Class 12 Chemistry Unit 6: Transition Metals Notes (NEB New Syllabus) | List, Properties, Oxidation States, Complex Ions, CFT | Transition Metals PDF
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1. Syllabus

Unit 6: Transition Metals

6.1 Characteristics, oxidation states
6.2 Complex ions and metal complexes
6.3 Shapes of complex ions and d-orbitals in complex ions
6.4 Colour of transition metal compounds
6.5 Catalytic properties

These notes follow the NEB Class 12 Chemistry syllabus (+2 Chemistry, Nepal). Practise the multiple choice questions first, then work through the solved numerical and subjective questions, and finish with the PDF notes for quick revision.

2. Multiple Choice Questions

Question 1. What is the maximum oxidation state shown by transition elements?

  • a) +5
  • b) +6
  • c) +7
  • d) +8

Correct Answer: d) +8

Reason: The highest oxidation state in the whole d-block is +8, shown by the 4d and 5d elements ruthenium and osmium in \(\mathrm{RuO_4}\) and \(\mathrm{OsO_4}\).


Question 2. What is the maximum oxidation state shown by the 3d series?

  • a) +9
  • b) +8
  • c) +7
  • d) +6

Correct Answer: c) +7

Reason: Manganese (\([\mathrm{Ar}]3d^54s^2\)) can use all seven valence electrons, as in the permanganate ion \(\mathrm{MnO_4^-}\). This is the highest state in the 3d series.


Question 3. Which of the following does not form colorful compound?

  • a) \(\mathrm{Cu^{2+}}\)
  • b) \(\mathrm{Fe^{2+}}\)
  • c) \(\mathrm{Cu^{+}}\)
  • d) \(\mathrm{Co^{2+}}\)

Correct Answer: c) \(\mathrm{Cu^{+}}\)

Reason: Colour comes from d-d transitions, which need a partly filled d subshell. \(\mathrm{Cu^+}\) is \(3d^{10}\) (completely filled), so no d-d transition is possible and it is colourless. \(\mathrm{Cu^{2+}}\) (\(3d^9\)), \(\mathrm{Fe^{2+}}\) (\(3d^6\)) and \(\mathrm{Co^{2+}}\) (\(3d^7\)) are coloured.


Question 4. What is the general electronic configuration of transition metals?

  • a) \(ns^{1-2}(n-1)d^{1-9}\)
  • b) \(ns^{1-2}(n-1)d^{10}\)
  • c) \(ns^{2}(n-1)d^{1-10}\)
  • d) \(ns^{1-2}(n-1)d^{1-10}\)

Correct Answer: d) \(ns^{1-2}(n-1)d^{1-10}\)

Reason: The (n-1)d subshell is being filled. Some elements have one ns electron (Cr, Cu), the others have two, and the d subshell holds 1 to 10 electrons across the series.


Question 5. Which of the following element shows maximum number of oxidation state?

  • a) Mn
  • b) Ni
  • c) Cr
  • d) Cu

Correct Answer: a) Mn

Reason: Manganese shows every oxidation state from +2 to +7 (six states). Ni shows +2 to +4, Cr shows +2, +3 and +6 as the common ones, and Cu shows +1 and +2.

Question 6. Find the number of unpaired electron in iron metal ion of \(\mathrm{[Fe(H_2O)_6]^{2+}}\)?

  • a) 2
  • b) 3
  • c) 4
  • d) 5

Correct Answer: c) 4

Reason: The oxidation state of Fe is \(x+6(0)=+2\), so Fe is \(\mathrm{Fe^{2+}}\) with configuration \(3d^6\). Water is a weak field ligand, so the splitting is small and the electrons occupy the orbitals singly first (high spin): \(t_{2g}^{4}e_g^{2}\). Five orbitals hold six electrons, so one orbital is paired and four are unpaired.


Question 7. What is the coordination number of octahedral complexes?

  • a) 2
  • b) 4
  • c) 5
  • d) 6

Correct Answer: d) 6

Reason: An octahedral complex has six ligand donor atoms around the metal ion, placed at the six corners of an octahedron.


Question 8. Which of the following ion has no d electron?

  • a) \(\mathrm{Cr^{3+}}\)
  • b) \(\mathrm{Cu^{+}}\)
  • c) \(\mathrm{Sc^{3+}}\)
  • d) \(\mathrm{Fe^{2+}}\)

Correct Answer: c) \(\mathrm{Sc^{3+}}\)

Reason: Sc is \([\mathrm{Ar}]3d^14s^2\). After losing three electrons, \(\mathrm{Sc^{3+}}\) is \([Ar]3d^0\). The others are \(\mathrm{Cr^{3+}}\) (\(3d^3\)), \(\mathrm{Cu^+}\) (\(3d^{10}\)) and \(\mathrm{Fe^{2+}}\) (\(3d^6\)).


Question 9. What do we call the energy difference between the two sets of d-orbitals in an octahedral field splitting?

  • a) Bond enthalpy
  • b) Crystal field splitting energy
  • c) Ligand field energy
  • d) Coordination energy

Correct Answer: b) Crystal field splitting energy

Reason: The gap between the lower \(t_{2g}\) set and the higher \(e_g\) set is the crystal field splitting energy \(\Delta_o\).


Question 10. Which of the following ligand produces a strong crystal field?

  • a) \(\mathrm{CN^-}\)
  • b) \(\mathrm{H_2O}\)
  • c) \(\mathrm{NH_3}\)
  • d) \(\mathrm{I^-}\)

Correct Answer: a) \(\mathrm{CN^-}\)

Reason: In the spectrochemical series, \(\mathrm{I^-}\) < \(\mathrm{Cl^-}\) < \(\mathrm{H_2O}\) < \(\mathrm{NH_3}\) < \(\mathrm{CN^-}\) < CO, the field strength increases to the right. Cyanide is a strong field ligand and gives a large splitting.

3. Numerical and Subjective Questions

Short Answer Questions

Question 1. Give reasons:

a. Zinc is non typical transition element.

A typical transition element has a partly filled d subshell, either in the atom or in its common ions. Zinc is \([\mathrm{Ar}]3d^{10}4s^2\) and its only common ion, \(\mathrm{Zn^{2+}}\), is \(3d^{10}\). The d subshell is completely filled in both, so zinc does not show the typical properties: it has a single oxidation state (+2), its compounds are colourless and diamagnetic, and it is softer with lower melting point. Zinc is therefore called a non-typical transition element.

b. Transition element exhibits variable oxidation state.

The (n-1)d and ns orbitals have almost the same energy. After the ns electrons are lost, the (n-1)d electrons can also take part in bonding, so the element can lose different numbers of electrons under different conditions. Example: manganese shows +2, +3, +4, +5, +6 and +7.

ElementConfigurationOxidation states
Ti\([\mathrm{Ar}]3d^24s^2\)+2, +3, +4
V\([\mathrm{Ar}]3d^34s^2\)+2, +3, +4, +5
Cr\([\mathrm{Ar}]3d^54s^1\)+1, +2, +3, +4, +5, +6
Mn\([\mathrm{Ar}]3d^54s^2\)+2, +3, +4, +5, +6, +7
Fe\([\mathrm{Ar}]3d^64s^2\)+2, +3 (also +4, +5, +6)
Cu\([\mathrm{Ar}]3d^{10}4s^1\)+1, +2

The half-filled (\(d^5\)) and filled (\(d^{10}\)) arrangements are specially stable, so \(\mathrm{Mn^{2+}}\) (\(3d^5\)) is more stable than \(\mathrm{Mn^{3+}}\) (\(3d^4\)). Lower states (+2, +3) are generally ionic and the higher states are covalent (CrO is ionic, \(\mathrm{CrO_3}\) is covalent).

c. Transition metals behave as good catalyst.

(i) Variable oxidation states: the metal can take part in the reaction by gaining or losing electrons, forms unstable intermediate compounds with the reactants and gives a new path with a lower activation energy; it returns to its original state at the end. (ii) Surface adsorption: the metals and their compounds have a large surface with vacant or partly filled d orbitals; the reactant molecules are adsorbed on it, their bonds weaken, and the reaction is faster. Examples: Fe in the Haber process and \(\mathrm{V_2O_5}\) in the Contact process.


Question 2. Write the characteristics of transition metals.

Position and configuration: They are the d-block elements in the middle of the periodic table, groups 3 to 12, with general configuration \((n-1)d^{1-10}ns^{1-2}\).

Metallic character: hard, malleable and ductile, with a metallic lustre.
Conductivity: good conductors of heat and electricity.
Density, melting and boiling points: high, because of strong metallic bonding (both ns and (n-1)d electrons take part).
Variable oxidation states: because the ns and (n-1)d energies are close.
Coloured compounds: caused by d-d electronic transitions.
Paramagnetism: due to unpaired d electrons.
Complex formation: small size, high charge density and vacant d orbitals favour complex ions.
Catalytic property: many act as catalysts in industry.
Alloy formation: similar atomic sizes let them form alloys such as brass, bronze and steel.

Question 3. What do you mean by crystal field splitting?

In a free metal ion all five d orbitals have the same energy (they are degenerate). When ligands approach, the negative charge of the ligands repels the d electrons, and the repulsion is not the same for every d orbital, because the orbitals point in different directions. The d orbitals therefore split into groups of different energy. This loss of degeneracy of the d orbitals in the field of the ligands is called crystal field splitting. In an octahedral complex the five orbitals split into a lower \(t_{2g}\) set and a higher \(e_g\) set.


Question 4. What are transition elements? Why are they called so?

Transition elements are the elements whose atoms, or common ions, have a partly filled d subshell. They are the d-block elements of groups 3 to 12, with the last electron entering the (n-1)d subshell.

Why they are called so: they lie between the s-block elements (strongly electropositive metals, on the left) and the p-block elements (on the right). Their properties change gradually from one block to the other, so they form a transition between them.

Question 5. Define:

a. Ligands

A ligand is an atom, ion or neutral molecule that donates a lone pair of electrons to the central metal ion to form a coordinate bond. It acts as a Lewis base. Examples: \(\mathrm{H_2O}\), \(\mathrm{NH_3}\), \(\mathrm{Cl^-}\), \(\mathrm{CN^-}\). Ligands are monodentate (one donor atom), bidentate (two, for example ethylenediamine and oxalate), tridentate or polydentate (for example EDTA).

b. Coordination sphere

The central metal ion with the ligands directly attached to it, written in square brackets, is the coordination sphere. In \(\mathrm{[Co(NH_3)_6]Cl_3}\), the coordination sphere is \(\mathrm{[Co(NH_3)_6]^{3+}}\), and the three \(\mathrm{Cl^-}\) ions lie outside it.

c. Complex ion

A complex ion is a charged species made of a central metal ion bonded to a definite number of ligands. Its charge is the sum of the metal ion charge and the ligand charges. Examples: \(\mathrm{[Cu(NH_3)_4]^{2+}}\) (complex cation) and \(\mathrm{[Fe(CN)_6]^{3-}}\) (complex anion). A complex with no net charge, such as \(\mathrm{Ni(CO)_4}\), is a neutral complex.


Question 6. Which of the \(\mathrm{Fe^{2+}}\) and \(\mathrm{Cr^{3+}}\) has more number of unpaired electrons?

\(\mathrm{Fe^{2+}}\): Fe is \([\mathrm{Ar}]3d^64s^2\); losing the two 4s electrons gives \(3d^6\). The six electrons fill five orbitals, one orbital is paired and four are single: 4 unpaired electrons.

\(\mathrm{Cr^{3+}}\): Cr is \([\mathrm{Ar}]3d^54s^1\); losing three electrons gives \(3d^3\), with three single electrons: 3 unpaired electrons.

So \(\mathrm{Fe^{2+}}\) has more unpaired electrons (4 against 3).


Question 7. What do you mean by degenerate orbitals?

Orbitals that have the same energy are called degenerate orbitals. The five d orbitals of a free metal ion (\(d_{xy},d_{yz},d_{xz},d_{x^2-y^2},d_{z^2}\)) are degenerate, and so are the three p orbitals of an atom. The ligand field of a complex removes the degeneracy of the d orbitals.


Question 8. What is crystal field splitting energy?

The crystal field splitting energy is the energy difference between the two sets of d orbitals formed in the field of the ligands. In an octahedral complex it is called \(\Delta_o\) and it is the gap between the \(e_g\) and \(t_{2g}\) sets. The \(t_{2g}\) orbitals lie \(0.4\Delta_o\) below the average energy and the \(e_g\) orbitals lie \(0.6\Delta_o\) above it.

The value of \(\Delta_o\) depends on the nature of the ligand (strong field ligands such as \(\mathrm{CN^-}\) give a large value), the oxidation state of the metal and the geometry of the complex.

Question 9. Give any two examples of complex ions.

1. Tetraamminecopper(II) ion, \(\mathrm{[Cu(NH_3)_4]^{2+}}\) (complex cation, square planar).

2. Hexacyanoferrate(III) ion, \(\mathrm{[Fe(CN)_6]^{3-}}\) (complex anion, octahedral).

Other examples: \(\mathrm{[Fe(H_2O)_6]^{2+}}\), \(\mathrm{[Ag(NH_3)_2]^+}\) and \(\mathrm{[Ni(CN)_4]^{2-}}\).


Question 10. Transition metal ions are paramagnetic, why?

A substance is paramagnetic when it has unpaired electrons, because each unpaired electron behaves as a tiny magnet due to its spin and the substance is weakly attracted by a magnetic field. Most transition metal ions have a partly filled d subshell (\(d^1\) to \(d^9\)). By Hund’s rule the electrons occupy the d orbitals singly before they pair, so the ions have unpaired electrons and are paramagnetic. Ions with \(d^0\) or \(d^{10}\) (\(\mathrm{Sc^{3+}}\), \(\mathrm{Zn^{2+}}\)) have no unpaired electrons and are diamagnetic.

Long Answer Questions

Question 1. What is CFT? Write its postulates.

Crystal Field Theory (CFT) was developed by Hans Bethe and John Van Vleck in the 1930s. It explains the bonding, colour and magnetic properties of complexes by treating the metal-ligand attraction as purely electrostatic.

Postulates:

The metal ion is surrounded by an electric field created by the ligands.
The bond between the metal cation and the ligands is purely electrostatic (ionic); the ligands are treated as point negative charges or as dipoles with the negative end toward the metal.
The transition metal ion is a positive ion with a charge equal to its oxidation state.
In the absence of ligands the five d orbitals of the metal ion are degenerate (have the same energy).
In the field of the ligands the degeneracy is lost and the d orbitals split into groups of different energy. In an octahedral complex, dxy, dyz and dxz form the lower set t2g, and dx2-y2 and dz2 form the higher set eg.
The extent of splitting depends on the nature of the ligands and their arrangement around the metal ion.
The energy difference between the split sets leads to the crystal field stabilisation energy, which affects the stability, colour and magnetism of the complex.
Weak field ligands (H2O, OH-, halides) cause small splitting and give high spin complexes with more unpaired electrons; strong field ligands (CN-, CO, NH3) cause large splitting and give low spin complexes with more pairing.

Question 2. Explain how CFT accounts for the splitting of d-orbitals in octahedral complexes.
Class 12 Chemistry - Transition Metals

Place the metal ion at the origin and let the six ligands approach along the +x, -x, +y, -y, +z and -z axes. The five d orbitals fall into two groups according to their direction:

\(d_{x^2-y^2}\) and \(d_{z^2}\) point along the axes, directly toward the ligands. Their electrons feel a strong repulsion, so their energy rises. These form the \(e_g\) set.
\(d_{xy},d_{yz},d_{xz}\) point between the axes, away from the ligands. They feel less repulsion, so their energy falls. These form the \(t_{2g}\) set.

If the ligand field were spherical, all five orbitals would rise equally to the average energy. In the octahedral field the \(e_g\) set lies \(0.6\Delta_o\) above this average and the \(t_{2g}\) set lies \(0.4\Delta_o\) below it, and \(\Delta_o\) is the crystal field splitting energy. The energy balance is \(2(0.6\Delta_o)=3(0.4\Delta_o)\).

Example, \(\mathrm{Cr^{3+}}\) (\(d^3\)): the three electrons occupy the three \(t_{2g}\) orbitals singly, so the crystal field stabilisation energy is \(3\times(-0.4\Delta_o)=-1.2\Delta_o\).

Question 3. What are transition metals? Write their general characteristics.

Transition metals are the d-block elements of groups 3 to 12, whose atoms or ions have a partly filled d subshell. The general valence shell configuration is \((n-1)d^{1-10}ns^{1-2}\). They lie between the s-block and the p-block elements, which is why they are called transition elements. Examples: Fe, Cu, Ni, Zn, Cr, Mn.

General characteristics:

They are hard, malleable and ductile metals with metallic lustre, high density, and high melting and boiling points.
They are good conductors of heat and electricity.
They show variable oxidation states because the ns and (n-1)d electrons have close energies (Mn: +2 to +7).
Their compounds and ions are often coloured, because of d-d electronic transitions (for example blue \(\mathrm{[Cu(H_2O)_6]^{2+}}\) and purple \(\mathrm{MnO_4^-}\)).
Many of them and their ions are paramagnetic, because of unpaired d electrons.
They form a large number of complex ions, because of small size, high charge density and vacant d orbitals.
They and their compounds are good catalysts (Fe, Ni, \(\mathrm{V_2O_5}\), \(\mathrm{MnO_2}\)).
They form alloys easily, because their atomic sizes are similar (brass, bronze, steel).

Question 4. Explain why transition metals and their compounds exhibit catalytic property.

A catalyst changes the rate of a reaction by giving it a new path with a lower activation energy, without being used up. Transition metals and their compounds are good catalysts for two reasons.

1. Variable oxidation state. A transition metal can gain or lose electrons easily, so it can pass through several oxidation states. It forms an unstable intermediate with a reactant, which then changes into the product, and the metal returns to its original state. The new path has a lower activation energy and the reaction becomes faster.

2. Surface adsorption. A solid transition metal, or its compound, offers a large surface with vacant or partly filled d orbitals. Reactant molecules are adsorbed on this surface, the concentration of the reactants on it increases and the bonds inside the reactant molecules weaken. The reactants come close in a favourable orientation, the intermediate forms and quickly turns into the products.

CatalystApplication
Fe-MoManufacture of ammonia by the Haber process
Ni or PdHydrogenation of vegetable oil (vanaspati ghee); \(\mathrm{CH_2=CH_2+H_2\rightarrow CH_3CH_3}\)
\(\mathrm{MnO_2}\)Decomposition of \(\mathrm{KClO_3}\) and of \(\mathrm{H_2O_2}\)
\(\mathrm{TiCl_4}\) (Ziegler-Natta catalyst)Preparation of polyethene from ethene
CuDehydrogenation of alcohols
\(\mathrm{V_2O_5}\) or PtManufacture of sulphuric acid by the Contact process

Question 5. Why do most transition metals form colorful compounds? What factors affect the color of transition metals?

Cause of colour. In a complex the ligands split the d orbitals into a lower set (\(t_{2g}\)) and a higher set (\(e_g\)). When the metal ion has a partly filled d subshell (\(d^1\) to \(d^9\)), an electron can absorb light of energy equal to the gap and move from the lower set to the higher set (a d-d transition):

\[ \Delta_o=h\nu=\dfrac{hc}{\lambda} \]

The wavelength absorbed is removed from white light, and the compound shows the complementary colour of the light absorbed. Example: in \(\mathrm{[Cu(H_2O)_6]^{2+}}\) (a \(d^9\) ion) the \(t_{2g}\) to \(e_g\) transition absorbs orange light, so aqueous copper(II) sulphate looks blue.

Ions with \(d^0\) (\(\mathrm{Sc^{3+}}\)) or \(d^{10}\) (\(\mathrm{Zn^{2+}}\), \(\mathrm{Cu^+}\)) have no d-d transition and are colourless. A strong charge transfer between the ligand and the metal ion (for example in \(\mathrm{MnO_4^-}\) and \(\mathrm{CrO_4^{2-}}\)) also gives an intense colour.

Factors affecting the colour:

Nature of the metal ion: its charge and the number of d electrons change the energy gap.
Oxidation state of the metal: \(\mathrm{FeCl_2}\) is pale green and \(\mathrm{FeCl_3}\) is yellow-brown; chromium gives violet, green, blue, yellow and orange compounds in different states.
Nature of the ligands: strong field ligands give a larger \(\Delta_o\) and absorb light of shorter wavelength (higher energy). \(\mathrm{[Cu(H_2O)_4]^{2+}}\) is pale blue while \(\mathrm{[Cu(NH_3)_4]^{2+}}\) is deep blue.
Geometry of the complex: the splitting in a tetrahedral complex (\(\Delta_t\approx\tfrac49\Delta_o\)) differs from that in an octahedral complex.

Question 6. Draw the shape of complex ions having coordination number 4 and 6.
Class 12 Chemistry - Transition Metals

The shape of a complex ion depends on its coordination number. A coordination number of 4 gives a tetrahedral or a square planar shape, and a coordination number of 6 gives an octahedral shape. (M is the central metal ion and L is a ligand; a solid wedge points toward you and a dashed line points away from you.)

Coordination numberShapeBond angleExamples
4Tetrahedral\(109.5^\circ\)\(\mathrm{[Zn(NH_3)_4]^{2+}}\), \(\mathrm{[CuCl_4]^{2-}}\), \(\mathrm{[Co(Cl)_4]^{2-}}\), \(\mathrm{Ni(CO)_4}\)
4Square planar\(90^\circ\)\(\mathrm{[Ni(CN)_4]^{2-}}\), \(\mathrm{[Pt(Cl)_4]^{2-}}\), \(\mathrm{[Cu(NH_3)_4]^{2+}}\)
6Octahedral\(90^\circ\)\(\mathrm{[Fe(H_2O)_6]^{3+}}\), \(\mathrm{[Co(NH_3)_6]^{3+}}\), \(\mathrm{[Fe(CN)_6]^{3-}}\)

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5. Frequently Asked Questions

What are transition metals? (Class 12 Chemistry)
Transition metals are the d-block elements, placed in the middle of the periodic table between the s-block and the p-block (groups 3 to 12 in the NEB textbook). Their last electron enters the (n-1)d subshell, so the general valence shell configuration is ns1-2 (n-1)d1-10. They are called transition elements because their properties lie between those of the highly electropositive s-block metals and the p-block elements. They are hard, have high melting points, show variable oxidation states, form coloured compounds and complexes, and act as catalysts.
What topics are included in Unit 6 Transition Metals of NEB Class 12 Chemistry?
6.1 Characteristics and oxidation states; 6.2 complex ions and metal complexes; 6.3 shapes of complex ions and d-orbitals in complex ions (crystal field theory); 6.4 colour of transition metal compounds; 6.5 catalytic properties.
Where can I find the transition metals list and the transition metals periodic table?
The d-block has four rows. 3d series: Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn (atomic numbers 21 to 30). 4d series: Y, Zr, Nb, Mo, Tc, Ru, Rh, Pd, Ag, Cd (39 to 48). 5d series: La, Hf, Ta, W, Re, Os, Ir, Pt, Au, Hg (57 and 72 to 80). 6d series: Ac, Rf, Db, Sg, Bh, Hs, Mt, Ds, Rg, Cn (89 and 104 to 112). Together they occupy groups 3 to 12 in the centre of the periodic table.
What are some transition metals examples and their uses?
Iron (steel, haemoglobin, Haber catalyst), copper (wires, alloys such as brass and bronze), nickel (hydrogenation catalyst, coins), zinc (galvanising, batteries), chromium (stainless steel, chrome plating), manganese (steel, MnO2 in dry cells), titanium (aircraft, implants), silver, gold and platinum (jewellery, catalytic converters), and vanadium (V2O5 catalyst in the Contact process).
What are the transition metals properties (characteristics)?
They are hard, malleable and ductile with a metallic lustre; they are good conductors of heat and electricity; they have high density and high melting and boiling points because of strong metallic bonding; they show variable oxidation states; they form coloured compounds because of d-d transitions; many are paramagnetic because of unpaired d electrons; they form complex ions; they act as catalysts; and they form alloys easily.
What are inner transition metals and post transition metals?
Inner transition metals are the f-block elements, the lanthanoids (4f series) and the actinoids (5f series), placed at the bottom of the periodic table; their last electron enters the (n-2)f subshell. Post-transition metals are the metals that follow the d-block, such as Al, Ga, In, Sn, Tl, Pb and Bi; they have a filled d subshell and do not show typical transition metal properties. Zinc, cadmium and mercury (d10) are sometimes also treated as non-typical transition elements.
Why does transition elements show variable oxidation states?
The (n-1)d and ns orbitals have very similar energies, so electrons from both subshells can take part in bonding. After the two ns electrons are lost (oxidation state +2), the (n-1)d electrons can also be lost, giving higher states. Manganese shows +2 to +7, and the highest oxidation state in the 3d series is +7 (MnO4-), while Ru and Os of the 4d and 5d series reach +8 (RuO4, OsO4).
What is a complex ion, ligand, coordination number and coordination sphere?
A ligand is an atom, ion or neutral molecule that donates a lone pair of electrons to the central metal ion through a coordinate bond (for example NH3, H2O, Cl-, CN-). A complex ion is the central metal ion with its ligands, written in square brackets, such as [Cu(NH3)4]2+. The coordination number is the number of ligand donor atoms attached to the metal ion, and the coordination sphere is the metal ion together with the ligands directly attached to it. Ligands may be monodentate, bidentate (ethylenediamine, oxalate), tridentate or polydentate (EDTA).
What are the shapes of complex ions?
Coordination number 2 gives a linear shape, such as [Ag(NH3)2]+. Coordination number 4 gives a tetrahedral shape, such as [Zn(NH3)4]2+ and [CuCl4]2-, or a square planar shape, such as [Ni(CN)4]2- and [Pt(Cl)4]2-. Coordination number 6 gives an octahedral shape, such as [Fe(H2O)6]3+, [Co(NH3)6]3+ and [Fe(CN)6]3-. The shape depends on the coordination number, the size and electronic configuration of the metal ion, the type and size of the ligands and ligand-ligand repulsion.
What is the crystal field theory (CFT)? What is crystal field splitting energy?
CFT, developed by Bethe and Van Vleck, treats the bond between the metal ion and the ligands as purely electrostatic. In an octahedral field, the five degenerate d orbitals split into a lower-energy t2g set (dxy, dyz, dxz, lowered by 0.4 delta-o) and a higher-energy eg set (dx2-y2, dz2, raised by 0.6 delta-o). The energy difference between the two sets is the crystal field splitting energy, delta-o. Weak field ligands (I-, Br-, Cl-, H2O) give high spin complexes and strong field ligands (NH3, CN-, CO) give low spin complexes.
Why do transition metal compounds show colour?
In a complex, the ligand field splits the d orbitals, and an electron in a partly filled d subshell can absorb visible light of energy equal to the splitting and jump from the t2g to the eg level (a d-d transition). The compound shows the colour complementary to the absorbed colour; for example, [Cu(H2O)6]2+ absorbs orange light and appears blue. The colour depends on the metal ion, its oxidation state, the ligands and the geometry. Ions with d0 (Sc3+) or d10 (Zn2+, Cu+) configuration are colourless.
Why are transition metals and their compounds good catalysts?
They have variable oxidation states, so they form unstable intermediates and give an alternative path with lower activation energy, and their surfaces adsorb the reactants, which weakens the bonds in the reactant molecules. Examples: Fe-Mo in the Haber process, V2O5 or Pt in the Contact process, Ni or Pd in the hydrogenation of vegetable oil, MnO2 in the decomposition of KClO3 and H2O2, TiCl4 in the Ziegler-Natta preparation of polyethene and Cu in the dehydrogenation of alcohols.
Where can I get Class 12 Chemistry Transition Metals notes in PDF?
This page includes handwritten notes and typed PDF notes of Unit 6 Transition Metals that you can view online free of cost, together with the solved multiple choice questions and the full exercise as per the NEB Class 12 Chemistry syllabus 2083.
Where can I get Class 12 Chemistry notes of NEB (handwritten, PDF, Nepal)?
The Class 12 Chemistry notes of the NEB syllabus 2083 are available on this website unit by unit, in handwritten and typed PDF form. Students who look for notes in the style of Note Mandir will find this unit in the same way: syllabus, solved exercise, handwritten notes and typed notes. These are NEB Class 12 Chemistry notes for Chapter 6, Transition Metals.

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