Chem 16 NotesUPD

Should only be a supplement to discussions

Table of Contents*[A] Lecture

[1] Thermodynamics[2]Energy[3]Enthalpy[4]Hess's Law[5]Determining Enthalpy[6]Heat Capacity[7]Calorimetry[8] Entropy[9] Gibb's Free Energy[10] Waves[11] Quantum Theory[12] Quantum Numbers[13] Electron Configuration[14] Periodic Table[15] Periodic Trends

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[16] Chemical Bonding[17] Formal Charge[18] Resonance Structure[19] Bonds[20] VSEPR[22] Valence Bond Theory[22] Molecular Orbital Theory

[B]Lab[1] Corrosion[2] Oxidation Reduction Reactions[3] Calorimetry[4] Qualitative Analysis[5] Flame Test[6] Molecular Model

*(Ctrl + F the roman numeral to skip to that part. Ex: [B.3])

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[A.1] ThermodynamicsThermodynamics

“thermo” = heat “dynamics” = movement It has 3 parts

system the object being observed

surroundings everything not part of the system

boundary the division between system and surrounding

Universe the sum of the 3 parts

Internal Energy energy found within the universe sum of all Potential Energy (PE), Kinetic Energy (KE), and energy is general

Δ E=E final−E init=E product−E reactant

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[A.2]EnergyEnergy

capacity for heat and/or work heat

symbolized by q movement of energy due to the difference in temperature when a system is “cold”, it means it has a higher temperature and the heat is leaving

the system When a system is “hot”, it means the surrounding has a higher temperature and the

heat is entering the system work

symbolized by w can done by or to the system When work is done by the system, there is a release of energy When work is done to the system, there is an absorption of energy

Δ E=q+w

sign depends on change +q = endo -q = exo

Properties of Energy 1st Law

energy is constant

Δ Euniv=Δ Esys+ΔE surr=0

SI unit is Joules (J)

1[(kg)(m)

2]

s2

1 cal = 4.184 J Capital Letter(Cal) = kilo (kilocal) State function

doesn't care on process, just endpoint work, not power

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[A.3]EnthalpyEnthalpy

symbolized by H heat gained/lost at a constant pressure Δ H=q=Δ E+PΔV

heat of reaction state function can be + or - magniture of heat = proportional to the amount of substance measured in kJ/mol

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[A.4]Hess's LawHess's Law

enthalpy change is the sum of the enthalpy of individual steps

Δ H total=Δ H 1+ΔH 2+...

In the case that the reaction needed is happening in the opposite direction (you want the product but it is in the reactant,etc.) To reverse the reaction, one must simply get the negative of the ∆H

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[A.5]Determining EnthalpyDetermining Enthalpy

Δ H rxn=ΣmΔH [ f ( product )]0

−ΣnΔ H [ f (reactant )]0

m/n number of moles

Standard Heat of Formation symbolized by ∆H0

f

enthalpy change for the formation equation when substances are in their standard state stable = neutral = 0 Determined at 1 atm, 25oC, 1M

Measuring Heat of Reaction heat is proportional to ∆T q=kΔT

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[A.6]Heat CapacityHeat Capacity

heat required to change temperature by 1K/10C

q(Δ T )

=JK

Specific Heat heat required to change the temperature of 1g by 1K/10C

c=

q[(mass)(Δ T )]

=JgK

Molar Heat heat required to change the temperature of 1 mol by 1K/10C

C=

q[(mole)(ΔT )]

=J

molK

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[A.7]Calorimetry

Calorimetry measures the heat types

constant pressure styro coffee cup

constant volume bomb

assumes no heat exits the universe adiabatic system

useful solutions

qsys=−qsurr

the heat released is the heat absorbed q=mcΔT

qcal=CcalΔ T

q=nLRΔH

qrxn=−(qcal+qH2O)

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[A.8] EntropySpontaneity

changes happening without continuing outside influence

Entropy measure of chaos, disorder, or randomness in a system Meaning

Δ S>0

increases favors spontaneity

Δ S<0

decrease does not happen spontaneously

Δ S=0

only occurs at 0K means nothing is occuring

Δ S univ=Δ S sys+Δ S surr>0

Δ S gas>Δ S liquid>Δ S solid

Predicting Entropy Change in # of Particles

if there is an increase

Δ S sys>0 Change in # of moles in Gaseous substance

if there are more gaseous substances

Δ S sys>0

Values reference state is 0K, Absolute Zero

JmolK

Solving

Δ S 2980

=Σ nS product0

−Σ nS reactants0

2nd Law Δ S>0

ΔG>0

Aspects exothermic (∆H) does not ensure spontaneity ∆S does not ensure spontaneity

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[A.9] Gibb's Free EnergyGibb's Free Energy

energy released to surrounding reliable indicator of spontaneity Meaning

negative(-) means spontaneous positive (+) means non spontaneous 0 means equillibrium

Stable

ΔG0=0

Solved at 250C, 1 atm same as Enthalpy

ΔG2980

=ΣmG product0

−ΣnG reactants0

don't SF, constant (same w/ Hess)

Free Energy Calculation ΔG=ΔH−T Δ S

Gatas Equation constant temperature and pressure goal is -∆G

since that would be spontaneous

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[A.10] WavesWave

frequency,f cycles/sec

length, λ distance from 2 crests

Visible Light 400-750 nm

Speed of Light, c 3 x 108 m/s λf

Wave-particle duality of light Light acts as both a wave and a particle

Notable Equation for solving with waves

E=mc2

=hf =hcλ

E is energy m is mass c is speed of light h is Planck's constant ( 6.626 x 10-34 Js or kgm^2/s) λ is wavelength

Heisenberg's Uncertainty Principle can't know speed AND location

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[A.11] Quantum TheoryQuantum Theory

atoms only exist in certain energy levels atoms emit or absorb light as they change states allowed energy elevels, described by number This serves as the basis for how we understand the quantum world

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[A.12] Quantum NumbersQuantum Numbers

n, principal level or shells +1,2,...

l, angular momentum subshell, shape 0,1,...,(n-1)

ml, magnetic moment orientation -l,..., 0,..., +l

ms, spin spin of the electron either + ½ or – ½

Pauli's Exclusion Principle no two electrons can have the same quantum numbers

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[A.13] Electron ConfigurationElectron Configuration

distribution of electrons in atom in orbitals Placement and Arrangement 2 electrons at rest per orbital for degenerate, place electron alone first

Aufbau Principle building up orbitals are arranged in increasing energy ↑ n value, ↑ energy

generally s < p < d < f

Hund's Rule when writing electrons, write the electrons alone first before pairing up

Writing Electron Configuration Shorthand

(n)(l)number of electrons

Orbital Diagram Drawing Boxes and Stuff Can be used to determine magnetism

Paramagnetic attract at least 1 unpaired e-

odd #e- mod value of last orbital

Diamagnetic repel all paired Group 2A/2, Group 2B/12, Group 8A/18

Condensed Noble Gas + EC

NOTE Cr and Cu

only have 4s1

remove if cation since it is still the outermost

higher stability less repulsion if inner is completed

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[A.14] Periodic TableDefinition of Terms

S-block, P-block, D-block, F-block how the periodic table is arranged

Group 1A Alkaline Metal

Group 2A Alkaline Earth

Group 6A Chalcogens

Group 7A Halogens

Group 8A Noble Gas

Isoelectronic same number of e-

Downwards group

Across period

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[A.15] Periodic TrendsPeriodic Trends

Size (Atomic and Ionic) Down a group

increase Across a period

decrease Ionic

negative is larger Ionization Energy

energy needed to remove one e-

>0 Down

decrease Across

increase Electron Affinity

energy needed to add an electron to an isolated gas energy released <0 Down

decrease Across

increase Electronegativity

attract e-

Most electronegativity F Down

decrease Across

increase Metallic Behavior

behavior to lose an electron Down

increase Across

decrease

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[A.16] Chemical BondingChemical Bonding

used by elements to help achieve 8 valence electrons Ionic

Metal + Non-metal Electron Transfer

Covalent Non-metal + Non-metal Electron Sharing

Metallic Metal + Metal Sea of Electrons

LEDS Drawing the valence electrons about a chemical's symbol bonding octet rule, usually

there is a duet rule, incomplete octet, and expanded octet rule if there is a charge, place in a bracket and write charge in upper right

Ionic just write the charge next to the elements no lines, just the symbols next to each other

Covelent H is an exception to the octet rule in that it needs only 2 electrons 2 shared electrons

shown by line can be bond, double bond, or triple bond

“Steps” for n < 3 determine the central atom

it is the least electronegative count e- needed to complete count total available e-

(Needed e−available e)2

=Number of Bonds=(e shared )

2

Distribute remaining

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[A.17] Formal ChargeFormal Charge

charge of molecule/ion number associated to each atom in the LEDS negative goes to the more elctronegative atoms (and so not the central)

though that would mean the central atom can become positive Solving for Formal Charge

Group # - (Dots + Lines) aim is zero for most, if not all

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[A.18] Resonance StructureResonance Structure

bonding all possible LEDS where there is a possibility for multiple different ways of bonding The changing of which atom the double bond is found is not a result of rotating the

molecule it is different atoms possibly double bonding with the central atom

shown by double-headed arrows leading to the different structures

Resonance Hybrid Average of all resonance structures dotted lines to denote incomplete/shared bonds no dots

Exceptions in LEDS Duet Rule

H and He Incomplete Octet

less than 8 valence electrons 3A, B, Be, Odd e-

Expanded Octet more than 8 valence electrons elements where n greater than or equal to 3

this is due to allowing the extra electrons to ending the d-orbital or f-orbital

Formal Charge > Octet- prioritize formal charge over octet rule in determining stability

Bond OrderαBond Energyα1

(bond length)

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[A.19] BondsBond Polarity

determined by EN Non Polar

=0 Polar

not equal to Zero Can simply be determined if same element or not

except for C and H since they have nearly the same electronegativity meaning when they bond, it is non polar

Δ H rxn=Σ BE

Positive endothermic

Negative exothermic

Bonds Breaking

positive there is a release of energy

Forming negative use energy to form the bond

When in gaseous states,HCl is bonded

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[A.20] VSEPRValence Shell Electron Pair Repulsion (VSEPR)

shapes arrangement of e- domains determines polarity Strongest repulsion is lone-lone, then lone-bond, then bond-bond Electron Group Geometry

e- pairs arrangement Molecule Group Geometry

molecule arrangement

Electron Group Geometry

Number of Pairs Shape Angles in degrees Look

2 Linear 180

3 Trigonal Planar 120

4 Tetrahedral 109.5

5 Trigonal Pyramidal 90, 120

6 Octahedral 90

Molecular Group Geometry

Number of Pairs Lone Pairs Shape Look

2 1 Linear

3 1 Bent

2 Linear

4 1 Trigonal Pyramidal

2 Bent/Angular

5 1 Seesaw

2 T-shaped

3 Linear

6 1 Square Pyramid

2 Square Planar

3 T-Shaped

4 Linear

When there are no lone pairs, EGG = MGG

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[A.21] Valence Bond TheoryValence Bond Theory

covalent bonds formed by overlap of atomic orbitals hybridization S orbit + P + D + …

needs to be single to be able to bond count e- groups

Basis of Hybrid = # of e- groups = EGG Sigma (σ) bonds are head-on Pi (Π) bonds are sideways Cheat sheet

1 bond = 1 sigma Double Bond = 1 Sigma, 1 Pi Triple Bond = 1 Sigma, 2 Pi

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[A.22] Molecular Orbital TheoryMolecular Orbital Theory

Atomic orbitals combine to become molecular orbitals

Number of Molecular Orbitals forward=Number of AtomicOrbitalscombined

σ = head on bonding; σ* = anti-bonding Π = sideways bonding; Π* = anti bonding

Star is higher energy Written = σ1sσ*1s

s = only sigma bonds p= 1 sigma, 2 pi Sigma bonds have 2 electrons while Pi bonds have 4 Still follow lowest energy to highest Can be

B,C,N σ1sσ*1sσ2sσ*2sΠ2pσ2pΠ*2pσ*2p

O,F,Ne σ1sσ*1sσ2sσ*2sσ2pΠ2pΠ*2pσ*2p

Order can change due to the similar levels of energy of pi 2p and sigma 2p

Bond Order:

(Number of Bonding e−Number of Anti bonding e )2

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[B.1] CorrosionExperiment 6“Corrosion”

observinig a form of redox reaction a reaction where the nail is attacked and lost in various reactions

Results Straight Nail

Center is pink Tip and Head are blue

Bent Nail Tip, point of stress, and head are blue pink around

Nail with Zinc white near Zinc

Nail with Copper Wire Pink center Tip and Head are blue Copper Wire is unaffected

Explanation Straight

Fe→ Fe+2+2e−1

oxidize, anode

O2+H 2O+4 e−1→ 4OH−1

reduce, cathode

Fe+2+Fe(CN )6

−3→ Fe3[Fe(CN )6]2

Turnbull's blue

OH−1+Phth→ Pink

Obviously, Phenolphthalein pink blue due to stress

Bent strained more active, more anodic

Nail with Copper same since Copper didn't affect redox Fe > Cu in reducing property

Nail with Zinc

Zn→ Zn+2+2e−1

oxidize, anode

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Zn+2+OH−1

→ Zn(OH )2

Zn+2+Fe (CN )6

−3→ Zn3[Fe(CN )6]2

both are white precipitate Zn > Fe in reducing property

In the end Zn > Fe > Cu in reducting property

Things to Learn Reducing Property Using indicators as a way to tell the reaction has occurred

Special Notes Chemists were also Painters

that's why they named the reactions as colors and so Prussian Blue = the blue the Prussians wore Red Lake = the red used to paint lakes...?

Anode Where oxidation occurs An Ox

Cathode where reduction occurs Red Cat

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[B.2] Oxidation Reduction ReactionsExperiment 5“Oxidation Reduction Reaction”

learning about Redox Reactions learning about oxidation or reducing property Since areaction will occur with a visible change, that is the basis for the follow

experiment Reduction

gain e-

lower/reduce Oxidation Number GEROA

Oxidation loss e-

increase in Oxidation Number LEORA

Results seen on the following pages

Things to Learn Oxidizing and Reducing Property Difference in Product based on Environment How to Solve Redox/Net Ionic Reactions

Special Notes Some products are more available depending on the medium Acidic

H+, H2O Basic

OH-, H2O

Cu2+ Fe2+ Zn2+ H+

Cu - No Change No Change No Change

- Cu + Fe2+ →NVR Cu + Zn2+ → NVR Cu + 2H+ → NVR

- Fe > Cu Zn > Cu H2 > Cu

Fe Metal – rustSolution - yellow

- No Change Metal – corrodesSolution - Bubbles

Fe + Cu2+ → Cu +Fe2+

- Fe + Zn2+ → Fe +Zn2+

Zn + 2H+ → Fe2++H2

Fe > Cu - Zn > Cu Fe > H2

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Zn Metal – from silverto black

Solution – Fromblue to greenish

Metal – turnedblack

Solution – lightyellow

- Bubbles

Zn + Cu2+ → Cu +Cu2+

Zn + Fe2+ → Fe +Zn2+

- Zn + 2H+ → Zn2++H2

Zn > Cu Zn > Fe - Zn > H2

Reactivity Series: Zn > Fe > H2 < Cu

Cl- Br- I-

Br2 Organic(org) – TurnedYellow

Aqueous(aq) –Colorless

- Org – YellowAq – Yellow

No Reaction - Br2 + 2I- → I2+ 2Br-

Cl > Br2 - Br2 > I

I2 Org – PinkAq - Colorless

Org – PinkAq - Yellow

-

No Reaction No Reaction -

Cl > I2 Br > I2 -

Reactivity Series: Cl2 > Br2 > I2

Cl- Aq is Colorless Cl2 Org is Colorless

Br- Aq is Yellow Br2 Org is Yellow

I- Aq is Yellow I2 Org is pink

SO32- + MnO4

-

Neutral 2MnO4

- + 3SO32- + H2O → 2MnO2+ 3SO4

2- + 2OH-

MnO2 is brown ppt Acidic

2MnO4- + 3SO3

2- + 6H+ → 2Mn2++ 5SO42- + 3H2O

Colorless Basic

2MnO4- + SO3

2- + 2OH- → 2MnO42- + SO4

2- + H2O auto dissociation

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[B.3] CalorimetryExperiment 8“Calorimetry

heat measurement system + surrounding = universe

seperated by a wall Exothermic = negative ∆H Endothermix = positive ∆ When Pressure is constant, this is a coffe cup calorimeter and solves for ∆H When Volume is constant, this is a bomb calorimeter and solves for ∆E Using a coffe cup calorimeter, seeing reactions and changes in temperature due to them Can be used to solve for heat, specific heat, temperature change,etc Notable Equations

q=(m)(c)(Δ T )=(Ccal )(ΔT )=(mwater)(4.18

JgC

)(ΔT )

or more specifically with the experiment

mcΔ T=−[C cal ΔT+mwater 4.18

JgC

ΔT ]

qrxn+qH20+qcal=0

Matter Energy

Open Permeable Permeable

Closed Impermeable Permeable Diathermal

Isolated Impermeable Impermeable Adiabatic

Things to Learn Solving Calorimetry problems

Special Notes Thermometer

part of the surroundings if an increase in temp is noted, it is exothermic

because the surround absorbed the energy released by the system(the reaction) Theoretical ∆H for Neutralization Reactions = -55.85 kJ/mol

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[B.4] Qualitative AnalysisExperiment 11“Qualitative Analysis”

to observe various reactions and use qualities to define elements doesn't care about quantity like how much substance but instead if the substance is

present or not Quantitative → numerical value/amount Qualitative → test of Presence

Classical has a limit of detection at .1M

Instrumental spectrometer

Tests Elimination

allows you to group the elements Confirmatory

allows you to identifyResults

seen in the following pages

Things to Learn- being able to tell the presence of elements based on qualities such as color and acidity

Cations Test

Add NaOH

Blue ppt Brown Ppt White Ppt Nothing

Copper Iron(Yellow Sol'n)

Calcium OR Zinc(Zn is sparingly soluble)

Ammonia

Cu2+ + 2(OH)- →Cu(OH)2

Fe3+ + 3(OH)- →Fe(OH)3

Ca2+ + 2(OH)- →Ca(OH)2

Zn2+ + 2(OH)- →Zn(OH)2

NH4 + + OH- → NH3 + H2O

Add Excess NaOH

PPT remains PPT dissolves

Zn(OH)2 + 2OH-

→ [Zn(OH)4]2-

Add NH3

Blue Ppt Brown Ppt White Ppt Nothing

Copper Iron Zinc Calcium OR Ammonia

Cu2+ + 2(OH)-

→ Cu(OH)2

Fe3+ + 3(OH)- →Fe(OH)3

Zn2+ + 2(OH)- →Zn(OH)2

CaWeak OH, soluble

NH4 + + OH- → NH3 +

H2O

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Add Excess NH3

Deep Blue White Solution

Cu(OH)2 + 4NH3

→ [Cu(NH3)4]2+Zn(OH)2 + 4NH3

→ [Zn(NH3)4]2+

Confirmatory Test

Copper Cu2+ + NH3 → [Cu(NH3)4]2+ Blue

Iron Fe + SCN Blood Red Complex

Calcium Ca + C2O4 White ppt

Zinc Zn2+ + Fe(CN)63- →

Zn3[Fe(CN)6]2

Yellow ppt

Zn2+ + Fe(CN)64- →

Zn2[Fe(CN)6]White ppt

Ammonia NH4+ + OH- → NH3 + H2O Evolution of Gas

Red Litmus to Blue

Anions Test

Add HNO3 and Ba(NO3)2

PptAdd Fe3+

Pink Organic Blood RedAqueous

Nothing

Carbonate OR Phosphate OR Sulphate Iodine Thiocynate Bromine OR Nitrate

Ba2+ + CO32-

→ BaCO3

Ba2+ + PO43-

→ Ba3(PO4)2

Ba2+ + SO42-

→ BaSO4

2I- → I2 + 2e Fe3+ + SCN→

[FeSCN]2+

Add Acetic Acid (Use new sol'n with HNO3 and Ba(NO3)2) Add KMnO4

Pptdisappears,

bubbles

Pptdisappears

Ppt remains Pink Organic YellowOrganic

Nothing

Carbonate Phosphate Sulphate Iodine Bromine Thiocynate OR Nitrate

BaCO3 +2OAc →

Ba(OAc) +H2CO3

Ba3(PO4)2 +6OAc →

Ba(OAc) +2H3PO4

BaSO4 +2OAc →Nothing

2I → I2 + 2e 2Br → Br2 +2e

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Confirmatory Test for Nitrate Sol'n + 1 drop 6M sulfuric + 3 drops Fresh Iron Sulfate + 2 drops 18 Sulfuric Evolution of Heat is NOT an indicator

Dilution of Sulfuric Acid always results in the evolution of heat

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[B.5] Flame TestExperiment 9“Flame Test”

igniting various compounds to see the color released a form of qualitative analysis in that the color released reflects the element contained a way of showing what happens when energy is absorbed and thus released Motion from the ground state to the excited state and back Useful Equations

E=

hcλ

E=mc2

f λ=c

Note

h=6.626 x10−34( J )s∨

[(kg)m2]

s

c=3 x 108m /s

Visible Light's wavelength is within 400-750 nm( 1 x 10-9)

ResultsSubstance Color of Flame

H 2O None

NaCl Red

CaCl2 Yellow

CuCl2 Blue Green

KCl Violet/Purple

BaCl2 Yellow-Green

Things to learn– Solving problems regarding the equations above

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[B.6] Molecular ModelExperiment 12“Molecular Model”

using sticks and clay to represent molecular compounds visualization of how these compounds are shaped application of VSEPR

Things to learn How to solve

Number of Valence Electrons group number

LEDS Formal Charge Resonance Structure EGG MGG Sigma Bonds Pi Bonds Bond Angles Bond Length Bond Polarity Hybridization Polarity of the Molecule

if charged, polar because it means it's soluble in water dissociation

OverlappingValence Orbitals

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