Hybridisation with s and p Orbitals

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Beryllium has two outer electrons, both paired up in its two s orbital. Zero unpaired electrons. So it should not form any bond at all, right? But in beryllium chloride, it forms two bonds, and the two bonds are completely identical. How? Today's idea, hybridisation, answers it. And in JEE, hybridisation questions come up often. So, let's begin.

0:31Beryllium's secret: promote, then mix

So, here is beryllium, B e, in its ground state: two s two. The small two up top means two electrons in the two s orbital, both paired. But a normal covalent bond needs an unpaired electron. So how can beryllium bond at all? It promotes an electron! One two s electron jumps into an empty two p orbital: two s one, two p one, where the small ones mean one electron each. Two unpaired electrons, so two bonds. But one bond from an s orbital and one from a p orbital would be different. And the real bonds are identical. So the two orbitals mix! They blend into two identical s p hybrids, pointing opposite ways, at one hundred and eighty degrees; the small circle means degrees. Each hybrid overlaps a three p orbital of one chlorine atom, written C l on screen. Two identical orbitals, two identical bonds. Puzzle solved! Clear? This mixing is called hybridisation. Orbitals of nearly equal energy, on the same atom, mix together. Two s and two p are close in energy, so they can. Okay? And the rule: n orbitals go in, n identical hybrids come out; the arrow means become. Hybrids have bigger lobes, so they form stronger sigma bonds; sigma is the Greek letter you see there. And they point in fixed directions. Got it?

1:55sp, sp2, sp3: shapes, angles and s character

Now look at the figure. s p: one s plus one p gives two hybrids at one hundred and eighty degrees. That is the first picture, a straight line. Okay? Next, s p two. The small two up top counts the p orbitals: one s plus two p, so three hybrids, flat in one plane, one hundred and twenty degrees apart. Right? And s p three: one s plus three p, so four hybrids, at one hundred and nine degrees, twenty eight minutes. The little tick after twenty eight means minutes of angle. With me? Quick question. Look under s p three in the figure: twenty five percent s character. s character means how much of each hybrid comes from the s orbital. But why exactly one quarter? Because four orbitals went into the mix, and only one of them was an s orbital! One out of four, twenty five percent. Yes! The same way, s p two is one s out of three, about thirty three percent, and s p is one out of two, fifty percent. Fewer p orbitals, more s character. Clear?

2:57Finding the hybridisation: count, don't guess

Alright, now the most useful part. To find the hybridisation, count the steric number. It equals the number of sigma bonds, plus the lone pairs on the central atom. Only sigma bonds, remember. Okay? Then read the answer off this line; each arrow means gives. Two gives s p, three gives s p two, four gives s p three. Five, six and seven give s p three d, s p three d two, and s p three d three. Those use d orbitals, and come in the next lesson. With me? Now try this. Carbon dioxide, C O two. Carbon makes two double bonds, four bonds in all. So the steric number is four, and carbon is s p three, right? No, no, careful! This is the most common trap. Each double bond is one sigma bond plus one pi bond; pi is the Greek letter on screen. Pi bonds use the unhybridised p orbitals, the ones left out of the mixing, or sometimes d orbitals. So never count them. Clear? So, O double bond C double bond O; the two short lines are the double bonds. Two sigma bonds, zero lone pairs: steric number two. Carbon is s p, not s p three! Got it? Now, the two pi bonds. s p uses only one of carbon's three p orbitals, so two p orbitals are left, at right angles to each other and to the axis. Each overlaps sideways with a p orbital of one oxygen: two pi bonds. Beautiful!

4:24More s character: what changes

Now, s character is not just a number. Look at the right side of the figure: more s character means shorter, stronger bonds. Okay? It also makes the atom more electronegative: it holds the bonding electrons more tightly. So tell me: an s p, s p two or s p three carbon, which is the most electronegative? s p, yes, exactly! Fifty percent s, the highest. So s p is greater than s p two, greater than s p three; these signs mean greater than. Right? So a hydrogen on a triple-bonded carbon is more acidic than one on a double-bonded or single-bonded carbon. The three lines are the triple bond, and the dash is the bond to hydrogen. That s p carbon holds the bond electrons tightly, so the hydrogen leaves most easily. Remember this one.

5:13Example 1: hybridisation of BeCl2, BCl3, CH4, NH3, H2O

Alright, example one. Give the hybridisation of the central atom in B e C l two, B C l three, C H four, N H three, and water, H two O; the small numbers below count the atoms. Pause the video, and try it first. Done? Okay, let's check. Beryllium chloride, our puzzle molecule: two sigma bonds, zero lone pairs, steric number two. The double arrow means so: s p, just as we found. Okay? Boron trichloride: boron's three valence electrons all go into the three bonds. Three sigma, zero lone pairs: s p two. Right? Methane: four sigma bonds, zero lone pairs. So, s p three. Easy! Now ammonia. Three bonds, just like boron trichloride. So ammonia is s p two as well, right? Oh no, careful! Count the lone pairs too. Nitrogen has five valence electrons; three go into bonds, and two are left: one lone pair. Three plus one is four: s p three. With me? And water: oxygen has six, two go into bonds, and four are left: two lone pairs. Two plus two is four: s p three again. Methane, ammonia and water: all s p three. Lone pairs always count!

6:34Example 2: hybridisation of S in SO2, SO3 and SO4 2-

Now a slightly harder one. Give the hybridisation of sulphur in S O two, S O three, and S O four two minus; the two minus up top is the charge on the ion. Pause and try. Hint: count the lone pairs on sulphur. Done? Okay, S O two first. It looks just like carbon dioxide: two oxygens, two double bonds. So sulphur is s p, like carbon, right? No, no, wait! Use the quick count from the VSEPR chapter: half the valence electrons, and double-bonded oxygens add nothing. The one over two means half. Sulphur has six, so three pairs. Two oxygens take two, and one lone pair is left. Two sigma plus one lone pair: s p two, not s p! Okay? S O three: half of six, three pairs again. Three oxygens use all three, so zero lone pairs: s p two. Right? And S O four two minus: add two for the negative charge. Six plus two is eight, and half of eight is four pairs. Four oxygens, zero lone pairs: s p three. So, s p two, s p two, and s p three. That one lone pair in S O two changed everything!

7:49Quick revision: steric number, pi bonds, s character

So, quick revision. One: steric number equals sigma bonds plus lone pairs; two is s p, three is s p two, four is s p three. Two: pi bonds use the leftover p orbitals, or d orbitals, so never count them. Three: more s character means shorter, stronger bonds, and a more electronegative atom. Now go and try the practice problems on this page. You can do them. All the best!