MCEN 1024 Exam 2

Ionic Compounds

formed between a metal and a non-metal, transferring electrons from one element to another, neutral b/c they contain equal number of positive and negative charges.

Covalent Compound

usually a bond between non-metals, they share valence electrons.

Ion

charged particles that form when atoms lose or gain electrons.

Cation

a positively charged Ion.

Anion

a negatively charged Ion.

What are the elements in Group 1A?

Alkali Metals

What are the elements in Group 2A?

Alkaline Earth Metals.

What are the elements in Group 8A?

Noble Gasses

What are the elements in Group 7A?

Halogens, they are highly reactive.

Polyatomic Ions

They are covalent bonds w/ Ions, Ionic compounds which consist of two or more atoms bond covalently.

How to name Ionic Compounds?

recognize which is non-metal and metal,

Place cation (metal) first, then the non-metal (anion),

"ide" is added to the non-metal,

Ex: Lithium Oxide, Copper II phosphide, Sodium sulfide

How to name Polyatomic "Oxyanions" with two in the family?

Non-metal bound with a oxygen atom,

The ion w/ more O atoms takes "ate".

The ion w/ less O atom takes "ite".

How to name Polyatomic "Oxyanions" with four in the family?

Non-metal bound with oxygen atoms,

The ion w/ most O atom takes "per" then "ate"

The ion w/ l fewer O atom takes just "ate"

Ion w/ 2 fewer O atoms takes "ite"

Ion w/ least O atom takes "hypo" then "ite"

Naming Hydrated Ionic Compounds

has specific number of water molecules in each formula unit,

Shows a center dot in the middle of the formula,

Use greek pre fix before word hydrate

CaSO4*2H2O - Calcium Sulfate dihydrate

How to name Binary acids?

a gaseous compound dissolved in water.

Use prefix "Hydro" + "non-metal root"+"ic"+acid

HCl-"Hydro" "Cholor""ic" "acid"

Naming Covalent compounds

two non-metals-

Element w/ lower group # comes first

Element w/ higher group# comes second

second element uses "Greek" prefixes and "ide" at the end

If both in the same group, the element with a higher period# comes first.

Exceptions: Oxygen and Halogens - Halogens first

EX: Sulfur hexafloride

Principle Quantum # (n)

specifies the energy level, relative size of orbital and distance from the nucleus

Angular momentum Quantum # (L)

is from (0 to n-1), related to shape of orbital and sets limits on L.

Magnetic Quantum # (M↓l)

from (-L to +L), prescribes 3-d orientation

Quantum Spin # (M↓S)

(+1/2 or -1/2) describes the spin of the electron

S orbital

L=0, spherical shape w/ the nucleus at its center (M↓l = 0)

P orbital

L=1, w/ n=2 or higher - lowest level is 2p

D orbital

L=2, w/ n=3 of higher - lowest level 3d

F orbital

L=3, n=4 or higher

Shielding

the effect of electron repulsion on each electron from other sub levels.

Valence electrons

outer electrons, are going to be shared or given, involved in forming compounds.

Bond order

the # of electron pairs being shared by a given pair of atoms. (number of bonds)

Ex: single bond, double bond, triple bond.

Oxidation # (O.N.)

#valence electrons -(#of shared electrons + #of unshared electrons)

Bond polarity

when ever atoms of different EN for a bond the bonding pair is shared unequally.

Non-polar covalent bond

When atoms are identical, the bonding pair is shared equally.

Electronegativity Difference (DeltaEN)

the difference between EN values of the bonded atoms is directly related to bond polarity.

Drawing Lewis structures

1)Place central atom in center (atom w/ lowest EN)

2)write tentative lewis structure

3)check valance electrons and Formal charge

Resonance Structures

have the same relative placement of atoms but different locations of bonds and lone electron pairs. A ↔ is placed between both lewis structures drawings.

Electron pair delocalization

in a single, double, or triple bond, each electron pair is localized between the bonded atom.

Formal Charge

The charge a compound has if the bonding electron were share equally.

Determining Formal Charge

=(#valence e)-(#of bonds)-(# of unshared e)

Choosing resonance structures

1)Smaller formal charge (+or-) are preferable to large one.

2)the same non-zero formal charge on adjacent atoms are not preferred.

3)A more negative formal charge should be on a more EN atom.

Electron deficient atoms

is an exception to octet rule, they have fewer than 8 e⁻ around the central atom.

Ex: Beryllium (Be) and Boron (Br)

Odd e⁻ Atoms

is an exception to octet rule, species that contain one unpaired e⁻(free radicals) which makes them parametric and extremely reactive.

Ex; Groups 5A and 7A(Halogens)

Free radical

one unpaired e⁻ and react w/ each other to pair up their lone e⁻.

Expanded Valance Shell

is an exception to octet rule, occur only w/ non-metal from period 3 or higher b/c d orbitals are available. More than 8e⁻ around central atom, expands its valance shell for more bonds and releases energy.

VSERP

Valence Shell Electron Repulsion

VSERP theory

to minimize repulsion, each group of valence e⁻ around a central atom is located as far as possible form each other.

Molecular shape

3-D arrangement of nuclei joined by the bonding groups.

Linear, Trigonal Planer, Tetrahedral, Trigonal bipyramidal, and Octahedral.

Bond angle

the angle formed by the nuclei of the surrounding atoms w/ the nucleus of the central atom as the vertex.

Linear Arrangement

Two e⁻ groups attached to a central atom in opposing direction w/ 180° bond angel (AX₂)

Trigonal Planer Arrangement

3 e⁻ groups around a central atom point to the comers of an equilateral triangle, w/ 120° angles (AX₃)

Trigonal Planer one lone pair e⁻

bent or V shape, lone pair repels bonding pair more than bonding pair repels each other, w/ 95° angles

Tetrahedral Arrangement

base with 2 or 3 e⁻ groups lies in a plane, requires 3-D dimensions to maximize separation, w/ 109.5° angles, all molecule or ions w/ 4 e⁻ groups around a central atom adopt this arrangement. (AX₄)

Tetrahedral one lone pair

Trigonal pyramidal shape w/ 107.3° angles (AX₃E)

Tetrahedral two lone pair

bent or V shaped (AX₂E₂) 104.5° angles

Ex: Water molecule

Molecular Polarity

an imbalance charge over the whole molecule or large portion of it.

Dipole moment

measurement of molecular polarity, units are (D) comes form SI units (Coulomb C) and Length (m)

1 D = 3.34X10⁻³ Cm

Polar Bond trends

The present of a polar bond does not always result in a polar molecule. We must consider the shape of the atom surrounding the central atom.

SO₄²⁻

Sulfate

NO₃⁻

Nitrate

ClO₃⁻

Chlorate

ClO₂⁻

Chlorite

ClO⁻

Hypocholrite

ClO₄⁻

Perchlorate

PO₄⁻

Phosphate

CO₃⁻

Carbonate

NO₂⁻

Nitrite

SO₃⁻

Sulfite

OH⁻

Hydroxide

Ammonium

NH₄

Trigonal Bipyramidal

5 or 6 e⁻ groups, central atom must be from n=3 or higher b/c D orbitals. Ideal bond angle is 90° and 120°. Has variations of 1,2,3 lone pairs of e⁻, and bond angles of 86.8° and 101.5°.

Octahedral

6 e⁻ groups, ideal bond angel is 90°, 90°, comes in variations of 1 and 2 lone pairs, w/ bond angles 81.5°.

Valance bond theory

basic principle is that a covalent bond forms when orbitals of two atoms overlap and a pair of e⁻ occupy the overlap region.

Maximum overlap of bond orbitals

Bond strength depends on the attraction between nuclei and shared e⁻, the greater the orbital overlap the stronger the bond.

Hybridization of atomic orbitals

the valence atomic orbitals in the isolated atoms become different when they are in the molecule.

Hybridization

The process of orbital mixing.

Hybrid orbitals

When orbitals mix they form new atomic orbitals, the # of orbitals = the # of atomic orbitals mixed, the shape and orientation of a hybrid orbital maximizes overlap w/ the orbital of the other atom in the bond, they correspond to the 5 e⁻ group arrangements in VSEPR theory.

SP Hybridization

Linear arrangement, two regions, one S and one P mix and form two hybrid orbitals, oriented 180° apart.

SP² Hybridization

Trigonal planer arrangement, 3 regions, mixing one S and two P orbitals gives 3 hybrid orbitals, axes are 120° apart.

SP³ Hybridization

Tetrahedral arrangement, 4 regions, Ideal bond angle 109.5°. Lone pair Trigonal pyramidal shape w/ 107.3° angles.

SP³d

Trigonal bipyramidal arrangement, 5 regions, have central atom from period 3 or higher, ideal bond angle is 90° and 120°/ Has variations of 1,2,3 lone pairs of e⁻, and bond angles of 86.8° and 101.5°.

SP³d²

Octahedral arrangement, 6 regions ideal bond angle is 90° and 90°,comes in variations of 1 and 2 lone pairs, w/ bond angles 81.5°

Sigma Bonds

End to end overlap, which has its highest e⁻ density along the bond axis, all single bonds.

Pi bonds

side to side overlap, has two regions (lobes) of e⁻ density, one above the sigma bond and one below the sigma bond axis, the two e⁻ in one pi bond occupy both lobes, any double bond consists of one sigma bond and one pi bond.

Percent yield %

Actual product formed / theoretical product formed

isomers

two or more compound with same molecular formula but different molecular structures / properties.

Molarity M =

moles of solution / L of solution

Avogadro's # =

6.022 x 10²³ units/atoms

Moles to mass

Mass (g) = moles ∙ #of grams / mole

Mass to mole

Moles = mass (g) ∙ moles / # of grams

Moles to atoms

Atoms = moles ∙ 6.022x10²³ atoms / 1 mole

Atoms to moles

Moles = atoms ∙ 1 mole / 6.022x10²³

Mass percent of X of a compound =

(atoms of X ∙ atomic mass of X) / molecular mass of compound x100

Mass % moles of a compound =

(moles of X ∙ molar mass of X) / mass (g) of 1 mole of compound x100

Mass of Element from mass % =

mass of compound ∙ mass of element in 1 mole of compound / mass of 1 mole of compound

Empirical formula

derived from mass analysis, show lowest whole number of moles and the relative # of atoms of each element in the compound.

Structural fomula

show the relative placement and connection of atoms in the molecule.

Determining Empirical formula

1) Find the mass (g) of each component element.

2) Convert mass(g) into moles and write preliminary formula.

3) Convert preliminary subscripts to whole integers

i)divide each subscript by smallest subscript

ii)If necessary, multiply through by the smallest integer that turns all subscripts into integers.

Determining Molecular formula

Whole # multiplier = molar mass (g) / empirical formula mass (g/mol)

Then multiply the empirical formula by whole # multiplier to get molecular formula.

Mass % to molecular formula

1)assume 100.0(g) of compound to express each mass percent directly as mass(g).

2)convert each mass(g) to moles.

3)Derive empirical formula

4)Then find whole # multiplier and molecular formula.

Limiting reaction

1)Balance equation

2)convert to moles

3)Multiply amount of moles to use mole ratio of reactants / product

4)Use limiting reactant to find moles of product the convert to grams if necessary.

Solute

A small quantity of substance in a solution.

Solvent

a large quantity of substance in a solution.

Concentration

the quantity of solute dissolved in a given quantity of solution.