M5-Transition Elements And Quantitative Analysis Flashcards by Leo Cheveau

d block element

Highest energy electron in a d orbital

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Writing electron configurations for copper and chromium

Exceptions to normal as when the 4s and 3d subshell can exist full or exactly half full, this is the preferred more stable arrangement of electrons when filling shells
So copper- 1s2 2s2 2p6 3s2 3p3 4s1 3d5
Chromium- 1s2 2s2 2p6 3s2 3p6 4s1 3d10

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Where are electrons lost from

Highest energy shell
Furthest from nucleus so less nuclear attraction

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When taking electrons from d black elements you must

Taking electrons from 4s than 3d

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Transition element

An element that forms at least one ion with a partially filled d-sub-shell

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Which d block elements ARE NOT transition metals

Sc- empty d subshell when sc3+
Zn- filled d subshell when Zn2+
Ag- filled subshell when Ag+

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Characteristics of transition elements

Have variable oxidation states
Form coloured compounds
They are good catalysts

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Iron oxidation states, colours and catalyst

Fe2+- pale green
Fe3+- yellow
Catalyses the harder process for producing ammonia

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Manganese oxidation states, colours and catalyst

Mn2+- pale pink
Mn7+ ( in MnO4-)- purple
MnO2 Catalyses the decomposition of hydrogen peroxide as
2H2O2 —> 2H2O

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How to identify aqueous transition metals

Add NaOH and coloured precipitate forms

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How to identify alkaline gasses

Ammonia will turn damp red litmus paper blue
As NH3(g) + H2O —> NH4+(aq) + OH-

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Oxidation of Fe2+ —> Fe3+

Redox titration
Add MnO4-
Colour change from purple to pale pink, it’s said MnO4- is decolourised by Fe2+

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Reduction of Fe3+ —> Fe2+

React with iodide ions
Colour change from yellow to green/ brown due to iodine and Fe2+
Pale green colour difficult to see as iodine is produced is brown do view of Fe2+ is obscured

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The reduction of Cr2O72- —> Cr3+

Used to oxidise alcohols
Colour change from orange to green
Can be reduced by reacting with zinc

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Oxidation of Cr3+ —> CrO42-

HOT ALKALINE HYDROGEN PEROXIDE
Used to oxidise chromium 3+ ions to CrO4-
Half equation for OH- - H2O2 + 2e- —> 2OH-
Cr3+ + 8OH- —> CrO42- + 3e- + 4H2O
Redox equation- 3H2O + 2Cr3+ + 10OH- —> 2CrO42- + 8H2O
Colour change from green due to Cr3+ —> yellow due to CrO42- ion

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Reduction of Cu2+ —> Cu+

Add iodide ions
Iodide ions converted to iodine
Cu2+ converted into Cu+
Also forms a white precipitate due to CuI
The precipitate is hard to see as CuI is also formed as a white precipitate obscuring the view of the brown I2

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Disproportionation of Cu+ ions

Cu+ with hot dilute H2SO4
Forms a blue solution due to Cu2+ ions and brown precipitate due to Cu(s)

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Complex ion

Central metal ions surrounded by ligands

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Ligand

A molecule or ion that donates a lone pair of electrons to the central metal ion to form a coordinate bond (dative covalent bond)

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Coordination number

The number of dative covalent (coordinate) bonds to the central metal ion

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Monodentate ligand

Donates only one pair of electrons forming one coordinate bond

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Bidentate ligands

Donates two lone pairs of electrons forming two coordinate bonds

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Tow common bidentate ligands

Ethan-1,2-diamine ( en) charge= 0
Ethandioate charge= -2

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Shapes of complex ions

Coordiantion number 6- octahedral- bond angle 90°
Coordination number 4- tetrahedral- bond angle 109.5°
- square planar- bond angle 90°

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Stereoisomerism

Same structural formula, different arrangement of atoms in space

Two types of stereoisomerism

Cis-trans isomerism Optical isomerism

Cis-trans stereoisomerism

Cis- same group on same side Trans- same group on different side

What shape of transition elements can cis-trans isomerism be used with ?

Square planar

Complex PtCl2(NH3)2]

Cis-platin or trans-platin Used as a cancer chemotherapy drug as binds to the DNA in cancer cells and prevents cell division

Equalibira in ammonia solution

NH3(g) + H2O(l) <=> NH4+(aq) + OH-(aq)

Ligand substitution- copper with water complex with NH3

Step 1- adding DROP WISE ammonia Cu2+(aq) + 2OH-(aq) —> Cu(OH)2(s) Forms a blue precipitate Step 2- Reagent- Dilute ammonia solution in EXCESS Equation- [Cu(H2O)6]2+ + 4NH3 —> [Cu(H2O)2(NH3)4]2+ Observation- pale blue solution to a dark blue solution

Ligand substitution- copper with water and HCL

Reagent- CONCENTRATED HCl(aq) Equation- [Cu(H2O)6]2+ + 4Cl- —> [CuCl4]2- + 6H2O Observation- pale blue solution to a yellow solution

Ligand substitution- chromium with water and NH3

Step 1- adding DROP WISE ammonia Cr3+(aq) + 3OH-(aq) —> Cr(OH)3(s) From green solution to grey-green precipitate Step 2- Reagent- Dilute ammonia solution in EXCESS Equation- [Cr(H2O)6]3+ + 6NH3 —> [Cr(NH3)6]3+ + 6H2O Observation- green solution to a purple solution

Ligand substitution- chromium with water and NaOH

Step 1- adding DROP WISE NaOH Cr3+(aq) + 3OH-(aq) —> Cr(OH)3(s) From green solution to grey-green precipitate Step 2- Reagent- EXCESS NaOH Equation- [Cr(H2O)6]3+ + 6OH-—> [Cr(OH)6]3- + 6H2O Observation- green solution to a darker green solution

Haemoglobin and oxygen

Hb + 4O2 <=> HbO8 Forms oxyheamaglobin

Heamaglobin and CO

CO binds to Fe2+ but doesn’t unbind so heamaglobin now unfunctional

Explain how ligand substitutions allow haemoglobin to transport oxygen in the blood

Oxygen binds to Fe2+ using lone pair of electrons making a coordinate bond Then when needed, the oxygen is replaced by CO2 or water

AgCl colour and with NH3

White precipitate Soluble in dilute NH3

AgBr colour and NH3

Cream precipitate Soluble in concentrated NH3

AgI colour and with NH3

Yellow precipitate Insoluble in concentrated NH3

Cu2+ colour

Light blue

Fe2+ colour

Pale green

Mn2+ colour

Pale pink

Fe3+ colour

Yellow

Cr3+ colour

Green (violet when first made)

[Cu(H2O)6]2+

Pale blue From CuSO4 and 5H2O

Cu(OH)2

NH3 drop wise to Cu2+ Pale blue precipitate

[Cu(NH3)4(H2O)]2+

Excess NH3 to Cu2+ Dark blue

CuCl42-

Excess HCL to Cu2+ Yellow

Cu+

Reduction of Cu2+ with I- White precipitate And brown I2

Cu2+/Cu

Disproportionation of Cu+ Brown solid (Cu) and blue solution (CuSO4)

CoCl42-

From Cocl2(6H2O)(s) plus water and excess HCl Blue

[Cr(H2O)6]3+

Pale purple Heat —> green due to impurities in solution

Cr(OH)3

NH3 drop-wise to Cr3+ Dark Green precipitate

[Cr(H2O)6]3+

Excess NH3 to Cr2+

[Cr(OH)6]3-

Excess OH- to Cr(OH)3 Dark green

Cr2O72-

From potassium dichromate in 0.1mol-1DM3 of H2SO4 Orange

CrO4-

Oxidation of Cr3+ with hot alkaline H2O2 Yellow

[Mn(H2O)6]2+

From MnSO4 and 4H2O Pale pink

Mn(OH)2

NaOH drop wise to Mn2+ Pale brown precipitate

[Fe(H2O)6]2+

From FeSO4 and 7H2O Pale green

Fe(OH)

NaOH drop wise to Fe2+ Dark green precipitate

[Fe(H2O)6]3+

From FeCl3 and 6H2O Yellow

Fe(OH)3

NaOH drop wise to Fe3+ Rusty brown precipitate

MnO42- with Fe2+ —> Mn2+ with Fe3+

Purple to pale pink so dilute in practicals basically colourless

I- with Fe3+ —> I2 with Fe2+

Orange brown (Fe3+) to brown (I2)

M5-Transition Elements And Quantitative Analysis Flashcards

(66 cards)