Transition Metals Flashcards

(46 cards)

1
Q

Transition Metals

A

D block elements that can form one or more stable ions with incompletely filled d orbitals

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2
Q

When ions are formed

A

S electrons are removed first

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3
Q

Sc and Zn aren’t transition metals

A

Sc- Only forms one ions- Sc3+ Which has an empty d sub shells (Ar)
Zn- Only forms one ion, Zn2+ which has a full d sub shell (Ar) 3d104s2

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4
Q

Transition Metlas

A

Have variable oxidation states
To form a compound or complex containing an ion with a certain oxidation number, the energy given out when the ion forms a compound or a complex needs to be greater than the energy taken to remove the outer electrons and form the ion (ionisation energy)

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5
Q

Transition metals- Why variable ox states

A

Form by losing electrons from both their 4s and 3D au shells- Similar energy levels so it takes a similar amount of energy to remove an electron from the 4s au shell

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6
Q

The energy released when ions from a complex or compound increase

A

With ionic charge- Therefore the increase in the energy required to remove outer electrons to form transition metal ions with higher oxidation numbers is usually counteracted by the increase in energy released

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7
Q

Complex ions

A

Metal ion surrounded by dative covalent bonded Kiyan’s

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8
Q

Ligand

A

Atom. Ion or molecule that donates a pair of electrons to a central metal atom or ion

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9
Q

Bidentate

A

1,2-diaminoethane

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10
Q

Multidentate

A

Ligands with more than 2 lone pairs- EDTA4- has six lone pairs

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11
Q

Coordination no

A

No. Dative covalent bonds formed with the central metal ion

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12
Q

Complexes with four fold coordination

A

Tetrahedral- [CuCl4]2-
109.5 degrees

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13
Q

Square planar

A

Cis platin
90 degree angles
Pt
Ni

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14
Q

Complex ions and colour - y?

A

Usually 3D orbitals have same energy
Ligand binds to ions
3D orbitals split into 2 different energy levels
Electrons tend to occupy lower
To jump, need energy equal to energy gap- Visible light
Amount of energy needed to make electrons jump depends on central metal ion and oxidation number of ligands

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15
Q
A
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16
Q

Colour

A

Rest of frequencies of light are transmitted
These transmitted combine to make complement go colours absorbed

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17
Q

MnO4-

A

Purple

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18
Q

Cr2O72-

A

Orange

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19
Q

Cr3+

20
Q

YBGV

21
Q

Fe3+

22
Q

Fe2+

23
Q

CO2+

24
Q

Cu2+

25
Chromium Ions
6+- Cr2O7^2- Orange
26
Chromium (3) ions- [Cr(H2O)6]3+
React with alkaline or acid- Both get chromium hydroxide precipitate [Cr(OH)3(H2O)3] - Grey green precipitate
27
Grey green precipitate - [Cr(OH)3(H2O)3
Amphoteric- React with acid or with base
28
Add excess sodium hydroxide to a chromium hydroxide precipitate,
H2O ligands deprotonate and solution containing [Cr(OH)6]3- forms- Dark green solution
29
Add acid to chromium hydroxide precipitate,
OH- ligands protonate and a solution containing [Cr(H2O)6]3+ forms
30
Add excess ammonia to chromium hydroxide precipitate- Ligand reaction occurs.
Grey green precipitate + 6NH3-> [Cr(NH3)6]3+- Purple solution + 3OH- + 3H2O
31
Catalysts
Good because they can change oxidation number by gaining or losing electrons within their d orbitals - They can transfer electrons to speed up reactions
32
Contact Process
SO2 is oxidised to SO3 V2O5 + SO2-> V2O4 + SO3 V2O4 + 1/2O2-> V2O5
33
Autocatalysis
34
Homogeneous Catalysts
35
Positive Entropy- Makes more stable compound
Ligand échange réaction Dative bonds are broken and formed Strength of the bonds broken is often very similar to the strength of the new bonds being made- Enthalpy change for this reaction is very small
36
Reactions of ligands
37
Ligands exchanging
38
Monodentate ligands
Substituer with bidentate or muktidentate- no particles in solution increases- greater entropy- more likely to occur
39
Hexadentate ligand
Replaces monodentate or bidentate ligands, complex formed is more stable
40
CrO4^2-
Yellow 6+
41
Cr^3+
3+ Green (Violet)
42
Cr^2+
2+ Blue
43
Chromium 6+ reduction
Reduced using a reducing agent such as zinc and dilute acid Cr2O7^2- + 14H+ + 3Zn-> 3Zn^2+ + 7H2O + 2 Cr^3+
44
Cr^3+ Reduction
2Cr^3+ + Zn -> Zn^2+ + 2Cr^2+
45
Oxidation of Cr3+ to Cr6+
2Cr^3+ + 10 OH^- + 3 H2O2-> 2CrO4^2- + 8 H2O Alkaline solution Hydrogen Peroxide- Oxidation
46
Adding acid to Yellow solution- CrO4^2-
2CrO4^2- + 2H+ ->Reversible reaction -> Cr2O7^2- + H2O Gives orange solution