Transition Elements

Question 1

Transition element

Answer 1

A transition element is a d-block element that forms one or more stable ions with an incomplete d subshell.

Question 2

Why are scandium (Sc) and (Zn) not defined as transition element?

Answer 2

Scandium, with the electronic configuration [Ar] 3d1 4s2, forms only one ion, Sc3+. This 3+ ion has no electrons in its 3d sub-shell: the electronic configuration of Sc3+ is just its argon core, [Ar].
Zinc, with the electronic configuration [Ar]3d10 4s2, forms only one ion, Zn2+. This 2+ ion has a complete 3d sub-shell: the electronic configuration of Zn2+ is [Ar]3d10.
So neither scandium nor zinc forms an ion with an incomplete d sub-shell.

Question 3

Pattern of electron filling and removal in subshells of transition elements (along with pattern of exceptional transition elements)?

Answer 3

Filling
In atoms of the transition elements, the 4s sub-shell is normally filled and the rest of the electrons occupy orbitals in the 3d subshell. However, chromium and copper atoms are the exceptions.
Chromium atoms have just one electron in the 4s sub-shell. The remaining five electrons are arranged in the 3d sub-shell so that each of its five orbitals is occupied by one electron.
Copper atoms also have just one electron in the 4s sub-shell. The remaining ten electrons are arranged in the 3d sub-shell so that each orbital is filled by two electrons.
Removal
When transition elements form ions, their atoms lose electrons from the 4s sub-shell first, followed by 3d electrons.

Question 4

Variable oxidation state of Transition elements

Answer 4

Each transition metal can form more than one positive ion. For example, the common ions of copper are Cu+ and Cu2+. We say that the transition metals have variable oxidation states. The resulting ions are often different colours.
It is due to the similarity in the energy of the 3d and 4s atomic orbitals that make these variable oxidation numbers possible. Because there are variable oxidation states, the names of compounds containing transition elements must have their oxidation number included, e.g. manganese(IV) oxide, MnO2, and cobalt(II) chloride, CoCl2.
.

Question 5

Why does higher oxidation states vary from element to element (trend)?

Answer 5

The maximum oxidation number of the transition elements at the start of the row involves all the 4s and 3d electrons in the atoms. For example, vanadium’s maximum oxidation state is +5, involving its two 4s electrons and its three 3d electrons. At the end of the row, from iron onwards, the +2 oxidation state becomes most common as 3d electrons become increasingly harder to remove as the nuclear charge increases across the period. The higher oxidation states of the transition elements are found in complex ions or in compounds formed with oxygen or fluorine. Common examples are the chromate(VI) ion, CrO42−, and the manganate(VII) ion, MnO4−.

Question 6

Redox reactions and titration (IMPT)

Answer 6

Example 1, 2 and 3; Practical Activity 24.1; Worked Example 1, 2; Self-Assessment Q 2; End of Ch Q 5 (must).

Question 7

Define Ligand, Complex ion, Co-ordination number, Bidentate ligand & Monodentate ligand

Answer 7

Ligand: a molecule or ion with one or more lone pair of electrons which form dative covalent bond to a central transition element atom or ion.
Complex ion: a central transition metal ion surrounded by ligands bonded to the central ion by dative (also called coordinate) covalent bonds.
Co-ordination number: the number of coordinate bonds formed by ligands with a transition element ion in a complex.
Bidentate ligand: a ligand that forms two coordinate bonds or dative bonds to the central transition metal ion in a complex.
Monodentate ligand: a ligand forming one coordinate bond with a transition element ion in a comlpex.

Question 8

Ligands and their complexes (shapes and coordination no.s)

Answer 8

Page 526, 527.

Question 9

Geometric isomers / cis-trans isomers

Answer 9

Geometric isomerism are possible in transition metal complexes, where no double bond exists. In this case, the term ‘geometric isomerism’ refers to complexes with the same molecular formula but different geometrical arrangements of their atoms. Examples are any square planar complexes with the general formula [M(A)2(B)2], where M is the transition element, and A and B are its surrounding ligands.

Question 10

Properties of geometric / cis-trans isomers.

Answer 10

Differences in electronegativity of the atoms in ligands forming the dative bonds to the central transition metal ion can result in polar and non-polar complexes. For example, the cis-isomer will have two identical groups on one side of any square planar complex. So any difference in electronegativity between the two pairs of isomers will cause an imbalance of charge, making a polar complex. The atoms with the higher electronegativity have a stronger pull on the electrons in the dative bonds and will carry a partial negative charge. However, the trans-isomer will have identical ligands directly opposite each other at the corners of the square planar complex. So the pull on the electrons in the dative bonds to the central metal ion are perfectly balanced. This means that the charge is balanced and results in a non-polar complex.

Question 11

EXAMPLE 1.
(Examples of geometric / cis-trans and stereo / optical isomers)

Answer 11

In cis-platin, the chlorine atoms are next to each other in the square complex but in trans-platin, they are diagonally opposite. Cis-platin has been used as an anti-cancer drug. It acts by binding to sections of the DNA in cancer cells, preventing cell division (see the ‘Fighting cancer’ section at the start of Chapter 24). Trans-platin does not have the same beneficial medical effects.

Question 12

EXAMPLE 2.
(Examples of geometric / cis-trans and stereo / optical isomers)

Answer 12

Octahedral complexes, with the general formula [M(A)4(B)2], can also display geometric isomerism. An example is the cobalt(II) complex ion, [Co(NH3)4(H2O)2]2+ (see Figure 24.9). Again, the cis-isomer is a slightly polar complex, whereas the trans-isomer is non-polar. The imbalance of charge in cis-isomer results from the asymmetric shape of the cis-[Co(NH3)4(H2O)2]2+ isomer and the difference in electronegativity between the oxygen and nitrogen atoms bonded to the central cobalt(II). The side of the complex with the water ligands will be partially negative as oxygen is more electronegative than the nitrogen in the ammonia ligands. The differences in electronegativity still apply in the trans-[Co(NH3)4(H2O)2]2+ isomer but its symmetrical arrangement ensures the charge in the complex is spread evenly around the complex.

Question 13

EXAMPLE 3.
(Examples of geometric / cis-trans and stereo / optical isomers)

Answer 13

Stereoisomerism is also commonly shown by octahedral (six co-ordinate) complexes associated with bidentate ligands. An example is the complex containing nickel as the transition metal and 1,2-are bidentate?diaminoethane (NH2CH2CH2NH2) as the bidentate ligand (Figure 24.10). The two isomers are stereoisomers because the two different molecules are mirror images of each other and cannot be superimposed. They are optical isomers, differing only in their ability to rotate the plane of polarised light in opposite directions.

Question 14

EXAMPLE 4.
(Examples of geometric / cis-trans and stereo / optical isomers)

Answer 14

The complex ion consisting of nickel(II) bonded to two bidentate 1,2-diaminoethane (en) ligands and two monodentate ligands, such as water or chloride ions, can form both geometric (cis/trans) isomers and optical isomers. Look at the cis- and trans-isomers of is [Ni(H2NCH2CH2NH2)2(H2O)2]2+ in Figure 24.11, simplified to [Ni(en)2(H2O)2]2+. Of the two cis- and trans- isomers of [Ni(en)2(H2O)2]2+ only the cis- isomer is optically active, with its two non-superimposable mirror images. The symmetrical nature of the trans-isomers of [Ni(en)2(H2O)2]2+ means that its mirror images can be superimposed so they do not display optical isomerism. Look at the optical isomers of [Ni(en)2(H2O)2]2+ in Figure 24.12.

Question 15

Ligand exchange reactions forming Copper complexes

Answer 15

[Cu(H2O)6]2+ gives a solution of copper sulfate its blue colour. On adding sodium hydroxide solution, we see a light blue precipitate forming. Two water ligands are replaced by two hydroxide ligands in the reaction:

[Cu(H2O)6]2+(aq) + 2OH–(aq) —–> Cu(OH)2(H2O)4(s) + 2H2O(l)

If you now add concentrated ammonia solution, the pale blue precipitate dissolves and we get a deep blue solution:

Cu(OH)2(H2O)4(s) + 4NH3(aq) —–> Cu(H2O)2(NH3) 4]2+(aq) + 2H2O(l) + 2OH(aq)

(The first reaction can also be achieved by adding concentrated ammonia solution to copper sulfate solution drop by drop or by adding a dilute solution of ammonia. The pale blue precipitate formed will then dissolve and form the deep blue solution when excess ammonia is added)
Water ligands in [Cu(H2O)6]2+ can also be exchanged for chloride ligands if we add concentrated hydrochloric acid drop by drop. A yellow solution forms, containing the complex ion [CuCl4]2−:

[Cu(H2O)6]2+(aq) + 4Cl–(aq) —–> [CuCl4]2–(aq) + 6H2O(l)

Question 16

Ligand exchange reactions forming Cobalt complexes

Answer 16

This ion gives an aqueous solution of cobalt(II) sulfate its pink colour. On adding sodium hydroxide solution, we see a blue precipitate of cobalt(II) hydroxide forming, which turns red when warmed if the alkali is in excess.

[Co(H2O)6]2+(aq) + 2OH–(aq) —->
Co(OH)2(H2O)4+(s) + 2H2O(l)

Water ligands in [Co(H2O)6]2+ can also be exchanged for ammonia ligands if we add concentrated aqueous ammonia drop by drop. The brown cobalt(II) complex ion is oxidised by oxygen in the air to [Co(NH3)6]3+(aq), a cobalt(III) complex ion.

[Co(H2O)6]2+(aq) + 6NH3(aq) —–>
[Co(NH3)6]2+(aq) + 6H2O(l)

Adding concentrated hydrochloric acid drop by drop to an aqueous solution of cobalt(II) ions results in the formation of a blue solution containing the tetrahedral complex [CoCl4]2−(aq).

[Co(H2O)6]2+(aq) + 4Cl–(aq) —–>
[CoCl4]2–(aq) + 6H2O(l)

Question 17

Stability constant, K stab

Answer 17

‘The equilibrium constant for the formation of the complex ion in a solvent from its constituent ions or molecules.’
Different ligands form complexes with different stabilities. The position of equilibrium lies in the direction of the more stable complex.
Example:
As we increase the concentration of ammonia, this process continues until four of the water molecules are replaced by ammonia molecules. The solution formed is a deep blue colour. The complex with ammonia as a ligand is more stable than the complex with just water as a ligand. If we dilute the complex with water, the position of equilibrium shifts to the left and a complex with more water molecules as ligands forms.

Question 18

How is stability constant, k stab calculated and what do the values refer?

Answer 18

An overall stability constant, Kstab, is given rather than the stepwise constants. The method for writing equilibrium expressions for stability constants is similar to the one we used for writing Kc.
K stab = products’ conc. / reactants’ conc.
Note :
- Water does not appear in the equilibrium expression because it is in such a large excess that its
concentration is regarded as being constant.
- The units for the stability constant are worked out in the same way as for the units of Kc.
Stability constants are often given on a log10 scale. When expressed on a log10 scale, they have no units (see Table 24.4).
Stability constants can be used to compare the stability of any two ligands. The values quoted usually give the stability of the complex relative to the aqueous ion where the ligand is water. The higher the value of the stability constant, the more stable the complex.

Question 19

Examples regarding to K stab values

Answer 19

1. The addition of excess ammonia to the complex [CuCl4]2−(aq) should result in the formation of a dark blue solution of the ammonia complex. That is because the stability constant of the ammonia complex is higher than that of the chloride complex. The position of equilibrium is shifted to the right in the direction of the more stable complex.
2. Addition of excess 1,2-dihydroxybenzene to the dark blue ammonia complex results in the formation of a
   green complex with 1,2-dihydroxybenzene. This is because the stability constant with the 1,2-dihydroxybenzene is much higher than that for ammonia. These values of Kstab predict that there will be hardly any of the copper complex with ammonia left in the equilibrium mixture.

Question 20

Example on finding k stab by deriving equil. conc. from before before reaction reaches equil.

Answer 20

Worked example 3 (e) pg. no. 533

Question 21

How are colors generally observed from transition elements?

Answer 21

White light is made up of all the colours of the visible spectrum. When a solution containing a transition metal ion in a complex appears coloured, part of the visible spectrum is absorbed by the solution. The colour we see is called the complementary colour, made up of light with frequencies not absorbed. For example, copper(II) ions absorb light from the red end of the spectrum, so the complementary colour seen is a pale blue (called cyan).

Question 22

Set of complementary colors

Answer 22

red||cyan
yellow||blue
green||magenta

Question 23

Define degenerate orbitals and non-degenerate orbitals

Answer 23

Degenerate orbitals : Atomic orbitals at the same energy level in a given subshell.
(The five d orbitals in an isolated transition metal atom or ion are described as degenerate orbitals, meaning they are all at the same energy level)
Non-degenerate orbitals : Groups of atomic orbitals in the same subshell that have slightly different amounts of energy is (splitting of orbitals).

Question 24

Explain the coloured nature of transition element

Answer 24

The ligands in a complex cause the d orbitals to split, forming two sets of non-degenerate orbitals. The difference in the energy (ΔE) between the non-degenerate d orbitals corresponds to the energy of part of the visible spectrum of light. So when light travels through a solution or a solid containing the complex, an electron from one of the three lower non-degenerate orbitals absorbs that amount of energy (ΔE) and jumps into one of the two higher non-degenerate orbitals. This leaves the transmitted light coloured.

Question 25

Difference in splitting pattern of d orbitals in an octahedral complex vs tetrahedral complex

Answer 25

In a complex with six ligands, the ligands are arranged in an octahedral shape around the central metal ion. The lone pairs donated by the ligands into the transition metal ion repel electrons in the two dx2-y2 and dz2 more than those in the other three d orbitals. This happens because these two d orbitals line up with the dative (coordinate) bonds in the complex’s octahedral shape.As the electrons in the dx2-y2 and dz2 orbitals are closer to the bonding electrons in the octahedral arrangement, the repulsion between electrons increases. Therefore, the d orbitals are split, with these two d orbitals at a slightly higher energy level than the dyz, dxz and dxy orbitals.
In tetrahedral complexes, such as [CoCl4]2−, the splitting of the d orbitals is different (Figure 24.18). The bonding pairs of electrons from four ligands line up with the dyz, dxz and dxy orbitals of the transition metal ion. Unlike the octahedral arrangement, the dx2-y2 and dz2 orbitals in a tetrahedral complex lie between the metal–ligand bonds. So there is less repulsion between electrons in dx2-y2 or dz2 orbitals and the lone pairs of bonding electrons donated by the ligands. Therefore, when the d orbitals split in this case, the dx2-y2 or dz2 are at a lower, more stable energy level than the dyz, dxz and dxy orbitals.