15.1 transition metals & complex ions

Question 1

where in the periodic table are transition metals found?

Answer 1

in the d-block.

Question 2

what is a d-block element?

Answer 2

an element which has its highest energy electron in its d sublevel.

Question 3

give the definition of a transition metal.

Answer 3

transition metals are d-block elements that can form one or more stable ions with incompletely filled d-orbitals.

Question 4

explain an advantage of transition elements having similar atomic radii.

Answer 4

transition elements have similar atomic radii, which make it possible for the atom of one element to replace that of another element in the formation of an alloy.

Question 5

s-block elements are also metals. explain why d-block metals are generally denser than s-block metals, and why the densities generally increase across the first series of d-block metals.

Answer 5

• d-block metals are generally denser than s-block metals because most d-block metals have close-packed lattice structures whilst most s-block metals do not.
• the densities generally increase across the first series of d-block metals due to the decrease in atomic radius across the series.

Question 6

when transition metals form positive ions, which electrons are removed first?

Answer 6

outer s electrons.

Question 7

explain why the first ionisation energy of d-block metals are greater than those of s-block metals in the same row of the periodic table.

Answer 7

the d-block metals are smaller in size than the s-block metals in the same row of the periodic table, so have greater effective nuclear charges.

Question 8

explain why the first ionisation energies of the d-block transition metals increase slightly and irregularly across the first series.

Answer 8

• across the first series of transition metals, the nuclear charge of the elements increases and additional electrons are found in the inner 3d subshell.
• the screening effect of the additional 3d electrons is so significant that the effective nuclear charge of the elements increases only very slightly across the series.

Question 9

with reference to electronic configuration, explain why the second ionisation enthalpies of both chromium and copper are higher than those of their next elements respectively.

Answer 9

• the second ionisation enthalpy of chromium involves the removal of an electron from a half-filled 3d subshell, which has extra stability. therefore, this second ionisation enthalpy is relatively high.
• the second ionisation enthalpy of copper involves the removal of an electron from a fully-filled 3d subshell, which also has extra stability. therefore, the second ionisation enthalpy for copper will also be relatively high.

Question 10

with reference to electronic configuration, explain why the third ionisation enthalpy of magnesium is higher than that of its next element.

Answer 10

• the third ionisation enthalpy of magnesium involves the removal of an electron from a half-filled 3d subshell, which has extra stability.
• therefore, the third ionisation enthalpy of magnesium is relatively high.

Question 11

all transition elements with the exception of chromium and copper have complete 4s subshells. with reference to the stability of the elements, explain why chromium and copper have incomplete 4s subshells.

Answer 11

• the electronic structure of chromium, with incomplete 3d and 4s subshells, and an equal distribution of charge around the nucleus is more stable than an electronic structure with a complete 4s subshell.
• the electronic structure of copper with a complete 3d subshell and an incomplete 4s subshell is more stable than an electronic structure with a complete 4s subshell.

Question 12

scandium and zinc are d-block elements but are not transition metals. explain why scandium and zinc are not transition metals.

Answer 12

• scandium only forms one ion, Sc³⁺, which has an empty d subshell.
• when scandium loses three electrons to form Sc³⁺, it obtains the electronic configuration of [Ar], the nearest noble gas configuration.
• zinc only forms one ion, Zn²⁺, which has a full d subshell.
• when zinc forms Zn²⁺, it loses 2 electrons, both from the 4s subshell. this means that zinc keeps its full 3d subshell.

Question 13

explain why the electronegativity of d-block metals are generally higher than those of s-block metals.

Answer 13

• generally, d-block metals have smaller atomic radii than s-block metals.
• the nuclei of d-block metals can attract the electrons in a bond more tightly towards themselves.

Question 14

explain why the electronegativity of d-block metals generally shows a slight increase with increasing atomic numbers across the series.

Answer 14

• there is a gradual increase in effective nuclear charge and a decrease in atomic radius across the series.
• the closer the electron shell to the nucleus, the more strongly the additional electron in a bond is attracted.
• this results in a higher electronegativity.

Question 15

give two reasons why the melting temperatures of the d-block metals are much higher than those of the s-block metals.

Answer 15

• d-block metal atoms are small in size and are closely packed in a metallic lattice. all group I and some group II metals do not have close packed structures.
• both 3d and 4s electrons of d-block metals participate in metallic bonding by delocalising into the electron sea. therefore, the metallic bond is very strong.
• s-block metals only have one to two valence electrons per atom delocalising into the electron sea, resulting in a weaker metallic bond.

Question 16

most transition metals can form multiple stable ions. explain the difference in energy transfer required for a transition metal to form a complex or compound containing an ion with a certain oxidation number.

Answer 16

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

Question 17

transition metals form ions by losing electrons from both their 3d and 4s subshells. with reference to the energy levels of both subshells, explain why multiple electrons can be removed from these subshells to form ions with different oxidation numbers.

Answer 17

• the 4s and 3d subshells are at similar energy levels, meaning that it takes a similar amount of energy to remove an electron from the 4s subshell as it does to remove an electron from the 3d subshell.
• there is not a large increase between the ionisation energies of removing successive electrons, so multiple electrons can be removed from these subshells to form ions with different oxidation numbers.

Question 18

explain the relationship between the energy released when ions form a complex or compound, and ionic charge.

Answer 18

• the energy released when ions form a complex or compound increases ionic charge.
• therefore, the increase in energy required to remove the outer electrons to form transition metal ions with higher oxidation numbers is usually counteracted by the increase in the energy released.

Question 19

define the term ‘ligand’.

Answer 19

an atom, ion, or molecule that donates a lone pair of electrons to a central metal atom or ion.

Question 20

give the three types of ligand.

Answer 20

• monodentate - donates one lone pair.
• bidentate - donates two lone pairs.
• multidentate - donates more than two lone pairs.

Question 21

transition metals can form complex ions. define the term ‘complex ion’.

Answer 21

a complex ion is a metal ion surrounded by dative covalently bonded ligands.

Question 22

what is a dative covalent bond?

Answer 22

a covalent bond in which both electrons in the shared pair come from the same atom.

Question 23

give the equation used to calculate the oxidation number of a complex metal ion.

Answer 23

oxidation number of the metal ion = total oxidation number - sum of the charges of the ligands.

Question 24

define the term ‘coordination number’.

Answer 24

the number of dative covalent bonds formed around the central metal ion.

Question 25

explain what causes complexes with different coordination numbers to form distinctive shapes.

Answer 25

• the bonding electrons in the dative covalent bonds of a complex repel each other.
• as a result, ligands are generally positioned as far away from each other as possible.
• this causes complexes with different coordination numbers to form distinctive shapes.

Question 26

give the shape and bond angle formed from complexes with six-fold coordination.

Answer 26

six-fold coordination forms an octahedral shape, with 90° bond angles.

Question 27

give the shapes and bond angles formed from complexes with four-fold coordination.

Answer 27

• four-fold coordination generally forms a tetrahedral shape, with 109.5° bond angles.
• occasionally, four-fold coordination can result in a square planar shape, with 90° bond angles.

Question 28

explain the type of isomerism shown by square planar and octahedral complex ions that have at least two pairs of identical ligands.

Answer 28

• square planar and octahedral complex ions that have at least two pairs of identical ligands show cis/trans isomerism.
• cis isomers have the same groups on the same side.
• trans isomers have the same groups on opposite sides.

Question 29

explain how the platinum (II) metal ion can form a cis-platin complex.

Answer 29

• cis-platin is a complex of platinum (II) with two chloride ions and two ammonia molecules, which form a square planar shape.
• the two chloride ions are next to each other, so the complex formed is cis-platin.

Question 30

give one use of cis-platin.

Answer 30

cis-platin is used as an antineoplastic (chemotherapy) drug.

Question 31

explain why only the cis form of the platinum (II) complex can be given to patients being treated for cancer.

Answer 31

• if the two chloride ions found in the complex were on opposite sides, the complex formed would be trans-platin.
• trans-platin is toxic, so only the cis form of the complex can be given to patients being treated for cancer.

Question 32

explain what happens to the 3d orbitals of transition metal ions when bonded with ligands.

Answer 32

• normally, the 3d orbitals of transition metal ions all occupy the same energy level.
• however, when these ions bond with ligands, the 3d orbitals split into two different energy levels.

Question 33

electrons tend to occupy the lower energy orbitals, known as the ‘ground state’. in order to ‘jump’ to higher energy orbitals, these electrons need energy equal to that of the energy gap, ΔE. where is this energy obtained from?

Answer 33

visible light.

Question 34

describe the relationship between the energy gap and the frequency of visible light absorbed.

Answer 34

the larger the energy gap, the higher the frequency of light that is absorbed.

Question 35

give the three factors which affect the size of the energy gap, ΔE.

Answer 35

• the oxidation number of the central metal atom.
• the ligands involved in the complex.
• the coordination number.

Question 36

give the product formed when a solid containing a transition metal is dissolved in water.

Answer 36

the transition metal will form an aqueous complex in solution (the metal ion will be surrounded by water ligands)

Question 37

the colour of this aqueous solution can be used to identify what?

Answer 37

the transition metal ion that is present.

Question 38

vanadium is a transition metal which can exist in up to four oxidation states. determine the colour change(s) observed when zinc metal is added to an acidified solution containing VO²⁺ (aq) ions. Zn²⁺ (aq) + 2e⁻ ⇌ Zn (s) E⦵ = - 0.76V

Answer 38

• E⦵ for the first three reactions will be positive, so zinc will reduce vanadium (V) to vanadium (IV), which will then be reduced to vanadium (III), and eventually to vanadium (II).
• the reduction potential for the reaction of vanadium (II) with zinc is negative, so under standard conditions, vanadium (II) will not be reduced to vanadium metal.
• the colour of the solution will change from yellow, to blue, to green, to violet.

Question 39

chromium is a transition metal which can exist in up to five oxidation states. chromium forms two ions with oxygen - chromate (VI) ions, and dichromate (VI) ions. explain why these ions are good oxidising agents.

Answer 39

chromate (VI) ions, and dichromate (VI) ions are good oxidising agents because they are easily reduced to Cr³⁺.

Question 40

when Cr³⁺ ions are surrounded by six water ligands, the colour of the ions appear violet. however, the resulting the solution appears green. explain why the resulting solution appears green.

Answer 40

the resulting solution appears green because because the water ligands are usually substituted with impurities in the water, such as Cl⁻.

Question 41

give the reagent used to reduce dichromate (VI) ions to Cr³⁺.

Answer 41

a reducing agent, such as zinc and dilute acid.

Question 42

explain why an inert (chemically inactive) atmosphere must be used when Cr³⁺ is reduced to Cr²⁺ using metal zinc.

Answer 42

an inert atmosphere must be used because Cr²⁺ is unstable, and would otherwise oxidise back to Cr³⁺ upon contact with the air.

Question 43

give the reagent used to oxidise Cr³⁺ to chromate (VI) ions.

Answer 43

hydrogen peroxide in an alkaline solution.

Question 44

an aqueous solution of chromium (III) ions mixed with aqueous sodium hydroxide or aqueous ammonia produces a chromium hydroxide precipitate. this precipitate is amphoteric. define the term ‘amphoteric’.

Answer 44

amphoteric - able to react with both acids and bases.

Question 45

explain what happens to the H₂O ligands when excess sodium hydroxide is added to a chromium hydroxide precipitate.

Answer 45

the H₂O ligands deprotonate (lose a proton), and a solution containing chromium (III) hydroxide forms.

Question 46

explain what happens to the OH⁻ ligands when a dilute acid is added to a chromium hydroxide precipitate.

Answer 46

the OH⁻ ligands protonate (gain a proton), and a solution containing hexaaquachromium (III) chloride forms.

Question 47

explain why these reactions are acid-base reactions and not ligand exchanges.

Answer 47

these reactions are acid-base reactions and not ligand exchanges because the ligands are chemically modified by the acid or alkali through the addition or removal of a proton.

Question 48

the addition of which reagent to a chromium hydroxide precipitate would cause a ligand exchange reaction.

Answer 48

excess aqueous ammonia.

Question 49

explain why the enthalpy change for a ligand exchange reaction is usually small.

Answer 49

• when a ligand exchange reaction occurs, dative covalent bonds are broken and formed.
• the strength of the bonds being broken is very similar to the strength of the new bonds being formed.
• therefore, the enthalpy change for a ligand exchange reaction is usually small.

Question 50

explain why an increase in entropy change occurs when monodentate ligands are exchanged with bidentate or monodentate ligands.

Answer 50

• when monodentate ligands are substituted with bidentate or monodentate ligands, the number of particles in the solution increases.
• an increase in particles is directly proportional to an increase in entropy change.

Question 51

explain why transition metals and their compounds make good catalysts.

Answer 51

• transition metals and their compounds make good catalysts because they can change oxidation number by gaining or losing electrons in their 3d orbitals.
• as a result, transition metals and their compounds can exchange electrons to increase the speed of reactions.

Question 52

in the ‘Contact Process’, SO₂ is oxidised to SO₃. give the catalyst used in this process, including the type of catalyst.

Answer 52

vanadium (V) oxide, a type of heterogeneous catalyst, is used in the Contact Process.

Question 53

explain why transition metals make good heterogeneous catalysts.

Answer 53

transition metals make good heterogeneous catalysts because they can lose electrons from their partially filled d-orbitals to form weak bonds with the reactant molecules.

Question 54

what is autocatalysis?

Answer 54

when a reaction is catalysed by a product of the reaction.

Question 55

catalytic converters are used in cars to reduce emissions of nitrogen monoxide and carbon monoxide produced by internal combustion engines. explain how a catalytic converter, such as platinum, converts these emissions into less harmful substances.

Answer 55

• the reactant molecules are attracted to the surface of the heterogeneous catalyst and adhere to it (adsorption)
• the surface of the catalyst activates the molecules, weakening the bonds between the reactants’ atoms, which makes them easier to break and reform as products.
• the product molecules leave the surface of the catalyst, making way for new reactants to take their place (desorption)