Effective nuclear charge increases in which direction across the periodic table?
Effective nuclear charge increases going from left to right in a period.
An ionic bond is the Coulombic attraction of two oppositely charged ions. Compounds between __________ and __________ are ionic.
Metals; nonmetals
When does a covalent bond occur?
A covalent bond results when atoms share one or more pairs of electrons. The electrons that are shared in a covalent bond are called bonding electrons or bonding pairs.
What is the number of electron regions around the sulfur atom in the following?

3
A double bond is single electron region. Therefore, a single bond, a double bond, and a lone pair form 3 electron regions.
Atomic radii increase in which direction across the periodic table?
Atomic radii increase going down a group and decrease going across a period.
Nonmetals form __________ by gaining the number of electrons required to fill their valence shell (outermost s and p sublevels). Metals form __________ by losing the outermost electrons in their valence shell.
Anions; cations
The Sn–Cl bond is which of the following?
Purely Covalent
Polar Covalent
Ionic
Polar covalent
Although tin is a metal, it lies well on the right side of the periodic table and is one of the metals that is an exception to the rule that metal-nonmetal bonds are ionic. The Sn–Cl bond is a polar covalent bond with only about 25% ionic character.
Which molecule has the larger bond angle?
NO3-
OR
BrO4-
NO3- > BrO4-
NO3- = 120o
BrO4- = 109o
Use only a Periodic Table to determine the atom in with the greatest ionization energy.
Si
P
S
P
Half-filled sublevels are usually unstable
Determine the oxidation state of Mn in MnO4-.
+7
O is –2 by Rule 5, and the charge on the MnO41– ion is –1, so x + 4(–2) = –1, or x = +7. The oxidation state of Mn is +7 in MnO41–. This very high oxidation state is stabilized by the highly electronegative oxygen atoms.
What is the name of the following compound?
Dinitrogen tetroxide
The 'a' at the end of a prefix is usually not used when the name of the element begins with a vowel, so N2O4 is dinitrogen tetroxide.
What is the hybridization around O3?

Sp3
O3 has four electron regions, so it is sp3 hybridized. The four hybrids are used in two σ bonds (O–H and C–O) and two lone pairs that are not shown. Oxygen always obeys the octet rule, so lone pairs must frequently be added to give oxygen an octet.
Which atom has the higher electronegativity?

X
Write the formula for a compound forming between Copper and Sulfur.
Cu2S
Copper is a transition element, and it commonly adopts the +2 oxidation state. Sulfur is a 6A nonmetal, so its oxidation state is –2. The formula is CuS. However, copper also adopts the +1 oxidation state, in which case, the formula would be Cu2S.
How many lone pairs exist in a Sulfite ion?
SO32- ;1
The sulfur atom has 3 shared pairs around it, which is 6 electrons. Thus, it needs two more electrons (1 lone pair) to obtain an octet.
Lewis structure and bonding in tetrafluoroethylene (C2F4):
ER = 6(8) = 48 electrons required with no sharing
VE = 2(4) from C + 4(7) from F = 36 valence electrons
SP = 1/2(48 - 36) = 6 shared pairs

How many π bonds are there?
P has a Zeff = 4.89, Pb, has a Zeff = 12.39. Which element has the lowest energy outermost orbitals given their effective nuclear charge?
Pb < P
Use:
E = -( Zeff2 / n2 ) to solve for the energy of the orbital
EPb = -4.26
EP = -2.66
What is the formula between compounds X and Z?

X2Z
X is +1 and Z is –2, so the compound is X2Z
Based on the bonding regions, how many resonance structures can be drawn for SO?
2
There are 3 shared pairs (bonds) but there are only 2 bonding regions. One bonding region must contain two of the bonds while the other contains a single bond. Since the double bond can be placed in either bonding region, there are two resonance forms of SO2.
Below is the MO diagram for He2. What type of MO does the He2 form?

Nonbonding MO
Each He atom has two electrons in its 1s orbital, so the two-atom system has four electrons. Two electrons would be placed in the σ orbital and two would be placed in the σ*. However, ΔE* > ΔE, so the energy of four electrons in He2 is greater than in two separated He atoms. The bond order of the He-He bond would be 1/2(2 − 2) = 0, no bond. Consequently, He2 does not form.