Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. To make calcium oxide, calcium has to turn into a gas, and then give away two electrons. Both cost energy. Oxygen has to split into atoms, and each atom has to take two electrons; that costs energy too. Step after step, we keep paying. And yet, forming calcium oxide from calcium and oxygen gives out six hundred and thirty five kilojoules per mole! So who pays the bill? We will settle the account at the end of this class. This is Part A of our unit problem bank: JEE Main level problems from all of chemical bonding. Questions like these come up often. So, let's begin.

0:45Tool 1: ionic compounds and polarising power

So, first tool: ionic compounds. Look at these four: calcium carbide, sodium peroxide, potassium superoxide and magnesium chloride. The small numbers below the symbols are subscripts; they count the atoms. All four are ionic. Now tell me: which of them contain a covalent bond as well? If you said none, careful! Look at the anions. The carbide ion, C two, two minus; the peroxide ion, O two, two minus; and the superoxide ion, O two minus, with the charges on top. Each holds two like atoms, joined by a covalent bond. So the first three qualify, and magnesium chloride has only simple ions. Clear? Now, polarising power: how strongly a cation pulls the electron cloud of an anion. Roughly, it goes as charge over size; the wavy equals sign means approximately. Here are four cations: lithium and sodium with one plus, beryllium and magnesium with two plus. Which one polarises the most? Beryllium, yes, exactly! Tiny and doubly charged, so by far the highest. Now lithium and magnesium are almost the same size, but magnesium carries double the charge, so its charge over size is about twice as big. So magnesium is next after beryllium, and sodium, big and singly charged, is the lowest. So the less than signs read: sodium, then lithium, then magnesium, then beryllium. Got it?

2:12Tool 2: count lone pairs with VSEPR

Alright, second tool: VSEPR, and the quick count from the course. Take V plus M, minus c, plus a, and halve the whole thing: that gives the electron pairs on the central atom. V is its valence electrons, M the number of monovalent atoms attached, like hydrogen or a halogen, c the positive charge and a the negative charge. Subtract the attached atoms, and you have the lone pairs. With me? Now a problem from the bank. Count the lone pairs on the central atoms of S F four, B r F three, I F five, X e F four and I three minus, and add them up. Okay? The first four are quick. Sulphur: six plus four is ten, so five pairs; minus four atoms, one lone pair. Bromine: five pairs, three atoms: two. Iodine in I F five: six pairs, five atoms: one. Xenon: six pairs, four atoms: two. Got it? Now the tricky one, I three minus. The central iodine has seven valence electrons, and two iodine atoms are attached. So, half of seven plus two, right? Oh no, careful! Half of nine is four and a half, and pairs never come in halves. You forgot the minus charge, so a is one. Seven plus two plus one is ten: five pairs, minus two attached atoms, three lone pairs. So the total is one plus two plus one plus two plus three: nine. Very good!

3:41Tool 3: forces between molecules

Now the third tool: forces between molecules. Arrange these by boiling point: C H three F, methanol and methane. Fluorine is the most electronegative atom. So C H three F forms hydrogen bonds, right? No, no, wait! For a hydrogen bond, the hydrogen itself must be bonded to fluorine, oxygen or nitrogen. In C H three F, every hydrogen sits on carbon. So, no hydrogen bonds there. Okay? So methane has only dispersion forces. C H three F is polar, which helps, but it still has no hydrogen bonds. Clear? And methanol has an O H group, so its molecules are hydrogen bonded. So the order is methane, then C H three F, then methanol. Hydrogen bonding wins. Got it?

4:29Example 1: most stable of He2+, He2, Li2, Be2 by bond order

Now, example one. Which is the most stable: H e two plus, H e two, L i two or B e two? Pause the video, and try it first. Done? Okay, let's check. Bond order equals half of N b minus N a, okay? N b, with the small subscript b, counts the electrons in bonding molecular orbitals, and N a counts those in antibonding ones. Higher bond order means more stable, and a bond order of zero means the species does not exist. Right? Helium first. H e two plus has three electrons: two bonding, one antibonding. Half of two minus one: zero point five. H e two has four: two and two. Bond order zero, so H e two does not exist. Clear? Now L i two and B e two. B e two has the most electrons of all four. So more bonding, and the most stable, right? No, no, careful! L i two has six electrons: four bonding and two antibonding, so the bond order is one. B e two has eight, and the two extra electrons go into an antibonding orbital. Four and four: bond order zero! More electrons does not mean more bonding. With me? So the bond orders are zero point five, zero, one and zero. The highest is one, so L i two is the most stable. Perfect!

5:55Example 2: lattice enthalpy of CaO by Born-Haber cycle

Now the harder one, and it settles our opening puzzle. Find the lattice enthalpy of calcium oxide. The capital delta, the triangle, means a change, H is enthalpy, and the small subscripts name each step. Sublimation of calcium: one hundred and seventy eight. First plus second ionisation enthalpy: one thousand seven hundred and thirty five. Half the dissociation enthalpy of oxygen: two hundred and forty nine. First plus second electron gain enthalpy of oxygen: plus seven hundred and three. And the enthalpy of formation of calcium oxide: minus six hundred and thirty five. All in kilojoules per mole. Pause and try. Hint: use the Born Haber cycle. Done? Okay, let's check. The Born Haber cycle: delta f H equals delta sub H, plus delta i H, plus half of delta diss H, plus delta e g H, minus U, okay? U is the lattice enthalpy: the energy needed to pull one mole of the solid apart into gaseous ions. Building the lattice gives this energy out, so it comes in with a minus sign. Clear? Now the electron gain term. Electron gain usually gives out energy, so we put minus seven hundred and three, right? No, no, wait! The question says plus seven hundred and three. The second electron gain of oxygen is strongly endothermic, so the sum is positive. So, minus six hundred and thirty five equals the four step values added, minus U, okay? Add the four steps: two thousand eight hundred and sixty five. Every one of them costs energy, just as we said at the start. Right? So U equals two thousand eight hundred and sixty five, plus six hundred and thirty five: three thousand five hundred kilojoules per mole! Puzzle solved. The huge lattice enthalpy pays for every step, with six hundred and thirty five to spare. That is why calcium oxide forms. Very good!

7:52Quick revision: polarising power, lone pairs, bond order

So, quick revision. Ionic compounds: look inside the anion for covalent bonds, and polarising power goes roughly as charge over size. Lone pairs come from the quick count, and a hydrogen bond needs the hydrogen on fluorine, oxygen or nitrogen. Bond order is half of N b minus N a, and in the Born Haber cycle, the lattice enthalpy pays the bill. Now go through all eighteen problems of Part A on this page, then the four practice problems. Try each one yourself first. You can do it. All the best!