Protons and neutrons in a nucleus are held together by the strong nuclear force. It's the strongest of the four fundamental forces because it must overcome the electrical repulsion between the protons.
Unstable nuclei are said to be radioactive, and they undergo a transformation to make them more stable--they do this by altering the number and ratio of protons and neutrons. This is called radioactive decay. There's 3 types: alpha, beta, and gamma. The nucleus that undergoes radioactive decay is called the parent, and the resulting (more stable nucleus) is called the daughter.
Alpha: When a large nucleus wants to become more stable by reducing the number of protons and electrons it emits an alpha particle--it contains 2 protons and 2 neutrons. This reduces the parent's atomic number by 2 and the mass number by 4.
Beta: there are 3 types: Beta (-), Beta (+), and electron capture. Each type involves the transmutation of a neutron into a proton (and vice versa) through the action of the weak nuclear force; beta particles are less massive than alpha particles and therefore less dangerous
Beta (-): Unstable nucleus contains too many neutrons--> it converts a neutron into a proton and an electron (Beta (-) particle that is ejected; the resulting atomic number is increased by 1 but the mass number remains the same. This is the most common type of beta decay so when the MCAT mentions it, it means this.
Beta (+): Unstable nucleus contains too few neutrons--> it converts a proton into a neutron and a positron (ejected). The positron is like an electron, only positive. The resulting atomic mass is 1 less than the parent but the mass number remains the same.
Electron Capture: unstable nucleus capture an electron from the closest electron shell (n=1) and uses it to convert a proton into a neutron--> causes the atomic number to be reduced by 1 while the mass number remains the same
Gamma Decay: is simply an expulsion of energy; a nucleus in an excited energy state (which is usually the case after a nucleus has undergone alpha or any type of beta decay) can "relax" to its ground state by emitting energy in the form of one or more photons. These photons are called gamma photons. They have neither mass nor charge. Their ejection from a radioactive atom changes neither the atomic number nor the mass number of the nucleus (i.e. does not change the identity of the nucleus like alpha or beta decay).
Quick note on nuclear binding energy: every nucleus that contains protons and neutrons has this. It is the energy that was released when the individuals nucleons were bound together by the strong force to form the nucleus. It's also equal to the energy that would be required to break up the intact nucleus into its individual nucleons. In short, the greater the binding energy per nucleon, the more stable the nucleus.
Mass defect: when nucleons bind together to form a nucleus, some mass is converted to energy, so the mass of the combined nucleus is less than the sum of the masses of all its nucleons individually. The difference, deltaM, is the mass defect and will always be positive.
DeltaM=total mass of separate nucleons - mass of nucleus
Saturday, January 3, 2009
Gchem Lecture 4: Electron Quantum Numbers
Electrons held by an atom can exist only at discrete energy levels--that is electron energy levels are quantized. This quantization is described by a unique "address" for each electron, consisting of four quantum numbers designating the shell, subshell, orbital, and spin.
The First Quantum Number: the "principle" quantum number is the shell number, n. It's related to the size and energy of an orbital. The value of n can be any whole number starting with 1, and generally, the greater the value of n, the greater the electron's energy and average distance from the nucleus.
The Second Quantum number (shape): the subshell number, is denoted by letter L. It describes the shape (and energy) of an electrons orbital. The possible values of L depend on the value of n as follows, L=0, 1, 2,..., n-1. For example, if the principal quantum number is 3, then L could be 0, 1, or 2.
L=0 s subshell
L=1 p subshell
L=2 d subshell
L=3 f subshell
The Third Quantum Number: orbital number, is denoted by m(l). It describes the three-dimensional orientation of an orbital. The possible values of m(l) depend on the value of L as follows: m(l)=-L, -(L-1),..., -1, 0,...(L-1), L. For example, if L=2, the m(l) could be -2, -1, 0, 1, 2
If L=0, then m(l) can only be equal to 0 (one possibility), so each S subshell has just 1 orbital
If L=1, then m(l) can equal -1, 0, 1 (three possibilities), so each P subshell has 3 orbitals
If L=2, then m(l) can equal -2, -1, 0, 1, 2 (five possibilities), so each D subshell has 5 orbitals
If L=3, then m(l) can equal -3, -2, -1, 0, 1, 2, 3 (seven possibilities), so the F subshell has 7 orbitals
The Fourth Quantum Number: spin number, denoted m(s); designates the electron's intrinsic magnetism; spin can either be (+1/2) spin up or (-1/2) spin down. Every orbital can accommodate 2 electrons, one spin up and one spin down. When an orbital is full, we say the electrons are "spin paired".
Here's the alternative/shorthand method to learning this:
Shells-->subshells-->orbitals-->spin number
Shells are the energy levels. The shells have subshells, which in turn have atomic orbitals. The orbitals are within the subshell. The azimuthal quantum numbers s, p, d and f designate the subshells. Consider the p subshell. It has 3 orbitals: px, py and pz. Each orbital can hold a total of 2 electrons, making 6 the total number of electrons a p subshell can house. The d subshell has 5 orbitals, and can hold a total number of ten electrons. The f subshell has 7 orbitals, and can house a total number of 14 electrons. Think of a shell as a neighborhood, a subshell as a house, and the orbitals as the different bedrooms within the house where the electrons stay.
The First Quantum Number: the "principle" quantum number is the shell number, n. It's related to the size and energy of an orbital. The value of n can be any whole number starting with 1, and generally, the greater the value of n, the greater the electron's energy and average distance from the nucleus.
The Second Quantum number (shape): the subshell number, is denoted by letter L. It describes the shape (and energy) of an electrons orbital. The possible values of L depend on the value of n as follows, L=0, 1, 2,..., n-1. For example, if the principal quantum number is 3, then L could be 0, 1, or 2.
L=0 s subshell
L=1 p subshell
L=2 d subshell
L=3 f subshell
The Third Quantum Number: orbital number, is denoted by m(l). It describes the three-dimensional orientation of an orbital. The possible values of m(l) depend on the value of L as follows: m(l)=-L, -(L-1),..., -1, 0,...(L-1), L. For example, if L=2, the m(l) could be -2, -1, 0, 1, 2
If L=0, then m(l) can only be equal to 0 (one possibility), so each S subshell has just 1 orbital
If L=1, then m(l) can equal -1, 0, 1 (three possibilities), so each P subshell has 3 orbitals
If L=2, then m(l) can equal -2, -1, 0, 1, 2 (five possibilities), so each D subshell has 5 orbitals
If L=3, then m(l) can equal -3, -2, -1, 0, 1, 2, 3 (seven possibilities), so the F subshell has 7 orbitals
The Fourth Quantum Number: spin number, denoted m(s); designates the electron's intrinsic magnetism; spin can either be (+1/2) spin up or (-1/2) spin down. Every orbital can accommodate 2 electrons, one spin up and one spin down. When an orbital is full, we say the electrons are "spin paired".
Here's the alternative/shorthand method to learning this:
Shells-->subshells-->orbitals-->spin number
Shells are the energy levels. The shells have subshells, which in turn have atomic orbitals. The orbitals are within the subshell. The azimuthal quantum numbers s, p, d and f designate the subshells. Consider the p subshell. It has 3 orbitals: px, py and pz. Each orbital can hold a total of 2 electrons, making 6 the total number of electrons a p subshell can house. The d subshell has 5 orbitals, and can hold a total number of ten electrons. The f subshell has 7 orbitals, and can house a total number of 14 electrons. Think of a shell as a neighborhood, a subshell as a house, and the orbitals as the different bedrooms within the house where the electrons stay.
Friday, January 2, 2009
Gchem Lecture 3: Strong Acids/Bases
It's a good idea to go ahead and learn these.
Strong Acids:
-H2SO4
-HNO3
-HCl
-HBr
-HI
-HClO3
-HClO4
*note that HF is NOT a strong acid, and its omission from the list is not an accident!!!
**only the first proton on H2SO4 is strong (i.e., completely dissociates); the second one is weak.
Strong Bases:
-Group I elements with hydroxide (ex. NaOH, KOH)
-Group II elements with hydroxide [ex. Ca(OH)2]
-some organic bases, including alkoxide ions, sodium hydride, Grignard reagents, and LDA
Strong Acids:
-H2SO4
-HNO3
-HCl
-HBr
-HI
-HClO3
-HClO4
*note that HF is NOT a strong acid, and its omission from the list is not an accident!!!
**only the first proton on H2SO4 is strong (i.e., completely dissociates); the second one is weak.
Strong Bases:
-Group I elements with hydroxide (ex. NaOH, KOH)
-Group II elements with hydroxide [ex. Ca(OH)2]
-some organic bases, including alkoxide ions, sodium hydride, Grignard reagents, and LDA
Gchem Lecture 2: Period Trends Part II
If you have read and understood the previous post about Z, you will now find that the other four trends we mentioned are all quite intuitive. Let's consider them one at a time.
Atomic Radius: Remember that the size of the nucleus of an atom is extremely small relative to the size of the entire atom. So when we talk about an atom's size, what we are really considering is the size of its electron cloud.
Radius of Neutral Atoms: As we move from left to right across the periodic table, we find that Z is increasing. So what exactly does this mean? Well, as Z increases, the net positive pull of the nucleus increases. (Z is, after all, the number of protons that are not needed to cancel out with the core electrons.) As we increase the net positive charge of the nucleus, we will also increase its ability to pull on the electron cloud, even the valence electrons. (This should make sense to you from Coulomb's Law, also; if you increase the magnitude of Q, you also increase the magnitude of F between the two charges.) So as Z increases from left to right, those nuclei will pull their electrons in more and more tightly to the nucleus, and this will decrease the overall size of the atom. Thus, the smallest atoms are found on the right side of the table, and the largest are on the left.
As we go down a group, Z doesn't change very much. But what does change? Well, we are adding a new principle quantum number each time we move down a row, which means that we've added more core electrons. So even though we aren't changing the net pull of the nucleus, we ARE changing the distance between the nucleus and the valence electrons. In fact, we are greatly increasing it. (From Coulomb's Law, we know that increasing the distance between two charges decreases the force between them.) So as we move down the periodic table, we will continue to add more layers of electrons without significantly changing Z, making the atom grow larger and larger.
Radius of Cations: You may be asked on your test to compare the radius of a cation with that of a neutral atom. The most important thing that removing an electron does is that it decreases electron-electron repulsion within a shell. Recall from the previous post that we are assuming that valence electrons don't repel one another, but in reality, they actually do a little bit. So when you remove an electron to form a cation, you don't change your value of Z, but you do decrease repulsion among the remaining valence electrons, which allows the nucleus to pull them in even tighter. Thus, cations are always smaller than the neutral atom.
This effect is especially pronounced if you remove an electron from a Group I element, since you lose an entire shell when you do this. Thus, Li+ cation is MUCH smaller than neutral Li.
Radius of Anions: The effect of adding an electron to a neutral atom is the opposite of the effect of removing one. Again, you are not changing Z, but you ARE increasing repulsion among the valence electrons, and so the net effect is to increase the size of the atom. If you add an electron to a group VIII element, this effect is especially pronounced, since you will be adding another whole quantum level to the atom.
Ionization Energy (IE): IE is the energy required to remove an electron from a lone atom in the gaseous state. Note that the atom MUST be in the gas phase, and it MUST be alone (i.e., not bonded to anything else).
As we have seen, the value of Z increases from left to right as we go across the periodic table. This means that the atoms are able to hold on to their valence electrons more and more tightly. Thus, it gets harder to remove an electron as the value of Z increases, and IE increases from left to right.
Going down a group, Z doesn't change much. But, the size of the atom increases greatly, and because of this, so does the distance between the nucleus and the valence electrons. That decreases the magnitude of the force of attraction between the nucleus and the outermost electrons. Thus, IE decreases going down a group from top to bottom.
You may also get asked on your test to compare first and second IE values. In general, it always takes more energy to remove a second electron than it does to remove the first. This is because you are trying to remove an electron from an already positively charged species, and make it have a positive charge of even higher magnitude. So second IE is always greater than first IE, third IE is greater than second IE, and so on.
Electron Affinity (EA) and Electronegativity (EN): These two concepts are similar to one another, and they follow the same trends. EA measures the energy released when a lone gaseous atom accepts an electron. It is the opposite of IE, and as with IE, the atom must not be in a bond, and it must be in the gas phase. Note that EA values are given as the absolute values of the energy released. (That is, we usually report exothermic reactions, where energy is being released, as having a negative energy value. However, EA is always reported as the absolute value of that negative energy magnitude, and so EA values are always positive.) EN measures the ability of an atom in a bond with another atom to pull the bond's electron density toward itself.
As we travel from left to right across the periodic table, and Z increases, the ability of the atom to accept an electron will also increase. This is because the pull of the nucleus on the valence electrons increases from left to right. Thus, both EA and EN increase from left to right.
As we go down a group, Z doesn't change much, but the valence electrons get farther and farther from the nucleus. Because of this, the atom is less and less able to "handle" having another electron added to it. Thus, both EA and EN will decrease as we go from top to bottom down the group.
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Summary of Periodic Trends
Effective nuclear charge increases from left to right, and does not significantly change from top to bottom. The left to right trend occurs because the number of core electrons remains constant, while the number of net protons left over after cancelling out the core electrons increases. The top to bottom trend occurs because the net number of protons left over after cancelling out the core electrons remains constant.
Atomic radius decreases from left to right, and increases from top to bottom. The left to right trend occurs because Z is increasing, and the top to bottom trend occurs because each row adds another layer of core electrons without significantly changing Z.
Ionization energy increases from left to right, and decreases from top to bottom. The left to right trend occurs because Z is increasing, and the top to bottom trend occurs because each row adds another layer of core electrons without significantly changing Z, making the valence electrons farther from the nucleus.
Electron affinity increases from left to right, and decreases from top to bottom. The left to right trend occurs because Z is increasing, and the top to bottom trend occurs because each row adds another layer of core electrons without significantly changing Z, making the valence electrons farther from the nucleus.
Electronegativity increases from left to right, and decreases from top to bottom. The left to right trend occurs because Z is increasing, and the top to bottom trend occurs because each row adds another layer of core electrons without significantly changing Z, making the valence electrons farther from the nucleus.
This post is compliments of QofQuimica from SDN.
Atomic Radius: Remember that the size of the nucleus of an atom is extremely small relative to the size of the entire atom. So when we talk about an atom's size, what we are really considering is the size of its electron cloud.
Radius of Neutral Atoms: As we move from left to right across the periodic table, we find that Z is increasing. So what exactly does this mean? Well, as Z increases, the net positive pull of the nucleus increases. (Z is, after all, the number of protons that are not needed to cancel out with the core electrons.) As we increase the net positive charge of the nucleus, we will also increase its ability to pull on the electron cloud, even the valence electrons. (This should make sense to you from Coulomb's Law, also; if you increase the magnitude of Q, you also increase the magnitude of F between the two charges.) So as Z increases from left to right, those nuclei will pull their electrons in more and more tightly to the nucleus, and this will decrease the overall size of the atom. Thus, the smallest atoms are found on the right side of the table, and the largest are on the left.
As we go down a group, Z doesn't change very much. But what does change? Well, we are adding a new principle quantum number each time we move down a row, which means that we've added more core electrons. So even though we aren't changing the net pull of the nucleus, we ARE changing the distance between the nucleus and the valence electrons. In fact, we are greatly increasing it. (From Coulomb's Law, we know that increasing the distance between two charges decreases the force between them.) So as we move down the periodic table, we will continue to add more layers of electrons without significantly changing Z, making the atom grow larger and larger.
Radius of Cations: You may be asked on your test to compare the radius of a cation with that of a neutral atom. The most important thing that removing an electron does is that it decreases electron-electron repulsion within a shell. Recall from the previous post that we are assuming that valence electrons don't repel one another, but in reality, they actually do a little bit. So when you remove an electron to form a cation, you don't change your value of Z, but you do decrease repulsion among the remaining valence electrons, which allows the nucleus to pull them in even tighter. Thus, cations are always smaller than the neutral atom.
This effect is especially pronounced if you remove an electron from a Group I element, since you lose an entire shell when you do this. Thus, Li+ cation is MUCH smaller than neutral Li.
Radius of Anions: The effect of adding an electron to a neutral atom is the opposite of the effect of removing one. Again, you are not changing Z, but you ARE increasing repulsion among the valence electrons, and so the net effect is to increase the size of the atom. If you add an electron to a group VIII element, this effect is especially pronounced, since you will be adding another whole quantum level to the atom.
Ionization Energy (IE): IE is the energy required to remove an electron from a lone atom in the gaseous state. Note that the atom MUST be in the gas phase, and it MUST be alone (i.e., not bonded to anything else).
As we have seen, the value of Z increases from left to right as we go across the periodic table. This means that the atoms are able to hold on to their valence electrons more and more tightly. Thus, it gets harder to remove an electron as the value of Z increases, and IE increases from left to right.
Going down a group, Z doesn't change much. But, the size of the atom increases greatly, and because of this, so does the distance between the nucleus and the valence electrons. That decreases the magnitude of the force of attraction between the nucleus and the outermost electrons. Thus, IE decreases going down a group from top to bottom.
You may also get asked on your test to compare first and second IE values. In general, it always takes more energy to remove a second electron than it does to remove the first. This is because you are trying to remove an electron from an already positively charged species, and make it have a positive charge of even higher magnitude. So second IE is always greater than first IE, third IE is greater than second IE, and so on.
Electron Affinity (EA) and Electronegativity (EN): These two concepts are similar to one another, and they follow the same trends. EA measures the energy released when a lone gaseous atom accepts an electron. It is the opposite of IE, and as with IE, the atom must not be in a bond, and it must be in the gas phase. Note that EA values are given as the absolute values of the energy released. (That is, we usually report exothermic reactions, where energy is being released, as having a negative energy value. However, EA is always reported as the absolute value of that negative energy magnitude, and so EA values are always positive.) EN measures the ability of an atom in a bond with another atom to pull the bond's electron density toward itself.
As we travel from left to right across the periodic table, and Z increases, the ability of the atom to accept an electron will also increase. This is because the pull of the nucleus on the valence electrons increases from left to right. Thus, both EA and EN increase from left to right.
As we go down a group, Z doesn't change much, but the valence electrons get farther and farther from the nucleus. Because of this, the atom is less and less able to "handle" having another electron added to it. Thus, both EA and EN will decrease as we go from top to bottom down the group.
****************
Summary of Periodic Trends
Effective nuclear charge increases from left to right, and does not significantly change from top to bottom. The left to right trend occurs because the number of core electrons remains constant, while the number of net protons left over after cancelling out the core electrons increases. The top to bottom trend occurs because the net number of protons left over after cancelling out the core electrons remains constant.
Atomic radius decreases from left to right, and increases from top to bottom. The left to right trend occurs because Z is increasing, and the top to bottom trend occurs because each row adds another layer of core electrons without significantly changing Z.
Ionization energy increases from left to right, and decreases from top to bottom. The left to right trend occurs because Z is increasing, and the top to bottom trend occurs because each row adds another layer of core electrons without significantly changing Z, making the valence electrons farther from the nucleus.
Electron affinity increases from left to right, and decreases from top to bottom. The left to right trend occurs because Z is increasing, and the top to bottom trend occurs because each row adds another layer of core electrons without significantly changing Z, making the valence electrons farther from the nucleus.
Electronegativity increases from left to right, and decreases from top to bottom. The left to right trend occurs because Z is increasing, and the top to bottom trend occurs because each row adds another layer of core electrons without significantly changing Z, making the valence electrons farther from the nucleus.
This post is compliments of QofQuimica from SDN.
Gchem Lecture 1: Periodic table Trends
The worst way to learn periodic trends (or anything else in the PS portion of the MCAT) is to memorize them. This is because memorization without understanding will only allow you to solve problems that are asked in the same format that you have memorized. However, the MCAT is infamous for forcing you to take information that you've learned in your college classes and apply it to new scenarios with which you have no previous experience.
The periodic trends that you should know for the MCAT are atomic radius, ionization energy, electron affinity, and electronegativity. We will consider each one in turn in a second post, but first, we must consider a fifth trend upon which the other four all depend, which is the effective nuclear charge, Z.
Effective Nuclear Charge (Z or Zeff): This trend is probably the hardest for students to understand, but once you do understand it, all of the other trends will seem very logical and intuitive to you. Z is basically a book-keeping method, where we cancel out each core electron with one of the protons in the nucleus. (Core electrons are the ones in the inner shells; in other words, they are the non-valence electrons.) We do not cancel the valence electrons. (Remember that valence electrons are the ones in the outer shell.) When we have completed this exercise, we will find that, for a neutral atom, we are left with the same number of uncancelled protons in the nucleus as we have valence electrons. The number of uncancelled protons is equal to the value of Z, and it will be the same as the group number for a neutral atom.
The reason why we only cancel out core electrons is that these electrons are considered to be shielding the valence electrons. This means that they block the valence electrons from easily having access to the nucleus. Remember that the nucleus is positive, and all electrons are negative. Since unlike charges attract, the valence electrons are attracted to the nucleus. But the problem is that there are some core electrons in the way, and they too are negatively charged. Since like charges repel, core electrons will repel the valence electrons, somewhat countering their attracting to the nucleus.
Let's look at some actual atoms to make this concept clearer.
Calculating Z for Li: We'll start with lithium, which has three protons and three electrons as a neutral atom. If you begin to write the electron configuration of lithium, where do the electrons go? Well, the first two go into the 1s orbital. These are two core electrons. The third goes into the 2s orbital, and it's a valence electron. Doing our book-keeping, we find that we have cancelled two protons for our two core electrons, and this leaves us one proton left over. Thus, the value of Z for Li is approximately 1.
Calculating Z for Be: The next element after Li is beryllium, which has four protons and four electrons as a neutral atom. Its configuration is 1s2, 2s2, which means that again we have two core electrons as with Li. However, we now have two valence electrons, and when we have finished cancelling our core electrons with two of the protons, we now have two protons left over. Thus, the value of Z for Be is approximately 2. (Note that in doing this, we are assuming that electrons in the same shell do not shield one another. That is, we are assuming that the two valence electrons in Be do not shield one another. Technically that is not exactly true, but for our purpose, the effect of electron-electron repulsion by electrons in the same shell can be ignored.)
Calculating Z for B: After Be comes boron, which has five protons and five electrons. Its configuration is 1s2, 2s2, 2p1. As we have already seen, we have two core electrons from the 1s-orbital, which we can cancel with two of the five protons. This leaves us with three protons left over, and three valence electrons. (Again, we are assuming that the 2s electrons do not shield the 2p electron, but technically this is not completely true.) So now our Z value is 3.
Calculating Z for the rest of the elements in row 2: As we continue to fill the 2p orbitals, one electron at a time, we will find that the Z value continues to increase by one as we move from left to right. Here are the Z values for the rest of the elements in row 2:
Element Z-value
C.............4
N.............5
O.............6
F.............7
Ne............8
Calculating Z for the Group I Metals: Now, let us consider what happens to Z when we go down a group. We'll use Group I, the alkali metals. We've already seen that Li has a value of 1 for Z. What about Na, right below Li? Well, sodium has 11 protons and 11 electrons as a neutral atom. Its configuration is 1s2, 2s2, 2p6, 3s1. All of the electrons in the second shell are now core electrons; only the one electron in the 3s orbital is a valence electron. If we cancel out the ten core electrons with ten of the protons, we are left with one proton, just as with Li. So the value of Z for Na is also approximately 1, just as it is for Li. What about K, element 19? Its configuration is 1s2, 2s2, 2p6, 3s2, 3p6, 4s1 as a neutral atom. Just as before, there is only one valence electron. The other 18 electrons are all part of the core. So they each cancel a proton, leaving us with 1 proton left over, and a Z value of about 1. Clearly, then, ALL of the alkali metals have a Z value of approximately 1. This same pattern will hold true for each of the other main groups in the periodic table.
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Here is a summary of the trends for Z.
As we travel from left to right across the periodic table, Z increases. This occurs because we are adding more protons as we move from left to right, but we are not adding more shielding core electrons.
As we move down a group from top to bottom, Z stays approximately equal. This occurs because we are not changing the net number of protons left over after we have cancelled out all of the shielding core electrons.
This post was compliments of QofQuimica on SDN.
The periodic trends that you should know for the MCAT are atomic radius, ionization energy, electron affinity, and electronegativity. We will consider each one in turn in a second post, but first, we must consider a fifth trend upon which the other four all depend, which is the effective nuclear charge, Z.
Effective Nuclear Charge (Z or Zeff): This trend is probably the hardest for students to understand, but once you do understand it, all of the other trends will seem very logical and intuitive to you. Z is basically a book-keeping method, where we cancel out each core electron with one of the protons in the nucleus. (Core electrons are the ones in the inner shells; in other words, they are the non-valence electrons.) We do not cancel the valence electrons. (Remember that valence electrons are the ones in the outer shell.) When we have completed this exercise, we will find that, for a neutral atom, we are left with the same number of uncancelled protons in the nucleus as we have valence electrons. The number of uncancelled protons is equal to the value of Z, and it will be the same as the group number for a neutral atom.
The reason why we only cancel out core electrons is that these electrons are considered to be shielding the valence electrons. This means that they block the valence electrons from easily having access to the nucleus. Remember that the nucleus is positive, and all electrons are negative. Since unlike charges attract, the valence electrons are attracted to the nucleus. But the problem is that there are some core electrons in the way, and they too are negatively charged. Since like charges repel, core electrons will repel the valence electrons, somewhat countering their attracting to the nucleus.
Let's look at some actual atoms to make this concept clearer.
Calculating Z for Li: We'll start with lithium, which has three protons and three electrons as a neutral atom. If you begin to write the electron configuration of lithium, where do the electrons go? Well, the first two go into the 1s orbital. These are two core electrons. The third goes into the 2s orbital, and it's a valence electron. Doing our book-keeping, we find that we have cancelled two protons for our two core electrons, and this leaves us one proton left over. Thus, the value of Z for Li is approximately 1.
Calculating Z for Be: The next element after Li is beryllium, which has four protons and four electrons as a neutral atom. Its configuration is 1s2, 2s2, which means that again we have two core electrons as with Li. However, we now have two valence electrons, and when we have finished cancelling our core electrons with two of the protons, we now have two protons left over. Thus, the value of Z for Be is approximately 2. (Note that in doing this, we are assuming that electrons in the same shell do not shield one another. That is, we are assuming that the two valence electrons in Be do not shield one another. Technically that is not exactly true, but for our purpose, the effect of electron-electron repulsion by electrons in the same shell can be ignored.)
Calculating Z for B: After Be comes boron, which has five protons and five electrons. Its configuration is 1s2, 2s2, 2p1. As we have already seen, we have two core electrons from the 1s-orbital, which we can cancel with two of the five protons. This leaves us with three protons left over, and three valence electrons. (Again, we are assuming that the 2s electrons do not shield the 2p electron, but technically this is not completely true.) So now our Z value is 3.
Calculating Z for the rest of the elements in row 2: As we continue to fill the 2p orbitals, one electron at a time, we will find that the Z value continues to increase by one as we move from left to right. Here are the Z values for the rest of the elements in row 2:
Element Z-value
C.............4
N.............5
O.............6
F.............7
Ne............8
Calculating Z for the Group I Metals: Now, let us consider what happens to Z when we go down a group. We'll use Group I, the alkali metals. We've already seen that Li has a value of 1 for Z. What about Na, right below Li? Well, sodium has 11 protons and 11 electrons as a neutral atom. Its configuration is 1s2, 2s2, 2p6, 3s1. All of the electrons in the second shell are now core electrons; only the one electron in the 3s orbital is a valence electron. If we cancel out the ten core electrons with ten of the protons, we are left with one proton, just as with Li. So the value of Z for Na is also approximately 1, just as it is for Li. What about K, element 19? Its configuration is 1s2, 2s2, 2p6, 3s2, 3p6, 4s1 as a neutral atom. Just as before, there is only one valence electron. The other 18 electrons are all part of the core. So they each cancel a proton, leaving us with 1 proton left over, and a Z value of about 1. Clearly, then, ALL of the alkali metals have a Z value of approximately 1. This same pattern will hold true for each of the other main groups in the periodic table.
****************
Here is a summary of the trends for Z.
As we travel from left to right across the periodic table, Z increases. This occurs because we are adding more protons as we move from left to right, but we are not adding more shielding core electrons.
As we move down a group from top to bottom, Z stays approximately equal. This occurs because we are not changing the net number of protons left over after we have cancelled out all of the shielding core electrons.
This post was compliments of QofQuimica on SDN.
Organic Lecture 8: Hydrogen bonding and Acidity
We are going to talk about hydrogen bonding and acidity in the context of alcohols. Alcohols form intermolecular hydrogen bonds because they have hydroxyl (-OH) groups. This results from a strong dipole in which the hydroxyl group's proton acquires as substantial partial positive charge and the oxygen has a negative charge. The hydrogen can therefore interact electrostatically with a non-bonding pair of electrons on a nearby oxygen, resulting in a hydrogen bond.
Intermolecular hydrogen bonds between molecules can increase the boiling and melting points by holding the molecules together. Intermolecular means between individual molecules. Intramolecular hydrogen bonds are on the same molecule and decrease the amount of of intermoleculear hydrogen bonding interactions that can occur between molecules thereby decreasing the melting and boiling points.
The acidity of a compound is determined by the ease with which it can lose a proton (H+). Alcohols are relatively acidic functional groups for the same reason that they engage in hydrogen bonding: the large difference in electronegativity between oxygen and hydrogen. Phenols are more acidic than alcohols due to resonance stabilization.
Intermolecular hydrogen bonds between molecules can increase the boiling and melting points by holding the molecules together. Intermolecular means between individual molecules. Intramolecular hydrogen bonds are on the same molecule and decrease the amount of of intermoleculear hydrogen bonding interactions that can occur between molecules thereby decreasing the melting and boiling points.
The acidity of a compound is determined by the ease with which it can lose a proton (H+). Alcohols are relatively acidic functional groups for the same reason that they engage in hydrogen bonding: the large difference in electronegativity between oxygen and hydrogen. Phenols are more acidic than alcohols due to resonance stabilization.
Thursday, January 1, 2009
Organic Lecture 7: Sn1, Sn2, E1, E2 (All you need to know)
Most organic reactions occur between nucleophiles and electrophiles. Nucleophiles are species that have unshared pairs of electrons or pi bonds and frequently have a negative charge. They are "nucleus seeking" or "nucleus loving" and electron donors (aka Lewis Base).
Nucleophilicity is a measure of how "strong" a nucleophile is. Here are some general trends:
1. Nucleophilicity increases as negative charge increases. NH2- is more nucleophilic than NH3.
2. Nucleophilicity increases going down the periodic table within a particular group. This is related to polarizability (how easy it is for electrons surrounding an atom to be distorted--therefore, larger atoms are more polarizable and more nucleophilic). F < Cl< Br< I
3. Nucleophilicity increases going left on the periodic table. NH2 is more nucleophilic than OH-. This is related to electronegativity. The more electronegative the atom is, the better it is able to support its negative charge. Therefore, the less electronegative an atom is, the higher its nucleophilicity.
Electrophiles are electron-deficient species. They have a full or partial positive charge and "love electrons". Since electrophiles are electron pair acceptors, they are also known as Lewis acids.
Here's a brief synopsis of the substitution and elimination mechanisms:
Sn2: nucleophile attacks an electrophile (generally an alkyl halide); this is a backside attack. Alkyl halide acts as a leaving group and the nucleophile forms a bond with the electrophile. There is only one step in this mechanism--therefore, it's a concerted mechanism. Inversion occurs when the substrate is chiral (this means that if you have an absolute configuration of S, then it goes to R, and vice versa). This is a bimolecular reaction because the rate is determined by the concentrations of both the nucleophile and the electrophile.
Sn2 reaction rate = k[nucleophile][electrophile]
**Less substituted substrates react faster than more substituted ones
**To favor Sn2, avoid protic solvents (i.e. water and alcohols) becasue these strongly solvate the nucleophile and therefore hinder the backside attack
**Use aprotic (non-hydrogen bonding) solvents--acetone, DMF, DMSO
Sn1: occurs in two distinct steps. In the first step, a carbocation is formed--this occurs when the leaving group falls off. This is the slow step of the mechanism (rate limiting step). In the final step, racemization occurs as the nucleophile attacks on either side of the carbocation. Racemic mixture results. The rate of the reaction depends only upon the concentration of the electrophile
Sn1 rate = k[electrophile]
**The more substituted the carbocation intermediate the better. This stablizes it and therefore makes the reaction proceed faster.
**Use protic solvents (water and alcohols) to help stablize the forming carbocation and solvate the leaving group.
E1: leaving group falls off and then a weak base removes a proton, leaving behind its electrons to form a C=C double bond; E1 works best with tertiary substrates because they best support the positive charge after the leaving group leaves; most highly substituted alkene is formed; this mechanism is favored by a protic solvent
E2: proceeds via a 1-step mechanism; a strong base removes the beta hydrogen while the leaving group leaves; the carbon-carbon double bond forms at the same time; this mechanism only works with alkyl halides; must have ANTIPERIPLANAR geometry (H is anti to leaving group); small bases yield the most substituted alkene while bulky bases favor the least substituted one (because of steric hindrance)
I found this on SDN. It's just a different way to explain what I have above. Very good though.
1) Number of R Groups on the Substrate
This is a very important factor. Keep in mind that primary substrates (like alkyl halides) will not react via Sn1 or E1, because the carbocation formed would be too unstable. Likewise, tertiary substrates will not react via Sn2, because they are too hindered. To sum:
-Primary substrates may undergo Sn2 or E2, but not Sn1 or E1
-Tertiary substrates may undergo Sn1, E1, or E2, but not Sn2.
-Secondary substrates are tricky and may react by any of the four mechanisms, depending on other factors. Keep reading.
2) Strength of the Base/Nucleophile
This is another very important factor. Keep in mind that carbocations are very powerful Lewis acids. You may not be used to thinking of them this way, because the Bronsted-Lowry definition of an acid doesn't include carbocations. However, the same general principle applies to carbocations that applies to all strong acids: when you mix them with strong bases, you get a neutralization reaction. That is not what we are trying to accomplish in organic reactions! Thus, E1 and Sn1 reactions, both of which have carbocation intermediates, will not occur in the presence of a strong base. In contrast, E2 requires the presence of a strong base, and a strong base will promote E2, particularly if the base is also bulky. To sum:
-A species that is BOTH a good nucleophile and a strong base (ex. hydroxide, Grignard reagents, and alkoxide ions) will tend to promote E2 or Sn2, depending on other factors like substrate hindrance and base bulkiness
-A species that is a good nucleophile but NOT a strong base (ex. cyanide, sulfur nucleophiles, azide) will tend to promote Sn2 if the substrate is not too hindered.
-A species that is a strong base but is NOT nucleophilic (ex. NaH) will often just deprotonate an acidic hydrogen on the substrate.
-A species that is BOTH a poor nucleophile and a weak base (ex. water, alcohols) will tend to promote Sn1 and E1, assuming that the substrate is hindered enough.
3) Solvent Type
This is another important factor that is often overlooked by students. Generally the solvent will be polar, and it can either be protic or aprotic, which affects the mechanism. Thus, it is essential that you learn how to correctly identify whether a solvent is protic or aprotic. (Read the Intermolecular Interactions post in the General Chemistry Explanations thread about protic versus aprotic solvents if you are not familiar with these terms.) A protic solvent stabilizes charged species. For mechanisms that form carbocations (Sn1 and E1), this is a good thing, because it lowers the potential energy of the carbocations, making them easier to form. However, for mechanisms where the base or nucleophile is charged, a polar protic solvent will stablize the nucleophile and make it less reactive. In contrast, a polar aprotic solvent will destabilize it and make it more reactive. (This is sometimes referred to as making a "naked nucleophile.") To sum:
-Polar protic solvents favor E1 and Sn1 by lowering the energy of the carbocation intermediate and making the RDS occur more readily.
-Polar aprotic solvents favor Sn2 by increasing the energy of the nucleophile and making it more reactive.
-E2 reactions are often carried out in polar aprotic solvents or in a polar protic solvent that is the conjugate acid of the strong base (ex. t-butanol for t-butoxide).
4) Heat and Hindrance
It is relatively easy to force the reaction to go by E2 over Sn2 by manipulating the choice of base. A large, bulky base like t-butoxide or LDA (lithium diisopropylamide) is too hindered to serve as a nucleophile, and the presence of either of those bases should immediately tell you that the correct mechanism is E2. Forcing a reaction to go by E1 instead of Sn1 is very difficult, because these two mechanisms have the same RDS (formation of the carbocation). The one thing you can do to promote elimination over substitution is to heat the reaction. In organic chemistry, we use the Greek symbol delta (looks like a little triangle over the reaction arrow) to designate that the reaction is being heated. (Note that the delta symbol means "change in" for the physical sciences, and often precedes a variable: eg. delta G = delta H - T delta S.) To sum:
-A large, bulky base (t-butoxide or LDA) should immediately clue you in that the mechanism is E2.
-A heat symbol (the Greek letter delta) over the reaction arrow tells you that the reaction will probably go by elimination rather than substitution.
5) None of the Above
Don't get so caught up in trying to decide which mechanism will predominate for a given reaction that you don't consider the possibility that no reaction will occur at all. There are two common situations that you should watch out for where there will be no reaction.
One is if you have a primary substrate and no good nucleophile or strong base is present. You can't have E1 or Sn1 in this scenario because the substrate is primary. You can't have E2 because there isn't a strong base. And you can't have Sn2 because there isn't a good nucleophile. The correct response here is that no reaction will occur.
Another situation where you must be cautious is if you are asked about how a cyclohexane or other hindered sytem will react, and you see that you have E2-only conditions. Remember that for an E2 to occur, the base, proton, both carbons forming the new double bond, and the leaving group must all be in what is called the antiperiplanar geometry. This means that all five atoms must lie in the same plane of space, and that the proton and leaving group must be 180 degrees apart in orientation. If the ring cannot orient itself in such a way that the proton and the leaving group are coaxial (both in axial positions on the ring with one up and one down), then an E2 cannot occur. Again, the correct response here is that there will be no reaction.
Nucleophilicity is a measure of how "strong" a nucleophile is. Here are some general trends:
1. Nucleophilicity increases as negative charge increases. NH2- is more nucleophilic than NH3.
2. Nucleophilicity increases going down the periodic table within a particular group. This is related to polarizability (how easy it is for electrons surrounding an atom to be distorted--therefore, larger atoms are more polarizable and more nucleophilic). F < Cl< Br< I
3. Nucleophilicity increases going left on the periodic table. NH2 is more nucleophilic than OH-. This is related to electronegativity. The more electronegative the atom is, the better it is able to support its negative charge. Therefore, the less electronegative an atom is, the higher its nucleophilicity.
Electrophiles are electron-deficient species. They have a full or partial positive charge and "love electrons". Since electrophiles are electron pair acceptors, they are also known as Lewis acids.
Here's a brief synopsis of the substitution and elimination mechanisms:
Sn2: nucleophile attacks an electrophile (generally an alkyl halide); this is a backside attack. Alkyl halide acts as a leaving group and the nucleophile forms a bond with the electrophile. There is only one step in this mechanism--therefore, it's a concerted mechanism. Inversion occurs when the substrate is chiral (this means that if you have an absolute configuration of S, then it goes to R, and vice versa). This is a bimolecular reaction because the rate is determined by the concentrations of both the nucleophile and the electrophile.
Sn2 reaction rate = k[nucleophile][electrophile]
**Less substituted substrates react faster than more substituted ones
**To favor Sn2, avoid protic solvents (i.e. water and alcohols) becasue these strongly solvate the nucleophile and therefore hinder the backside attack
**Use aprotic (non-hydrogen bonding) solvents--acetone, DMF, DMSO
Sn1: occurs in two distinct steps. In the first step, a carbocation is formed--this occurs when the leaving group falls off. This is the slow step of the mechanism (rate limiting step). In the final step, racemization occurs as the nucleophile attacks on either side of the carbocation. Racemic mixture results. The rate of the reaction depends only upon the concentration of the electrophile
Sn1 rate = k[electrophile]
**The more substituted the carbocation intermediate the better. This stablizes it and therefore makes the reaction proceed faster.
**Use protic solvents (water and alcohols) to help stablize the forming carbocation and solvate the leaving group.
E1: leaving group falls off and then a weak base removes a proton, leaving behind its electrons to form a C=C double bond; E1 works best with tertiary substrates because they best support the positive charge after the leaving group leaves; most highly substituted alkene is formed; this mechanism is favored by a protic solvent
E2: proceeds via a 1-step mechanism; a strong base removes the beta hydrogen while the leaving group leaves; the carbon-carbon double bond forms at the same time; this mechanism only works with alkyl halides; must have ANTIPERIPLANAR geometry (H is anti to leaving group); small bases yield the most substituted alkene while bulky bases favor the least substituted one (because of steric hindrance)
I found this on SDN. It's just a different way to explain what I have above. Very good though.
1) Number of R Groups on the Substrate
This is a very important factor. Keep in mind that primary substrates (like alkyl halides) will not react via Sn1 or E1, because the carbocation formed would be too unstable. Likewise, tertiary substrates will not react via Sn2, because they are too hindered. To sum:
-Primary substrates may undergo Sn2 or E2, but not Sn1 or E1
-Tertiary substrates may undergo Sn1, E1, or E2, but not Sn2.
-Secondary substrates are tricky and may react by any of the four mechanisms, depending on other factors. Keep reading.
2) Strength of the Base/Nucleophile
This is another very important factor. Keep in mind that carbocations are very powerful Lewis acids. You may not be used to thinking of them this way, because the Bronsted-Lowry definition of an acid doesn't include carbocations. However, the same general principle applies to carbocations that applies to all strong acids: when you mix them with strong bases, you get a neutralization reaction. That is not what we are trying to accomplish in organic reactions! Thus, E1 and Sn1 reactions, both of which have carbocation intermediates, will not occur in the presence of a strong base. In contrast, E2 requires the presence of a strong base, and a strong base will promote E2, particularly if the base is also bulky. To sum:
-A species that is BOTH a good nucleophile and a strong base (ex. hydroxide, Grignard reagents, and alkoxide ions) will tend to promote E2 or Sn2, depending on other factors like substrate hindrance and base bulkiness
-A species that is a good nucleophile but NOT a strong base (ex. cyanide, sulfur nucleophiles, azide) will tend to promote Sn2 if the substrate is not too hindered.
-A species that is a strong base but is NOT nucleophilic (ex. NaH) will often just deprotonate an acidic hydrogen on the substrate.
-A species that is BOTH a poor nucleophile and a weak base (ex. water, alcohols) will tend to promote Sn1 and E1, assuming that the substrate is hindered enough.
3) Solvent Type
This is another important factor that is often overlooked by students. Generally the solvent will be polar, and it can either be protic or aprotic, which affects the mechanism. Thus, it is essential that you learn how to correctly identify whether a solvent is protic or aprotic. (Read the Intermolecular Interactions post in the General Chemistry Explanations thread about protic versus aprotic solvents if you are not familiar with these terms.) A protic solvent stabilizes charged species. For mechanisms that form carbocations (Sn1 and E1), this is a good thing, because it lowers the potential energy of the carbocations, making them easier to form. However, for mechanisms where the base or nucleophile is charged, a polar protic solvent will stablize the nucleophile and make it less reactive. In contrast, a polar aprotic solvent will destabilize it and make it more reactive. (This is sometimes referred to as making a "naked nucleophile.") To sum:
-Polar protic solvents favor E1 and Sn1 by lowering the energy of the carbocation intermediate and making the RDS occur more readily.
-Polar aprotic solvents favor Sn2 by increasing the energy of the nucleophile and making it more reactive.
-E2 reactions are often carried out in polar aprotic solvents or in a polar protic solvent that is the conjugate acid of the strong base (ex. t-butanol for t-butoxide).
4) Heat and Hindrance
It is relatively easy to force the reaction to go by E2 over Sn2 by manipulating the choice of base. A large, bulky base like t-butoxide or LDA (lithium diisopropylamide) is too hindered to serve as a nucleophile, and the presence of either of those bases should immediately tell you that the correct mechanism is E2. Forcing a reaction to go by E1 instead of Sn1 is very difficult, because these two mechanisms have the same RDS (formation of the carbocation). The one thing you can do to promote elimination over substitution is to heat the reaction. In organic chemistry, we use the Greek symbol delta (looks like a little triangle over the reaction arrow) to designate that the reaction is being heated. (Note that the delta symbol means "change in" for the physical sciences, and often precedes a variable: eg. delta G = delta H - T delta S.) To sum:
-A large, bulky base (t-butoxide or LDA) should immediately clue you in that the mechanism is E2.
-A heat symbol (the Greek letter delta) over the reaction arrow tells you that the reaction will probably go by elimination rather than substitution.
5) None of the Above
Don't get so caught up in trying to decide which mechanism will predominate for a given reaction that you don't consider the possibility that no reaction will occur at all. There are two common situations that you should watch out for where there will be no reaction.
One is if you have a primary substrate and no good nucleophile or strong base is present. You can't have E1 or Sn1 in this scenario because the substrate is primary. You can't have E2 because there isn't a strong base. And you can't have Sn2 because there isn't a good nucleophile. The correct response here is that no reaction will occur.
Another situation where you must be cautious is if you are asked about how a cyclohexane or other hindered sytem will react, and you see that you have E2-only conditions. Remember that for an E2 to occur, the base, proton, both carbons forming the new double bond, and the leaving group must all be in what is called the antiperiplanar geometry. This means that all five atoms must lie in the same plane of space, and that the proton and leaving group must be 180 degrees apart in orientation. If the ring cannot orient itself in such a way that the proton and the leaving group are coaxial (both in axial positions on the ring with one up and one down), then an E2 cannot occur. Again, the correct response here is that there will be no reaction.
Organic Lecture 6: Ring Strain
We'll start the new year off with a short lesson, partly because I'm extremely hung over.
Ring strain arises when bond angles between ring atoms deviate from the ideal angle predicted by hybridization of the atoms.
Cyclopropane is very strained because the carbon-carbon bond angles are 60 degrees rather than the idealized 109 degrees for sp3 hybridized carbons.
Cyclobutane has carbon-carbon bond angles of 88 degrees.
**The deviation of the bond angles from the normal tetrahedral 109 degrees causes cyclopropane and cyclobutane to be unstable. The strain weakens the carbon-carbon bonds and increases the reactivity of them (C-C bonds in these highly strained cyclic molecules are significantly more reactive). Because of this, both cyclopronane and cyclobutane can undergo hydrogenation. However, cyclopentane and cyclohexane are more stable because they have near tetrahedral bond angles due to the conformations they adopt.
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