Oxidation States and Oxides

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Chromium forms a compound, C r O five, with five oxygen atoms. Each oxygen is minus two, five oxygens give minus ten, so chromium must be plus ten. Chromium at plus ten, can that really be true? Something is wrong here, right? By the end of this video, you will crack it yourself. Today we study the oxidation states of the transition elements, and their oxides. This topic comes up often in JEE, so let's begin.

0:33Why so many oxidation states?

Let us begin with the reason. In a transition metal, both the n minus one d electrons and the n s electrons can be lost. The n minus one in brackets means the d subshell just below the outer shell, and n s is the outer s subshell, okay? So these metals show many oxidation states, usually one unit apart, like steps on a staircase. Now, the highest state. The maximum oxidation state equals the number of n minus one d electrons plus n s electrons, and this works up to manganese. Clear? Let me ask you this. Manganese is three d five, four s two, with the five and the two written small and raised, as in the badge. So what is the highest oxidation state of manganese? Yes, exactly, it is plus seven! Five plus two makes seven, and the double arrow on screen means this gives plus seven. But beyond manganese, in the three d series, the maximum falls, okay? So in the three d series, never keep counting electrons past manganese. Got it?

1:40The most common state, and two exceptions

Now, a quick one for you. In the three d series, which oxidation state do you meet most often? Think about which electrons leave first. Yes, it is plus two! The two four s electrons are lost most easily, so plus two is the most common state in the three d series. Right? Now two exceptions. Scandium shows only plus three, because after losing three electrons it is left with the very stable argon core. And zinc shows only plus two. With me?

2:12Highest states: oxides versus fluorides

Now, the highest states show up in oxides. M n two O seven has manganese at plus seven; the small numbers written below count the atoms, okay? C r O three has chromium at plus six, and O s O four and R u O four, of osmium and ruthenium, reach plus eight! Now the fluorides. The highest ones are T i F four, V F five and C r F six, so plus four, plus five, plus six. But for manganese, only M n F four, just plus four. Right? So manganese reaches plus seven with oxygen, but only plus four with fluorine. What can oxygen do that fluorine cannot? Yes, exactly, oxygen can form multiple bonds! The two lines between M n and O on screen mean a manganese oxygen double bond. Fluorine forms only single bonds. So plus seven with fluorine would need seven fluorine atoms round one manganese, and they cannot fit. Clear?

3:11Low states, and the trend down a group

Now the other end, the low states, zero and plus one. They are stabilised by pi acceptor ligands; pi is the Greek letter before the word acceptor. Examples are N i C O four and F e C O five. The brackets hold C O, carbon monoxide, and the small number outside counts the C O groups, okay? Next, going down a group. In the p block, the lower states become more stable as we go down. So for transition metals too, high states should become less stable, right? No, no, careful, for transition metals it is the reverse! Down a group, the higher oxidation states become more stable. Mixing up these two trends is a common trap. Okay? Look at chromium, molybdenum and tungsten, with chromium at the top of the group. Molybdenum six and tungsten six, written M o and W with a Roman six in brackets for plus six, are stable. But chromium six is a strong oxidant, and readily falls to chromium three. Got it?

4:15Oxides: basic, amphoteric, acidic

Finally, the oxides. When the metal is in a low oxidation state, its oxide is basic, okay? Now push the oxidation state higher and higher. What do you think happens to the oxide? Yes, it turns more and more acidic! In an intermediate state the oxide is amphoteric, and in a high state it is acidic. Right? There is one case to remember exactly. V two O five, with vanadium at plus five, is amphoteric, but mainly acidic. Clear?

4:47Example 1: nature of MnO, Mn2O7, CrO, Cr2O3 and CrO3

Now, Example one. Classify M n O, M n two O seven, C r O, C r two O three and C r O three as basic, amphoteric or acidic. Pause the video, and try it first. Done? Okay, let's check. First, the oxidation state of each metal, with oxygen at minus two. M n O is plus two. In M n two O seven, minus fourteen is shared by two manganese atoms, so each is plus seven. In the same way, C r O is plus two, C r two O three is plus three, and C r O three is plus six. Okay? Now apply the rule: acidity rises with the oxidation state. So M n O, at plus two, is basic, and M n two O seven, at plus seven, is acidic. Right? Now chromium. Its three oxides here are at plus two, plus three and plus six. So where does C r two O three, at plus three, fit in? Yes, it is in the middle, so it is amphoteric! C r O is basic, C r two O three amphoteric, and C r O three acidic. So, in the order of the question: basic, acidic, basic, amphoteric, acidic. Clear?

6:09Example 2: oxidation state of Cr in CrO5 and Fe in Fe(CO)5

Now, Example two, and this one solves our puzzle from the start. Find the oxidation state of chromium in C r O five, and of iron in F e C O five. Pause the video, and try it first. Done? Okay, let's check. Chromium first. Five oxygens at minus two each give minus ten, so chromium is plus ten, right? Oh no, careful, that is the trap! Chromium is three d five, four s one, so it has only six electrons in its three d and four s orbitals, and its highest state is plus six. So plus ten is impossible, okay? So the oxygens are not all the same. C r O five has one oxo oxygen, that is, a single oxygen atom on its own, and two peroxo groups. Each peroxo group is O two, with two minus written small and raised: two oxygen atoms joined together, carrying minus two in total. One plus two times two makes five oxygens. With me? Now let x be the oxidation state of chromium. The oxo oxygen gives minus two, and each peroxo group gives minus two. So x, minus two, minus two times two, equals zero, which gives x equals plus six. Remember our plus ten from the start? It was never real; the peroxo groups were hiding the answer! And iron in F e C O five is easy. C O is neutral, so x plus five times zero equals zero, and iron is zero. That is a low state, stabilised by pi acceptor ligands, just as we saw. So the answers are plus six, and zero. Got it?

7:54Quick revision: maximum state, CO ligands, nature of oxides

So, quick revision. First, the maximum oxidation state equals the number of n minus one d plus n s electrons, up to manganese at plus seven. Second, the highest states appear in oxides, because oxygen forms multiple bonds, and low states appear with C O ligands. And third, oxides go from basic at low states, to amphoteric, to acidic at high states. Now go and try the two practice problems on this page. All the best!