Transition Metals

Transition metals are d-block elements that have their d-orbitals partially filled. Zinc and Scandium are not considered transition metals, even though they're d-block elements. They do not have partially filled d-orbitals. Zn has the electronic configuration [Ar]4s²3d¹⁰. Sc has the electronic configuration [Ar]4s²3d¹.

Properties of transition metals:
1. They have high melting and boiling temperatures.
In their metallic structure, 3d as well as 4s electrons are available for delocalisation. Since there are more electrons within the sea of electrons, there is a greater electrostatic attraction - hence the higher melting temperatures.
2. They have variable oxidation states - except Scandium (Sc) and Zinc (Zn)
3. They form coloured compounds and ions - except Sc and Zn form white compounds and colorless ions
4. They show catalytic activity

Sn can only form Sn ³⁺ ion which has the same electronic configuration as [Ar].
Zn loses both of its 4s electrons to form only Zn²⁺ ions (with the electronic configuration of
[Ar]3d¹⁰).

Electronic Configurations:

In d block, electrons are added to an inner d-oribital and this shields the outer 4s electrons from the increased nuclear charge (i.e. they're easier to lose). Therefore, the atomic radius only decreases slightly and electronegativity and ionisation energies increase only slightly.

Examples of some electronic configurations:
Sc ----> [Ar]4s²3d¹
Zn ----> [Ar]4s²3d¹⁰
Cu ----> [Ar]4s¹3d¹⁰
Cr ----> [Ar]4s¹3d⁵

In Cr and Cu, the 3d orbitals fill first before the 4s orbitals. This is due to the increased stability offered by full and half-filled shells.
Moreover,
Fe²⁺ (3d⁶) is readily oxidised to Fe³⁺ (3d⁵)
Mn²⁺ (3d⁵) is not readily oxidised to Mn³⁺ (3d⁴)

When transition metals form ions, they lose electrons from the 4s sub-shell first before 3d sub-shell. Fe²⁺ has the eletronic structure [Ar]3d⁶  rather than [Ar]4s²3d⁴.
This occurs because of the repulsion forces on the 4s electrons, repelling them further away from the nucleus, by the 3d electrons. Therefore, 4s electrons are pushed to a higher energy level, higher than 3d, thus electrons in 4s sub-shell are lost before 3d sub-shell.

What are complex ions and ligands?

Complex ion has a metal ion at its centre with a number of other molecules or ions surrounding it that are attached to the central ion by dative covalent (coordinate) bonds.

Ligands are the molecules or ions surrounding the central ion, for example: water, ammonia and chloride ions. These ligands have active lone pairs of electrons in their outer energy level which are used in the dative covalent bond with the metal ions.

All ligands are lone pair donors. Therefore, all ligands function as lewis bases.

Dative covalent bonds are:
covalent bonds in which both electrons come from the same atom, unlike in a simple covalent bond where each atom supplies one electron. Remember: in covalent bonds two atoms share a pair of electrons. Both atoms are held together due to the forces of attraction between the positive nuclei and the shared pair of electrons.

Transition Metals - Formation of complex ions

Today, we will discuss the fomation of complex ions, specifically aqua ions.

Aqua ions occur when d-block cations dissolve in water and become hydrated (i.e. surrounded by water molecules). The oxygen atom in the water molecule has a lone pair of electrons that forms a bond with an empty 3d or 4p orbital in the metal ion.

The theory behind the exact nature of the bonding is beyond the scope of A-level. All you need to know is that a dative covalent bond forms with the oxygen atom as the donor atom.
There are other theories that tend to explain the bonding such as: the electrostatic forces of attraction between the delta negative oxygen and the positive metal ion.

An example of an aqua ion is the hydrated chromium (III) ion, [Cr(H₂O)₆]³⁺
The water molecules are called ligands

One of the lone pairs of electrons on the oxygen atom of each water molecule forms a dative covalent bond with an empty orbital in the Cr³⁺ ion. Six dative bonds form so the hydrated ion has the coordination number 6.

The ion, with its water molecules bonded to the central metal ion, is called a complex ion.

The coordination number is the number of near neighbouring atoms that are bonded to the central ion.

Ligands are organic molecules that donate the necessary electrons to form dative (coordinate) covalent bonds with metallic ions. (e.g. H₂O, NH₃)

Note: When drawing the hydrated ions make sure that the dative covalent bonds all start from the O of the H₂O and not from the H.

There are six dative bonds, each containing a pair of electrons. These six pairs of bonding electrons repel each other to the position of maximum separation and minimum repulsion. The shape is therefore a octahedral.

All complex ions with coordination number 6 are octahedral.

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Copper:

In the solid state, hydrated copper (II) ions have four water molecules arranged in a plane around Cu²⁺ion. However, in aqueous solution, two more water molecules are weakly bonded at right angles, forming an octahedron. The two non planar water molecules are further from the copper ion than are the four planar water molecules. Because of this, the formula of the hydrated ion in solution is sometimes written as [Cr(H₂O)₄]²⁺ instead of [Cu(H₂O)₆]²⁺

4.7 Acid/base equilibria - Introduction

Ideas about the nature of acids and bases have developed over the centuries. Simple ideas about acids were:

·         Corrosive liquids

·         Have a sour taste

·         Give characteristic colors with indicators

Acids also were known to react with:

·         Bases (and alkali) to give a salt and water – neutralization reaction

·         Carbonates to give salt, carbon dioxide (observed as effervescence) and water

·         Some metals to  give a salt and hydrogen

Ideas have developed about acids and acidity:

 Arrhenius theory

The theory of electrolytic dissociation: it states that when acids, bases or salts dissolve in water they split up partially or completely into their ions.

·         Arrhenius acids give hydrogen ions in aqueous solution

HA _(aq) ⇌   H⁺ _{(aq) +} A^- _(aq)

·         Arrhenius bases give hydroxide ions in aqueous solution

BA _(aq) + H₂O _(aq) ⇌   BH⁺ _{(aq) +} OH^- _(aq)

Limitations of the theory: some acid-base reaction took place in solvents other than water or did not even need water to react. The theory restricts reactions to aqueous solutions only.

 BrØnsted-Lowry theory

The theory proposed that:

·      An acid is a proton (H⁺) donor

·         A base is a proton acceptor

Conjugate acid-base pairs

·         Acid-base Equilibria involve the transfer of protons

·         Each acid has a conjugate base

·         Each base has a conjugate acid

   HA _(aq) + H₂O _(l) ⇌   H₃O⁺ _{(aq) +} A^- _(aq)

The acid HA is transformed into A^- after donating a proton to the base, H₂O.  In the reverse reaction, the A^- ion acts as a base by accepting a proton to form the acid again. Therefore, HA and A^-   are referred to as conjugate acid-base pairs. Similarly, H₂O and H₃O⁺ are conjugate acid-base pairs.

Revision Check - Entropy: how far?

How far? – entropy

Revision check:

You need to be able to:

a) Demonstrate an understanding that, since endothermic reactions can occur spontaneously at room temperature, enthalpy changes alone do not control whether reactions occur

b) Demonstrate an understanding of entropy in terms of the random dispersal of molecules and of energy quanta between molecules

c) Demonstrate an understanding that the entropy of a substance increases with temperature, that entropy increases as solid → liquid → gas and that perfect crystals at zero Kelvin (absolute zero) have zero entropy

d) Demonstrate an understanding that the standard entropy of a substance depends mainly on its physical state but also on its complexity

e) Demonstrate an understanding that reactions occur due to chance collisions, and that one possible ordered arrangement, e.g. in a crystalline solid, can be rearranged into many possible disordered arrangements, e.g. in a solution, so the probability of disorder is greater than order

f) Interpret the natural direction of change as being in the direction of increasing total entropy (positive entropy change), e.g. gases spread spontaneously through a room

g) Carry out experiments and relate the results to disorder and enthalpy changes including:

i.             Dissolving a solid, e.g. adding ammonium nitrate crystals to water

ii.            Gas evolution, e.g. reacting ethanoic acid with ammonium carbonate

iii.           Exothermic reaction producing a solid, e.g. burning magnesium ribbon in air

iv.           Endothermic reaction of two solids, e.g. mixing solid barium hydroxide, Ba(OH)2.8H2O with solid ammonium chloride

h) Demonstrate an understanding that the entropy change in any reaction is made up of the entropy change in the system added to the entropy change in the surroundings, summarized by the expression: ΔStotal = ΔSsystem + ΔSsurroundings

i) Calculate the entropy change in the system for a reaction, ΔSsystem, given entropy data

j) Use the expression ΔSsurroundings = -ΔH/T to calculate the entropy change in the surroundings and hence ΔStotal

k) Demonstrate an understanding that the feasibility of a reaction depends on the balance between ΔSsystem and ΔSsurroundings, and that at higher temperatures the magnitude of ΔSsurroundings decreases and its contribution to ΔStotal is less. Reactions can occur as long as ΔStotal is positive even if one of the other entropy changes is negative

l) Demonstrate an understanding of and distinguish between the concepts of thermodynamic stability and kinetic inertness

m) Understand that endothermic reactions can occur spontaneously at room temperature

n) Define the term enthalpy of hydration of an ion and use it and lattice energy to calculate the enthalpy of solution of an ionic compound

o) Demonstrate an understanding of the factors that affect the values of enthalpy of hydration and the lattice energy of an ionic compound

p) Use entropy and enthalpy of solution values to predict the solubility of ionic compounds.

Melting & Boling temperature - entropy: how far?

At 0C°, the freezing point of water, equilibrium between ice and water is made. At that temperature, neither the ice melts nor the water freezes, and neither direction is thermodynamically feasible. Also, there’s an equilibrium made when water reaches its boiling temperature.

At these melting and boiling points, where equilibrium is made, the value of ΔS° _(total) is zero.

ΔS° _(system) + (-ΔH /T) = 0

To calculate the temperature at which the reaction is at equilibrium:

T = ΔH / ΔS° _(system)

The expression above explains why boiling and melting temperatures depend on the strength of the forces between the particles. As, the melting & boiling temperatures depend on the amount of energy required for the change of state.

Strong force = large amount of energy needed to separate the particles = high melting or boiling temperatures.

Note: The reactants are thermodynamically unstable relative to the products if ΔS° _(total) for the change is positive. That means that the reaction is thermodynamically feasible.

AS Practical Chemistry

Test for halides ions. Acidify with dil. nitric acid and add silver nitrate The following silver halides form
F-	Colourless	No ppt to dissolve
Cl-	White ppt produced (AgCl)	Re-dissolves in dil NH3
Br-	Cream ppt produced (AgBr)	Re-dissolves in conc NH3
I-	Yellow ppt produced (AgI)	Does not dissolve in NH3

Test for unsaturated carbon-carbon bonds C=C Add bromine water to organic compounds and shake

Result: brown color disappear and solution becomes colorless

Additionally When burned on a watch glass they produce a sooty flame (incomplete combustion)

Test for sulphite anion SO32-

Add acid and test the gas evolved is SO2

Test for sulfur dioxide

Add potassium chromate paper goes from orange to green

Test for carbonate and hydrogen carbonate (CO32- and HCO3-)

Add universal indicator and if the solution is blue you have carbonate (pH 12), if it is dark green you have hydrogen carbonate (pH 9)

Additionally you can add Phenolphthalein

This goes dark pink with carbonate and light pink with hydrogen carbonate

Add calcium chloride to CO32- or HCO3-

Carbonate will form a white ppt

Add MgSO4, then heat

Cloudy solution

Flame Tests

Salt	Colour
Lithium	Red
Sodium	Yellow
Barium	Apple green
Magnesium	Colorless
Strontium	Crimson (dark red)
Calcium	Brick red (orange)
Potassium	Lilac

Ammonium compounds

Unlike ionic solids they sublime at low temperature

Test for Halogenalkanes

Add sodium hydroxide and acidified silver nitrate. Refer to group 7 test for color of precipitate formed.

Test for alcohols –OH group

Add Phosphorous Pentachloride PCl₅, misty fumes are produced which are acidic (hydrogen chloride HCl_(g))

Test for gases

Gas	Test and result
ammonia	Turns damp red litmus paper blue (alkaline gas)
Hydrogen chloride gas (same for hydrogen bromide/ iodide)	Misty fumes turns blue litmus red
Carbon dioxide	Turns limewater milky
Chlorine (same for Br2 and I2)	Turns damp blue litmus paper red then bleaches
Hydrogen	‘pops’ with a lighted splint
Oxygen	Relights a glowing splint
Nitrogen dioxide	Brown gas turns blue litmus red (acidic gas test in fume cupboard)
Sulfur dioxide	Yellow gas turns blue litmus red (acidic gas tested in fume cupboard) Turns potassium chromate paper from orange to green)

Test for water

Test	Result
Cobalt chloride paper	Blue paper turns pink with water
Copper sulfate (anhydrous)	White compound goes blue with water

Test for cations

Cation	Effect of aqueous NaOH	Effect of aqueous NH3
Aluminum (Al3+)	White ppt., soluble in excess giving colorless solution	White ppt., insoluble in excess
Ammonium (NH4+)	Ammonia produced on warming
Calcium (Ca2+)	Slight white ppt., insoluble in excess	No ppt or very slight white ppt.
Copper (Cu2+)	Light blue ppt., insoluble in excess	Light blue ppt., soluble in excess giving a dark blue solution
Iron (II) (Fe2+)	Dark green ppt., insoluble in excess	Dark green ppt., insoluble in excess
Iron (III) (Fe3+)	Brown/orange ppt., insoluble in excess	Brown/orange ppt., insoluble in excess
Zinc (Zn2+)	White ppt., soluble in excess giving colorless solution	White ppt., soluble in excess giving colorless solution

Test for anions

Anion	Test	Test result
Carbonate (CO32-)	Add dil. acid	Effervescence, CO2 produced
Iodide in solution (I-)	Acidify with dil. Nitric acid and add lead (II) nitrate	Bright yellow ppt produced PbI2)
Nitrate in solution (NO3-)	Add sodium hydroxide and aluminium foil and heat gently	Ammonia gas produced
Sulfate in solution (SO42-)	Acidify with dil nitric acid and add barium nitrate	White precipitate, BaSO4

Indicator tests – Titration information

Indicators: Phenolphthalein

Goes from colorless to pink (acid to alkali)

Methylorange

Goes from yellow to orange

Concordance: two titres need to be within 0.20 cm3 of each other, take an average and then multiply by the concentration to get the moles

Sources of error: misreading meniscus, parallax error, bubbles in the burette, not rinsing with reactant before start, leaving the funnel at the top

Points to consider: ratio of reactants in the equation e.g. 2:1 for NaOH + H2SO4

Organic compounds test

Organic molecules	Test
Alkenes	·        Shake with bromine water, goes from orange/brown to colorless Alkanes don’t react and the solution remains orange/brown ·        Also alkaline KMnO4 can be added, it goes from purple to colorless and a diol is made ·        Also when burned alkenes make a sooty flame
Alcohols	·        Acidified K2Cr2O7 is added, heated under reflux ·        Primary go from orange to green and aldehyde forms then carboxylic acid is made ·        Secondary go from orange to green and ketones are made ·        Tertiary do not oxidize and it stays orange Additionally you can add PCl5 and misty fumes of HCl are produced
Aldehyde	Addition of Benedict’s solution causes a red ppt of Cu2O to be produced (Ketones do not react)
Organic acids	Add a carbonate, effervescence of CO2 is observed
Halogenalkanes	Add sodium hydroxide and heat, then acidified silver nitrate (see halides tests for results)