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Y12 Organic Chemistry > Atomic Structure > Flashcards

Atomic Structure Flashcards

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1
Q

Mass of proton

A

1

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2
Q

Mass of neutron

A

1

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3
Q

Mass of electron

A

1/1800

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4
Q

Charge of proton

A

+1

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5
Q

Charge of neutron

A

0

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6
Q

Charge of electron

A

-1

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7
Q

Atomic number

A

number of protons in the nucleus

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8
Q

Mass number

A

Total number of protons and neutrons

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9
Q

Isotopes

A

Atoms with the same number of protons but different number of electrons

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10
Q

Reason mass spec is done under a vacuum

A

Otherwise air particles would ionise and be detected

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11
Q

Steps of mass spec

A

Ionisation, acceleration, drift, detection

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12
Q

Electron impact ionisation

A

High energy electrons are fired at a vaporised sample removing an outer electron forming positive ions

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13
Q

Electrospray ionisation

A

Sample is dissolved in a volatile, polar solvent and injected through a fine needle. The tip of the needle has a high voltage adding a proton to the sample. M(g) + H= –> MH+

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14
Q

Use of electron impact

A

elements and substances with low formula mass. Larger molecules may fragment

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15
Q

Use of electrospray ionisation

A

larger organic molecules as fragmentation is less likely

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16
Q

Acceleration

A

Positive ions are accelerated by an electric field to a constant kinetic energy = 1/2 m v2

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17
Q

Flight Tube/ Drift region

A

Lighter ions move down the tube quicker than heavier ones and are distinguished by flight times

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18
Q

Detection

A

positively charged ions are attracted to a negatively charged plate generating a current, which a computer analyses. The more ions of a particular size, the larger the current produced

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19
Q

Relative atomic mass

A

weighed average of all isotopes of an atom compared to 1/12th mass of a C-12 atom

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20
Q

Relative atomic mass calculation

A

(mass 1 x abundance 1) + (mass 2 x abundance 2) + … / total abundance

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21
Q

What is the relative atomic mass for Mg?
Mg-24 = 78.70%, Mg-25 = 10.13%
Mg-26 = 11.17%

A

24.3

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22
Q

Key features to include when asked for species for a peak on a mass spectrum

A

charge and mass number. e.g. 24Mg+

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23
Q

Calculate the relative atomic mass of Te. Te-124 = 2, Te-126 = 4, Te-128 = 7, Te-130 = 6

A

127.8

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24
Q

Copper has 2 isotopes 63-Cu and 65-Cu. The RAM is 63.5. What is the abundance of each isotope

A 
63-Cu = 72.5%
65-Cu = 27.5%
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25
Q

Molecular Ion Peak

A

Shows the relative molecular mass and is the peak furthest to the right

26
Q

Order sub shells are filled –> 5p

A

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p

27
Q

Electronic structure for Ca

A

1s2, 2s2, 2p6, 3s2, 3p6, 4s2

28
Q

s block element

A

element with highest energy electron in an s sub shell (first two groups of PT)

29
Q

p block element

A

element with highest energy electron in an p sub shell (end 6 groups of PT)

30
Q

d block element

A

element with highest energy electron in an d sub shell (central section of PT)

31
Q

electronic configuration for Sc

A

[Ar] 4s2, 3d1

32
Q

electronic configuration for Ti

A

[Ar] 4s2, 3d2

33
Q

electronic configuration for V

A

[Ar] 4s2, 3d3

34
Q

electronic configuration for Cr

A

[Ar] 4s1, 3d5

35
Q

electronic configuration for Mn

A

[Ar] 4s2, 3d5

36
Q

electronic configuration for Fe

A

[Ar] 4s2, 3d6

37
Q

electronic configuration for Co

A

[Ar] 4s2, 3d7

38
Q

electronic configuration for Ni

A

[Ar] 4s2, 3d8

39
Q

electronic configuration for Cu

A

[Ar] 4s1, 3d10

40
Q

electronic configuration for Zn

A

[Ar] 4s2, 3d10

41
Q

electronic configuration for Ti3+

A

[Ar] 3d1

42
Q

electronic configuration for V3+

A

[Ar] 3d2

43
Q

electronic configuration for Cr3+

A

[Ar] 3d3

44
Q

electronic configuration for Mn2+

A

[Ar] 3d5

45
Q

electronic configuration for Cu2+

A

[Ar] 3d9

46
Q

electronic configuration for Zn2+

A

[Ar] 3d10

47
Q

Rules for filling orbitals

A

Fill lowest energy orbitals first. Singly fill orbitals before doubly filling

48
Q

Rules for filling and emptying d block element sub-shells

A

Fill 4s before 3d. Empty 4s before 3d.

49
Q

First Ionisation Energy

A

Energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions

50
Q

Equation for first ionisation energy

A

X(g) –> X+(g) + e

51
Q

Second Ionisation Energy

A

Energy required to remove one mole of electrons from one mole of gaseous 1+ ions to form one mole of gaseous 2+ ions

52
Q

Second ionisation energy equation

A

X+(g) –> X2+(g) + e

53
Q

Factors that affect ionisation energy

A

Amount of protons, distance of electrons from nucleus and amount of shielding

54
Q

Why are successive ionisation energies always larger?

A

2nd IE is larger than 1st IE because the electron is being removed from a 1+ ion with greater attraction to the outgoing electron.

55
Q

What does a big jump in successive ionisation energy show?

A

A change of shell that the electron is being removed from. E.g. a big jump from electron 4 to 5 indicated the element is in group 4 with 4 electrons on the outer shell

56
Q 
What group is this element in?
1st IE = 590
2nd IE = 1150
3rd IE = 4940
4th IE = 6480
5th IE = 8120
A

Group 2

57
Q

How does IE change down a group?

A

Decreases - despite having more protons, the elements have a greater distance from the nucleus to the outgoing electron with more shielding

58
Q

What is the general trend in IE across a period

A

Increases - successive element have more protons but the electrons are added to the same shell with the same amount of shielding and slightly less distance between the outer shell and the nucleus

59
Q

Why is there a drop in IE from Mg to Al?

A

The outermost electron in Al is in a 3p orbital compared to 3s in Mg so is slightly further away from the nucleus with more shielding so easier to remove.

60
Q

Why is there a drop in IE from P to S?

A

The outermost electron in S is in a paired 3p orbital compared to singly filled 3p in P. This means there is some repulsion from the other electron making it easier to remove.

Y12 Organic Chemistry (3 decks)

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