Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, one quick question before we start. Fluorine is more electronegative than chlorine; it pulls electrons harder. So in B F three, boron should be left hungrier for electrons than in B C l three, right? And so B F three should be the stronger Lewis acid. Sounds perfect. But are you sure? Keep your answer ready; example two will settle it. Today is a concept check class: short true or false statements from the whole of chemical bonding. And look, in JEE these short statements look easy, but one word can flip the answer. So, let's begin.

0:38Watch the big words: all, always, every

So, first, a simple method for true or false. Look for the big words: all, always, every. Such a statement must hold in every single case. First one on screen: all ionic compounds are soluble in water. Common salt dissolves easily, so true, right? Hmm, careful! Think about what dissolving costs. First the crystal must be pulled apart into separate ions; that cost is the lattice enthalpy. Water pays it back as the hydration enthalpy, the energy released when water molecules surround the ions. So a salt dissolves only if the hydration enthalpy beats the lattice enthalpy. With me? And for some salts, it does not. Barium sulphate, B a S O four; the small four below means four oxygen atoms. Then calcium fluoride, C a F two, and silver chloride, A g C l: all three are almost insoluble. So the statement is false. One counterexample is enough to break the word all. Got it? Next statement: a molecule with a lone pair on its central atom always has a dipole moment. The word always, again! Hmm. The lone pair makes the molecule lopsided, so it must be polar, right? Not always! Look at xenon difluoride, X e F two, and xenon tetrafluoride, X e F four. Xenon carries three lone pairs in the first, and two in the second. But X e F two is linear, and X e F four is square planar: both completely symmetric. So the bond dipoles cancel, and the dipole moment is zero. False! Clear?

2:15Is every pi bond weaker than every sigma bond?

Now, the next statement is right there in the heading: every pi bond is weaker than every sigma bond. Pi and sigma are the two Greek letters on screen. You learnt that a sigma bond, made by head-on overlap, is stronger than a pi bond, made by sideways overlap. So, true? Look at the word every, again! The rule, sigma stronger than pi, compares bonds between the same pair of atoms. Change the atoms, and the rule says nothing at all. Right? So, let's compare different atoms. The pi bond inside a carbon carbon double bond, that equals sign between the two carbons, is worth about two hundred and seventy kilojoules per mole. The wavy equals sign means approximately, and k J mol to the minus one means kilojoules per mole. Okay? And the sigma bond in fluorine, the single F F bond: just one hundred and fifty eight. It is weak because the lone pairs on the two small fluorine atoms repel each other. So, which one is stronger here? The pi bond! Two hundred and seventy is greater than one hundred and fifty eight. So this pi bond beats this sigma bond, and the statement is false as a general statement. The rule, sigma beats pi, holds only for the same pair of atoms. Got it?

3:33Paramagnetic O2: explained by VB theory?

Now, a statement about oxygen, in the heading: the paramagnetism of O two is explained by valence bond theory. We will use the figure on the right in a minute. Paramagnetic means the molecule is drawn into a magnetic field, because it has unpaired electrons. So, true or false? Let's check with valence bond theory, or V B theory. It gives the Lewis structure O double bond O, the two lines on screen. Every electron in it is paired. So, no unpaired electrons, and no explanation for the magnetism. With me? Now molecular orbital theory, M O theory, in the figure. On the left and right, the three short lines are the two p orbitals of each oxygen atom; O two p four, with the small four up top, means four electrons in two p. In the middle, from the bottom: sigma two p z, with a small z below, then pi two p, then pi star two p, and sigma star two p z at the top. The star marks an antibonding orbital. Now look at pi star two p: two orbitals of equal energy, one electron in each, both arrows pointing up. Okay? So O two has two unpaired electrons, and that is exactly why it is paramagnetic. The line under the figure gives B O, the bond order: one half of ten minus six. That counts all sixteen electrons, not just the ones drawn: ten bonding, six antibonding. So the bond order is two, the same double bond as before. Right? So the statement is false. Valence bond theory cannot explain it; only M O theory explains the two unpaired electrons. Remember this one; it comes up often.

5:17Example 1: is a hydrogen bond stronger than a covalent bond

Now, example one. True or false: hydrogen bonds are stronger than covalent bonds. Pause the video, and try it first. Done? Okay, let's check. Hydrogen bonds are worth about ten to forty kilojoules per mole. That is the range on screen: ten to forty. And a typical covalent bond? About four hundred kilojoules per mole. So, which is stronger? The covalent bond, by far! Even the top value, forty, is just one tenth of four hundred. So, false. Hydrogen bonds are usually stronger than other intermolecular forces, but much weaker than covalent bonds. Don't mix the two up. Clear?

6:03Example 2: why BCl3 is a stronger Lewis acid than BF3

Now example two, the harder one, and it is our opening puzzle. True or false: B C l three is a stronger Lewis acid than B F three. Remember, a Lewis acid accepts a pair of electrons. Pause and try. Hint: think about back bonding. Done? Alright, first the quick answer from the start. Fluorine is more electronegative, so it pulls more electron density away from boron. So B F three is the stronger acid, and the statement is false, right? No, no, careful! Look at boron first. In both molecules, boron has only six electrons around it, and one empty two p orbital. That empty orbital is what accepts the electron pair. Okay? Now, fluorine has a filled two p orbital, and it overlaps sideways with the empty two p of boron; the arrow shows electron density flowing back to boron. Same size orbitals, so the overlap is efficient. This is back bonding, and it partly fills boron's vacancy. Now, chlorine uses a three p orbital, which is larger. So, will its back bonding be stronger or weaker? Weaker, exactly! Chlorine's large three p does not match boron's small two p, so the back bonding is weaker. Boron gets much less electron density back. With me? So boron in B C l three stays more electron deficient, hungrier for an electron pair. B C l three is the stronger Lewis acid. True! Our opening puzzle, solved: the electronegativity argument completely misses back bonding. And bromine uses an even larger four p, so its back bonding is weaker still. So, Lewis acid strength: B F three, less than B C l three, less than B B r three; those are the less than signs on screen. Remember this order; it is a favourite in concept checks.

8:01Quick revision: big words, paramagnetic O2, back bonding

So, quick revision. One: with all, always or every, a single counterexample makes the statement false. Two: O two is paramagnetic, and only M O theory explains its two unpaired electrons. Three: back bonding makes B F three the weakest Lewis acid of the three: B F three, less than B C l three, less than B B r three. Now go and try the practice problems on this page. You can do them. All the best!