Issue 34: Chemistry: Bonding (Hoyle's Freshman Honor Chem Class)

Metallic Bonds

Bonding between two metals

A crystalline lattice is formed

Mobile Sea of Electrons

           properties + explanation of properties

Covalent Bonds

dot structure

polarity

properties

shape

bond+molecules

Ionic Bonds

naming

dot structure

properties

def.

In a metallic bond the electrons of the metal are released by the metal because the metal doesn't want the electrons.  When 2 or more metals bond they all release the electrons but they still are bonded because the free electrons still have nowhere to go and the metal who now have a positive charge are attracted to the electrons.  the metals still don't want it but are attracted.  with this each electron and metal atom are each attracted to each other forming the bond. this set of electrons are movable because they aren't stuck to any particular set of metals(called sea of electrons).  this lets it conduct electricity because the current of electrons are able to move through it by pushing one in on one end and one has to come out on the other.  the metal also becomes malleable because the positive metals cannot touch each other and repel each other  the metal atoms are cushioned by the sea of electrons so it becomes malleable.  it also has a very high melting point because the attraction of the electrons and metals are so high it is very hard to overcome.

https://lh6.googleusercontent.com/Uk_12vEHSH7J5sCoRi_177gwWe7lPWmCOVk2wgJZTLT7iihiVeya45Ck5Mxd6FwXyO7mrZzcKhfYWWtTJTJnWWJMwGGRC4BKtRt-umrtsFgCcqrxVtIp

Covalent bonds are bonds between nonmetals. The nonmetals share electrons.

Diatomic Molecules: Elements that are bonded to themselves. They are nonpolar bonds.

Dot Structures:

Polarity:

• Polar bond: covalent bonds in which atoms share electrons unevenly due to a difference in electronegativity
• Polar molecule: has polar bonds and is asymmetric

Properties:

• Lower melting and boiling points than Ionic compounds
• Can’t conduct electricity when dissolved in water
• Formed by sharing electrons
• Can be either soft, hard, or flexible

Shape:

# of lone pairs on “central” atom	# of bonding groups/domains on “central” atom	Electron-pair geometry	Molecular Geometry	Bond angle
0	2	Linear	Linear	180
0	3	Trigonal Planar	Trigonal Planar	120
1	2	Trigonal Planar	Bent	Less than 120
0	4	Tetrahedral	Tetrahedral	109.5
1	3	Tetrahedral	Trigonal Pyramidal	Less than 109.5 about 107
2	2	Tetrahedral	Bent	Less than 109.5
0	5	Trigonal Bipyramidal	Trigonal Bipyramidal	90, 120, and 180
1	4	Trigonal Bipyramidal	Seesaw	90, 120, and 180
2	3	Trigonal Bipyramidal	T-shaped	90 and 180
3	2	Trigonal Bipyramidal	Linear	180
0	6	Octahedral	Octahedral	90 and 180
1	5	Octahedral	Square Pyramidal	90 and 180
2	4	Octahedral	Square Planar	90 and 180

Notes for Molecular geometry and Electron Geometry:

Total lone pairs and bonding groups/domains	Electron Geometry
2	Linear
3	Trigonal Planar
4	Tetrahedral
5	Trigonal Bipyramidal
6	Octahedral

*If there are no lone pairs, the electron geometry and molecular geometry are the same

Naming:

• Add a prefix for the number of atoms
• 1-mono
• 2-di
• 3-tri
• 4-tetra
• 5-pent
• 6-hex
• Drop mono if it is the first atom
• Ends in -ide
• Drop vowels to make it sound nicer

Valence Shell Electron-Pair Repulsion Theory (VSEPR): VSEPR uses the Lewis structures to predict the molecular geometry of a covalently bonded molecule. It states that 3D arrangements of atoms surrounding the central atom is determined by the repulsions between the bonding and nonbonding electron pairs in the central atom’s valence shell.

Here are the steps to predicting the geometrical structure of a molecule:

1. Draw the Lewis Dot Structure of the molecule
2. Count the total number of bonding and nonbonding electron pairs in the central atom’s valence shell
3. Arrange the electron pairs so that they are as far apart from each other as possible

http://intro.chem.okstate.edu/1314f97/chapter9/vsepr.html

LINK HAS INFO ON VSEPR

Covalent Bonding Examples:

Phosphorus Pentachloride:

Nitrate (NO3−) (Resonance Structure):

Hydrogen Cyanide (HCN):

Formal Charge:

Atoms want to have the smallest possible formal charge

Formal Charge Link:

http://cosm.georgiasouthern.edu/chemistry/general/molecule/fc.htm

Resonance Structures Link:

http://cosm.georgiasouthern.edu/chemistry/general/molecule/resonan.htm

Ionic Bonding:

Basics: Ionic bonding consists of metals bonding with non-metals. Metals tend to lose electron to achieve a full shell of electrons because they usually have 4 or less electrons (not including transition metals). To achieve this full shell metals give away electrons to non-metals therefore achieving a full shell that complies with the octet rule. Non-metals on the other hand have 5,6,7 electrons and will receive electrons rather than give away them to achieve a full shell. Non-metals when they receive electrons from metals can achieve a full shell.

Example:

Magnesium (2 valence electrons), Oxygen (6 valence electrons), Magnesium gives away 2 electrons and gets a full shell, and oxygen receives the 2 electron and gets a full shell.

Ions: Ions are formed when an atom in a an ionic bond either gives away or receives them. When an atom loses electrons it becomes a positive ion, because now the protons outnumber the electrons in the atom. A positive ion is called a cation. On the other hand when an atom receives electrons it gets a negative charge because now its electrons outnumber the protons. A negative ion is called an anion.

Examples:

Magnesium (2 valence electrons), Oxygen (6 valence electrons), Magnesium gives away 2 electrons and gets a full shell, and oxygen receives the 2 electron and gets a full shell. Magnesium now has 2 less electrons than protons which means it formed a 2+ ion. Oxygen has received 2 electrons which means it has 2 more electrons than protons therefore forming a 2- ion.

Octet Rule: Atoms become stable when they have a full valence shell.

Ionic-Bonding: When the ions are formed they will have opposite charges. Opposite charges attract so the two atoms bond.

Examples:

Magnesium (2 valence electrons), Oxygen (6 valence electrons), Magnesium gives away 2 electrons and gets a full shell, and oxygen receives the 2 electron and gets a full shell. Magnesium now has 2 less electrons than protons which means it formed a 2+ ion. Oxygen has received 2 electrons which means it has 2 more electrons than protons therefore forming a 2- ion. Since 2+ and 2- are opposite charges magnesium and oxygen bond.

Naming: The first term will be the positive ion (Magnesium) unaltered. Then you attach the second term (Oxygen) however, you add an -ide suffix (Oxide). If the naming includes a transition metal, the charge is denoted by Roman Numerals in parenthesis next to the transition metal. The transition metals Silver, Zinc, and Aluminum do not need their charges denoted with Roman Numerals since they can only have one possible charge.

Example:

Magnesium Oxide

Iron(II) Sulfate

Chemical Symbol

To represent an ionic bond put the positive ion symbol first then attach the negative ion symbol second. If there are multiple amount of atoms indicate it with subscripts

Examples:

Li2O

Lewis dot structures with ionic bonds: This is a more complex way of showing an ionic bond compared to the chemical symbol format, and is simpler than Bohr notation. First you put the positive ion including the ion number (Mg 2+), and then you put the negative ion in brackets with 8 electrons around it ([O]2-).

Really Good Example:

https://lh3.googleusercontent.com/UE_UxcmWZIdFRjuRRxnA8wfhFawmdRRBYs_7-fklRAhyA_CwAFyVamOR_Yi6psATj5TKlR6y0N9iy4j-5toaprhcNJPd_Xe6iQ7j0HeQfhyyu3rV3DKihttps://lh3.googleusercontent.com/UE_UxcmWZIdFRjuRRxnA8wfhFawmdRRBYs_7-fklRAhyA_CwAFyVamOR_Yi6psATj5TKlR6y0N9iy4j-5toaprhcNJPd_Xe6iQ7j0HeQfhyyu3rV3DKi

Electronegativity:

Electronegativity is the force an atom has for absorbing electrons. Usually non-metals tend to have high electronegativity because they “want” electrons. The higher the electronegativity, the easier it is for an atom to receive an electron

Properties of Ionic Compounds and Ionic Bonds

• They are formed through a transfer of electrons
• They are hard
• They are brittle
• They have high melting points and boiling points
• They are formed between a metal and nonmetal
• They conduct electricity when they are dissolved in water (aqueous) or in a molten state; not when they are in a solid form
• Electrolytes dissociate in water

Hybridization

Hybridization is when the orbitals are shared so that an element can bond with more elements. When the two orbital combine they form a new one which is usually a hybrid of the two orbitals.

Valence Bond Theory-covalent bonds are overlap of orbitals

*Hybridization is an expansion on this theory

The first bond is called a sigma bond.

The second and third bonds are called pi bonds.

Method to find hybridized orbitals:

1) Draw dot structure

2) Count # of unique electron sites around the central atom (steric #)

3) Steric # is the number of hybridized orbitals

4) Start with s orbital & hybridize until you reach steric #https://lh3.googleusercontent.com/5qdEQBzQYaqMBnvfuelRLGsbJsxtLOZKmSFuJdCyALTx8WVEd5o_ZgjBgYj29r1uvE9nsxZI7l8miXhYdBmJgbQ_5CU24-OmeATVnhrxF9Mdtq6iaml7

Ammonia Example:

Since all of the hydrogen atoms bond equally with the carbon (CH4), the bonds need to be equal. The orbitals need to be the same, so the 2s and 2p would hybridize, meaning they would form sp3. You can tell it would be sp3 because of the method described above.

Link about hybridization:

https://chemistry.boisestate.edu/richardbanks/inorganic/bonding%20and%20hybridization/bonding_hybridization.htm

Issue 33: Chemistry: Exothermic/Endothermic (Hoyle's Freshman Honor Chem Class)

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Exothermic: Heat released. Heat is product.

Endothermic: Heat absorbed. Heat is reactant.

Calorimetry can be used to find out how much heat is released:

1.Using q(water)=-q(rxn)

2.Then use q=mCΔT

Heat is in proportion to coefficients

Enthalpy-heat of a rxn(ΔH)

Sign of ΔH shows if a reaction is endothermic or exothermic. If ΔH is positive, the rxn is endothermic, but if ΔH is negative, the rxn is exothermic.

Potential Energy is stored in bonds, so heat in rxns comes from the breaking of bonds.

Potential Energy Diagram

If reactants are higher than products, the reaction is exothermic. If the products are higher than the reactants, the reaction is endothermic.

ΔH=potential energy of products-potential energy of reactants

Universe favors lower energy systems

Activation energy-energy needed to get a reaction started

Activation energy is the potential energy of the highest point on the potential energy diagram

Activated complex-unstable transition state

Catalyst-activation energy decreases

Entropy-measure of disorder (ΔS)

Systems tend towards higher disorder so systems tend towards higher entropy

Gases have high entropy, solids have low entropy

Higher entropy is probably because there are more ways to be disordered than to be ordered

How is entropy measured

To start off, entropy is the measure of system’s disorder. The higher is the disorder, the higher is entropy, and vice versa. Although determining entropy seems quite impossible, in fact, scientists derived formula for measuring change in disorder.

It is   , in which S stands for increase in disorder, q is the heat absorbed by substance (In J, of course).To understand this, we must understand that solid has lower entropy (less disorder) than gas, because solid’s atoms have less combinations, compared to gas. Also, the higher is the temperature, the more energy is applied to the atoms, thus canceling out the IMFs between the atoms, and pulling them apart, thus creating a gas. So, at higher temperatures, the substance has higher change in entropy.

On the other hand, T is the temperature at which the absorption takes place (again, in K). This characteristic of entropy is a little bit more confusing. To understand this, let’s make an analogy:

If a 10-year-old boy is allowed to play in his bedroom for half an hour, the increase in disorder is  lower, because the room is already disordered (i.e., at a higher “temperature”). By contrast, if the boy is let loose for half an hour in a neat and tidy living room (i.e., at a lower “temperature”), the effect is much more dramatic.

So, the higher is the temperature at which the absorption is taking place, the lower is the change in disorder, because the substance is already in disorder (again, gas has higher entropy than solid).

Now, let’s see this equation in action!

In this problem, we have an a substance that absorbed 1 kJ (1000 J), at different temperatures. First is 3 K.

Second is 300 K.

Third is 3000 K.

Ready? Set. Chemistry!

a) At 3 K we have:

           

As you can see, at lower temperatures, the change in disorder is extremely high.

b) At 300 K, similarly,

           

c) At 3000 K

           

Now, we notice that change in disorder is extremely low. That is due to the temperature at which this absorption is taking place. The substance is already at disorder, thus it is pretty hard for it to become more disorderly.

Calorimetry

To start off, let’s define what do we mean by calorimetry. Usually, it is the energy that is located inside the substance. However, in the classroom, most of the time, we don’t really talk about calorimetry. Rather, when we use the word “calorimetry,” we mean heat of capacity. Heat of capacity is the amount of heat (energy, usually in J or kJ, sometimes in Kilocalories) required to raise the temperature of one gram of substance by 1 K/C. Remember that during these reactions, volume and pressure must remain constant (thus, if this is not stated in the problem, assume it does). Now, let’s go from theory to actual Math!

Necessary Equations:

           q water = -q rxn

           q=mC△T    for water, C=4.18J=1 Cal     

First of all, we must remember the heat capacity of water. It is 4.18 J per gram/mL of water. Thus, to raise a gram of water by 1 C/K, we need to apply 4.18 J of energy on it. This is arch-important! Now, if the amount of water doubles (2 mL instead of 1, let’s say) then the energy that is required to raise its temperature by 1 mL doubles. Now, let’s put it to practice!

Let’s say we have 10 mL of water. It is at 20 C. However, we don’t want to drink water that cold! No, we want water of room temperature (33 C)! So, how much heat should we apply to it in order to heat it to 33 C?

Remember:

q=mC△T   

So, we substitute the mass of the sample for m. 10 mL = 10 g

For C, as it was mentioned before, we put 4.18 J

△T is the change in temperature. Now, here is the tricky part. Sometimes, when we use Celsius, one temperature value is negative. Thus, in order to not get confused, we should transfer it into Kelvin.

C = K - 273.

So 20 C = 293 K, and 33 C = 306 K

So, △T is = 306 - 293, which is 13 K

Now, let’s substitute!

q = (10. g)(4.18 J)(13 K)

q = 543.4 J (Don’t forget to round to sig. figs!)

q = 540 J

Now, this is where is gets tricky! Time to introduce calorimeter itself. Calorimeter is used to determine heat of capacity of any substance. Sometimes, it is even used to determine heat of solidification and fusion. In order to use calorimeter, we might want to introduce new equation, which comes out of first 2 equations. Here it is:

mC△T (for water) = mC△T (for substance, heat of capacity of which we want to know)

Although it looks quite complicated, with lots of variables, it is actually quite simple. In order to fully explain this, let’s put it to practice first.

Let’s say we had 10 g metal at 50 C (room temperature, let’s say). We then place it into a cup filled with 100 mL of water at 30 C. Then, we record the final temperature of water. It is 40 C. Lots of info, right. So, let’s plug it in!

Let’s start with water!

mC△T

What is the mass of water? Volume is 100 mL, so mass must be 100 g.

What is the Heat of capacity? Well, we know it, It’s 4.18 J

What is the change in temperature? 30 - 40 must be 10 (remember, change in temperature is an absolute value)!

Let’s plug it in and solve it:

(100 g)(4.18 J)(10 C)

You can do this in your head, you say!?

4180 J, right?

Now, it’s time for the right side of the equation, the metal itself

We know the mass, we know the initial temperature, we know the final temperature (which is always the temperature of the water). The only thing we have to figure out is the heat of solidification. Let’s plug the values in now!

(10 g)(x J)(50 C - 40 C)

X joules x 100

Now, let’s equate those

4180 = 100 multiplied by X joules. Divide by 100 on both side, and we get

41.8 = J/g *Now, I know that sig. figs are so wrong, but you get the idea, right?

1. Which one of the following pairs of samples has higher entropy?

a. Br2 (l) or Br2 (g)

b. KOH (s) or KOH (aq)

2. Predict the entropy change for the following processes:

a. O2 (g) → 2O (g)

b. 2O3 (g) → 3O2 (g)

c. CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

d. NaCl (s) → Na+ (aq) + Cl- (aq)

e. C2H5OH (l) → C2H5OH (g)

f. Ag+ (aq) + Cl- (aq) → AgCl (s)

3. Of the following reaction,

which are spontaneous at any temperature,

which are never spontaneous regardless of the temperature,

which are spontaneous only at high temperature

and which are spontaneous only at low temperature

*Note: I won’t add the reactions themselves (too lazy). Just going to add the △H & △S

△H △S Temp △G Spontaneous

a. - + any - yes

b. + - any + no

c - - high - yes

d. - - low + no

e. + + high + no

f. + + low - yes

Ammonium nitrate spontaneously dissolves in water at room temperature and the process causes the solution to become quite cold. Which of the following is TRUE about the dissolution of ammonium nitrate?

a) The process is exothermic. False

b) Its solubility will be greater in warmer water. True

c) △S for the reaction is negative. False

d) All the solutions of ammonium nitrate are supersaturated. False

e) All the solutions of ammonium nitrate are cold. False

Examples

Thermodynamic Equation: CH4(g) + 2O2 (g) ---> CO2(g) + 2H2O(l) –890.4

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Exothermic: CH4(g) + 2O2 (g) ---> CO2(g) + 2H2O(l)          delta H= -890.4

Endothermic: N2(g) + O2(g)-----> 2NO(g)                             delta H=182.6

Created by Hoyle's Chem Class 2012-2013  (Freshman Honor Chem)