Mixed Problems: Set 1 (Groups 13 and 14)

Transcript

0:00Introduction

Hello everyone! Let me start with a small question. Boron forms four trihalides, with fluorine, chlorine, bromine and iodine. Which of them is the strongest Lewis acid? Most students say boron trifluoride, because fluorine is the most electronegative element, right? Hold that thought. Today we solve mixed problems from groups thirteen and fourteen, the kind that come up often in JEE Main. So, let's begin.

0:30Idea 1: count per atom, then divide by sharing

Idea number one is counting. Find the number of carbon carbon double bonds in one C sixty molecule. C sixty, with the small sixty below the C, is buckminsterfullerene, a hollow cage. Each carbon is s p two, with a small raised two, and bonded to three other carbons. Okay? Now, each carbon forms exactly one double bond; the two short lines between the letters C mark it. Sixty carbons, one double bond each, so sixty double bonds, right? No, no, that is the trap, because every double bond joins two carbons. Counting per carbon counts each bond twice. So sixty over two gives thirty double bonds. Clear? Let us check. Each carbon has three neighbours, so sixty times three, over two, gives ninety edges. Thirty double plus sixty single makes ninety. So count per atom, then divide by how many atoms share each bond. Got it?

1:31Idea 2: boron trihalides and back bonding

Now idea number two, our opening question. B X three, with X for any halogen, is planar, s p two, and electron deficient, because boron has an empty two p orbital. With me? But the halogen has filled p orbitals, and one of them can donate into that empty two p orbital, shown by the arrow. This is back bonding, p pi, p pi, with pi the Greek letter. Okay? So, which halogen gives the best overlap with the two p orbital of boron? Fluorine, yes, exactly! Two p with two p overlaps best, while two p with the five p of iodine is the poorest. So in B F three, back bonding relieves the electron deficiency of boron the most. Right? So B F three is the weakest Lewis acid, and B I three, boron triiodide, the strongest. Read the less-than signs: B F three, then B C l three, B B r three and B I three. It is the reverse of what electronegativity suggests, and that answers our opening question. Clear?

2:38Idea 3: silicates, count the shared oxygens

Idea number three is silicates, built from S i O four units, in square brackets, with a four minus charge up top. Each unit is a tetrahedron. In the figure, each triangle is one tetrahedron seen from above, with silicon at the dot. The pale circles are terminal oxygens; the grey ones are bridging oxygens, shared by two tetrahedra. Okay? Now say each tetrahedron shares some oxygen atoms, and call that number s, a small letter s, for shared. Each shared oxygen counts only half, so the composition is S i O, four minus s over two, written below the oxygen. Got it? Now the problem. Take the silicate anion S i two O five, in round brackets, with a small n below and two n minus up top. How many oxygen atoms of each tetrahedron are shared? First, the oxygen per silicon. Five over two gives two point five. But careful, that is not the answer; it counts oxygens per silicon, not shared ones. Right? So four minus s over two equals two point five. Then s over two equals one point five, so s equals three; the double arrow on screen reads, which gives. Three shared oxygens per tetrahedron make a sheet silicate. Clear?

4:02Idea 4: a statement pair on tin and lead

Idea number four is a statement pair. Statement one, the Roman numeral one, says P b O two gives H two O two, hydrogen peroxide, with dilute acids. P b O two has two oxygens, so is it a peroxide? No, and that is the trap! P b O two is lead four oxide, the Roman four in brackets, with lead in the plus four state, not a peroxide. So statement one is false. With me? Statement two, the Roman two, says S n C l two reduces mercury two chloride, H g C l two. Tin in the plus two state is a good reducing agent. Okay? S n C l two plus two H g C l two give H g two C l two, a white precipitate, the down arrow, plus S n C l four. So statement one is false, and statement two is true. Got it?

4:58Example 1: 3c-2e bonds in B2H6 plus sp3 borons in borax

Now, example one, an easy one. Find the sum of two numbers: the three-centre two-electron bonds in diborane, written three c, two e, and the s p three boron atoms in the borax anion. Pause the video and try it first. Done? Okay, let's check. On the left of the figure is diborane, B two H six, with four terminal hydrogens and two bridging hydrogens on curved orange bonds. Each bridge, BHB, is one three-centre two-electron bond, so the double arrow gives two. The aluminium chloride dimer on the right is not needed today. Right? Now the borax anion: B four O five, O H four, in square brackets, with a two minus charge up top. It has four boron atoms. So all four of them are s p three, right? No, no, wait, that is the trap here! Only two borons are tetrahedral, s p three, and the other two are trigonal, s p two. So only two of them count. Clear? So the sum is two plus two, which is four. Small counting questions like this come up often, so picture the structure before you count. Got it?

6:12Example 2: volume of HCl to titrate a borax solution

Now the harder one, example two. One point nine zero seven grams of borax, molar mass capital M equal to three hundred and eighty one point four, is dissolved to make one hundred millilitres, written m L, of solution. What volume of zero point one zero zero molar H C l, the M after the number, neutralises twenty point zero millilitres of it? Pause the video and try it first. Done? Okay, let's check. The molar mass three hundred and eighty one point four is for borax with its ten waters of crystallisation; the equation leaves that water out, and the mole ratio stays the same. Borax with H C l: N a two B four O seven, plus two H C l, plus five H two O, give two N a C l and four H three B O three, boric acid. So one borax needs two H C l. With me? Now the moles of borax in the one hundred millilitres. One point nine zero seven over three hundred and eighty one point four gives zero point zero zero five zero zero moles, or five point zero zero millimoles, written m mol. Okay? So five millimoles of borax needs ten millimoles of H C l, and that is one hundred millilitres of acid, right? Oh no, careful, that is the trap here! We titrate only twenty point zero millilitres out of one hundred, which is one fifth of the solution. Twenty over one hundred, times five point zero zero, gives one point zero zero millimole of borax. Clear? Each borax needs two H C l, so two times one point zero zero gives two point zero zero millimoles of H C l. Right? Finally, capital V, the volume, is millimoles over molarity. Two point zero zero over zero point one zero zero gives twenty point zero millilitres. Equation first, then only the portion you titrate. Got it?

8:17Quick revision: counting, back bonding, borax titration

So, quick revision. For counting, count per atom, then divide by sharing. C sixty has thirty double bonds, and silicates are S i O, four minus s over two. B F three has the best back bonding, so it is the weakest Lewis acid, and B I three the strongest. In titrations, balanced equation first, then only the portion you titrate. Now go and try the practice problems on this page. All the best!