Set 1 (Ionic Bonding, Lewis Structures and Polarity)

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Two carbon chlorine bond moments, each of one point six debye, act at an angle of one hundred and ten degrees. What is the resultant? One point six plus one point six, three point two debye, right? Hmm, no. The real answer is much smaller. So where did the rest go? By the end of this class, you will know. This is our first mixed problem set in chemical bonding: ionic bonding, Lewis structures and polarity. Questions like these come up often in JEE Main. So, let's begin.

0:38Tool 1: ionic compounds can hide covalent bonds

So, first tool: ionic bonding. Look at these five compounds: sodium nitrate, potassium cyanide, calcium chloride, ammonium bromide and magnesium oxide. The small numbers below are subscripts: they count the atoms. All five are ionic compounds. Now tell me: how many of them contain covalent bonds as well? If you said zero, careful! Calcium chloride and magnesium oxide are made of simple ions, single atoms carrying a charge. So in these two, ionic bonds only. Okay? But look inside the other three. Nitrate, cyanide and ammonium are polyatomic ions; the minus and plus signs on top are their charges. Inside each of them, the atoms are joined by covalent bonds. So sodium nitrate, potassium cyanide and ammonium bromide have both kinds. The answer is three. Clear?

1:35Tool 2: Fajans' rules decide covalent character

Now, second tool: Fajans' rules. The cation pulls the electron cloud of the anion towards itself. This is polarisation, and it gives the ionic bond some covalent character. So the bond is more covalent when the cation is small and highly charged, and when the anion is large. Okay? Now arrange these four chlorides by ionic character: beryllium, magnesium, calcium and barium chloride. From beryllium to barium, the cation keeps growing. So which one is the most ionic? Barium chloride, yes, exactly! Polarising power falls as the cation grows, so the ionic character rises. The less than signs on screen show the order: beryllium chloride is the least ionic. Right? Your turn now. Sodium fluoride, sodium iodide, potassium fluoride and potassium iodide. Which one is the most covalent? Sodium iodide, correct! The smallest cation, sodium, with the largest anion, iodide. Both rules push the same way. Got it?

2:39Tool 3: formal charge, V minus L minus half B

Alright, third tool: formal charge. Formal charge equals V, minus L, minus half of B, okay? V is the number of valence electrons of the free atom, and L is the number of lone-pair electrons. Half of B is one for each shared pair, so a double bond counts two and a triple bond counts three. With me? Take sulphur dioxide, drawn with two S O double bonds, and one lone pair on sulphur; the equals sign is a double bond. For sulphur, V is six. One lone pair, so L is two. Two double bonds, so half of B is four. Six minus two minus four: zero. Okay? Now the thiocyanate ion, in square brackets, with the minus charge outside. Nitrogen, a triple bond, carbon, a single bond, then sulphur. Each pair of dots marks a lone pair; nitrogen has one, and sulphur actually has three, though only one is drawn. Let's start with sulphur. Three lone pairs, so L is three, right? No, no, careful! L counts lone-pair electrons, not lone pairs. Three pairs means six electrons. So for sulphur: six, minus six, minus one for its single bond. Minus one! Nitrogen: five, minus two for its lone pair, minus three for the triple bond. Zero. Carbon: four, minus zero, minus four for the triple bond and the single bond. Zero. So in this drawing, the minus charge sits on sulphur. Clear?

4:14Tool 4: polarity, symmetric shapes cancel

Now the fourth tool: polarity. Mu, the Greek letter, is the dipole moment. Here are six fluorides: B F three, N F three, P F five, S F four, X e F two and C l F three. We want to count the ones with mu equal to zero. Remember, the bond moments cancel only when the shape is symmetric. Okay? Now look at X e F two. Xenon carries three lone pairs. With so many lone pairs, it must be polar, right? No, no, wait! The three lone pairs go to the equatorial positions, so X e F two is linear. Its two bond moments point in opposite directions and cancel. So mu is zero. Clear? B F three and P F five are symmetric too. N F three, S F four and C l F three are not, so they are polar. So the answer is three. And when the moments do not cancel, we add them as vectors. For two moments at an angle theta, the Greek letter on screen, mu squared equals mu one squared, plus mu two squared, plus two times mu one times mu two times cos theta. Mu one and mu two, with the small subscripts, are the two moments. Keep this ready; we need it soon. Right?

5:30Example 1: lone pairs in the perchlorate ion ClO4-

Now, example one. How many valence electrons are there in total in the perchlorate ion, C l O four minus? And how many lone pairs does it carry in all, when it is drawn with three C l O double bonds? Pause the video, and try it first. Done? Okay, let's check. Chlorine brings seven. Four oxygens bring six each, that is twenty four. And the minus charge adds one more. Thirty two electrons, so sixteen pairs. First answer: thirty two. Got it? Now the lone pairs. Chlorine is bonded to four oxygens, so four bonds. Sixteen minus four: twelve lone pairs, right? Oh no, careful! Three of those bonds are double bonds, and a double bond is one sigma bond plus one pi bond: two pairs. Sigma and pi are the Greek letters on screen. So four sigma plus three pi: seven bonding pairs. Okay? Sixteen minus seven: nine lone pairs. So the answers are thirty two electrons and nine lone pairs. Always count the pi bonds too!

6:41Example 2: resultant of two 1.6 D moments at 110 degrees

Now the harder one, and it is our opening puzzle. Two carbon chlorine bond moments of one point six debye each, with capital D for debye, act at one hundred and ten degrees; the small circle is the degree sign. Estimate the resultant. You are given that cos one hundred and ten degrees is minus zero point three four two. Pause and try. Hint: use the formula from tool four. Done? Okay, let's check. One point six plus one point six: three point two debye, right? No, no, wait! Dipole moments are vectors, so the angle matters. Use the formula: mu squared equals one point six squared, plus one point six squared, plus two times one point six squared, times minus zero point three four two. Okay? One point six squared is two point five six, and two of them make five point one two. The last term is minus one point seven five; it is negative because the angle is more than ninety degrees. So mu squared is three point three seven. Take the square root: about one point eight four debye; the wavy equals sign means approximately. Not three point two! Puzzle solved: at more than ninety degrees, the two moments partly work against each other. That's the power of vectors!

8:03Quick revision: Fajans rules, formal charge, polarity

So, quick revision. Ionic bonding: look inside polyatomic ions, and a small cation with a large anion means more covalent character. Formal charge is V minus L minus half of B, and L counts electrons, not pairs. Polarity: symmetric shapes give zero dipole moment; otherwise, add the moments as vectors. Now go through all fifteen problems of this set on this page. Try each one yourself before you read its solution. You can do it. All the best!