would the formula for the phosphide (P^{3-}) of sodium be?

Na_{3}P

- If the formula for the fluoride of of zinc is ZnF
_{2}, what would the formula for the fluoride of cadmium, Cd, be?

CdF_{2}

- If the formula for the oxide of titanium is TiO
_{2}, what would the formula for the oxide of zirconium, Zr, be?

ZrO_{2}

- If the formula for the oxide of aluminum is Al
- Are the compounds below ionic or covalent?

- CH
_{4}

covalent

- Na
_{2}S

ionic

- SO
_{3}

covalent

- Na
_{2}SO_{3}

ionic

- MgH
_{2}

ionic

- P
_{2}S_{3}

covalent

- BF
_{3}

covalent

- AlBr
_{3}

ionic

- SeO
_{2}

covalent

- CO
_{2}

covalent

- CH
- Determine the formal oxidation state of the underlined element in the compounds and ions below:

__Cl__O_{4}^{-}

cl + 4(-2) = -1

cl - 8 = -1

cl -8 + 8 = -1 + 8

cl = +7

- P
_{2}__O___{3}

oxygen wants two electrons, so o = -2

__Fe___{2}(SO_{4})_{3}

2(fe) + 3(-2) = 0

2(fe) - 6 = 0

2(fe) - 6 + 6 = 0 + 6

2(fe) = +6

fe = +6/2 = +3

__P__PF_{5}

p + 5(-1) = 0

p - 5 = 0

p - 5 + 5 = 0 + 5

p =+5

__Cl__O_{2}^{-}

cl + 2(-2) = -1

cl - 4 = -1

cl - 4 + 4 = -1 + 4

cl = +3

__Sn__S_{2}

sn + 2 (-2) = 0

sn - 4 = 0

sn - 4 + 4 = 0 + 4

sn = +4

__P__O_{4}^{3-}

p + 4(-2) = -3

p - 8 = -3

p -8 + 8 = -3 + 8

p = +5

__Cl__O^{-}

cl + -2 = -1

cl + -2 + 2 = -1 + 2

cl = +1

- Li
_{3}__N__

N would like to gain three electrons, so -3.

__C__O_{3}^{2-}

c + 3(-2) = -2

c - 6 = -2

c - 6 + 6 = -2 + 6

c = +4

- Label all the columns and parts of the periodic table on the blank periodic table below: