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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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