Covalent and Ionic Radii, Effective Nuclear Charge, Orbital and Promotion Energies

Question 1

What is the Covalent radius?
How can we calculate the distance for new bond?
However what must we take into account?

Answer 1

• Covalent radius is the defined as half the length of a symmetrical homonuclear bond (X-X)
• Knowing the radii of two elements we can calculate distances for new bonds
• Generally works well, but we must take into account electronegativity differences

Question 2

What is the metallic radius?

Answer 2

Is the equivalent distance between ions in a metal lattice

Question 3

For some elements like helium and neon the covalent radius is difficult to measure due those elements not having any bonds
How is this overcome?

Answer 3

through measuring the van der Waals radius

Question 4

What is the ionic radii?

Answer 4

equivalent distance between positive cations and negatie ions
But additionally increase with increasing -ve charge and decrease with increasing +ve charge

Question 5

How does the size of the atomic radii change across the p-block from left to right?

Answer 5

• from left to right, the size of the elements decreases across a period (angstrom radii numbers shown)
• so flourine is smaller than boron

Question 6

What is 1 Angstrom (Å) in pm?

Answer 6

1 Angstrom = 100ppm

Question 7

What do Slater’s rules (Zeff) recognise?

Answer 7

the outermost electrons feel a nuclear charge which is less than the actual nuclear charge, because of shielding effects from other electrons

Question 8

When using Slater’s rules to work out effective nuclear charge, for a particular electron in a ns or np orbtial
(i) Each of the other electrons with the same principle quantum number in the same (ns, np)(nd) group contributes =
(ii)Each of the electrons in the (n-1) shell contributes =
(iii) Each of the electrons in the (n-2) or lower shells contributes =

Answer 8

(i) 0.35
(ii) 0.85
(iii) 1.00

Question 9

When using Slater’s rules to work out effective nuclear charge, for an electron in a nd or nf orbital
(i) how much does each of the other electrons in the (nd, nf) group contribute
(ii) how much does each of the electrons in a lower group than the one being considered contribute

Answer 9

(i) 0.35
(ii) 1.00

Question 10

Calculate the shielding constant, s, for the 2p electron in a boron atom, and the effective nuclear charge, Zeff, that this electron experiences
B: 1s² 2s² 2p¹

Answer 10

Zeff = Z - S
S = 2(0.35) + 2(0.85) = 2.4
5-2.4 = 2.6

Question 11

Calculate the shielding constant, s, for the 4p electrons in an atom of germanium, and the effective nuclear charge, Zeff, that this electron experiences
Ge: [Ar] 3d¹⁰ 4s² 4p²

Answer 11

= 3(0.35) + 18(0.85) + 10(1)
s = 26.35
zeff = 32 - 26.35
= 5.65

Question 12

Calculate the shielding constant, s, for the 3d electron in an atom of nickel, and the effective nuclear charge, Zeff, that this electron experiences
Ni: [Ar] 3d⁸ 4s²

Answer 12

= 7(0.35) + 18(1.00)
S = 20.45
= 28-20.45
= 7.55

Question 13

How does effective nuclear charge change going accross the periodic table from left to right?

Answer 13

• Effective nuclear charge increases substantially across the periodic table
• Hence explaining why F has such a high ionisation energy (hard to remove that first proton due to high amounts of nuclear charge felt) and also a smaller atom

Question 14

Slater’s rules predict an increase in nuclear charge going down the group
Why is this incorrect

Answer 14

• Because we know the ionisation energy decreases on descending a group
• Slater’s rules are simplistic: they do not account for distance from nucleus or penetration (nodes, varying likelihood of finding an electron etc)

Question 15

What are the s- vs p-electrons energies like

Answer 15

• s-electron energy are lower in energy due to being closer to the nucleus hence more stabilised by positive attraction force
• p-electron energies are higher due to being less pentrating and hence less stabilised by positie attractive force

Question 16

Within a group, the highest occupied molecular orbital (HOMO) energies increases (become less negative/less stable) from top to bottom and the s-p separation decreases
Why is this?

Answer 16

Higher principal quantum number shells so electrons further away from the nucleus

Question 17

s and p energies decrease across a period, but more for s
Why?

Answer 17

• Zeff for s increases, resulting in the s-electrons becoming much more stable than p
• (this is why the sigma and pi MOs swap around from N₂ to O₂ (s-p mixing)

Question 18

the 4s orbital energies for As, Se, Br and Kr are lower than expected
Why?

Answer 18

related to d-block contraction
Increased Zeff from 3rd period which has a strong affect on 4s (penetration to inner core)

Question 19

Describe the trend in Promotion Energies

Answer 19

General increase in energies required to promote e- from ground state to excitied state going down a group
e.g. s²p² → s¹p³

Question 20

What two thing affect promotional energies

Answer 20

• The energy difference between orbitals (this increases down a group)
• The effect on the energies of other electrons in the shell which have not been promoted and now experience less electrostatic repulsions

Question 21

Why do Ga and Ge deviate from the trned of promotional energies

Answer 21

• 4p row immediately proceded by full 3d row so lower than expected s orbital energy - d-block contraction

Question 22

Why are the promotional energies for Tl and Pb so unexpectly high?

Answer 22

Tl and Pb have highest promotional energies indicating that the reorganisation energies are substantial (relates to rehybridiation)
(when one electron moves the others do not stay static but their energies change due to electron correlation)

Question 23

Why can the first row not expand its octet?

Answer 23

Because there are only s and p orbtials which are available

Question 24

How do the properites of the p-block compound affect the structure of the oxide it forms

Tl(I), Pb(II) and Bi(III)

Answer 24

• All metals (minus Sn) will form an ionic structure with some covalent character
• All metalloids (plus Se) will form a covalent network structure
• All non-metals (minus nobel gases) will form a discrete covalent structure

Question 25

How do the properties of the p-block compounds affect the structure of the fluorides it forms

Answer 25

* All metals (minus Sn) will form an ionic structure with some covalent character * There is no pattern after that * This is because the increase in electronegativity (F vs O) shifts the boundary between covalent, polymeric/macromolecular and ionic compounds * (atoms in flouride compounds have higher covalency parameters - average electronegativity)

Question 26

What is the inert pair effect?

Answer 26

* The tendency of the **electrons in the outermost** atomic s-orbitals to **remain unionised** or unshared in compounds of the group **13-16** elements * OR * The observation that as a group is descended, the **n-2 oxidation state become more favoured** (where n=group oxidation state) * There is no simple explanation for this effect BUT the **strength of covalent bonds decreases down a group** due to poor orbital overlap which results in bond enthalpy not offsetting the rehybridiation energy cost + relativist effect which stabilise 6s orbitals

Question 27

CH₄ is tetrahedral: needs sp³ hybridisation for 109.5° bond angles Formation of the bonds is exothermic (offsets endothermic cost) What does this mean in terms of rehybridisation and overall energy change

Answer 27

* In order to make a tetrahedral molecule of methane, we need 4x singularly occuped sp³ hybrids * This requires the promotion of an electron from the 2s to 2p orbital * However for hybridiation there is no net change in energy * The C-H bonds are strong and promotion energy is low so CH₄ formation is favourable (C⁴⁺) * (this is not always the case of other group 14 hydrides e.g PH₄)

Question 28

What is the relativistic effect?

Answer 28

The heavier the element, the faster the outer electrons go The theory of Special Relativity states that objects moving near the speed of light gain mass (basically the weight of an electron is not the same while it is moving as it would be stationary)

Question 29

For Bi, the 1s electron reach a speed of 60% c (speed of light) resulting in m=1/26 What affect does this have on the 1s orbital

Answer 29

The 1s orbital contracts by 20% and its energy is lowered - resulting in stabilisation Called the direct relativistic orbital contraction

Question 30

How does the direct relativistic orbital contraction affect the s- vs the p-orbital

Answer 30

* The s-orbitals are stabilised the most (less available for bonding) * Due to best penetration * p-orbitals contract but not as much * due to poorer penetration

Question 31

How are the d and f-block affected by the direct relativistic orbital constraction

Answer 31

* d- and f-orbitals experience indirect relativistic orbital expansion (destabilised) * poorer penertration and the contraction of s and p leaves them more shielded and less affected by Zeff * As a result, the s electrons of heavy elements are more stable and less likely to be ionised or involved in bonding (hence, the inert pair effect observation)

Question 32

Comment on some of the general trends

Answer 32

* From left to right and from bottom to top there is an increase in electronegativity * The more electronegative an element is the more likely they will form a covalent molecule * The less electronegative an element is the more likely they will form an ionic compound * Tl, Pb and Bi are all impacted by relativistic and intert paur effects so can form ionic compounds