General Trends in Group 17

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Fluorine is the most electronegative element, so surely it releases the most energy when it gains an electron, and surely that makes it the strongest oxidising agent among the halogens. Well, one of these is false, and the other is true, but for a different reason! Group seventeen trends come up often in JEE, so let's find out.

0:26Group 17 at a glance

Group seventeen is the halogen family, F, C l, B r and I on screen, fluorine to iodine. The atomic radius, in picometres, the p m after the slash, goes sixty four, ninety nine, one hundred and fourteen, and one hundred and thirty three. Okay? Electronegativity goes four point zero for fluorine, the highest of all, then three point two, three point zero, and two point seven. Now, electron gain enthalpy, the Greek capital delta with a small e g, then H, on the badge. Fluorine is the most electronegative, so its value must be the most negative, right? Oh no, careful, that is exactly the trap many students fall into! In kilojoules per mole, where mol to the minus one means per mole, fluorine is minus three hundred and thirty three, but chlorine is minus three hundred and forty nine. Then bromine is minus three hundred and twenty five, and iodine is minus two hundred and ninety six. So chlorine, not fluorine, has the most negative value. With me? The reason is size. The added electron enters the small two p shell of fluorine, and the electrons already there repel it strongly. So the first half of our puzzle is false. Clear?

1:42Why the halogens are coloured

Now let us look at their colours. F two, where the small two means two atoms, is a pale yellow gas, and C l two is a greenish-yellow gas. Got it? B r two is a red-brown liquid, and I two is a violet-black solid. So the colour deepens down the group. Right? They are coloured because they absorb visible light, which lifts an electron from pi star to sigma star, the Greek letters with a small star; the arrow means to. This gap decreases down the group, so the colour deepens. Clear?

2:12Oxidising power and displacement

Now, oxidising power, with X for any halogen. E naught, the E with a small circle, is the standard electrode potential for X two, slash, X minus, in volts; the raised minus marks the halide ion. It goes plus two point eight seven, plus one point three six, plus one point zero nine, and plus zero point five four. Okay? The higher the E naught, the stronger the oxidising agent. So F two is greater than C l two, greater than B r two, greater than I two; the pointed sign means greater than. Fluorine is the strongest, the true half of our puzzle. Right? Now, each halogen displaces the heavier halide ions. C l two plus two B r minus gives two C l minus plus B r two; this arrow means gives. So, will bromine water free iodine from potassium iodide, K I? Yes, it does, because bromine is above iodine. B r two plus two I minus gives two B r minus plus I two. But the reverse does not happen: iodine cannot free bromine from bromide.

3:15Why fluorine is different

Got it? Now, fluorine is the odd one out. It has a small size, the highest electronegativity, and no d orbitals at all. So tell me, can fluorine ever show a positive oxidation state? No, it never can, because nothing is more electronegative than fluorine. So it shows only minus one. Chlorine, bromine and iodine show minus one, plus one, plus three, plus five and plus seven. Clear? Fluorine is such a strong oxidising agent that it oxidises water itself. Two F two plus two H two O gives four H F plus O two, oxygen gas. Okay? Chlorine and bromine only disproportionate, one atom oxidised and one reduced: X two plus H two O gives H X plus HOX, two acids. And iodine does not react with water at all. Right? No oxidising agent is stronger than F two, so none can make fluorine from fluorides. F two is made by electrolysis of K F in anhydrous H F, H F with no water. Got it?

4:21The hydrogen halides

Next, the hydrogen halides. In water, acid strength rises from H F to H I; this pointed sign means less than. The reason is in the brackets: the bond enthalpy of H X, written with a dash, falls. Okay? Thermal stability goes the other way, with H F the most stable and H I the least. Reducing power is the reverse, so H I is the strongest reducing agent. Clear? Now let us check boiling points. H F has the smallest molar mass of the four. So it should boil the lowest, right? No, no, wait, H F boils the highest of all! The order is H C l, H B r, H I, then H F, because H F molecules are hydrogen bonded. With me?

5:04Example 1: bond enthalpy order Cl2, Br2, F2, I2 and why

Now example one, an easy one. Arrange F two, C l two, B r two and I two in decreasing bond dissociation enthalpy, and explain the position of F two. Pause the video and try it first. Done? Okay, let's check. The delta with bond below it means bond enthalpy. In kilojoules per mole, it is one hundred and fifty eight point eight for fluorine, two hundred and forty two point six for chlorine, one hundred and ninety two point eight for bromine, and one hundred and fifty one point one for iodine. Bond enthalpy normally falls as atoms grow. So where should F two be, and where is it? Yes, exactly, F two should be at the top, but it sits only third! The decreasing order is C l two, then B r two, then F two, then I two. Right? The reason is this. The F F bond, the dash on screen, is very short, so the lone pairs on the two small fluorine atoms come close and repel each other. That repulsion weakens the bond. Clear?

6:05Example 2: why F2 is a stronger oxidising agent than Cl2

Now example two, which settles our puzzle. Fluorine has a less negative electron gain enthalpy than chlorine, yet it is the stronger oxidising agent in water. Can you explain why? Pause the video and try it first. Done? Okay, let's check. In water, oxidising power depends on the whole cycle. One half of X two gas, the g in brackets, splits into atoms, each atom gains an electron, and the ion is hydrated, the a q meaning in water. So in which steps does fluorine win? First, bond breaking. The F F bond enthalpy is one hundred and fifty eight point eight kilojoules per mole, versus two hundred and forty two point six for the C l C l bond. This step is cheaper for fluorine, so fluorine wins it. Okay? Next, electron gain gives minus three hundred and thirty three for fluorine, versus minus three hundred and forty nine for chlorine. So chlorine wins this step, but only slightly. Right? Third, in hydration, the F minus ion is very small, and its hydration enthalpy is very high. So fluorine wins this step, and by a lot. With me? So the low bond enthalpy and the high hydration enthalpy outweigh the slightly smaller electron gain enthalpy. That is why E naught is plus two point eight seven volts for fluorine, versus plus one point three six for chlorine. And that solves our puzzle. Got it?

7:36Quick revision: halogen trends and the hydrogen halides

So, quick revision. Chlorine has the most negative electron gain enthalpy, and bond enthalpy goes C l two, B r two, F two, I two. Oxidising power runs from F two down to I two, thanks to the weak F F bond and the high hydration of F minus. And acid strength rises to H I, while H F boils the highest. Now try the practice problems on this page. All the best!