Transition Metals Flashcards
Transition metal definition
forms one at least stable ion with a partially filled d sub shell
2 d block elements that are not transition metals
Zinc as Zn 2+ is the only ion for dc and its d sub shell is full
Scandium as sc3+ is the only ion formed and its d sub shell is empty
Chromium and copper electronic configuration
Chromium = more stable with a half d sub shell as 4s1 and 3d5
Copper = more stable with a full d subshell as 4s1 and 3d10
When writing electronic configuration
Key features of transition metals CCCV
C= act as a catalyst
C= form complex ions
C = coloured compounds
V = variable oxidation states ( different oxidation states)
Coordinate bond
shared pair of electrons that cone from the sane atom
Ligand
ion or a molecule with a lone pair that forms a coordinate bond with a transition metal
Bidentate ligand
ligand with 2 atoms with lone pairs that form 2 coordinate bond with a transition metal
Coordination number
number of coordinate bonds that a transition metal forms
Complex ion
central atom surrounded by ligand by coordinate binds
Monodente and examples
each ligand forms 1 coordinate bond
water, ammonia, cyanide , chlorine
Bidentate and multidentate and examples
forms 2 or more Coordinate binds
ethane-1,2-diamine and ethandioate ion
Skeletal formula of ethane-1,2-diamine and ethandioate ion
Displayed formula of ethane-1,2-diamine and ethandioate ion
NH2CH2CH2NH2
C2O42-
4 shapes transition metals form
Octahedral (cu(H2O)6)2+
Tetraderdral = only molecules with chlorine can form as they are large (cucl4)2-
Square planar= get rid of lone pair and platinum forms (Pt(NH3)2cl2)
Linear =only sliver forms (ag(nh3)2)+
What can chlorine only form
Tetrahedral as it’s a large molecule
EDTA4-
form 6 coordinate bonds 2 from n atoms and 4 from o atoms
Chelate effect
mono dente substituted by bi/multidentate. If moles increase from left to right then entropy increases eg which means reaction is feasible
Explain why this reaction is thermodynamically feasible ?
Delta h is negligible as make abd break same number if bonds
Delta S = increase as number of miles from reactants to products increases
Delta G =delta H-Tdelta s is negative as t delta s is negative than delta H
Heamoglobin structure
Fe2+ forms 4 coordinate bonds with n atoms in the haem group forming a square planar
One coordinate bond is formed between fe2+ and protein globin and another with oxygen
When oxygen has been transported, it breaks the quad not bond releasing the oxygen. Then carbon dioxide forms a coordinate bond to iron to be transported back into the lungs.
Effect of carbon monoxide on haemoglobin
Carbon monoxide forms are stronger permanent coordinate bond to iron which means no longer transport oxygen around the body
cis / trans
Trans = opposite / cis(sis=same)
Transition metals as coloured compounds
Transition metals are coloured because of the partially filled d subshell
They all have equal energy, which is known as the ground state
Presence of other atoms causes the d orbitals to have different energies
This allows electrons to be excited from one d orbital to another so they can move from a lower energy to a higher energy which is called the excited state. Energy to make this transition is taken from white light the colour of the light which is absorbed is missing from the light and the one that is missing reflects from the substance giving it its colour
2 equation to work out delta E which the the change from ground to excited state
Delta e = hf
H=plan is constant
F= frequency of light
Delta e = hxc/f
H = plancks constant = 6.624x10-34
C =speed of light
L=/ lambda wavelength of light
Red
low energy light
Violet
high energy light
Large difference between ground state and excited state
Large energy energy difference between ground stay and excited state for the subshell
High energy of spectrum is absorbed and excite the electron to excited state
Low energy is transmitted
Small difference between ground state and excited state
Small energy difference between ground stay and excited state
Low energy part of visible light spectrum is absorbed
High energy part is transmitted
change in the excited state between the D sub shall meaning it can change the colour of the compound you need to remember
LOCOS , ONE OF THESE CAN ALTER THE COLOUR
l= change in ligand
O= change in oxidation state
Co= change in coordination number
S= change in shape of complex
Calorimetery
Add an appropriate legend to intensify the colour
A range of colours of the same complex iron are made. of known concentration
These are tested in a colorimeter and the transmission is measured
A calibration curve is plotted for concentration v transmission and draw a line of best fit
The transmissions of a unknown solution is measured in a colorimeter and concentration is determined by reading of calibration curve
Key points in colorimetery
The length of the container affects the amount of absorbent so you would use the same path length, which is the distance that light travels through
A filter is used to allow one colour of light through the sample
Two types of catalyst
Heterogeneous= catalyst is in a different phase to the reactant
Homogeneous catalyst is in the same phase as the reactants
Heterogeneous
catalyst is in a different phase to the reactant
Homogeneous
catalyst is in the same phase as the reactants
Heterogeneous catalytic action on surface
Reactants adsorb ( stick to the surface )on the surface of the catalyst of an active site
Reaction occurs on the surface and weak bonds by stress and disorder
Products desorb ( unstick ) from the surface
To make catalyst as efficient as possible you can
Increase the surface area
Spread the catalyst over an inner support medium
They don’t last forever as poisoning can occur
So you
Impurities can block the active site
This prevents reactant from adsorbing
Purifying the reactant is the best way to prevent poisoning
Examples of heterozygous catalysts
Making ammonia in the haber process. Catalysed by iron N2 + 2H2 = 2NH3
Making sulphuric acid in the contact process. Catalysed by solid vanadium oxide.= V2O5. Step one.= SO2 + V2O5=SO3+V2O4. Step two.= 2V2O4 + O2=2V2O5. SO3 + H2O = H2SO4
Manufacture of methanol. Step one.= CH4 + H2O=CO + 3H2. Step two= catalysed by chromium oxide. Co +H2 =CH3OH
Making ammonia in the haber process.
Catalysed by iron N2 + 2H2 = 2NH3
contact process.
Catalysed by solid vanadium oxide.= V2O5. Step one.= SO2 + V2O5=SO3+V2O4. Step two.= 2V2O4 + O2=2V2O5. SO3 + H2O = H2SO4
Manufacture of methanol
Step one.= CH4 + H2O=CO + 3H2. Step two= catalysed by chromium oxide. Co +H2 =CH3OH
Homogenous example
between s2O82- and I- ions where fe2+ has to be used as a catalyst. Without a catalyst these negative ions would naturally repel each other and never react.
Autocatalysis
one of the products can act as a catalyst.
This means that overtime as the amount of product increases the rate of reaction also increases as it becomes catalysed
MNO4- AND C2O42- an acidic conditions where mn2+ ions produced would act as a catalyst
Overall equation
Mn2+ to mn3+ and combine with mnO4- to mn2+ ( purple colour )
Mn3+ to mn3+ and combine with c2O42- to CO2
Explain
Initially, the reaction is slow as there is high activation energy as no catalyst and both reactants having a negative charge so repel
As rate begins to increase mn2+ slowly being produced and catalyse the reaction
Rate decreases as all reactants are used up
Why can s block / period 1 not be used can a catalyst
same oxidation state
Fe 2+ goes to (green )
Fe 3+(brown)
MnO4- goes to (purple )
Mn2+( colourless )
Cr2O72- goes to ( orange )
Cr3+ green
C2O42- goes to
CO2
Zn goes to
Zn2+
S2O82- goes to
SO42-
I- goes to
I2
Mn3+ goes to
Mn2+
