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Do Now When should we do our next assessment – After chapter 3 (like Mondayish….) – After chapter 4 (chapter 4 is BIG….5 sections…much work to be done)

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Presentation on theme: "Do Now When should we do our next assessment – After chapter 3 (like Mondayish….) – After chapter 4 (chapter 4 is BIG….5 sections…much work to be done)"— Presentation transcript:

1 Do Now When should we do our next assessment – After chapter 3 (like Mondayish….) – After chapter 4 (chapter 4 is BIG….5 sections…much work to be done) – OR Do section 4.5 Metallic bonding then test then do the rest of chapter 4 (1-4 sections) This would push test to later next week.

2 Topic 3: Periodicity 3.2 Periodic Trends

3 Understandings: Vertical and horizontal trends in the periodic table exist for atomic radius, ionic radius, ionization energy, electron affinity, and electronegativity.

4 Periodic Trends Trends are primarily based on a few factors – Distance of outer most electrons from the nucleus – Effective nuclear charge.

5 Effective Nuclear Charge The positive charge that an electron experiences from the nucleus, equal to the nuclear charge but reduced by any shielding or screening from any intervening electron distribution (core electrons- shielding electrons). Down a group the increase in nuclear charge is largely offset by the increase in number of inner electrons so the effective nuclear charge remains approx +1 all the way down the group. ElementNaMgAlSi Nuclear Charge11121314 Electron configuration [Ne]3s 1 [Ne]3s 2 [Ne]3s 2 3p 1 [Ne]3s 2 3p 2 Effective nuclear charge ≈ 11-10 ≈ +1 ≈ 12-10 ≈ +2 ≈ 13-10 ≈ +3 ≈ 14-10 ≈ +4

6 Atomic Radius Atomic Radius – one-half the distance between neighboring nuclei. Increases Decreases

7 Atomic Radius The radius increases down a family or group – As the principle quantum number (primary energy level) increases, the size of the orbitals increases therefore increasing the distance of the electrons from the nucleus. The radius decreases across a period – The principle quantum number stays constant so the size of the shells doesn't increase – The effective nuclear charge increases (more protons are added to the nucleus but the core electrons stay constant) so the electrons are pulled in more strongly.

8 Atomic radius Top to Bottom – increases – Electrons are added to shells farther from the nucleus Left to right – decreases – The effective nuclear charge increases pulls more strongly on the valence electrons so radius gets smaller

9 Ionic radius Cations are smaller than parent atoms – Remove all valence electrons so fewer shells are occupied and there are more p+ than e- so a stronger pull from nucleus occurs Anions are larger than parent atoms – Electrons are added but no protons are added. This increases electron repulsion and the electrons spread out further increasing the size

10 Ionization Energy (IE) Ionization Energy (IE)- The energy required to remove one electron from a mole of gaseous atoms or ions. Decreases Increases

11 Ionization Energy (IE) The IE decreases down a family or group – The electrons being removed are further from the positive charge of the nucleus and are easier to remove. The IE increases across a period – The effective nuclear charge increases (more protons are added to the nucleus but the core electrons stay constant) so it is harder to pull an electron away. – The electrons are held closer to the nucleus (smaller radius)

12 Guidance Only examples of general trends across periods and down groups are required. For ionization energy the discontinuities in the increase across a period should be covered.

13 Group 3 elements with e- configurations ns 2 np 1, have lower 1 st IE than group 2 with the configuration ns 2, as p orbitals have higher energy than s (it is easier to remove an e- at higher energy)

14 The drop between group 15 and 16 occurs as the electron removed from group 16 is taken from a doubly occupied p orbital (easier to remove because it is repelled by it’s partner)

15 Electron Affinity Electron affinity- The energy change when one mole of electrons is added to one mole of gaseous atoms to form one mole of gaseous ions Decreases Increases

16 Electron Affinity (these trends are less dramatic than the two previous examples) The e- affinity decreases down a family or group – The electron shells are farther from the nucleus so it is harder to add an electron to this level. The e- affinity increases across a period – The effective nuclear charge increases (more protons are added to the nucleus but the core electrons stay constant) so the atom more readily accepts an electron.

17 Electronegativity Electronegativity- a measure of the ability of its atoms to attract electrons in a covalent bond. Decreases Increases

18 Electronegativity The electronegativity decreases down a family or group – The shared electrons are further from the positive charge of the nucleus in atoms at the bottom of groups The electronegativity increases across a period – The effective nuclear charge increases (more protons are added to the nucleus but the core electrons stay constant) so it will attract the electrons more strongly then elements at the beginning of a period. – The electrons are held closer to the nucleus (smaller radius)

19 Trends in metallic and non-metallic behavior are due to the previously discussed trends

20 Metals conduct electricity – Due to valence e- moving away from nucleus – Related to low IE and electroneg – Transition from metal to metalloid to non-metal occurs as these values increase Metals get oxidized – Low IE Non-metals get reduced – High electron affiniy

21 Guidance Discussion of the similarities and differences in the properties of elements in the same group (1 and 17).

22 Melting point trends in group 1, 17 Alkali Metals (group 1) – Have metallic structures – Held together by forces b/t delocalized outer e- & nucleus – Attraction decreases with distance – Melt at lower temperatures Halogens (group 17) – Have molecular strucures (diatomic) – Held together by London dispersion forces – These increase with # e- in molecule – Melting points increase

23 Reactivity Alkali metals – Increase reactivity as you move down group (electrons easier to remove to make ions and new compounds) – All form a +1 ion – With water form MOH + H2 (MOH is basic or alkaline…giving the group its name) – https://www.youtube.com/watch?v=uixxJtJPVXk https://www.youtube.com/watch?v=uixxJtJPVXk

24 Halogens – Similarities Colored gases – Trends Gradual trend from gas to liquid to solid at room temp Reactivity decreases down the group – Adding electrons gets more difficult Reactivity includes – Reactions with alkali metals » Bottom metals react most with top halogens – Displacement reactions ie top halogen will replace lower » KBr (aq) + Cl 2(aq)  2KCl (aq) + Br 2(aq)

25 Oxides change from basic through amphoteric to acidic across a period. – Guidance: Construction of equations to explain the pH changes for reactions of Na 2 O, MgO, P 4 O 10 and the oxides of nitrogen and sulfur with water.

26 Oxides Oxide compounds – From left to right oxide compounds transition from giant ionic, to giant covalent to molecular covalent (changing their “Ionic character”) – Type of bonding is dependent on difference in electronegativity b/t the element electroneg and oxygen’s of 3.4 – Oxides become more ionic when oxygen is bonded with elements farther down the P.T.

27 Acid-base properties of oxides are linked to their bonding. Na2O + H2O(l)  2NaOH (aq) Formula of oxideNa 2 O (s) MgO (s) Al 2 O 3(s) SiO 2(s) P 4 O 10(s) / P 4 O 6(s) SO 3(l) / SO 2(g) Cl 2 O 7(l) / Cl 2 O (g) Acid-base characterbasicamphotericacidic

28 Basic oxides Dissolve in water to form alkaline solutions Na 2 O + H 2 O (l)  2NaOH (aq) React with acids to produce salt and water Li 2 O + 2 HCl  2 LiCl (aq) + H 2 O (l)

29 Acidic oxides React with water to produce acidic solutions P 4 O 10(s) + 6 H 2 O (l)  4H 3 PO 4(aq)

30 Amphoteric oxides Aluminum oxide oxide does not affect the pH whuen it is added to water b/c it is not very soluble. It shows acid and base behavior Al 2 O 3 (s) + H 2 SO 4  Al 2 (SO 4 ) 3 (aq) + 3 H 2 O (l) Al 2 O 3(s) + 3H 2 O + 2NaOH (aq)  2NaAl(OH) 4(aq)


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