0.77 + 1.14

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0.77 + 1.14 Approximate bonding atomic radii for the elements have been tabulated. The distance between bonded nuclei can be pproximated by adding radii from both atoms. e.g., Bonding atomic radii are as follows: So the approximate distance between bonded C and Br nuclei = = 1.91 A C = 0.77 A, Br = 1.14 A r r d

description

C = 0.77 A, Br = 1.14 A. = 1.91 A. r. r. d. Approximate bonding atomic radii for the elements have been tabulated. The distance between bonded nuclei can be approximated by adding radii from both atoms. e.g., Bonding atomic radii are as follows:. - PowerPoint PPT Presentation

Transcript of 0.77 + 1.14

Page 1: 0.77 + 1.14

0.77 + 1.14

Approximate bonding atomicradii for the elements havebeen tabulated.

The distance between bonded nuclei can beapproximated by adding radii from both atoms.

e.g., Bonding atomic radii are as follows:

So the approximate distance betweenbonded C and Br nuclei = = 1.91 A

C = 0.77 A, Br = 1.14 A

r rd

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more p+, but no new (i.e., farther away) energy levels

Atomic Radius

As we go down a group,atomic radius…

-- principal quantum numberincreases (i.e., a newenergy level is added)

increases.

As we go from left to right across theTable, atomic radius… decreases.

-- effective nuclear charge increases, but principal quantum number is constant

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Coulombic attraction depends on…

2– 2+

2+ 2– 1–

2–

1+

2+

amount of charge distance between charges

+ + – –

H

He

+ –

+ – + –

As we go , more coulombic

attraction, no new energy level, more pull, smaller size

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Sr < Ba < Cs

Arrange the following atoms in order of increasingatomic radius: Sr, Ba, Cs

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Ionization Energy: the minimum energy needed to remove an e– from an atom or ion

M(g) + 1st I.E. M+(g) + e–

M+(g) + 2nd I.E. M2+(g) + e–

M2+(g) + 3rd I.E. M3+(g) + e–

Successive ionization energiesare larger than previous ones.

-- (+) attractive force remains the same, but there is less e–/e– repulsion

I.E. e –

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The ionization energy increases sharply whenwe try to remove an inner-shell electron.

e.g., For Mg, 1st IE = 738 kJ/mol 2nd IE = 1,450 kJ/mol 3rd IE = 7,730 kJ/mol

As we go down a group, 1st IE… decreases.

-- more e–/e– repulsion and more shielding

(strong evidence that only valence e– are involved in bonding)

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Generally, as we go from left to right, 1st IE…

Exceptions: e.g., B < Be

Be: 1s2 2s2

B: 1s2 2s2 2p1 B doesn’t like

2p

(easier to remove B’s single 2p e– than one of Be’s two 2s e–s)

N: 1s2 2s2 2p3

O: 1s2 2s2 2p4

More stable to have

than to have

2p

Subshells prefer to beeither completely filled

OR half-filled.

This e– is easier to remove…

…than any of these.

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Firs

t

acro

ss a

per

iod…

across a perio

d…

down a group…

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released.

Electron affinity: the energy change that occurs when an e– is added to a gaseous atom

For most atoms, adding an e–

causes energy to be…

Exceptions:

noble gases: the added e– must go into a new, higher energy level

group 2 metals: the added e– must go into a higher-energy p orbital

group 15 elements: the added e– is the first one to double-up a p orbital

eq. for e– affinity: A + e– A–

I.E. e –

e– energy

ener

gy

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more (–)e– affinity

The halogens have the most (–) electronaffinities, meaning that they become verystable when theyaccept electrons.

Electron affinities don’t varymuch going down a group.

more willing toaccept an e–=

–328F

Cl

Br

I

O

S

Se

Te

Ne

Ar

Kr

Xe

He

–349

–325

–295

–141

–200

–195

–190

+

+

+

+

+