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PHYSICAL CHEMISTRY 3.1 > 3.1.1 - ATOMIC STRUCTURE > Flashcards

3.1.1 - ATOMIC STRUCTURE Flashcards

1
Q

state the definition of an orbital

A
  • a region of space that an electron is likely to be found
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2
Q

how many orbitals (therefore how many electrons can fit) in each sub level?

A
  • s has 1 orbital and can fit 2 electrons
  • p has 3 orbitals and can fit 6 electrons
  • d has 5 orbitals and can fit 10 electrons
  • f has 7 orbitals and can fit 14 electrons
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3
Q

state electronic configuration for Copper, Cu

A

[Ar] 4s1 3d10

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4
Q

what is time of flight mass spectrometry used for?

A
  • abundance and mass of each isotope in an element sample
  • allows for its relative atomic mass to be found (r.a.m or Ar)
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5
Q

state the stages of TOF mass spectrometry

A

1 - ionization
2 - acceleration
3 - flight tube
4 - detection

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6
Q

state the 2 forms of ionization and what type of substances they are used for in TOF mass spec

A
  1. electron impact ionistion - substances with a low formula mass, e.g. small molecules and individual atoms
  2. electrospray ionisation- substances with a higher molecular mass, e.g. proteins
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7
Q

describe electron impact ionisation

A
  • sample is vapourised
  • high energy electrons fired at it (from electron gun)
  • these knock of an electron from each particle, forming a +1 ion.
    Ex) X(g) + 2e —> X+(g) + 2e
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8
Q

describe electrospray ionistion

A
  • sample dissolved in volatile solvent
  • injected through a fine hypodermic needle to give a fine mist/aerosol
  • needle attached to positive terminal of power supply
  • particles gain a proton (H+ ion) from the solvent as they leave the needle
    Ex) X(g) + H+ —> XH+(g)
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9
Q

describe the acceleration stage of TOF mass spec

A
  • positive ions are accelerated using an electric field
  • ensures all particles have the same kinetic energy
  • lighter particles have faster velocities
  • heavier particles have slower velocities
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10
Q

describe the flight tube stage in TOF mass spec

A
  • positive ions travel through hole in the negatively charged plate into the flight tube
  • the lightest ions reach the detector first, as they have the fastest velocities
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11
Q

describe the ionisation stage in TOF mass spec

A
  • particles either lose an electron or gain a proton to form a +1 ion
  • positive ions are then attracted to a negatively charged electron plate, where they are accelerated
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12
Q

describe the detection stage in TOF mass spec

A
  • positive ions hit the negatively charged electron plate
  • ions are discharged by gaining electrons from the plate
  • generates a movement of electrons, therefore a current can be measured
  • size of current is proportional to abundance
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13
Q

what does the size of current tell us in TOF mass spec?

A
  • measures the number of ions hitting the plate
  • is proportional to the abundance of that isotope
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14
Q

how is a mass spectrum determined?

A
  • a computer shows the mass to charge ratio (m/z) and abundance of each isotope
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15
Q

what does the peak on the greatest mass to charge (m/z) value tell us?

A
  • the Mr, molecular mass
  • e.g. 84 for Kr84
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16
Q

what are peaks bigger than the Mr caused by?

A

isotopes of atoms in molecules

17
Q

what are peaks smaller than the Mr caused by?

A
  • fragmentation
  • occurs during electron impact ionisation only
18
Q

what can be calculated from mass spectrum graphs?

A
  • the relative atomic mass (R.A.M or Ar)
  • due to the relative abundance and mass being provided
19
Q

what are the isotopes of chlorine and their abundances?

A
  • Cl 35 = 75% abundance
  • Cl 37 = 35% abundance
20
Q

why are there peaks on the lower m/z end of the chlorine mass spectra?

A
  • due to fragmentation of Cl2 molecules during impact ionisation
21
Q

state the electron configuration of Chromium, Cr

A

[Ar] 4s1 3d5

22
Q

explain why the electron configuration for Copper is NOT [Ar] 4s2 3d9, but instead [Ar] 4s1 3d10

A
  • it reduces repulsion
  • as pairs of electrons in the 4s orbit cause more repulsion than pairs of electrons in 3d orbits
23
Q

explain why the electron configuration for Chromium is NOT [Ar] 4s2 3d4, but instead [Ar] 4s1 3d5

A
  • reduces repulsion
  • as pairs of electrons in the 4s orbit cause a high level of repulsion, so it is easier to move an electron from the 4s orbit to an empty 3d orbit
24
Q

Sc’s electronic configuration is [Ar] 4s2 3d1.
state the electronic configuration for its ion, Sc 2+

A

[Ar] 4s0 3d1

25
Q

when writing electronic configurations of ions, where are electrons removed from first and why?

A
  • the 4s orbit
  • as once filled with electrons, the 4s orbit becomes higher in energy than the 3d orbit
26
Q

state the 2 elements that have anomalous electronic configurations

A
  • Chromium, Cr
  • Copper, Cu
27
Q

define ionisation energy

A

the minimum energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of gaseous +1 ions

28
Q

state 3 factors that determine ionisation energy

A
  1. atomic radius
  2. nuclear charge
  3. shielding
29
Q

state the effect of atomic radius on ionisation energy

A
  • distance between outermost e- and the nucleus
  • larger atomic radius, lower force of attraction
  • e- is easier to remove
  • LOWERS IE
30
Q

state the effect of nuclear charge on ionisation energy

A
  • no. of protons in the nucleus
  • larger nuclear charge, greater attraction between e- and + nucleus
  • INCREASES IE
31
Q

state the effect that shielding has on ionisation energy

A
  • when outershell e-‘s are repelled by inner shell e-‘s
  • reduces attraction between outer e- and +nucleus
  • LOWERS IE
32
Q

why do successive IE’s increase?

A
  • due to overall decrease of repulsion
  • e-‘s are more attracted to + nucleus
33
Q

how do successive IE graphs provide evidence of electron shells?

A
  • sudden large increases indicate a change in energy level
  • e-‘s closer to nucleus require more energy to remove, due to greater attraction
34
Q

describe the trend in IE across period 3

A
  • generally increases
  • dips from Mg to Al
  • dips from P to S
35
Q

why does IE generally increase across a period?

A
  • increase in atomic number
  • therefore INCREASES NUCLEAR CHARGE
  • no. of electrons shells stay the same, so no effect on shielding
36
Q

why does IE decrease from Mg to Al

A
  • Mg has outermost e- in 3s sub shell
  • Al has outermost e- in 3p sub shell
  • electrons in 3p sub shell have more energy, so easier to remove
  • LOWER IE FOR AL
37
Q

why does IE decrease from P to S?

A
  • P has no pairs of electrons in an orbit
  • S has 1 pair of electrons in its 3p orbit
  • e-‘s in same orbit repel each other, so require less energy to remove
  • S HAS LOWER IE THAN P
38
Q

explain the trend in IE going down group 2

A
  • IE DECREASES
  • shielding increases, due to more electron shells
  • atomic radius increases, greater distance between + nucleus and e-

PHYSICAL CHEMISTRY 3.1 (4 decks)

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