Need To Know Flashcards

1
Q

electronegativity

A

the ability of an atom to attract the pair of electron in a covalent bond

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2
Q

what factors affect electronegativity

A

CRAM

charge raidus attraction and shielding ie

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3
Q

Cl2 AND WATER HYDROCHLORIC ACID and chloric acid

A

Cl2 + H2O –> 2H+ + CL- + HClO + ClO-/HCl

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4
Q

relative atomic mass

A

Mean mass of one atom /

1/12 mass of one atom of C12

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5
Q

whats makes a tranistion metal and what it defing feutures?

A
incomplete d subshell
forms colour ions
variable oxidation state
catlystic abilities
complex formations
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6
Q

whats a ligand (bidentate lingand)

A

(2) atom dontates a (2) lone pair and forms (2) co-od bond

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7
Q

why is reaction feable for transition metals changing chelate effect/.

A

more moles in products = increasing entropy
same no. bond broken and formed enthalpy change small
there gibs freee energy is negative as small number ets

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8
Q

4 shapes of transition metals

A

any comon octohedral
cl- ligand tetrahedral
pt as metal square [lanar
ag metal linear

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9
Q

trans isomerism in cisplatin and an octahdral

A

basic knowldege hoe

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10
Q

optical isomers tm

A

octerhedral with 3x bidentate ligand

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11
Q

what causes colour change in anion metal thing

A

change in oxidation state
co oc no.
ligand
change in shape

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12
Q

equations which link the colour, wavelength and
frequency of the light absorbed with the energy
differ beteeen split orbitals

A
ΔE = hv or ΔE = hc/λ
v = frequency of light absorbed (unit s-1 or Hz)
H = Planck’s constant 6.63 × 10–34 (J s)
E = energy difference between split orbitals (J)
c = speed of light 3.00 x 108 (m s–1)
λ = wavelength of light absorbed (m)
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13
Q

VO2 + Oxidation state +5
VO 2+ Oxidation state + 4
V3+ Oxidation state + 3
V2+ Oxidation state + 2

A

Yellow
blue
green
violet

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14
Q

what do u need to know about EDTA4 as a ligand

A

EDTA4- can form six coordinate bonds; two coordinate bonds form from the N atoms, and four coordinate bonds form the O- atoms.

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15
Q

What do you need to know about haemoglobin

A

O2 forms a co-ordinate bond to Fe2+ in haemoglobin enabling oxygen to be transported in the blood.

CO is toxic because CO bonds more strongly to the Fe2+ in haemoglobin.
This prevents O2 from bonding to the Fe2+, causing suffocation.

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16
Q

big e biv abs and small e roy abs

hint- roygibiv

A

Big E BIV abs
If a transition metal compound has LARGE ∆E between d sub-shells:
- High energy light (i.e. Violet, Indigo and Blue)
- will be absorbed to excite the electrons
- Red, Orange, Yellow light will be reflected
- This means the compound will look Red/Orange

Small E ROY abs
If a transition metal compound has SMALL ∆E between d sub-shells:

  • Low energy light (i.e. Red, Orange, Yellow)
  • will be absorbed to excite the electrons
  • Blue, Indigo, Violet) will be reflected by the compound
  • This means the compound will look Blue/Purple
17
Q
Colour of
[Fe(H2O)6]2+
[Fe(H2O)6]3+
[Cr(H2O)6]2+
[Cr(H2O)6]3+
[Co(NH3)6]2+
[Co(NH3)6]3+
A

[Fe(H2O)6]2+ Green [Fe(H2O)6]3+ Pale Brown
[Cr(H2O)6]2+ Blue
[Cr(H2O)6]3+ Red-Violet
[Co(NH3)6]2+ Brown [Co(NH3)6]3+ Yellow

18
Q

How can colorimetry be used to determine unkown conc of a tm

A

(1) An appropriate ligand (such as -SCN) is added to the solution in order to intensify the colour.
(2) A range of solutions of the same complex ion are made of known concentrations.
(3) One at a time these are tested in a colorimeter and the transmission (or absorbance) is measured.
(4) A graph is plotted of Conc vs Transmission (or absorbance) and a line of best fit drawn.
(5) The transmittance/ absorbance of the unknown solution is measured in a colorimeter, and its concentration is determined by reading off the calibration curve.

19
Q

WHAT IS CTALYST

HOW DOES IT WORK

A
  1. -A substance which increases the rate of a chemical reaction AND Without being used up
  2. Provide an alternative reaction pathway AND With a lower activation energ
20
Q

Heterogenous catalyst vs homogeneous catalysts

A

Heterogeneous – where the catalyst is in a different phase to the reactants
Homogeneous – where the catalyst is in the same phase as the reactants

21
Q

how does Heterogeneous Catalysis work?

why more effiecient?

A

(1) Reactants adsorb onto the surface of the catalyst on an active site.
(2) Reaction occurs on the surface of the catalyst
(3) Products desorb from the surface if the catalyst

  • Increase the surface area
  • Spread the catalyst over an inert support medium.
22
Q

catalyst poising?

A

poisoning can occur:

  • Impurities can block the active sites.
  • This prevents the reactants from adsorbing
  • Purifying the reactants is the best way to prevent poisoning.
23
Q

heterogenous catalyst process you need to know?

A

(1) Making Ammonia in the Haber Process Catalysed by solid IRON

N2(g) + 3N2(g) 2NH3(g)

(2)Making Sulfuric Acid in the Contact Process. Catalysed by solid Vanadium (V) Oxide – V2O5
Step 1 – Sulfur Dioxide is oxidised to Sulfur Trioxide

SO2(g) + V2O5(s) SO3(g) + V2O4(s)

Step 2 – The Vanadium (IV) Oxide is then converted back to Vanadium (V) Oxide with oxygen.

2V2O4(s) + O2(g) 2V2O5(s)

This means the overall reaction is:
2SO2(g) + O2(g) -> 2SO3(g)
Sulfuric Acid is then formed by reacting SO3 with H2O.

(3)Manufacture of Methanol
Firstly this reaction forms a mixture known as synthesis gas
CH4(g) + H2O(g) -> CO(g) + 3H2(g)

Then the following reaction is catalysed by solid Chromium (III) Oxide – Cr2O3
CO(g) + 2H2(g) -> CH3OH(g)

24
Q

Homogeneous catalyst example you MUST know?

S2O82-(aq) + 2I-(aq) -> 2SO42-(aq) + I2(aq)

A

The uncatalyzed reaction has a high Ea as the two negative ions repel each other:
S2O82-(aq) + 2I-(aq) -> 2SO42-(aq) + I2(aq)

This reaction is catalysed by Fe2+ ions in a two-step process.
Step 1: S2O82-(aq) + 2Fe2+(aq) -> 2SO42-(aq) + 2Fe3+(aq)
Step 2: 2I-(aq) + 2Fe3+(aq) –> I2(aq) + 2Fe2+(aq)

Both steps now involve reactions between positive and negative ions
In the second step the Fe2+ catalyst is regenerated.

Fe3+ could also be used as a catalyst, as step 2 could occur before

25
Q

How does Fe 2+ act as a catalyst in S2O82-(aq) + 2I-(aq) -> 2SO42-(aq) + I2(aq)

A

The uncatalyzed reaction has a high Ea as the two negative ions repel each other:
S2O82-(aq) + 2I-(aq) -> 2SO42-(aq) + I2(aq)

This reaction is catalysed by Fe2+ ions in a two-step process.
Step 1: S2O82-(aq) + 2Fe2+(aq) -> 2SO42-(aq) + 2Fe3+(aq)
Step 2: 2I-(aq) + 2Fe3+(aq) –> I2(aq) + 2Fe2+(aq)

Both steps now involve reactions between positive and negative ions
In the second step the Fe2+ catalyst is regenerated.

Fe3+ could also be used as a catalyst, as step 2 could occur before

26
Q

Balance and coulour change of?

Fe2+ -> Fe3+
MnO4- -> Mn2+
Cr2O72- _> Cr3+

A

Fe2+ -> Fe3+ +E-
Green brown

5E- + 8H+ + MnO4- -> Mn2+ + 4H2O
Purple Colourless

Cr2O72- _> 2Cr3+
Orange green

27
Q

AUTOCATALYSISI reaction to learn

A

The uncatalyzed reaction has a high Ea as the two negative ions repel each other:

2MnO4-(aq) + 16H+(aq) + 5C2O42-(aq)  2Mn2+(aq) + 8H2O(l) + 10CO2(g)

The Mn2+ ions produced can catalyse the reaction in a two step process as shown below. Initially the rate is slow, but as more catalyst is produced the rate increases.

4Mn2+(aq) + MnO4-(aq) + 8H+(aq)  5Mn3+(aq) + 4H2O(l)

In the first step the Mn2+ is oxidised to MnO4-.

2Mn3+(aq) + C2O42-(aq)  2CO2(g) + 2Mn2+(aq)

28
Q

what happens when Vanadate(V) is reduced by Zinc in acidic conditions.

A

Zn + 4H+ + 2VO2 + -> Zn2+ + 2VO 2+ + 2H2O
yellow to blue

Zn + 4H+ + VO2+ -> Zn2+ + V3+ + 2H2O
yellow to green

3Zn + 8H+ + 2VO2+  3Zn2+ + 2V2+ + 4H2O
yellow to purple

29
Q

What does silver do in the tollens reagent to test for aldehydes?

A

2[Ag(NH3)2]+ + RCHO + 3-OH  2Ag + 4NH3 + RCOO- + 2H2O

silver in [Ag(NH3)2]+ to reduce from Ag(I) to Ag in its zero oxidation state. This causes a layer of solid silver = silver mirror test

30
Q

the trend of electronegativity in 7

A

down group decrease as more shielding so nuclear attraction so less able to draw electrons in covalent bond.
Flourine is most electronegative

31
Q

Making Sulfuric Acid in the Contact Process. hint: Catalysed by solid Vanadium (V) Oxide – V2O5
Step 1 – Sulfur Dioxide is oxidised to Sulfur Trioxide

SO2(g) + V2O5(s) SO3(g) + V2O4(s)

Step 2 – The Vanadium (IV) Oxide is then converted back to Vanadium (V) Oxide with oxygen.

2V2O4(s) + O2(g) 2V2O5(s)

This means the overall reaction is:
2SO2(g) + O2(g) -> 2SO3(g)
Sulfuric Acid is then formed by reacting SO3 with H2O.
what catlst exmpl?

A

Step 1 – Sulfur Dioxide is oxidised to Sulfur Trioxide

SO2(g) + V2O5(s) SO3(g) + V2O4(s)

Step 2 – The Vanadium (IV) Oxide is then converted back to Vanadium (V) Oxide with oxygen.

2V2O4(s) + O2(g) 2V2O5(s)

This means the overall reaction is:
2SO2(g) + O2(g) -> 2SO3(g)
Sulfuric Acid is then formed by reacting SO3 with H2O.

heterozydous

32
Q

Chlorine with cold water

A
33
Q

tst for ammoniumion

A

+naoh warmed = damp red litmus ppr blue