Counting Lone Pairs; Isostructural and Isoelectronic Species

Transcript

0:00Introduction

Hello everyone! Okay, a quick question to start. How many lone pairs are there in B r F five, in the whole molecule? One, on the bromine? That is what many students write. But the answer is sixteen! So where are the other fifteen hiding? Today we learn to count every lone pair in a molecule. Then we meet chemical twins: isoelectronic and isostructural species. Both ideas come up often in JEE, so let's begin.

0:32Counting lone pairs: start at the centre

So, step one: the central atom. Take the electron pairs from our quick count, minus the number of attached atoms. That gives the lone pairs on the central atom. Okay? Take B r F five. The small five below the F, the subscript, means five fluorine atoms. Bromine has seven valence electrons, and the five fluorines add five. Half of seven plus five, half of twelve: six pairs. Five atoms attached, so six minus five: one lone pair on B r, the bromine. Right? But our question was about the whole molecule. Each fluorine makes one bond, and completes its octet. So how many lone pairs sit on each fluorine? Three, exactly! An octet is four pairs, as the brackets say. One pair is the bond, so three are lone pairs. So every fluorine or chlorine atom bonded once carries three lone pairs. Clear? Now add them all. One on bromine, plus five times three on the fluorines; the cross means times. One plus fifteen: sixteen lone pairs! Puzzle solved. The trap is stopping at the central atom, so always check what the question asks for. Got it?

1:48Oxygen, nitrogen, and a quick check

Now oxygen, with the same octet idea. Suppose an oxygen atom makes one double bond. How many lone pairs are left on it? Two, correct! The double bond uses two of the four pairs, so two lone pairs are left. Okay? But an oxygen with just one single bond uses only one pair, so it keeps three lone pairs, like a fluorine. And nitrogen with three bonds uses three pairs, so it keeps one lone pair. With me? Now, a quick check for any molecule. Add all the valence electrons and halve them: those are the valence pairs. Subtract the bond pairs, counting each double bond as two, as the brackets say. Whatever is left must be lone pairs. Clear? Try it on B r F five. Bromine seven, plus five fluorines at seven each, thirty five. Half of forty two: twenty one pairs. Five bonds, so twenty one minus five: sixteen again. Perfect!

2:47Isoelectronic and isostructural

Now the second half: chemical twins. Two species are isoelectronic when they have the same number of electrons. Often we compare just the valence electrons; that is the word in brackets. Okay? Isostructural simply means the same shape. The figure shows the basic shapes: two pairs linear, three trigonal planar, four tetrahedral, and so on. The thick wedge comes towards you, the dashed line goes away, the small circle means degrees, and the little dash after twenty eight means minutes. Right? Let's test it. Count the valence electrons in C O two, and in N three minus; the minus at the top right is the charge on the ion. Sixteen each, very good! Carbon four, plus two oxygens at six each, twelve: sixteen. Three nitrogens at five each make fifteen, plus one for the negative charge: sixteen. So they are isoelectronic. Clear? Now the useful fact. Isoelectronic species with the same number of atoms are usually isostructural. Same electrons and same atoms, so usually the same shape. Got it? So here is the first family: C O two, N two O, N O two plus, N three minus, and OCN minus. The plus at the top of N O two plus is a positive charge. Three atoms and sixteen valence electrons each, and all of them linear, like the first drawing in the figure. With me?

4:15More families, and one trap

Now more families, and the number in brackets is the valence electron count. Trigonal planar, twenty four each: B F three, N O three minus, C O three two minus, and S O three. The two minus at the top means a charge of minus two. Right? Tetrahedral, eight each: C H four, N H four plus, and B H four minus. And tetrahedral again, thirty two each: C l O four minus, S O four two minus, and P O four three minus. Okay? Linear, twenty two each: X e F two, I three minus, and I C l two minus. And square planar, thirty six each: X e F four, and I C l four minus. Clear? Now a trap. S O three is trigonal planar. S O three two minus has the same four atoms. So it is trigonal planar too, right? No, no, careful! Count first. Sulphur six, the oxygens add nothing, and the charge adds two: half of eight, four pairs. Three oxygens attached, so one lone pair; l p means lone pair. So it is pyramidal. Same atoms, but two extra electrons, so it is not isoelectronic with S O three, and the shape changes. Got it?

5:40Example 1: total lone pairs in PCl5 and SF4

Now let's use all this. Example one. Find the total number of lone pairs in P C l five, and in S F four. Pause the video, and try it first. Done? Okay, let's check. P C l five: phosphorus has five valence electrons, plus five chlorines: half of ten, five pairs. Five chlorines attached, so zero lone pairs on P, the phosphorus. Okay? Zero on phosphorus. So the answer for P C l five is zero, right? No, no, wait! That is our trap again. Five chlorines, each bonded once, carry three lone pairs each. Zero plus five times three: fifteen. With me? Now S F four. Sulphur six, plus four fluorines: half of ten, five pairs. Four fluorines attached, so one lone pair on S, the sulphur. Then the fluorines: four times three, twelve. One plus twelve: thirteen. So the answers are fifteen and thirteen. Got it?

6:46Example 2: neutral species isostructural with ClO4-, ICl2-

Now a harder one. Example two. Name a neutral molecule that is isoelectronic and isostructural with C l O four minus, and one with I C l two minus. Pause and try. Hint: count the valence electrons first. Done? Let's check. C l O four minus: chlorine seven, four oxygens at six each, twenty four, plus one for the charge: thirty two valence electrons. And it is tetrahedral. Okay? Now, S O four two minus is in the same family: thirty two electrons, tetrahedral. So that is our answer, right? Oh no, careful! Read the question again. It asks for a neutral molecule, and S O four two minus is an ion. Clear? So we need thirty two electrons with no charge. Xenon has eight valence electrons, one more than chlorine, so it makes up for the charge. X e O four: xenon eight, plus twenty four from the oxygens, thirty two. Five atoms like C l O four minus, tetrahedral, and neutral. Here isoelectronic means the same number of valence electrons; the total electron counts differ, because xenon has many more inner electrons than chlorine. Got it? Same idea for I C l two minus: iodine seven, two chlorines at seven each, fourteen, plus one for the charge: twenty two, and linear. The arrow points to our answer, with xenon again: X e F two, eight plus fourteen, twenty two, linear, and neutral. So the answers are X e O four, and X e F two. See how the electron count did all the work?

8:32Quick revision: counting lone pairs, isoelectronic shapes

So, quick revision. Total lone pairs: the central atom first, then three on each fluorine or chlorine, and two or three on each oxygen. Check: valence pairs minus bond pairs gives the lone pairs. And same electrons with the same atoms usually means the same shape. Now go and try the practice problems on this page. Count every atom, not just the centre. All the best!