S2.2 Covalent Model - Hybridisation & Bonds

Question 1

Orbital Hybridisation

Answer 1

concept of mixing atomic orbitals to form new hybrid orbitals that can accommodate the bonding and lone pair electrons in a molecule

Question 2

SP3

Answer 2

creates 4 equal length bonds (exmaple carbon)
* ground state electron configuration changes = excitation
* 2s orbitals are promoted to higher energy
* creates 4 new bonding orbitals of equal energy
* make uo tetradhedral shape = 109.5 degrees

Question 2

SP2

Answer 2

when 3 bonds of equal length & shape want to be formed
* 2s orbital joins with p to make SP2 and 1 remaining 2p orbital
* sp3 orbitals form sigma bond
* unhybridised p orbital forms pi bond
* trigonal planar = 120 degrees

Question 3

SP

Answer 3

• 2s orbitals joins with p orbital and leaves 2 unhybridised p orbitals and 2sp orbitals
• sigma bond between sp orbitals
• 2 pi bonds

Question 4

Benzen hybridisation

Answer 4

• all 6 carbons in benzene are sp2 hybridised
• each C has sigma bond between adjacent C
• each C has delocalised pi bond with overlapping unhybridised p orbitals

Question 5

How are electron pairs arranged

Answer 5

Electron pairs are arranged in a way that they will maximise seperation and minimize repulsion

Question 6

Electron domain geometry

Answer 6

the shape of the arrangement of electron domains surrounding a central atom in a molecule or ion

Question 7

electron domain

Answer 7

a region in which bonding & non bonding pairs of electrons are most likely to be found

Question 8

Rules for bond angles

Answer 8

each lone pair reduces the bond angle by 2.5 degrees

Question 9

Trend in decreasing repulsion strength

Answer 9

bonding pair & bonding pair< non bonding pair & bonding pair< non bonding pair & non bonding pair

Question 10

How does repulsion affect bonding angle

Answer 10

the higher the repulsion, the smaller the bonding angle

Question 11

expanded octets/hypervalent

Answer 11

when an atom can accept more than 8 electrons using their available d orbitals (3rd period atoms can)

Question 12

linear

Answer 12

no lone pairs

180 degrees

CO2

Question 13

3 electron pairs

bent/angular (one lone pair)

Answer 13

<120 degrees

Question 13

trigonal planar

Answer 13

no lone pairs

120 degrees

BCl3

Question 13

bent or angular (two lone pairs)

Answer 13

two lone pairs
<109.5 degrees
H2O

Question 14

Octahedral

Answer 14

no lone pairs

90 degrees

SF6

Question 15

Trigonal planar

Answer 15

no lone pairs

90 degrees and 120 degrees

PCl5

Question 16

Seesaw/Bisphenoidal

Answer 16

<90 degrees and 117 degrees

5 electron domains

1 lone pair

SF4

Question 17

T shape/ triangular bypyramidal

Answer 17

TBP= 90 degrees (full bonded)
5 electron domains

T shape = <90 degrees

2 lone pairs

BrF3

Question 18

linear (3 lone pairs)

Answer 18

180 degrees

3 lone pairs

I3

Question 19

square planar

Answer 19

90 degrees

2 lone pairs

XeF4

Question 19

square pyramidal

Answer 19

90 degrees

1 lone pair

BF5

Question 20

formal charge

Answer 20

useful to determine possible lewis structures and which minimizes the charges

based on finding how many e belong to the atom and the difference between how many it actually has

=> most likely structure will always have charge closest to zero

Question 20

bond strength

Answer 20

measure of the energy required to break a bond

Question 21

bond length

Answer 21

distance between the nuclei of two bonded atoms

Question 21

How bond strength works

Answer 21

* double and triple bonds have greater strength * more shared e- = higher electrostatic attraction between e and nuclei * bond is harder to overcome

Question 21

How bond length works

Answer 21

* double and triple bonds have shorter bond length * the more shared electrons the stronger the electrostatic attraction * two nuclei are pulled closer together

Question 22

Evidence contradicting benzene

Answer 22

Bond lengths= if structure consisted of alternate double 6 single bonds = some shorter than others => X Ray Crystallygrophy showed that all bonds are euqal in length Enthalpy of hydrogenation= should be 3x that of cyclyhexane but it isnt = difference is due to extra stability from delocalisation of electrons = less exothermic Chemical evidence= test for double bond; YET benzene does not decolorise bromine water => resonance; electrons are delocalised in pi bond throughout ring= give greater stability= lack of reactivity

Question 23

What causes the proof of benzene to be so extroardinary

Answer 23

* lack of expected nethalpy of combustion * lack of reactivity towards electrophiles

Question 24

How the Ozone layer protects from UV

Answer 24

Formation * Oxygen absorbes UV light (240nm) shining on O2= 2 oxygen radicals * radical reacts with 02 = o3 Decomposition * Ozone absorbs Uv light (330nm) = Oxygen radical and 02 * Ozone reacts with =xygen radicla = 2O2 => reduced UV concentration reation earth

Question 25

Why O2 and O3 are dissociated by different wavelengths

Answer 25

* covalent molecules absord UV radiation when radiation has sufficient energy to break bonds within molecule = dissociates * Ozone has weaker O-O bonds than oxygen (double bond) * bonds are stronger than a single bond but weaker than a double bond * require a longer wavelength of light to break their bonds than O=O double bond in oxygen molecules * but a shorter wavelength than oxygen–oxygen single bond.

Question 26

Resonance

Answer 26

occurs when there is more than one possible position for a double bond in a molecule by delocalising a pair of electrons

Question 27

Delocalisation

Answer 27

occurs when an electron is not fixed to one position in a molecule but can move freely across bonds

Question 28

How can you identify resonnace structures

Answer 28

* dashed line * when multiple version of the lewis structure can be drawn

Question 29

Bond order

Answer 29

the number of bonds between 2 atoms

Question 30

Equation showing NO removing ozone

Answer 30

NO∙(g)+O 3 (g)→NO2∙(g)+O 2 (g) 1 mark NO2∙(g)+O∙(g)→NO∙(g)+O2(g) NO∙(g)+O∙(g)→NO∙(g)+O 2(g)