Lewis Structures, Introduction, Formal Charge, Molecular Geometry, Resonance, Polar or Nonpolar

Summary

Highlights

• Dive into electron configurations and their role in chemical bonding 🔬.
• Learn to draw Lewis structures for common molecules like O2, N2, and F2 ✏️.
• Explore the concept of molecular geometry with practical examples such as water and methane 🔎.
• Understand the significance of resonance in molecules like CO2 and SO2 🌌.
• Discover how polarity affects a molecule's behavior and interactions 🧲.
• Master the art of deducing hybridization states for different molecules 🔄.

Key Takeaways

• Drawing Lewis structures becomes easier once you understand the periodic table basics and valence electrons 💡.
• Determining molecular geometry is key in predicting the shape and angle of molecules 🔍.
• Formal charge calculation helps in identifying the most stable structure ➕.
• Resonance structures reveal the versatility and stability of molecules ➿.
• Recognizing polar or non-polar molecules depends on the symmetry and charge distribution ⚖️.
• Organic chemistry is a mix of rules and creative drawing; practice makes perfect 🧠.

Overview

Chapters

• 00:00 - 01:00: Introduction and Basics The chapter introduces the various topics and skills that will be covered, including drawing Lewis structures of simple molecules, organic compounds, and polyatomic ions. It also discusses determining molecular geometry, calculating formal charges, identifying resonance structures, and assessing the polarity of molecules.
• 01:00 - 02:00: Molecular Geometry The chapter provides an introduction to the periodic table, focusing on the second row elements such as boron, carbon, nitrogen, oxygen, and fluorine. These elements are categorized into different groups based on their valence electrons: boron in group 3A with three valence electrons, carbon in group 4A with four valence electrons, nitrogen with five, oxygen with six, and fluorine with seven. The information applies similarly to elements positioned below these in the periodic table.
• 02:00 - 03:00: Polarity and Bond Order Discusses the valence electrons of various elements such as chlorine, bromine, beryllium, and lithium.
• 03:00 - 04:00: Oxygen, Sulfur, and Phosphorus Molecules The chapter focuses on the properties of non-metals, specifically those located at the upper right corner of the periodic table, such as oxygen, sulfur, and phosphorus. These elements tend to acquire electrons to achieve a stable electron configuration, often aiming for eight electrons in their outer shell, a state known as the octet rule. Florine is used as an example; it is highly electronegative and needs one additional electron to complete its octet, leading it to form one bond. The discussion highlights the electron affinities of these elements and how they achieve stability through bonding.
• 04:00 - 05:00: Molecular Shapes and Bond Angles Oxygen, with six valence electrons, aims to acquire two more electrons to achieve a full valence shell of eight electrons. Consequently, oxygen typically forms two covalent bonds. In comparison, nitrogen, which aims for the same valence shell configuration, forms three bonds, while carbon forms four. Conversely, elements positioned on the left side of the periodic table, often metals like Boron, are predisposed to lose electrons. For instance, Boron more readily gives away its three valence electrons than attempts to gain additional electrons to complete its shell. This behavior of donating electrons is a common trait among metals in contrast to the electron-accepting tendencies of nonmetals.
• 05:00 - 06:00: Resonance Structures and Formal Charge This chapter discusses the concept of resonance structures and formal charges in chemistry. It highlights how different elements tend to form bonds based on their valence electrons. Boron typically forms three bonds due to its three valence electrons. Beryllium forms two bonds, while lithium usually forms one ionic bond. Notably, carbon forms the most number of bonds among the elements in the second row of the periodic table.
• 06:00 - 07:00: Advanced Examples and Organic Molecules In this chapter, we explore advanced examples of organic molecules with a focus on elements capable of expanding their octet. Typically, atoms in the second row, like carbon, are limited to an octet rule, meaning they cannot have more than eight electrons in their valence shell. However, elements in the third row and beyond, such as phosphorus, sulfur, and chlorine, are capable of expanding their octet. This means they can form more than four bonds, due to the presence of d-orbitals which can accommodate extra electrons. Understanding this concept is crucial for comprehending the behavior and bonding potential of more complex organic molecules. The chapter also delves into the implications of this capability in various chemical reactions and molecular formations.
• 07:00 - 08:00: Polyatomic Ions and their Structures This chapter provides an overview of polyatomic ions and their electron structures. It emphasizes the electron configuration in atoms, stating that S can only accommodate two electrons. In the second energy level, there are two sub-levels: 2s and 2p. The P sub-level can hold up to six electrons, resulting in a total of eight electrons for the second row elements such as carbon, nitrogen, oxygen, and fluorine. The chapter suggests prior knowledge of electron configuration for better understanding.
• 08:00 - 09:00: Radicals and Specific Ions This chapter discusses the bonding behavior of carbon, nitrogen, oxygen, and fluorine, focusing on the concept that these elements will not form more than their typical number of bonds. It uses examples with oxygen in different compounds to illustrate how formal charges are distributed. In water, oxygen has two bonds and is neutral. In hydroxide (OH-), oxygen has one bond and a negative charge, whereas in the hydronium ion (H3O+), oxygen carries a positive formal charge. The discussion centers around understanding charges and bond formation.
• 09:00 - 10:00: Conclusion and Exam Preparation The chapter 'Conclusion and Exam Preparation' discusses the concept of formal charges in molecules with an example using the ammonium ion, NH4+. It explains how the nitrogen atom in NH3 is neutral with three bonds, but when it forms NH4+, it gains an additional bond and carries a formal charge, highlighting the deviation from its typical bonding number. Additionally, the chapter touches on the electron configuration of elements in the third row of the periodic table, elucidating the presence of three sublevels (3s, 3p, and 3d) and the total number of electrons each can accommodate. This explanation is part of the larger context of understanding molecular structures and properties, essential for exam preparation.

Lewis Structures, Introduction, Formal Charge, Molecular Geometry, Resonance, Polar or Nonpolar Transcription

• 00:00 - 00:30 in this video we're going to go over how to draw Le structures of simple molecules organic compounds polyatomic ions and a lot of other stuff as well we're going to talk about how to determine the molecular geometry also how to calculate the formal charge determine if there's any resonance structures and also determine if the molecule is polar or non-polar so we're going to cover all of that in this video so let's talk about the
• 00:30 - 01:00 Basics on the periodic table on the second row you'll see elements like boron carbon nitrogen oxygen and Florine Boron is found in group 3A of the periodic table so it has three valence electrons carbon is found in group 4 a so it has four valence electrons nitrogen has five oxygen has six Florine has seven now the same is true for or any elements below uh these
• 01:00 - 01:30 elements for example chlorine and bromine also have seven veence electrons brillium has two veence electrons and lithium has one now using this information you can determine or you can get a good estimation of how many bonds these elements like to form
• 01:30 - 02:00 now it's important to understand that elements on the upper right corner of the periodic table mostly the non-metals they like to acquire electrons they like to have eight electrons in their in their auto shell so to speak so Florine wants to have eight electrons it's electronegative it needs one more electron to get to eight so Florine is going to form one bond to get that one extra electron it needs to satisfy its oxide requirements
• 02:00 - 02:30 oxygen has six valence electrons it's a nonmetal it likes to acquire electrons it needs two more to get to eight so oxygen likes to form two bonds to get the eight electrons that it needs nitrogen likes to form three bonds carbon likes to form four now the elements on the the left side typically the metals they like to give away electrons it's easier for Boron to give away three electrons instead of trying to acquire five
• 02:30 - 03:00 electrons so Boron likes to form three bonds buril has two valence electrons and it likes to give away those two electrons so it's going to form two bonds lithium if you if you see lithium in a structure it's probably going to form one ionic bond so it's going to be just one for lithium so as you can see carbon forms the most number of bonds for the most part at least in the the second row for all of the elements in the
• 03:00 - 03:30 second row now once you go to the third row you can have what is known as an expanded Octan elements such as phosphorus sulfur chlorine they can have more than four bonds they can have more than eight electrons and here's why in the first row all you have is the 1s suble
• 03:30 - 04:00 and S can only hold two electrons now in the second row or in the second energy level you have two Su levels 2 s and 2 p hopefully at this point you've studied electron configuration so you should be familiar with these numbers and symbols now P can hold six electrons 2 + 6 is 8 so second row elements like carbon nitrogen oxygen and Florine they cannot have more than eight
• 04:00 - 04:30 electrons so carbon will never form more than four bonds same is true for nitrogen oxygen or Florine oxygen for instance likes to form two bonds in water if you draw the L structure for water oxygen has two bonds and oxygen is neutral in that particular molecule if you draw the structure hydroxide it has one Bond and it has a negative charge if you draw H2O plus the hydronium ion oxygen Bears a positive a formal charge so anytime an
• 04:30 - 05:00 element deviates from its Ido Bond number it typically contains a formal charge nitrogen is neutral in NH3 it has three bonds but if you draw the nh4 plus ammonium ion it's going to have four bonds and it's going to deviate from the ID number of three now in the third row you have three sublevels 3 S 3 p and 3D if you add up 2 six and 10 you will
• 05:00 - 05:30 get 18 so elements in the third row like phosphorus sulfur chlorine they can have more than four bonds four bonds equates to eight electrons so they can have what is known as an expanded oxide now these elements they have access to the 3D orbital the 3D Su and so that's why they
• 05:30 - 06:00 can have an expanded oxide they can have more than eight electrons or more than four bonds now in the course of this video you'll see that chlorine sometimes it's going to form one Bond sometimes it's going to form seven bonds or up to seven I should say chlorine has seven veence electrons it can acquire one electron to get to eight in which case it's going to form one bond to do that or it can give away all of it seven veence electrons and form up
• 06:00 - 06:30 to seven bonds sulfur has six veence electrons so sulfur can either acquire two to get to eight in which case it'll form two bonds or it can give away all six of its valence electrons to form six bonds if you draw the L structure for H2S sulfur has two bonds but if you draw the sulfate structure in this case it's going to have six bonds
• 06:30 - 07:00 phosphorus has five valence electrons it can acquire three electrons to get to eight in which case it's going to form three bonds or it can give away its five valence electrons and form five bonds so in ph3 phosphorus is neutral and it has three bonds it doesn't have a charge and in p43 minus the most stable resonance structure for the phosphate ion
• 07:00 - 07:30 you'll see that phosphorus will have five bonds so I want to just give you some general Trends in terms of the number of bonds that certain elements like to form so keep this information in mind as we continue to draw Le structures so let's start with simple structures how can we draw the leis structure for Florine Florine is a diatomic molecule but what can we do to draw the Le structure now the first thing you can do
• 07:30 - 08:00 is add up the number of valence electrons Florine has seven valence electrons and there's two of them so the structure that we need to draw should contain a total of 14 valence electrons now as we mentioned before Florine which is a hogen likes to form one Bond so knowing that it's safe to start with a single bond between the two Florine atoms this single bond is a a non-polar coent bond in the coent bond electrons are shared between the two
• 08:00 - 08:30 elements in a non-polar equalent Bond they're shared equally because the two flowing elements are the same the distribution of electrons will be the same so this is a non-polar molecule and the bond is also non-polar now typically whenever an element has one Bond attached to it it usually contains three lone pairs not always but usually when an
• 08:30 - 09:00 element has two bonds attached to it it usually contains two lone Pairs and when it has three bonds attached to it it usually contains one lone pair so that it can have a total of eight electrons around it to satisfy its oet requirement and that's just the general Trend now that General Trend typically is true for elements that are on the outside of the
• 09:00 - 09:30 molecule not the center element sometimes it's true for the center element except when it has an expanded octet but for the outer elements attached to the center element that Trend usually is or holds true now Florine has one Bond so it's going to have three lone pairs so that each Florine atom have eight electrons around itself so this is 2 4 6 eight electrons every bond equates to two electrons so each Florine atom has eight electrons
• 09:30 - 10:00 around it but two electrons are shared so the total is 14 this is 2 4 6 8 10 12 14 so if you can draw a structure where every atom that's in the second row have eight electrons typically the elements in a second row starting from carbon and to the right of carbon and if the number of electrons add up to this number it's safe to say that you have the correct
• 10:00 - 10:30 structure by the way elements to the left of carbon like Boron buril and lithium they will have what is known as an incomplete Octan they will have less than eight electrons so this is the L structure for F2 now let's draw the Le structure for O2 oxygen gas which is also a diatomic molecule so let's begin by adding the number of valence electrons so oxygen is a calogen in group 6A and
• 10:30 - 11:00 it has six valence electrons so it has a total of 12 and as you mentioned before because oxygen has six valence electrons and needs two more to get to eight oxygen likes to form two bonds so with that information it's good if we start with a double bond now when an element has two bonds typically it has two lone pairs so that it can have eight electrons around it so let's begin by putting two lone pairs on each oxygen atom so if you notice each
• 11:00 - 11:30 oxygen atom has eight electrons Around It 2 4 6 eight a double bond represents four electrons and the total number of electrons is 12 2 4 6 8 10 12 so this is the Le structure for the oxygen molecule that's how you draw it now what about nitrogen gas
• 11:30 - 12:00 N2 which is also diatomic what would you do to draw the Le structure for this molecule by the way for each of these examples it would be wise if you pause the video and work out these examples yourself to see if you're able to get the right answer the best way to learn is by practice and as you work out these examples you're going to get a better understanding of how to draw Le structures so nitrogen has five valence electrons and there's two of them so the nitrogen molecule has a total of 10 uh
• 12:00 - 12:30 valence electrons now because nitrogen has five electrons and it wants to have eight nitrogen likes to form three bonds so let's begin by putting a triple bond between the two nitrogen atoms and anytime an element has three bonds generally speaking it's going to have one lone pair so let's put one lone pair on each nitrogen
• 12:30 - 13:00 atom so the nitrogen on the left has 2 4 6 eight electrons so its octet requirement is satisfied and this entire molecule has a total of 2 4 6 8 10 electrons which is also correct so this is the Le structure for the nitrogen molecule this molecule has a linear geometry as you can see it's
• 13:00 - 13:30 straight the bond angle of a straight line is 180° for a full circle it's 360 now this molecule would you say it's polar or nonpolar anytime you have a substance that consists of only one element that substance is non-polar the electrons are shared equally in a substance like that for a substance to be polar one side has to
• 13:30 - 14:00 have a positive charge or a partial positive charge and the other side needs to be negatively charged this is a polarized object but since the elements are the same the electrons are shared equally and so it's a non-polar molecule so N2 O2 F2 even H2 all of these are non-polar molecules Now what is the bond order for
• 14:00 - 14:30 the nitrogen molecule it turns out that the bond order is basically equal to the number of bonds you see so for N2 the bond order is three since it has a triple bond for O2 the bond order is going to be two since it has a double bond between the two oxygen molecules now what about for F2 for F2 it contained a single Bond so the bond order is
• 14:30 - 15:00 one now what about this U molecule H2 hydrogen gas how can we draw the Le structure for it now hydrogen only has one valence electron it's in the first row we have two of them so this molecule has a total of two electrons elements in the first row like hydrogen or helium they can only have up to two electrons in their energy
• 15:00 - 15:30 level so hydrogen doesn't require an oxide of eight electrons so therefore hydrogen will always form one single Bond because that single Bond contains all the electrons that it needs which is two electrons you should never add any lone pairs to a hydrogen atom this is it that's the L structure for H2 now our next example is bh3
• 15:30 - 16:00 boring how can we draw the Le structure for this molecule now as we mentioned before we said that Boron is going to have an incomplete octet and Boron likes to form three bonds boron has three valence electrons hydrogen has one but there's three of them so the total is six uh valence electrons now we know that hydrogen will only form one Bond so we can start with a single Bond this is 2 four six electrons we already
• 16:00 - 16:30 have the number of electrons that we need which is six so we don't need to add any more electrons this is it that's the Le structure for bh3 it has six electrons Which is less than eight so boron has an incomplete octet but it has its desired number of bonds as you can see which is three bonds so this is the L structure for bh3 that's how you draw now what is the molecular geometry for this molecule
• 16:30 - 17:00 whenever you have a molecule where the center atom is attached to three atoms and if the center atom has no lone pairs this molecule has a trigonal planer shape now there's two things you need to be familiar with electron pair geometry and molecular geometry these two are going to be the same if this Center atom has no lone
• 17:00 - 17:30 pair if the center atom has a lone pair then the electron pair geometry and the molecular geometry will uh differ now what I recommend you doing or that you should do at this point is go to Google images and type in molecular geometry worksheet or something like that so you could be familiar with the different shapes and the names of those shapes so make sure sure you go ahead and do that when you get a
• 17:30 - 18:00 chance now this particular molecule would you say it's polar or non-polar anytime you have a molecule where the center atom doesn't have any lone Pairs and if all of the outer elements are the same basically if this molecule is symmetrical it's going to be non-polar all of the diap moments will cancel I'll go into more detail throughout this video video but for now
• 18:00 - 18:30 anytime all of the outer elements are the same it's going to be a non-polar molecule the bond angle that corresponds to a trigonal planer shape is 120° now what can help you remember that Bond angle is if you think of a circle a full circle is 360° so if you split a circle into three parts through 360 / 3 is
• 18:30 - 19:00 120 so the bond angle between two hydrogen atoms is going to be 120 that can help you to remember the bond angle for any trigonal planer shape now let's consider the Le structure for these three substances water de hydronium ion h3o
• 19:00 - 19:30 plus and the hydroxide ion o minus now earlier in this video we said that oxygen likes to form two bonds it has six valence electrons it needs two more to get to eight so it likes to form two bonds that's the ideal number of bonds however oxygen can form three bonds or one Bond but in a situation like that it's going to have a formal charge but whenever it has its desired number of bonds it's going to be stable which means the formal charge is zero
• 19:30 - 20:00 the equation to calculate the formal charge of an atom at least this is the equation I use it's the number of veence electrons minus the bonds and dots on that element so let's take water for example in water H2O hydrogen has one valence electron and there's two of them and oxygen has six so it has a total of eight so let's start with oxygen so we know hydrogen can only form a single Bond so that means hydrogen I mean oxygen has
• 20:00 - 20:30 two bonds Each Bond represents two electrons so we have four plus the two non the two lone pairs or also called non-bonding electrons so we can see oxygen has a total of eight 2 4 6 8 which agrees with that number now if we calculate the formal charge on oxygen it's going to be the
• 20:30 - 21:00 six veence electrons that oxygen has in this structure it has two bonds four dots two and four so 2 + 4 is 6 6 - 6 is 0 so as we can see whenever oxygen has two bonds and two lone peers it has a formal charge of zero which means oxygen is very stable in this configuration Now what is the molecular geometry of this
• 21:00 - 21:30 molecule whenever you have a molecule that has two atoms connected to the center atom and two lone pairs on a center atom if you look at that worksheet on Google Images this is going to have a bent molecular geometry or molecular shape the electron pair geometry is going to be tetrahedral anytime an element has four groups total if you count atoms and lone
• 21:30 - 22:00 pairs as groups not including the center atom so 1 2 3 4 anytime an atom has four groups it's going to the electron pair geometry is going to be tetrahedral if it has two groups it's going to be linear three trigonal planer four tetrahedral five trigonal bip peral six octahedral that's for the electron pair geometry so we have two lone pairs two atoms
• 22:00 - 22:30 attached to the central atom so we have a total of four groups the electron pair geometry is a tetrahedra for the structure now the bond angle for water you should basically memorize this number it's going to be 104.5 simply commit that to memory now let's move on to uh h2+ so how many valence electrons is is found in the hydronium ion
• 22:30 - 23:00 h2+ so oxygen has six hydrogen has 1 * 3 and whenever you see a plus charge you need to subtract one from the total number of veence electrons so 6 + 3 is 9 - 1 is 8 so let's start with oxygen so oxygen is attached to three hydrogen atoms
• 23:00 - 23:30 those three bonds represent six electrons we need to get to eight so that means oxygen has one L here so that's the Le structure for the hydronium ion now we know the net charge is+ one let's calculate the formal charge of the central oxygen atom so oxygen has six veence electrons it has three bonds and one lone pair
• 23:30 - 24:00 which is equal to two dots 3 + 2 is 5 6 - 5 is one so therefore oxygen has a plus one formal charge and that's why the total charge is plus one it's because of the oxygen atom as we can see because oxygen does not have its ideal number of two bonds it's going to have a formal charge now what is the molecular geometry and the electron pair geometry
• 24:00 - 24:30 of h3o+ how can we figure that part so oxygen has three atoms attached to it and it has one L paer for situation like that the molecular geometry is trigonal
• 24:30 - 25:00 pyramidal so that is the molecular geometry for h3o+ now the electron pair geometry because it has three atoms one one pair it has a total of four groups anytime an element has four groups the electron pair geometry it's going to be tetrahedral the same is true for water the total number of groups is four the electron here geometry will always be
• 25:00 - 25:30 tetrahedral whenever you have a total uh number of four groups so that's it for h2+ let's move on to hydroxide in hydroxide oxygen has six valence electrons hydrogen has one but because we have a negative charge we need to add one to the total number of valence electrons which once again we have eight now because it's only one hydrogen atom
• 25:30 - 26:00 this is going to be one Bond hydrogen will always form just a single Bond so right now we have only two electrons in that single Bond so we need to add three L peers 2 4 6 8 and as we mentioned before anytime an atom has one Bond generally speaking it's going to have three lone piers in the case of h2+ whenever an atom has three bonds it usually usually has one L pair and in a case of water
• 26:00 - 26:30 whenever an atom has two bonds it usually has two L pairs KN these General Trends will help you to quickly draw a molecule so we know that the total charge is going to be NE 1 for the hydroxide ion and we can see that oxygen does not have its ideal number of bonds which is two so that's a good indication it's
• 26:30 - 27:00 going to have a charge probably ne1 but let's use the equation to calculate it the formal charge equation is the number of veence electrons minus the bonds and the dots so oxygen has six veence electrons in this structure it only has one Bond but it has three lone pairs which is six dots 2 4 six Now 1 + 6 is 7 and 6 - 7 7 is equal to1 therefore oxygen has A1 formal
• 27:00 - 27:30 charge so as we can see whenever an element deviates from its ideal number of bonds it typically has a formal charge now consider the next two molecules CO2 also known as carbon dioxide and Co carbon monoxide now what I want you to do is I want you
• 27:30 - 28:00 to draw the Le structures for these two molecules and explain why carbon monoxide Co is polar but CO2 is non-polar when they both have polar bonds so let's analyze the co Bond carbon has an electro negativity value of 2.5 and oxygen has electro negativity value of
• 28:00 - 28:30 3.5 whenever the electro negativity difference between two elements if it's 05 or more the bond is considered to be a polar bond or polar coal Bond the reason why it's still called a calent bond is because the electrons between carbon and oxygen are being shared between those two elements but it's a polar equalent bond because they're being shared unequally because oxygen is more electr
• 28:30 - 29:00 negative than carbon it's going to pull the electrons toward itself and so oxygen is going to acquire a partial negative charge and carbon is going to be partially positive so whenever you have an unequal distribution of electrons one side of the molecule is going to be positive and the other side is going to be negative and when you have this separation of charge you have a polarized substance therefore we could
• 29:00 - 29:30 say that the carbon oxygen bond is polar one side is positive the other side is negative so if both molecules contain polar bonds how is it that CO2 is nonpolar but Co is polar so let's begin by drawing the Le structure of CO2 and then we'll draw the Le structure for Co now in CO2 carbon has four valence
• 29:30 - 30:00 electrons oxygen has 6 * 2 6 * 2 is 12 + 4 is 16 now there is a certain technique that will help you to draw the lres with ease and it's basically called the multiple of eight rule whenever a molecule has a multiple of eight electrons 8 16 2432 and if if there's no hydrogen atoms
• 30:00 - 30:30 attached to it then it's going to have no lone pairs on a center atom so in CO2 we have 16 valence electrons 16 is a multiple of eight so on the central carbon atom there should be no lone Pairs and I want to emphasize that this is only true if an only if there's no hydrogen atoms attached to this molecule
• 30:30 - 31:00 if there is then this multiple of eight technique will not work now we know that carbon likes to form four bonds and oxygen likes to form two bonds knowing that fact will help us to draw this Le structure with a so since oxygen likes to have two bonds let's start with a double bond in this situation carbon has four bonds each oxygen atom has two so everything is
• 31:00 - 31:30 good right now and as you mentioned before whenever an outer element has two bonds typically it has two lone pairs so let's put two lone pairs on each oxygen atom so every element has its desired number of bonds and also the octet requirement of each element is satisfied oxygen has eight electrons Around It 2 4 68 and carbon also has eight electrons
• 31:30 - 32:00 Around It 2 4 68 Each Bond is two electrons so whenever carbon has four bonds it's basically happy so this is the Le structure of carbon dioxide as you can see it's a linear molecule so it has a bond angle of 180° the hybridization at the center carbon atom is sp or SP hybridized
• 32:00 - 32:30 now what about Co let's draw the leou structure for that so carbon has four and oxygen has six so the total is 10 which is not a multiple of eight but then in this particular molecule we don't know what the center atom is either case there's going to be a lone paer on each element now carbon likes to form four bonds but oxygen likes to form two if we
• 32:30 - 33:00 put two bonds carbon is not going to be happy it wants four if we put four bonds oxygen is not going to be happy because it wants two so what can we do in this situation carbon wants four oxygen wants two so these two when they get together they need to form some sort of agreement so they're going to settle on three it turns out that in this molecule it's going to have three bonds which is not the ideal number for oxygen and it's not the ideal number for carbon it's simply
• 33:00 - 33:30 an average of 4 and two which is three so since each element does not have their ideal number of bonds both elements will have a formal charge now right now we have six electrons and as we mentioned before whenever an element has three bonds typically it has one lone paer so since each element has three bonds let's put a lone pair on each now we have a total of
• 33:30 - 34:00 10 electrons two 4 6 8 10 and each element has eight electrons around it which satisfies its octet requirement 2 4 6 8 so carbon and oxygen has eight electrons around it due to the uh six electrons that are being shared in the middle now let's calculate the formal charge on the carbon atom and on the
• 34:00 - 34:30 oxygen atom in Co so let's start with carbon so carbon has four valence electrons and in that structure it has a triple bond which is three bonds and it has a lone here which equates to two dots so 3 + 2 is 5 and 4 - 5 is1 now what about oxygen oxygen has six
• 34:30 - 35:00 valence electrons and this structure has three bonds two dots so 6 - 5 is + one so the carbon has a negative formal charge and oxygen has a positive formal charge when you add the charges the total charge is zero that's why overall carbon monoxide is a neutral molecule but the elements with in it contain formal
• 35:00 - 35:30 charges now why is it that Co is polar and CO2 is nonpolar we need to look at something called dipole moments to draw the dipole moment is basically an arrow that points towards the the negatively charged element oxygen is more electronegative than carbon so it's going to Bear the partial negative charge and carbon is partially positive so if we draw the dip moment towards the
• 35:30 - 36:00 partial negative oxygen atom notice that it cancels for cl2 due to the symmetry of this molecule but for Co it doesn't cancel so for CO2 the net dipole moment is zero that's why it's a non-polar molecule but for Co it's polar because it has a net dipole moment now let's move on to hybd
• 36:00 - 36:30 ization now let's say if you have an atom that's attached to two other atoms and let's say it has two lone pairs how can you calculate or determine the hybridization of the central atom if the number of groups on that atom let's say it's two where the number of groups is the number of atoms attached to it and lone pairs the hybridization is going to be S1
• 36:30 - 37:00 P1 if the total number of groups is three it's going to be S1 P2 it's sp3 for four S1 P3 D1 for 5 notice that the exponents add up to the number of groups and for six groups it's going to be S1 P3 D2 so in this particular example the hybridization at the central is going to be let's see 1 2 3 4 so we have four
• 37:00 - 37:30 groups it's going to be sp3 hybridized so knowing that what is the hybridization of boron in bh3 so Boron is attached to three atoms so it has a total of three groups therefore the hybridization is going to be S1 P2 or
• 37:30 - 38:00 simply SP2 hybridized now what about for water what's the hybridization of the central oxygen atom so in water oxygen is attached to two atoms and it has two lone pairs so it has four groups It's s P3
• 38:00 - 38:30 hybridized now try this one draw the Le structure for methane determine the molecular geometry and if it's polar or non-polar and also determine the hybridization carbon has four valence electrons hydrogen has 1 time four the total is eight now we know that hydrogen can only form one Bond and
• 38:30 - 39:00 carbon likes to form four bonds so this is the Le structure for CH4 this molecule has a tetrahedral molecular geometry and since there's no lone pairs on the central carbon atom the electron pair geometry is the same as the molecular geometry tetrahedral the bond angle at the central carbon atom is 109.5 as for the hybridization it has four groups so the
• 39:00 - 39:30 carbon atom is S3 hybridized now is it polar or is it nonpolar anytime you have a molecule that is composed of only carbon and hydrogen atoms it will always be non-polar and there's two reasons why first carbon hydrogen bond is non-polar carbon has an electr
• 39:30 - 40:00 negativity value of 2.5 and for hydrogen it's 2.1 so the difference or the electro negativity difference it's less than 0.5 which means it's a non-polar Bond nevertheless there is a small electro negativity difference so you still do have a small dipo moment now carbon is more electronegative than hydrogen so car carbon Bears the partial negative charge and hydrogen Bears the partial
• 40:00 - 40:30 positive charge so if you were to draw the diap moment it's going to point towards the carbon atom so even though these dipole moments are small they still exist nevertheless notice that they all cancel due to the symmetry of this molecule these two are in opposite directions and the same is true for those two to so the net dipole moment for this
• 40:30 - 41:00 molecule is zero therefore CH4 is a non-polar molecule now what about NH3 how can we draw the L structure for this molecule nitrogen has five valence electrons plus the three from the three hydrogen atom so the total is eight and we know that nitrogen likes to form one I mean three bonds hydrogen can only form one
• 41:00 - 41:30 Bond so three bonds equates to six electrons so nitrogen's going to have a lone Pier that's the Le structure for NH3 it has a trigonal pyramidal molecular geometry and the bond angle is about 107° now for the electron pair geometry the total number of groups is four nitrogen has three atoms and one lone
• 41:30 - 42:00 paer so it has a total of four groups so it has a a tetrahedral electron pair geometry now since it has four groups what is the hybridization at the central nitrogen atom for four groups the hybridization is going to be S1 P3 which adds up to four now what is the hybridization of the hydrogen at atom hydrogen only has one atom attached
• 42:00 - 42:30 to it so the hybridization for hydrogen is simply s now the NH3 molecule is it polar or is it nonpolar what would you say well let's analyze the NH Bond let's make some space first
• 42:30 - 43:00 nitrogen has an electro negativity value of 3.0 and for hydrogen it's 2.1 so the electro negativity difference is .9 which is much greater than 0.5 so the bond is indeed polar nitrogen Bears the partial negative charge since it's more electronegative than hydrogen so if we draw all of the diap moments they're going to point towards the center nitrogen
• 43:00 - 43:30 atom and as you can see the arrows are poting in the upward Direction and due to the presence of this lone PA the dipole moments do not cancel so there's a net dipole moment which means that NH3 is a polar molecule now let's consider two mole molecules CO2 and2 carbon dioxide and sulfur
• 43:30 - 44:00 dioxide carbon dioxide is non-polar but sulfur dioxide is polar the question is why why is SO2 a polar molecule where they both contain two oxygen atoms CO2 and SO2 they appear to be similar but why is it that 2 is polar and CO2 is
• 44:00 - 44:30 not we know that the Le structure for CO2 looks like this it has a linear molecular geometry and the carbon atom is sp hybridized since the carbon has two groups attached to it S1 P2 I mean S1 P1 adds up to two and the dipole moments cancel so that's why CO2 is nonpolar but
• 44:30 - 45:00 let's go ahead and draw the L structure for SO2 sulfur has six valence electrons and the same is true for oxygen it has six as well both elements are in the same group that is group 6A and so they all have six electrons sulfur and oxygen are known as cogens so if you add the electrons it's 18 now 18 is not a multiple of eight which means that there's going to be a
• 45:00 - 45:30 lone pair on a central sulfur atom to find out the number of Lone pairs subtract 18 by the highest multiple of eight just under 18 so the highest multiple of 8 is going to be 8 16 24 is too much so let's subtract 18 by 16 this will give us two this tells us that there's going to be two2 which equates to one L on a central
• 45:30 - 46:00 sulfur atom now we have two oxygen atoms now how should we draw the Le structure should it be like this with a single Bond or should we use double bonds and how can you tell well let's get into that later but let's say that this is the L structure for sulfur
• 46:00 - 46:30 dioxide is it polar or non-polar due to the presence of the lone pair the molecule has a bent shape and we know that oxygen is more electronegative than sulfur the electro negativity for oxygen is 3.5 for sulfur is 2.5 so the difference is about one which means that the so bond is polar if you draw the dip moments it's going going to point towards the oxygen
• 46:30 - 47:00 atom so notice that these two arrows they do not cancel that's why SO2 is a polar molecule they partially cancel but there's a net diap moment that goes this way now for those of you who have taken physics if you want to see why these dipole moments partially cancel but not completely see it this way so we have a dip moment going towards uh the left but
• 47:00 - 47:30 down at the same time and we have another one going this way so the first one and the second one has a y component that is directed in the negative y AIS and they both have an X component one is towards the right the others towards the left so the X components of the diap moments they cancel however the Y components they're parallel to each other they're going
• 47:30 - 48:00 down in the same direction so they add up and form a larger dipole moment in a negative y direction so that's why we could say this is a polar molecule now keep in mind a polarized object is simply an object where you have separation of charge if you were to draw the SO2 molecule it would look something like this it has three atoms the sulfur has a partial positive
• 48:00 - 48:30 charge and oxygen being more Electro negative Bears the partial negative charge so whenever you have a a separation of charge or an uneven distribution of electrons you have a polarized substance as you can see the top part of the molecule has a positive charge and the bottom part where the two oxygen are located has a negative charge and and this is a typical feature of a polar molecule one side is positive
• 48:30 - 49:00 and the other side is negative now I mentioned that there are other Le structures for SO2 and I'm going to talk about it when I go over resonance so stay tuned for that now earlier in this video I said that sulfur has six valence electrons and sulfur can either acquire two more electrons to get to eight or it can give away its six valence electrons so sulfur can form two bonds
• 49:00 - 49:30 sometimes it can form six bonds and still have a formal charge of zero consider the Le structure for H2S hydrogen has two electrons total since there's two of them plus six so the total number of electrons is eight so H2S has the same molecular geometry as H2O it has a bent molecular geometry and the electron pair geometry is
• 49:30 - 50:00 tetrahedral since it has two atoms two lone pairs which is four groups the sulfur hydrogen bond is relatively non-polar the electro negativity difference is only4 but nevertheless it does have a bent shape and H2S does dissolve in water so we should expect that this molecule is slightly polar due to the
• 50:00 - 50:30 the bend shape but as you can see in this structure sulfur only has two bonds sulfur Bears the partial negative charge and hydrogen is partially positive in this case sulfur is acquiring the two electrons it needs since it Bears the partial negative charge and that's why it has two bonds whenever an element AC requires an electron it gains a negative
• 50:30 - 51:00 charge whenever an element like a metal gives away an electron it acquires a positive charge so whenever sulfur is acquiring electrons it only needs to form two bonds to get to eight electrons since it has six and that's what we see here now sulfur can also give away all of its six electrons and it can form six bonds now this can happen when sulfur is
• 51:00 - 51:30 bonded to an element that's more electronegative than itself so in this molecule sf6 sulfur is going to have six bonds sulfur has six valence electrons Florine has 7 7 * 6 is 42 + 6 is 48 now 48 is a multiple of eight so the center element is not going to have any lone pairs
• 51:30 - 52:00 the Le structure of sulfur hexif floride looks like this it has an octahedral molecular geometry which is the same as the electron pair geometry and it's non-polar Florine is more electronegative than sulfur so if you draw the diap moments they will all cancel these two cancel each other these two will cancel each
• 52:00 - 52:30 other and the remaining two will cancel as well so due to the symmetry of this molecule it's going to be non-polar now the sulfur Florine bond is very polar the difference is huge Florine has an electro negativity value of 4.0 so the electro negativity difference is 1 .5 so the bonds are very polar but the molecule as a whole is
• 52:30 - 53:00 non-polar so whenever sulfur bonds to an element that has a higher electro negativity than itself it can form up to six bonds because sulfur has six valence electrons to give when sulfur is attached to an element that is less electronegative than itself like hydrogen sulfur only need to acquire two electrons to get to eight so it likes to
• 53:00 - 53:30 form two Bonds in that case so that's why in H2S it just it only has two bonds but in sf6 it can have up to six now it doesn't always have to have six if it's bonded to an element that is more Electro negative than itself for example an sf2 or sf4 sulfur can give away up to six electrons but it doesn't have to give all it can give two it can give four but it usually gives up to six
• 53:30 - 54:00 electrons so generally speaking sulfur will form anywhere between two and four bonds depending on what it's attached to but now let's talk about the bond angles for this molecule what is the bond angle for the octahedral
• 54:00 - 54:30 shape now the bond angle between these two floring atoms is 180 let's redraw the structure so we have a a sulfur atom at the center and between these two Florine atoms the bond angle is 180 now if you focus on the floing atoms at the
• 54:30 - 55:00 center the bond angles will be equal to 90° so the bond angles between these two are about 90 imagine if you have a square the angles in a square are 90° or imagine if you have a
• 55:00 - 55:30 circle and if you split the circle in four ways let's say this is the x axis this is the Y AIS these two are perpendicular the angle here is about 90 now the angle between these two Florine atoms is also
• 55:30 - 56:00 90° imagine if the four flooring atoms at the center is in the the XY plane the two floring atoms the one on top and on the bottom is on the z-axis so between the z-axis and the XY plane you have a 90° angle now try this one
• 56:00 - 56:30 pcl5 phosphorus pensac chloride go ahead and draw the Le structure for this molecule determine the hybridization the molecular geometry and the bond angles and if it's polar or non-polar so phosphorus has five valence electrons chlorine has 7 * 5 which is 35 + 5 is 40 since 40 is a multiple of eight there's going to be no lone pairs on the central phosphorus atom so we have five chlorine
• 56:30 - 57:00 atoms since each chlorine atom has one Bond it's going to have three lone pairs so if you add up all the electrons the total is 40 as we can see phosphorus has an expanded do 10 it has five bonds which is 10
• 57:00 - 57:30 electrons so as you mentioned before because phosphorus has five valence electrons it can try to acquire three electrons or give away all five electrons in this case it's giving away its five electrons to the chlorine atoms so that's why it's forming five bonds as we see here now what is the ization of the central phosphorus atom phosphorus is attached to five
• 57:30 - 58:00 chlorine atoms so it has five groups it's going to be S1 P3 D1 hybridize 1 + 3 + 1 adds up to five now what about the bond angles and what is the molecular geometry whenever you have an atom attached to five other atoms the molecular geometry which is going to be the same as the pair geometry is
• 58:00 - 58:30 trigonal by pyramidal when I hear the word trigonal I think of a triangle the three chlorine atoms in the center they form a triangle B pyramidal I think of two pyramids one at the top there's the top pyramid and one at the bottom here is the the bottom pyramid so trigonal by pyramidal that can help you to remember the name of this
• 58:30 - 59:00 shape now what about the polarity is it polar or non-polar whenever the center element if it doesn't have any lone Pairs and if all of the outer elements are identical typically it's going to be a non-polar molecule so PC L5 is non-polar the bonds are polar but the molecule is
• 59:00 - 59:30 nonpolar phosphorus has an electr negativity of 2.1 and for cl it's 3.0 so the difference is bigger than 0.5 so the bonds are polar now if you draw the dipole moments they're going to point towards the more electronegative chlorine atom so these two they cancel because they're anti parallel they're exactly opposite of each other now the other
• 59:30 - 60:00 three cancel as well so we have one arrow that goes towards left there's one that travels in the the Northeast Direction and one in the Southeast Direction so let's draw the vectors we have a vector towards the left another vector and another Vector so the vector that's in in the Northeast Direction it has an X
• 60:00 - 60:30 component and a y component the vector in a Southeast direction also has an X component and a y component notice that the two y components they're anti-parallel they cancel however the two x components is just enough to cancel with the vector that's in the the negative X Direction so these two cancel with the one in the left so overall this
• 60:30 - 61:00 is a non-polar molecule all of the vectors all of the dipole moments will cancel out so that the the sum of the dipo moments is zero which is typical of a non-polar molecule so anytime you have symmetry anytime all of the outer elements are the same just know it's going to be non-polar now one thing I forgot to do is go over the hybridization of sulfur hexif floride so what do you think the hybridization is for this
• 61:00 - 61:30 particular molecule it has an octaedro molecular geometry and it has six flowing atoms so the hybridization is going to be S1 P3 D2 1 + 3 + 2 adds up to six so I just want to make sure I went over that before I forget now when drawing difficult complicated molecules it helps to use the multiple of eight rule to determine the number of Lone pairs on the molecule or on the
• 61:30 - 62:00 center atom if you do this it's going to make it a lot easier to draw the Le structure so let's do a few examples of that so you can get used to the technique and remember this technique doesn't work if there's a hydrogen atom attached to the molecule try this one sf4 sulfur tetrafluoride so first first let's add up the number of valence electrons sulfur has six Florine has seven but
• 62:00 - 62:30 there's four of them 7 * 4 is 28 + 6 that's going to be 34 so 34 is not a multiple of eight which tells us that there's going to be at least one lone pair on a central sulfur atom to find how many lone peirs subtract 34 by the highest multiple of eight multiples of it are 8 16 24 32 40
• 62:30 - 63:00 is too much so let's subtract it by 32 so this gives us two electrons which equates to one lone pair so let's put two electrons on a central sulfur atom and we know Florine likes to form one Bond so this is the Le structure of sf4 now whenever Florine has one Bond it's going to have three lone pairs so every Florine atom has eight
• 63:00 - 63:30 electrons eight 16 24 32 34 so we have the correct number of electrons now what is the molecular geometry for this molecule it turns out whenever the central atom has or is attached to four other atoms and if it contains one here the molecular geometry is called
• 63:30 - 64:00 seesaw the electron pair geometry it has five groups which is four atoms one lone pair the electron pair geometry is going to be trigonal by pamal but the molecular shape is seesaw now is this molecule polar or or is it non-polar what would you say whenever an element if the central
• 64:00 - 64:30 element only has one lone pair not two but if it has one it's safe to say that it's going to be polar so let's analyze the diapo moments these two cancel however we don't have anything to cancel the dipole moments for the two arrows in purple so so the sf4 molecule is a polar molecule it has a net dipole
• 64:30 - 65:00 moment these two are pointed in the same general direction the Y components of those two vectors they they're additive but the X components are cancel because one is slightly to the left the other is slightly to the right oh by the way what is the hybridization on the central sulfur atom so if you add up the number of groups it has it has
• 65:00 - 65:30 four atoms one lone pair so it has five groups the hybridization is going to be S1 P3 D1 try this one sf2 go ahead and draw the molecular geometry for this substance so to draw the Le structure sulfur has
• 65:30 - 66:00 six valence electrons Florine has seven * 2 so that's 14 + 6 which is 20 20 is not a multiple of eight but the highest multiple of eight just under 20 is 16 so this gives us a difference of four which is two lone pairs or four dots now Florine likes to form one Bond and it's going to have three lone pairs each so this is the Le structure of the
• 66:00 - 66:30 sf2 molecule and like H2O it has a bent molecular geometry but a tetrahedral electron pair geometry and this molecule it's going to be polar which is typical of any bet molecule the dipole moments will Point towards the sulfur atom so if we draw the vectors the Y components of these two diap moments they're additive they're pointing in the same direction but the X components
• 66:30 - 67:00 cancel so therefore we have a net dipole moment that's in a negative y direction based on the way it's drawn so sf2 is a polar molecule and it has a Ben shape the hybridization at the central sulfur atom is going to be sp3 hybridized it has four groups 1 2 3 4 try this one
• 67:00 - 67:30 Xenon Tetra fluoride Xenon has eight valence electrons it's in group 8A which is the last column of the periodic table that's the noral gases and 7 * 4 seven for Florine atoms and it's four of them 7 * 4 is 28 + 8 is 36 if we subtract it by 3 um 32 excuse me which is the highest multiple of eight just under 36 you get four so Xenon has four dots or two lone
• 67:30 - 68:00 paers and it's attached to four Florine atoms so it looks like this Each of which will contain three lone pairs so what is the molecular geometry of this molecule what would you say and what is the electron pair geometry now we have four atoms attached
• 68:00 - 68:30 to the center atom and two lone pairs whenever you have that situation the molecular shape is going to be square planer the bond angles between the Florine atoms are 90° the electron pair geometry is going to be octahedral since you have a total of six groups you can treat the lone pairs as
• 68:30 - 69:00 if they were atoms so if you have an element with six atoms it's going to have an octahedral shape so that's the electron pair geometry now this particular molecule is it polar or is it non-polar molecules that have only one lone pair on a center atom are are usually polar molecules for example NH3 which has a trigonal pyramidal shape is
• 69:00 - 69:30 a polar molecule another example is SO2 it has only one lone pair it has a bent molecular geometry that's a polar molecule and also sulfur Tetra floride which has a seaw shape it too has only one L here on a center a at that's a polar molecule now you need to be careful with
• 69:30 - 70:00 elements that have two lone pairs because they may or may not be polar the key to focus on is symmetry if the electrons are distributed evenly it's usually nonpolar if they're distributed unevenly then it tends to be polar so in the case of water water usually looks like this it has a Ben shape but notice that the lone pairs I mean this molecule may seem symmetrical if you draw a line of
• 70:00 - 70:30 symmetry here but notice that the the lone pairs are not exactly opposite to each other they're on the same side as a result this is going to be a polar molecule but here notice that the lone pairs are on opposite sides of the molecule in that sense you can see that there's symmetry right here so so therefore XC F4 is a non-polar
• 70:30 - 71:00 molecule and also these two they cancel and these two Dio moments as well cancels so if you see the shape this Square planer molecular geometry it's going to be uh non-polar I guess if you think about the Symmetry and the case of water it appears to be symmetrical here but not
• 71:00 - 71:30 here but in XC F4 it's symmetrical on the Y AIS and on the x axis it's completely symmetrical so I guess you can use a symmetry test in both ways but either case this is a non-polar molecule oh by the way the the hybridization at the central Xenon atom is S1 P3 D2
• 71:30 - 72:00 hybridized it has six groups one 2 3 four and they two lone pairs try this one iodine pens floride so iodine has seven valence electrons the same is true for Florine 7 * 5 is 35 5 + 7 is 42 so to find the highest multiple of 8
• 72:00 - 72:30 it's going to be 8 16 24 3240 so we're going to have two dots or one lone pair on the central iodine atom each Florine atom is going to have a single Bond and they all have three lone pairs so this is the Le
• 72:30 - 73:00 structure for if5 now would you say it's polar or non-polar so notice that the center element only has one lone pair this molecule is going to be polar so the dipole moments in red will cancel they're opposite to each other and the dipole moments in purple will cancel as well however there's nothing
• 73:00 - 73:30 to cancel the dipole moment between the iodine and the Florine atom in yellow so this molecule has a net dipole moment which means that it's going to be polar the molecular geometry for this molecule is called Square pyramidal now let's talk about why that's the case so let's
• 73:30 - 74:00 uh redraw it now the base of the molecule has a shape of a square and we have a Florine atom on top and the lone pair in the bottom but looking at the atoms only you
• 74:00 - 74:30 could see that it has a base of a square and it forms a pyramid so it's a square pyramidal structure and if you can see that it's going to help you to remember the name of the molecular shape so it's Square pyramidal so therefore the bond angle between the Florine atoms in the plane of the square is about 90° and
• 74:30 - 75:00 also between these two since they're perpendicular to each other this is about 90 as well the hybridization at the central iodine atom is going to be S1 P3 D2 hybridized since it has a total of six groups five Florine atoms and one lone pair and the electron pair geometry is octahedral even though the molecular geometry is square peral the total
• 75:00 - 75:30 number of groups is six so whenever the number of atoms and lone pairs add up to six the electron pair geometry will be octahedral if they add up to five the EPG electron pair geometry is going to be trigonal by pidal if they add up to four it's going to be tetrahedral three trigonal planer and two linear now what about the triiodide ion I3
• 75:30 - 76:00 minus so let's draw the Le structure for it so iodine has seven valence electrons time 3 + one for the negative charge 7 * 3 is 21 + 1 is 22 so let's subtract it by 16 so this will give us six dots or three lone pairs on the central iodite atom typically iodine like any other
• 76:00 - 76:30 hogen chlorine bromine or Florine usually likes to form one Bond and this is going to be the case when it's not the center element for the examples that you've seen before whenever chlorine or Florine is not the center element it usually only has one Bond however when it's the center element it can have an expanded octad so it can have more than one Bond
• 76:30 - 77:00 just keep that in mind but when it's not the center element the halogens will usually have one bond this is the Le structure for the triiodide ion the overall charge is minus one let's calculate the formal charge of iodine so for this one it should be zero because it has its desired number of bonds the equation is it's the number of valence electrons which is seven minus
• 77:00 - 77:30 the number of bonds which is one plus the six dots so 7 - 7 is zero now let's calculate the formal charge of the iodine in the middle it's going to be S minus the two Bonds in six dots so 7 - 8 -1 so that's why the overall charge is negative because the one in the middle has a negative formal charge the molecular geometry is linear as you can see it's a straight line so
• 77:30 - 78:00 therefore the bond angle is 180 now the electron pair geometry is not linear because the total number of groups is five it has two atoms and three lone pairs so the electron pair geometry is trigonal by pyramidal the hybridization for five groups is S1 P3 D1 now I want to show you something else
• 78:00 - 78:30 that can also help you to identify the correct Le structure when there's multiple acceptable Le structures or at least how to find the most stable Le structure sometimes you'll have resonance forms where there's more than one way to draw Le structure and some Le structures are more stable than others generally speaking the most stable Le structure is the one where the formal
• 78:30 - 79:00 charge of the center element is zero now we said that the formal charge is equal to the veence electrons minus the bonds and the dots we know how to calculate the number of dots using the multiple of eight technique so for structures that have multiple acceptable Le structures we want the formal charge to be zero so let's solve for the number of
• 79:00 - 79:30 bonds so if we add B to both sides it turns out that the number of bonds to minimize the formal charge or to get a formal charge of zero it's going to be the valence electrons minus the number of dots now keep this equation in mind we're going to use it later if we add D to both sides notice that the number of bonds and dots adds up to the valence electrons let's focus on this expression
• 79:30 - 80:00 this is true if the net charge is zero if there's a formal charge of zero keep that in mind by the way so let's use sulfur as an example sulfur has six valence electrons notice that in most structures that we've come across so far the number of bonds and dots adds up to the number of valence electrons for example in a H2S notice that sulfur has four dots and two bonds which
• 80:00 - 80:30 adds up to the number of valence electrons in SO2 particularly this structure it has four bonds two dots which adds up to six in sf2 we had a structure that looks like
• 80:30 - 81:00 this so it has two bonds four dots which adds up to six in sf4 we had one lone pair four bonds so four bonds plus two dots adds up to six and finally in
• 81:00 - 81:30 sf6 we had six bonds zero dots which still adds up to six so generally speaking if the if you want to draw the most stable structure in which the formal charge is zero the number of bonds in Dot adds up to the number of valence
• 81:30 - 82:00 electrons now let's move on to structures that have resonance resonance occurs when you can draw a Le structure in multiple ways you can move around the electrons without moving the atoms so consider the carbonate structure CO3 2 minus let's draw the Le structure for this carbon has four valence electrons oxygen has six * 3 plus the min-2 charge 6 * 3 is 18 + 2 is 20 + 4 is 24 so 24 is a
• 82:00 - 82:30 multiple of eight and that tells us that there's going to be no lone pairs on a central carbon atom now we know that carbon likes to form four bonds so we have to draw it like this whenever oxygen has two bonds it's going to have two lone pairs when it has one Bond it's going to have three lone pairs this is the Le structure of the carbonate molecule so what is the formal charge on
• 82:30 - 83:00 this oxygen if you calculate the formal charge it's going to be the six valence electrons minus the two bonds and the four dots 6 - 6 is zero so it's neutral which is typical whenever oxygen has two bonds but when it has one Bond as in the case of hydroxide it's going to have a negative charge if you calculate the formal charge for this one it's it's going to be 6 minus the one bond in six dots 6 - 7
• 83:00 - 83:30 is-1 so each of these have a minus one formal charge thus we can see why the total charge is -2 so this is the Le structure for the carbonate molecule now if you want to you can enclose it in Brackets
• 83:30 - 84:00 now how can we draw the resonance structure of this molecule if you want to draw the resonance structure you can simply move the double bond from one oxygen atom to another but you can show how it's going to move and we can use arrows to represent the flow of electrons a four-headed Arrow represents the flow of two electrons a half Arrow repres represents the flow of one electron so we can take two electrons which equates to a lone pair and use it
• 84:00 - 84:30 to form a double bond and take the two electrons in this double bond and push it back on the oxygen on top so then we're going to get a structure that looks like this so the double bond is now here so we had three lone pairs on the oxygen and lower right now we only have two L Pi here now the oxygen in the lower left it remained the same now the oxygen on top it gained two
• 84:30 - 85:00 electrons so it has three lone pairs so these two initially had the negative charge now these two bear the negative charge so as you can see the electrons can move around creating another Le structure and so this is an example of resonance try this one the nitri ion go ahead and draw the Le structure and also one
• 85:00 - 85:30 resonance form so nitrogen has five valence electrons oxygen has 6 * 2 + 1 so that's 12 + 5 that's 17 + 1 which is 18 and let's subtract it by 16 so this tells us that the center or the central nitrogen atom has one lone pair or two dots now if we wish to calculate the number of
• 85:30 - 86:00 bonds on the cental nitrogen atom such that the formal charge will be zero it's going to be the valence electrons minus the dots nitrogen has five valence electrons minus two dots so it's going to want three bonds which is typical of a nitrogen atom so this is going to be 1 2 3 so one of these have to have a double bond so this is the Le structure of the nitri
• 86:00 - 86:30 ion whenever oxygen has one Bond it usually has a negative charge so thus we can see why the overall charge is minus one so draw the resonance form we can take a lone pair use it to form a double bond and break this double bond and push those two electrons on the other oxygen atom so now we have a Le structure that looks like
• 86:30 - 87:00 this so as you can see these two structures are equivalent they're the same and so that's how you can draw the resonance structure for the nitri ion now there's something called resonance hybrid the actual molecule is actually an average of the two the double bond is shared equally between these two oxygen atoms so if you
• 87:00 - 87:30 want to draw the resonance hybrid here's what you need to do so each oxygen atom will always have at least one Bond now because the double bond can be shared between both oxygen atoms you can draw it using dashes so this is the the resonance hybrid of NO2 now I'm missing two more electrons
• 87:30 - 88:00 because this one has two Ls this one has three so to distribute those two extra electrons equally let's put one on this oxygen atom and one on that one so it should look something like that now what about the BF3 molecule draw the Le structure for this molecule and also any resonance structures that you can think of boron has three Florine has 7 * 3 the
• 88:00 - 88:30 total is 24 which is a multiple of eight so there's no lone pairs on the central Boron atom NOW Boron likes to form three bonds and Florine likes to form one so the Le structure should look like this it has a trigonal plane of shape a bond angle of 120 and Boron is currently SP2 hybridized or at least it appears to be SP2
• 88:30 - 89:00 hybridized now can we draw any resonance structures for this molecule right now boron has an incomplete Octan it has three bonds but it can form four so the Florine atom can use one of its lone pairs to form a double bond in which case we're going to get a resin structure that looks like
• 89:00 - 89:30 this now which resonance structure is more stable the one on the left or the one on the right so far the other three the other two examples that we looked at the resonance structures were equal but now these two structures are not equal so one structure is more stable than the other which one is it so let's calculate the formal charge
• 89:30 - 90:00 of boron boron has three valence electrons minus three bonds zero dots so it has the formal charge of zero which is good now let's focus on Florine let's find the formal charge of this Florine atom which is is the same as the other two so Florine has seven veence electrons in this structure it has one
• 90:00 - 90:30 Bond six dots 7 - 7 is zero so the structure on the left is very stable because the formal charge of every element is zero now let's focus on the structure on the right side let's start with Boron so boron has three valence electrons it has four bonds zero dots so in this case it has a negative 1 formal charge 3 - 4 is1 now what about this Florine atom
• 90:30 - 91:00 because it has two bonds the formal charge is not going to be zero so Florine has seven veence electrons and in that structure it has two bonds four dots 7 minus 6 is 1 so it has a plus one charge so the total charge is still zero but do to the separation of charge this resonance form is less stable it's still an acceptable leis
• 91:00 - 91:30 structure because the octet of boron is satisfied boron has four bonds or eight electrons however the formal charge is ne1 Which is less stable than a formal charge of zero here it's neutral so due to the fact that this structure is completely neutral it doesn't have any form formal charge we should expect that this form is more stable however this form is still an
• 91:30 - 92:00 acceptable leis structure it's simply not the most stable Le structure now let's get back to our discussion on SO2 we said that it has multiple leis structures that you can represent it with now we know that the total number of electrons is 18 and if we subtract it by 16 we can see that it's going to have one Lo Pier or two dots now some of you
• 92:00 - 92:30 out there might be inclined to draw it uh this way you might put two Bonds on one oxygen and a single Bond on the other such that sulfur can have eight electrons to satisfy its oxit requirement so this is 2 4 6 8 so case sulfur has a complete octet which is good this structure is acceptable but is it the most stable
• 92:30 - 93:00 structure so we can draw the resonance form of this structure we can take a lone pair from oxygen and use it to form a double bond in which case we'll get the other structure so both Le structures are acceptable but they're not equally stable so remember if you want to draw the leis structure the most stable one where the
• 93:00 - 93:30 formal charge of the center atom is zero the number of valence electrons must equal the bonds plus the dots so if we look at the sulfur atom on the right it has four bonds two dots which adds up to the six valence electrons of the sulfur atom now for the structure on the left it has three bonds and two dots so therefore that tells us that
• 93:30 - 94:00 this sulfur atom has a formal charge so it's not the most stable structure therefore this one is the most stable structure if you calculate the formal charge on each starting with the one on the left it's going to be the six valence electrons of sulfur minus the three Bonds in two dots so 6 - 5 is+ one therefore the sulfur atom has a positive charge and the oxygen atom which has a single bond has a negative
• 94:00 - 94:30 charge for the structure on the right sulfur has six valence electrons four bonds two dots 6- 6 is zero so it's neutral therefore this is the most stable form of2 try this one draw the Le structure of the nitrite ion and also draw one resonance form for it nitrogen has five valence electrons oxygen has 6 * 3 plus the negative 1
• 94:30 - 95:00 charge 6 * 3 is 18 + 5 that's 23 + 1 is 24 now 24 is a multiple of eight so there's no lone pairs on a central nitrogen atom now nitrogen usually likes to form three bonds but in order for us to satisfy its octet requirement we need to put four bonds which means that it's going to have a formal charge it's not going to be neutral so 1 2 3
• 95:00 - 95:30 4 oxygen is going to have two lone pairs when it has a double bond and three lone pairs when it has a single Bond now we know that the nitrogen atom or the oxygen atom that has one bond is going to have a negative one formal charge to calculate the formal charge on the nitrogen atom it's going to be five valence electrons minus the four bonds zero dots so it has a positive formal charge -1 + 1 plus another negative 1
• 95:30 - 96:00 adds up to a net charge of negative one this is the Lis structure of the nitrate ion because it has a formal charge it doesn't have its desired number of bonds which is three bonds another example is nh4+ in this molecule nitrogen has four bonds so therefore it has a positive formal charge for the nitrate ion NO3 minus the molecular geometry is trigonal
• 96:00 - 96:30 planer so the bond angle is 120 for nh4+ it's a tetrahedral shape so we should expect the bond angle to be 109.5 for that now let's draw the resonance structure so let's take a l here from this oxygen let's use it to form a double bond and let's bring break this double bond and push a lone pair on that oxygen so this is going to be a single and this is going to be a
• 96:30 - 97:00 double and now there's going to be two lone pairs on this oxygen atom but three on this one so that's how you can draw the resonance structure for the nitr ion so basically this double bond can move anywhere among those three oxygen atoms so now let's try another problem how can we draw the Le structure for S F2 now let's put everything that we've
• 97:00 - 97:30 learned together to draw like these harder Le structures particularly when there's multiple elements so keep this in mind remember that oxygen likes to form two bonds and Florine likes to form one Bond whenever they're not the center element and when you have a molecule with three different elements how do you know which one is going to be the center atom it may not always be the middle one right now
• 97:30 - 98:00 oxygen is in the middle but doesn't mean that it's going to be the center element Florine likes to form one Bond oxygen likes to form two but sulfur can form anywhere between two to six in the presence of elements that are more electronegative than sulfur sulfur prefers to give away its six valence electrons than to acquire two the only way it's going to acquire two electrons is if it's attached to an element that's
• 98:00 - 98:30 less electronegative than itself but because oxygen and Florine really wants electrons they're going to pull it away from sulfur so in this case sulfur because it can form the most number of bonds it's going to be the atom in the center so let's add up the number of electrons in this molecule sulfur has has six oxygen has six as well and Florine has 7even but times 2 which is 14 + 6 is 20 plus another six is
• 98:30 - 99:00 26 so let's calculate the number of dots on a center sulfur atom so let's subtract 26 by the highest multiple of eight just under 26 that's 24 which is two now let's calculate the number of bonds that sulfur wants to have in order to minimize the formal charge so the number of bonds is going to be the valence electrons minus the number of dots
• 99:00 - 99:30 sulfur has six valence electrons and it has two dots so it wants four bonds so let's draw the sulfur atom and let's put the two lone Pairs and let's put the two Florine atoms and the oxygen atom in the middle now we need four bonds Florine
• 99:30 - 100:00 wants to have one Bond and oxygen wants to have two so this works out the number of bonds and dots at the central sulfur atom is six we have four bonds two dots which add up to six whenever fline has one Bond it's going to want three lone Pairs and oxygen has two bonds so it wants two lone pairs so that's the best way to draw the Le structure for S so F2 now let's try another example like this try this one p
• 100:00 - 100:30 cl3 actually just P so feel free to pause the video as you work on this example phosphorus has five valence electrons sulfur has six chlorine has seven so the total is 18 which means that phosphorus has one lone pair or two dots to calculate the number of bonds it's going
• 100:30 - 101:00 to be the valence electrons minus the number of dots if the formal charge can be zero so phosphorus has five valence electrons two dots so it wants to have three bonds oxygen likes to have two bonds chlorine likes to have one so this works out and this this is the Le structure of PCL let's try another
• 101:00 - 101:30 example s SO2 cl2 sulfur has six oxygen has 6times 2 and chlorine has seven so this is 6 + 12 + 14 6 + 14 is 20 + 12 that's 32 so 32 is a multiple of eight so there's no dots now the number of bonds it's going to be the number of valence electrons minus
• 101:30 - 102:00 the number of dots sulfur has six ve valence electrons and no dots so it wants to have six bonds so we have two oxygen atoms and two chlorine atoms chlorine likes to form one Bond oxygen likes to form two bonds so now we have six bonds whenever an outer element has only one Bond it's going to have three lone pairs when it has two bonds two lone pairs so
• 102:00 - 102:30 that's how you can draw the Le structure of so SO2 cl2 as you can see it has a tetrahedr molecular geometry and the hybridization at the central sulfur atom is going to be sp3 hybridized try this one x e o F2 Xenon has eight oxygen has six veence electrons Florine has seven 7 * 2 is 14
• 102:30 - 103:00 + 6 is 20 + 8 is 28 so if we take away 24 from it we're going to get four so that tells us that Xenon has two lone pairs or four dots so to calculate the number of bonds that will produce a formal charge of zero it's going to be the eight veence electrons of xenon minus four dots so Xenon wants to have four
• 103:00 - 103:30 bonds left over so let's put the two Florine atoms on a side and the oxygen atom in the middle just to make it symmetrical or to have some sort of symmetry so Florine likes to form one Bond oxygen likes to form two and then just add the appropriate number of Lone pairs so this is the L structure of XE o F2 so because Xenon
• 103:30 - 104:00 has eight veence electrons you want the number of bonds and dots to add up to eight so in this structure Xenon have four bonds four dots which adds up to the eight valence electrons that it has so this is the most stable and acceptable leis structure of this particular molecule now let's go over some polyatomic ions we're going to do a few of these
• 104:00 - 104:30 particularly the ones with expanded octadic so try this one sulfate so we're just going to focus on drawing the Le structures sulfur has six valence electrons oxygen has six but times four and let's add the let's add two electrons due to the minus two charge 6 * 4 is 24 4 + 4 is 30 + 2 is 32 so 32 is a multiple of 8 so there's no lone pairs
• 104:30 - 105:00 on the sulfur atom so to calculate the number of bonds to minimize the formal charge it's going to be the valence electrons minus the number of dots sulfur has six veence electrons zero dots so it wants to have six bonds the only way to do that is for two oxygens to be double bonded 1 2 3 4 four five six let's add two lone pairs to the double bonded oxygen atoms and three
• 105:00 - 105:30 lone pairs to the ones with a single Bond anytime oxygen has one Bond it's going to have a negative one formal charge so we can see why the total charge is minus two that's the Le structure for the sulfate ion as you can see the number of bonds and dots adds up to the valence electrons of sulfur whenever the form charge is zero try this one
• 105:30 - 106:00 phosphate phosphorus has five valence electrons and then oxygen is 6 * 4 + 3 this adds up to let's see 4 * 6 is 24 + 5 is 29 + 3 is 32 so that's a multiple of eight so there's no lone pairs on the central phosphorus atom which means that the number of valence electrons is going to be equal to the number of bonds since the number of dots is zero so phosphorus wants five Bonds
• 106:00 - 106:30 in this particular structure so 1 2 3 four five this is the Le structure for the phosphate ion so each of these single bonded oxygen atoms has a negative formal charge and we can see why the total is3 Let's uh enclose it in Brackets now what about the
• 106:30 - 107:00 phosphite ion try this one 3 * 6 is 18 + 5 that's 23 + 3 is 26 so 26 is not a multiple of eight if we subtract it by 24 that tells us that phosphorus it's going to have two dots or one lone paer so the number of bonds is going to be the valence electrons minus the number of dots so phosphorus
• 107:00 - 107:30 wants three remaining bonds so let's start with a lone pair and we have three oxygen atoms so each of them has to have a single Bond so that they each have a negative one formal charge so the total charge is still3 so this is the Le structure for the phosphite
• 107:30 - 108:00 ion now just to make sure that you've Master this topic I want you to draw the Le structure of the polyatomic ions that contain chlorine so draw the Le structure for the per chlorate ion the chlorate ion the chlorite ion and the hypochlorite ion if you can get these right then uh you basically mastered the topic so let's start with the per
• 108:00 - 108:30 chlorine ion cl4 minus so chlorine has seven valence electrons plus 4 * 6 for the four oxygen atoms and one for the negative 1 Charge 4 * 6 is 24 + 1 is 25 + 7 is 32 so since that's a multiple of eight there's no L pairs on the central chlorine atom now let's calculate the number of
• 108:30 - 109:00 bonds which is going to be the veence electrons minus the number of dots since there's no lone pairs it's just going to be 7 minus Z chlorine has seven veence electrons so we need seven bonds 1 2 3 4 five 6 7 so the oxygen which has a single bond is going to have three lone Pairs and a negative one charge the chlorine atom is going to be
• 109:00 - 109:30 neutral if you calculate the formal charge it's going to be seven valence electrons minus seven bonds and zero dots which is going to be zero so we can see why the total charge is minus one it's due to this oxygen atom so that's the lowest structure of the per Chlor ion now let's move on to the chlorate ion cl3 minus so it's 7 for chlorine 3 * 6 + 1 3 * 6 is 18 +
• 109:30 - 110:00 7 that's uh 25 + 1 is 26 so if we subtract it by 24 we're going to have two dots or one lone pair on the central chlorine atom now let's calculate the number of bonds which is going to be the valence electron minus the dots so chlorine has seven valence electrons two dots so it wants five bonds
• 110:00 - 110:30 remaining 1 2 3 four five so the only oxygen that has the negative charge is this one so that's the structure for the chlorate ion now let's move on to the chloride ion cl2 minus so it's going to be 7 + 2 * 6 + 1 2 * 6 is 12 + 7 that's 19 + 1 is 20
• 110:30 - 111:00 subtracted by 18 that gives us four so we have two lone pairs or four dots on the center chlorine atom now let's calculate the number of bonds so chlorine has seven veence electr r four dotts so we need three bonds so we're going to put a double
• 111:00 - 111:30 bond on one oxygen and a single Bond on the other so this is the one that has the negative charge so that's the lowest structure for the chloride ion and then the last one of the chlorine series is going to be the hypochlor ion which is 7 + 6 + 1 that adds up to 14 and 14 minus 8 is six so we're going to have six or six dots or three lone pairs
• 111:30 - 112:00 on the center chlorine atom so if you calculate the number of bonds it's seven valence electrons minus 6 it's just going to be one Bond and oxygen is going to have three lone period as well but it's going to Bear the negative charge so that's the Le structure of the the hypochlorite ion now let's move on to the
• 112:00 - 112:30 radicals let's draw the Le structure for NO2 a radical is basically a substance that have an odd number of electrons nitrogen has five valence electrons oxygen has 6 * 2 which is 12 + 5 is 17 whenever you get an odd number you're going to have a radical so basically instead of having a lone pair which is two dots you're going to have half of a lone pair which is a single
• 112:30 - 113:00 dot if you subtract this by 16 you're going to get one so this half Al here or a single dot on the nitrogen atom now we know that nitrogen cannot have more than eight electrons but we want to get to we want to get as close as to eight as
• 113:00 - 113:30 possible so what should we do in this case if we calculate the number of bonds which is valence electrons minus dots nitrogen has five valence electrons and it has one dot so ideally we would probably want four but we can't have four so this equ equ is not going to work for Radicals so it's good to know that so we're just going to have to just use our
• 113:30 - 114:00 understanding of L structures elements in a second row like carbon nitrogen oxygen and Florine cannot have more than eight so if we put four bonds this is not going to work nitrogen will have nine electrons that's a violation of the ACH rule it's not like phosphorus or sulfur where it can have an expanded oxide so therefore the most that we can get to is three bonds if we have three Bonds in one dot nitrogen has seven veence electrons 2 4 6 seven so we can't
• 114:00 - 114:30 exceed eight but we want to get to as close as eight as possible and this is as far as we can go and whenever oxygen has a double bond we need to add two lone Pairs and three when it has a single Bond so this is the Le structure for NO2 nitrogen dioxide now let's try another one n o nitrogen monoxide so nitrogen has five o has six
• 114:30 - 115:00 so this adds up to 11 now what can we do in this example so let's start by putting n and O we know that oxygen likes to form two bonds and nitrogen likes to form three so if we start out with uh well first let's subtract 11 by 8 if we do
• 115:00 - 115:30 that that's going to give us three that tells us that one of these elements is going to have three electrons which is probably the less electronegative element but now let's focus on the bonds let's get back to that three should we start with three bonds or two bonds well let's try three bonds let's see what's going to happen if we add three bonds oxygen can
• 115:30 - 116:00 only have one lone paer and the same is true for nitrogen otherwise they will have more than eight electrons and right now if you add up the electrons you only get a total of 10 2 4 6 8 10 so it can't be three bonds so let's go down to two bonds now if we have two bonds we need to add two lone pairs to oxygen so right now we have a total of eight
• 116:00 - 116:30 electrons and then we can add the other three to nitrogen or we can do it this way we can add three to oxygen and four to nitrogen so in both cases we have a total of 11 electrons and one element has eight electrons the other has seven so which structure is better which element prefers to have
• 116:30 - 117:00 eight oxygen is more electronegative than nitrogen the electro negativity for Florine is 4.0 for oxygen is 3.5 and for nitrogen is 3.0 because oxygen is more electronegative than nitrogen it's going to pull the electrons toward itself so oxygen is going to be the one that has of two lone pairs or four dots instead of nitrogen since it has a greater pole of electrons so therefore the most
• 117:00 - 117:30 acceptable Le structure is this one if you calculate the formal charge of each element it's going to be zero a quick way to tell is that or is to check that the number of bonds and dots to see if it's equal to the number of valence electrons in the case of oxygen oxygen has two bonds four dots which equates to six and oxygen has six
• 117:30 - 118:00 valence electrons nitrogen has two bonds three dots so that adds up to five which is good if you calculate the formal charge of the nitrogen atom it's going to be five valence electrons minus the two bonds and the three dots it's going to be zero but for the other structure that we had earlier that's not going to be the
• 118:00 - 118:30 case it's going to be five valence electrons for nitrogen two bonds four dots so in this structure nitrogen has a negative 1 formal charge which means oxygen is probably going to have a positive one formal charge it's going to be six valence electrons for oxygen two bonds three dots that's plus one so this particular Le structure is less stable so therefore this is the most acceptable leis structure for nitrogen
• 118:30 - 119:00 monoxide so anytime you have an odd number of electrons you're going to have a radical so we could see where the three comes from one of the elements have three dots around it now what about this structure SN minus the thiocyanate ion how can we draw the Le structure for that so we have a polyatomic ion with three different
• 119:00 - 119:30 elements so you want to identify which element is going to be the center atom so nitrogen likes to form three bonds carbon likes to form four bonds sulur could form anywhere from 2 to six so should we put sulfur in the middle more carbon in the middle what would you
• 119:30 - 120:00 say let's see what would happen if we put sulfur in the middle so carbon wants to have four bonds nitrogen wants to have three sulfur at most wants to have six so this is probably not going to be an ideal situation remember sulfur is going to be in between two to six bonds sometimes it could be one but it's r that sulfur is going to
• 120:00 - 120:30 have more than six bonds it's unlikely so let's try putting carbon in the middle so one possible structure that we can draw is put in a triple bond between carbon and nitrogen and a single bond with sulfur so if sulfur has one bond is probably going to have a formal charge but it's better than having seven bonds or we could do this we can give sulfur two bonds which it's going to be
• 120:30 - 121:00 okay with that and nitrogen two bonds so either case one of these elements is not going to be happy and we do have a negative charge so we should expect that it turns out both of these structures are acceptable whenever nit has three bonds it's going to have one lone pair when it has two bonds it has two lone pairs when sulfur has one Bond it's going to have
• 121:00 - 121:30 three lone Pairs and this one's going to have two if we calculate the formal charge on nitrogen it's going to be five minus the three bonds plus two dots so as we can see whenever nitrogen has three bonds it has a formal charge of zero but when it has two bonds it's going to be five minus the two bonds four dots 5 minus 6 is negative 1 now when sulfur has two bonds it has a
• 121:30 - 122:00 formal charge of zero but when it has one Bond it's going to be six minus the one Bond and six dots that's 6 - 7 which is-1 so these two structures are resonance forms of each other I can take a lone pair form a double bond which will give me this Bond here and break a triple bond and put a l here on this nitrogen giving me this structure so both fluid structures are
• 122:00 - 122:30 acceptable every atom has eight electrons around it so their oxy requirement is satisfied now the question is is it better to put a negative charge on nitrogen or on a sulfur atom so which is more important electron negativity or size on the periodic table we have elements like carbon nitrogen oxygen Florine and
• 122:30 - 123:00 sulfur if we were comparing nitrogen and oxygen the size between these two elements is roughly about the same oxygen is a little bit smaller than nitrogen but the atomic size doesn't vary that much for elements in the same row so between nitrogen and oxygen it's better to put the negative charge on oxygen because it's more electronegative it can stabilize the negative charge
• 123:00 - 123:30 better than the nitrogen atom now between nitrogen and sulfur sulfur is significantly bigger than nitrogen let's say if this is the size of nitrogen this would be the size of sulfur because sulfur is so much more bigger it can stabilize the negative charge better the negative charge is distributed over a larger surface area so it's less concentrated if you compare sh minus and
• 123:30 - 124:00 O minus hydroxide is a stronger base than the sh minus ion the reason being is water is less acidic than H2S H2S is more acidic than water so therefore hydroxide is a stronger base the stronger the acid the weaker the base the fact that hydroxide is a stronger base than sh minus means that
• 124:00 - 124:30 the negative charge or the oxygen atom that has the negative charge is less stable since the base is stronger here the base is weaker which means that it's more stable whenever a molecule is stable it's less reactive whenever it's more reactive it's less stable so as you can see the bigger atom can stabilize the
• 124:30 - 125:00 negative charge better than a smaller atom so this is going to be the best or most stable resonance form of the thio sin ion but both forms are acceptable it's just that this one is better it's more stable less reactive now let's go over organic molecules particularly those with mostly carbon atoms and hydrogen atoms for
• 125:00 - 125:30 these molecules you really don't need to add up the number of electrons all you need to do is simply put it together for example let's draw the L structure of ethane c2h6 for these compounds just remember that carbon likes to form four bonds hydrogen can only form one oxygen likes to form two and Nitrogen likes to form three bonds with that information it's enough to put this together so because carbon can form the most number of bonds we're
• 125:30 - 126:00 going to put that in the middle now what you want to do when drawing these types of molecules draw them evenly that is with symmetry so if there's six hydrogen atoms let's put three on each carbon atom so in this structure every hydrogen atom has one Bond every carbon atom has four bonds and that's how you draw ethane C2 H6 so now let's draw ethane C2
• 126:00 - 126:30 H4 so let's begin by putting the two carbon atoms in the center and since there's four hydrogen atoms let's spread them out equally so let's put two hydrogen atoms on each carbon atom so in order for carbon to have four bonds we need to put a double bond and so that's the Le structure for ethine now let's try acetylene or ethine c2h2 so each carbon atom is going to have one hydrogen atom and in order for
• 126:30 - 127:00 each carbon atom to have four bonds we need to put a triple bond and that's the low structure for acetylene so as you can see for the organic molecules you just got to connect them with the appropriate number of bonds for the organic molecules that we've just considered the first one where the carbon atoms only had single bonds that's called an alane and the one where we had a double bond between the two carbon atoms it's an Aline and for the triple bond it's an
• 127:00 - 127:30 alkine now the next molecule is going to be an alcohol ch3 this is called methanol so the first carbon atom has three hydrogens ch3 and it's attached to an oxygen atom that's attached to a hydrogen so in this structure carbon has four bonds and oxygen has two and we know that whenever oxygen has two bonds we need to add two L pairs so that's the L structure for
• 127:30 - 128:00 methanol so all you got to do is just put it together let's try this one ch3 CH if you see o it's typically an alcohol but if you see CH this is known as an alahh so how can we draw the Le structure we know that ch3 is going to look like this it's a carbon with three hydrogens and that's attached to another
• 128:00 - 128:30 carbon now we can't write it like this because hydrogen can't have two bonds and we can't draw it this way because this carbon atom will have two bonds and not four so the only way we can make this work is if we connect a carbon with a double bond to an oxygen since oxygen wants to have two bonds and with a hydrogen to the side this is the only way this structure can work so carbon has four oxygen has
• 128:30 - 129:00 two and every hydrogen atom has one Bond so that's how you can draw an alahh the next molecule that we're going to go over is the Ketone so go ahead and pause the video and draw the Lis structure so typically all of the carbon atoms are usually connected together the carbon atom on the left is a ch3 which means that it has three hydrogens attached to it and the carbon atom on
• 129:00 - 129:30 the right is also a ch3 so we have an oxygen in the middle and we know that oxygen likes to form two bonds so this is the best way to connect them together so that every carbon atom has four bonds and let's add the two lone pairs so that's the L structure for this particular molecule it's a ketone also known as two propanone or simply propanone the common name for this molecule is acetone which is found in nail
• 129:30 - 130:00 polish now let's try another example so this is known as an ether particularly dimethyl ether a ch3 group is known as a methyl group so let's start with the oxygen the oxygen is attached to two carbons which contain three hydrogens each oxygen likes to form two bonds which it already has it and it has two long pairs and that's it for The Ether structure
• 130:00 - 130:30 that's all there is to it and now let's draw the carboxilic acid so the first carbon is going to have three hydrogens and what about the second carbon how can we add the two oxygen and the hydrogen so we can't draw it like this because carbon is not going to have four bonds that means that we need to put an oxygen on
• 130:30 - 131:00 top we don't want to put it this way because the oxygen atoms won't have two bonds the best way to connect the structure is to put a double bond and to put the hydrogen on one of the oxygen atoms in this case every oxygen atom is going to have two bonds every carbon atom has four so this is the L structure for the carbic acid this is also known as acetic acid or ethanoic
• 131:00 - 131:30 acid this is found in vinegar what about this one this is known as an Esther so we have three carbon atoms but this time you got to be careful the first carbon atom has three hydrogens the second one is going to have two oxygen atoms but not all of the carbon atoms are connected to each other there's an oxygen in the
• 131:30 - 132:00 middle so this is the Lis structure of this particular Esther now what about this one this is known as an amine particularly this is called ethyl amine methyl is just a ch3 it's one carbon ethyl contains two carbons so the first carbon atom has
• 132:00 - 132:30 three hydrogen atoms the second one has two hydrogen atoms and it's attached to a nitrogen atom which has two hydrogen atoms and a lone Pier so that's the L structure for Etho Amine now this is known as an amide particularly ethanamide since it has two carbons to
• 132:30 - 133:00 draw the Le structure the first carbon is going to have three hydrogens it's a methyl group the second one is going to have a carbonal group which is a C1 o and it's going to have an nh2 group so whenever nitrogen has three bonds it's to have one lone pair and oxygen usually has two so that's how you can draw the Le structure for an
• 133:00 - 133:30 amide the last one is a Nitra whenever you see CN typically there's a triple bond between them so in order for this carbon atom to have four bonds you need to put a triple bond and so this is the leou structure for a Nitro so that is it for this video thanks for watching and I wish you well
• 133:30 - 134:00 on your exam on Le structures