First Ionization Energy Of Lithium

IONISATION ENERGY

This page explains what showtime ionisation energy is, and then looks at the manner it varies around the Periodic Tabular array - across periods and down groups. It assumes that y'all know most uncomplicated diminutive orbitals, and can write electronic structures for simple atoms. You will find a link at the bottom of the folio to a similar description of successive ionisation energies (2d, third and so on).

Of import!If you aren't reasonable happy about atomic orbitals and electronic structures you should follow these links before you go whatever farther.

Defining commencement ionisation energy

Definition

The first ionisation energy is the energy required to remove ane mole of the virtually loosely held electrons from one mole of gaseous atoms to produce i mole of gaseous ions each with a charge of 1+.

This is more easily seen in symbol terms.

Ten(g) 10⁺(yard) + e^-

It is the energy needed to conduct out this change per mole of X.

Worried virtually moles?Don't exist! For now, but take it as a measure of a item amount of a substance. It isn't worth worrying virtually at the moment.

Things to notice about the equation

The state symbols - (g) - are essential. When you are talking about ionisation energies, everything must be present in the gas state.

Ionisation energies are measured in kJ mol^-1 (kilojoules per mole). They vary in size from 381 (which yous would consider very depression) up to 2370 (which is very loftier).

All elements take a first ionisation energy - even atoms which don't form positive ions in test tubes. The reason that helium (1st I.E. = 2370 kJ mol^-1) doesn't ordinarily form a positive ion is because of the huge amount of energy that would be needed to remove one of its electrons.

Patterns of offset ionisation energies in the Periodic Table

The first 20 elements

Start ionisation free energy shows periodicity. That ways that it varies in a repetitive style as you move through the Periodic Tabular array. For example, look at the design from Li to Ne, and then compare it with the identical blueprint from Na to Ar.

These variations in outset ionisation energy can all be explained in terms of the structures of the atoms involved.

Factors affecting the size of ionisation free energy

Ionisation free energy is a measure out of the energy needed to pull a particular electron away from the attraction of the nucleus. A high value of ionisation free energy shows a high attraction betwixt the electron and the nucleus.

The size of that attraction volition exist governed past:

The charge on the nucleus.

The more protons at that place are in the nucleus, the more positively charged the nucleus is, and the more strongly electrons are attracted to it.

The altitude of the electron from the nucleus.

Allure falls off very rapidly with altitude. An electron close to the nucleus will be much more strongly attracted than one further away.

The number of electrons between the outer electrons and the nucleus.

Consider a sodium atom, with the electronic structure 2,8,1. (There'south no reason why you can't use this notation if information technology's useful!)

If the outer electron looks in towards the nucleus, it doesn't see the nucleus sharply. Betwixt it and the nucleus in that location are the two layers of electrons in the first and second levels. The 11 protons in the sodium'southward nucleus have their effect cutting down by the 10 inner electrons. The outer electron therefore only feels a cyberspace pull of approximately 1+ from the centre. This lessening of the pull of the nucleus by inner electrons is known as screening or shielding.

Warning!Electrons don't, of course, "look in" towards the nucleus - and they don't "see" annihilation either! Merely there's no reason why you can't imagine it in these terms if information technology helps you to visualise what's happening. Just don't employ these terms in an test! You may get an examiner who is upset by this sort of loose language.

Whether the electron is on its ain in an orbital or paired with another electron.

2 electrons in the same orbital feel a flake of repulsion from each other. This offsets the attraction of the nucleus, so that paired electrons are removed rather more easily than yous might expect.

Explaining the design in the get-go few elements

Hydrogen has an electronic structure of 1s¹. It is a very small atom, and the single electron is close to the nucleus and therefore strongly attracted. At that place are no electrons screening it from the nucleus and so the ionisation energy is high (1310 kJ mol^-1).

Helium has a structure 1s². The electron is being removed from the same orbital every bit in hydrogen'southward instance. It is close to the nucleus and unscreened. The value of the ionisation energy (2370 kJ mol^-1) is much higher than hydrogen, because the nucleus now has 2 protons alluring the electrons instead of 1.

Lithium is 1s²2s¹. Its outer electron is in the 2nd energy level, much more distant from the nucleus. You might debate that that would be offset by the boosted proton in the nucleus, but the electron doesn't experience the full pull of the nucleus - it is screened by the 1sⁱⁱ electrons.

Y'all can think of the electron as feeling a net 1+ pull from the eye (3 protons offset past the two 1s² electrons).

If y'all compare lithium with hydrogen (instead of with helium), the hydrogen'southward electron also feels a 1+ pull from the nucleus, just the distance is much greater with lithium. Lithium'southward first ionisation free energy drops to 519 kJ mol^-ane whereas hydrogen's is 1310 kJ mol^-1.

The patterns in periods two and iii

Talking through the next 17 atoms i at a time would take ages. We can exercise it much more neatly by explaining the main trends in these periods, so accounting for the exceptions to these trends.

The first thing to realise is that the patterns in the ii periods are identical - the difference being that the ionisation energies in period iii are all lower than those in menses 2.

Explaining the general trend across periods two and 3

The general trend is for ionisation energies to increase across a catamenia.

In the whole of period 2, the outer electrons are in 2-level orbitals - 2s or 2p. These are all the same sort of distances from the nucleus, and are screened past the same 1s² electrons.

The major difference is the increasing number of protons in the nucleus as yous go from lithium to neon. That causes greater allure between the nucleus and the electrons and and then increases the ionisation energies. In fact the increasing nuclear charge also drags the outer electrons in closer to the nucleus. That increases ionisation energies still more as you get beyond the period.

Notation:Factors affecting atomic radius are covered on a separate folio.

In catamenia three, the trend is exactly the same. This time, all the electrons being removed are in the third level and are screened by the 1s^two2s²2p^{half dozen} electrons. They all have the same sort of environment, but in that location is an increasing nuclear accuse.

Why the drib betwixt groups two and 3 (Be-B and Mg-Al)?

The caption lies with the structures of boron and aluminium. The outer electron is removed more easily from these atoms than the full general trend in their period would suggest.

Be		1s^two2s^two		1st I.E. = 900 kJ mol^-i
B		1s^two2s^two2p₁₀ ¹		1st I.E. = 799 kJ mol^-1

You might expect the boron value to be more than the beryllium value because of the extra proton. Offsetting that is the fact that boron's outer electron is in a 2p orbital rather than a 2s. 2p orbitals have a slightly higher energy than the 2s orbital, and the electron is, on average, to be found farther from the nucleus. This has two effects.

The increased altitude results in a reduced attraction so a reduced ionisation energy.
The 2p orbital is screened not only past the 1s² electrons but, to some extent, by the 2s^two electrons also. That also reduces the pull from the nucleus and so lowers the ionisation energy.

The explanation for the drop between magnesium and aluminium is the same, except that everything is happening at the 3-level rather than the two-level.

Mg		1s²2s²2p⁶3s²		1st I.Eastward. = 736 kJ mol^-1
Al		1s²2sⁱⁱ2p⁶3s²3p_x ^one		1st I.E. = 577 kJ mol^-1

The 3p electron in aluminium is slightly more afar from the nucleus than the 3s, and partially screened by the 3s² electrons besides as the inner electrons. Both of these factors offset the effect of the extra proton.

Warning!You might possibly run into a text book which describes the drib betwixt group 2 and grouping iii past saying that a full southward² orbital is in some way peculiarly stable and that makes the electron more than hard to remove. In other words, that the fluctuation is considering the group two value for ionisation energy is abnormally loftier. This is quite simply wrong! The reason for the fluctuation is because the group 3 value is lower than you might await for the reasons we've looked at.

Why the drib between groups five and six (N-O and P-Southward)?

Once more, you might await the ionisation energy of the group 6 chemical element to be higher than that of group five considering of the extra proton. What is offsetting it this time?

N		1s²2s²2p_ten ¹2p_y ^ane2p_z ¹		1st I.E. = 1400 kJ mol^-1
O		1s²2s²2p₁₀ ^two2p_y ¹2p_z ¹		1st I.E. = 1310 kJ mol^-one

The screening is identical (from the 1s² and, to some extent, from the 2s² electrons), and the electron is existence removed from an identical orbital.

The deviation is that in the oxygen case the electron existence removed is i of the 2p₁₀ ² pair. The repulsion between the two electrons in the same orbital means that the electron is easier to remove than it would otherwise be.

The drop in ionisation energy at sulphur is accounted for in the same mode.

Note:After oxygen or sulphur, the ionisation energies of the next two elements increase because of the boosted protons. Everything else is the same - the type of orbital that the new electron is going into, the screening, and the fact that it is pairing up with an existing electron.

Students sometimes wonder why the side by side ionisation energies don't autumn because of the repulsion caused by the electrons pairing up, in the same mode it falls between, say, nitrogen and oxygen.

Betwixt nitrogen and oxygen, the pairing up is a new factor, and the repulsion outweighs the effect of the extra proton. Just between oxygen and fluorine the pairing up isn't a new factor, and the only deviation in this case is the extra proton. Then relative to oxygen, the ionisation energy of fluorine is greater. And, similarly, the ionisation energy of neon is greater even so.

Trends in ionisation free energy down a grouping

As you go downward a group in the Periodic Table ionisation energies generally fall. You have already seen evidence of this in the fact that the ionisation energies in menses iii are all less than those in catamenia ii.

Taking Group 1 as a typical case:

Why is the sodium value less than that of lithium?

There are xi protons in a sodium cantlet but only three in a lithium atom, so the nuclear charge is much greater. You might have expected a much larger ionisation energy in sodium, just offsetting the nuclear charge is a greater distance from the nucleus and more screening.

Li		1s²2sⁱ		1st I.Eastward. = 519 kJ mol^-ane
Na		1s^two2s²2p⁶3s¹		1st I.Due east. = 494 kJ mol^-1

Lithium's outer electron is in the second level, and simply has the 1sⁱⁱ electrons to screen it. The 2s¹ electron feels the pull of three protons screened by 2 electrons - a net pull from the heart of ane+.

The sodium's outer electron is in the third level, and is screened from the 11 protons in the nucleus past a full of 10 inner electrons. The 3s^ane electron too feels a net pull of 1+ from the centre of the atom. In other words, the effect of the extra protons is compensated for by the effect of the extra screening electrons. The only factor left is the extra altitude between the outer electron and the nucleus in sodium's example. That lowers the ionisation energy.

Similar explanations hold as y'all become down the residual of this group - or, indeed, whatever other grouping.

Trends in ionisation energy in a transition series

Apart from zinc at the terminate, the other ionisation energies are all much the same.

All of these elements have an electronic construction [Ar]3d^north4s² (or 4s¹ in the cases of chromium and copper). The electron being lost always comes from the 4s orbital.

Notation:The 4s orbital has a higher energy than the 3d in the transition elements. That means that information technology is a 4s electron which is lost from the atom when it forms an ion. Information technology as well means that the 3d orbitals are slightly closer to the nucleus than the 4s - and then offer some screening.

Confusingly, this is inconsistent with what we say when we use the Aufbau Principle to piece of work out the electronic structures of atoms.

I have discussed this in detail in the folio well-nigh the club of filling 3d and 4s orbitals.

If you lot are a instructor or a very confident pupil and so yous might like to follow this link.

If yous aren't and so confident, or are coming at this for the offset fourth dimension, I propose that you ignore it. Recall that the Aufbau Principle (which uses the supposition that the 3d orbitals fill afterwards the 4s) is simply a useful way of working out the structures of atoms, but that in real transition element atoms the 4s is actually the outer, higher energy orbital.

As yous become from ane atom to the next in the serial, the number of protons in the nucleus increases, but so likewise does the number of 3d electrons. The 3d electrons have some screening upshot, and the actress proton and the extra 3d electron more or less cancel each other out as far as attraction from the centre of the atom is concerned.

The rise at zinc is like shooting fish in a barrel to explicate.

Cu		[Ar]3d^ten4s¹		1st I.E. = 745 kJ mol^-i
Zn		[Ar]3d¹⁰4s²		1st I.Due east. = 908 kJ mol^-ane

In each case, the electron is coming from the same orbital, with identical screening, but the zinc has one extra proton in the nucleus and so the attraction is greater. There will exist a degree of repulsion between the paired up electrons in the 4s orbital, but in this case it apparently isn't enough to outweigh the effect of the extra proton.

Note:This is really very similar to the increase from, say, sodium to magnesium in the third period. In that instance, the outer electronic structure is going from 3s^one to 3s². Despite the pairing-upward of the electrons, the ionisation energy increases because of the extra proton in the nucleus. The repulsion between the 3s electrons evidently isn't plenty to outweigh this either.

I don't know why the repulsion betwixt the paired electrons matters less for electrons in s orbitals than in p orbitals (I don't even know whether yous can make that generalisation!). I suspect that information technology has to do with orbital shape and possibly the greater penetration of s electrons towards the nucleus, only I haven't been able to find any reference to this anywhere. In fact, I haven't been able to find anyone who even mentions repulsion in the context of paired s electrons!

If you have whatsoever hard information on this, could yous contact me via the address on the well-nigh this site page.

Ionisation energies and reactivity

The lower the ionisation free energy, the more than easily this change happens:

X(one thousand) X⁺(g) + due east^-

Y'all can explain the increment in reactivity of the Group 1 metals (Li, Na, Grand, Rb, Cs) equally you go down the group in terms of the fall in ionisation energy. Whatever these metals react with, they have to course positive ions in the process, so the lower the ionisation energy, the more than hands those ions volition course.

The danger with this approach is that the formation of the positive ion is only i stage in a multi-step process.

For example, you wouldn't be starting with gaseous atoms; nor would you end upwards with gaseous positive ions - you would end upwardly with ions in a solid or in solution. The energy changes in these processes also vary from chemical element to chemical element. Ideally you need to consider the whole picture and non only one small-scale part of information technology.

Nonetheless, the ionisation energies of the elements are going to be major contributing factors towards the activation energy of the reactions. Think that activation energy is the minimum energy needed before a reaction will accept place. The lower the activation energy, the faster the reaction will be - irrespective of what the overall energy changes in the reaction are.

The fall in ionisation energy as y'all go down a group volition atomic number 82 to lower activation energies and therefore faster reactions.

Note:You will find a page discussing this in more detail in the inorganic section of this site dealing with the reactions of Group 2 metals with water.

Where would you like to go at present?: To look at 2nd (and successive) ionisation energies . . .; To the atomic properties menu . . .

To the atomic construction and bonding menu . . .

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