TRANSITION ELEMENTS

What are transition elements? Give the names of the metals of the first transition series.

The elements in which the differentiating electron enters into d – orbital are called d – block elements. The d – block-elements are arranged in 4 series and 10 groups. These elements are present between electropositive s – block and electronegative p – block elements in the periodic table. The d – block elements are called transition elements because they show a change from electropositive character of s – block elements to the electronegative” character of the p – block elements.

The transition elements exhibit, some characteristic properties such as 1) variable valency 2) formation of coloured compounds and ions 3) paramagnetic character 4) formation of complexes 5) formation of alloys etc. These characteristic properties of transition elements are due to incompletely, filled or partially filled d- orbitals.

Zinc, Cadmium and Mercury, the II B group elements, are not considered as transition elements because they do not exhibit the characteristic properties of transition elements. This is because in these elements the d – orbital is completely filled.

(Transition elements are those which have incompletely filled d-subshell either in its elemental state or in any of its stable oxidation states.)

The II B elements have  d10 configuration both in elemental state and also in their stable oxidation state. So they are not considered as transition elements. General outer electronic configuration of the transition elements is (n – 1) d1-10 ns* or   .      :

The names of the first transition series metals are

1)     Scandium     Sc

2)     Titanium       Ti

3)     Vanadium     V

4)     Chromium    Cr

5)     Manganese    Mn

6)     Iron               Fe

7)    Cobalt            Co

8)     Nickel           Ni

9)     Copper          Cu

Write the names and electronic configurations of first transition series.

 

Name of the element Symbol Electronic Configuration
Scandium Sc [Ar] 3d1 4s2
Titanium Ti [Ar]3d24s2
Vanadium V [Ar] 0 4s2
Chromium Cr [Ar] 3d5 4s1
Manganese Mn [Ar] 3d5 4s2
Iron Fe [Ar] 3d6 4s2
Cobalt Co [Ar]3d74s2
Nickel Ni [Ar] 3d8 4s2
Copper Cu [Ar] 3d10 4s1
Zinc Zn [Ar] 3d10 4s2

 

Define a “Transition element”. Into how many series the d-block elements are divided give the outer electronic configuration of at least one element in each series.

Transition elements are those which have incompletely filled d – subshell either in its elemental state or in any of its stable oxidation state.

The d – block elements are divided into four series.  , In the first series of ten elements starting from scandium (Z = 21) to zinc (Z= 30) the differentiate electron enters into 3d orbital. So they are called 3d series or first transition series.  .

In the second series of ten elements starting from yttrium (Z = 39) to cadmium (Z = 48) the differentiating electron enters into 4d orbital. So they are called 4d series or second transition series!

In the third series of ten elements starting from lanthanum (Z = 57) to mercury (Z = 80) except in fourteen lanthanides the differentiating electron enters into 5d orbital. So they are called 5d series or third transition series.

The fourth series is incomplete series. In these elements the differentiating electron enters into 6d orbital. So they are called 6d series or fourth transition series.

Outer electronic configuration of elements

 

21Sc                                          3d14s2

39 Y                                           4d15s2

57La                                         5d16s2

89 Ac                                        6d1 7s2

List the typical properties of transition elements. How are their properties correlated?

Transition elements exhibit typical characteristic properties.

  1. Electronic, configuration
  2. Variable oxidation states
  3. Para and ferromagnetic properties.
  4. Formation of coloured hydrated ions and salts
  5. Alloy forming ability
  6. Catalytic properties
  7. Complex forming ability
  8. Metallic character
  9. Ionisation energy
  10. Atomic and ionic radii
  11. Interstitial compounds.

These properties are correlated wherever possible with the electronic configurations or sizes or high nuclear charge and the presence of unpaired d – electrons.

Give the names of at least five d-block elements which have typical electronic configurations. Give their outer configurations.

 

 Scandium (Z = 21) 3d1 4s2
Titanium (Z = 22) 3d24s2
Vanadium (Z = 23) 3d3 4s2
Manganese (Z = 25) 3d5 4s2
Iron (Z = 26) 3d6 4s2

 

What is meant by variable oxidation states? How does this show its variable oxidation states? Explain.

Ans:- If the same element exhibit more than one oxidation state it is called variable valency. All the transition elements exhibit variable valency The energy difference between ns and (n-l)d orbitals is little. So the electrons in ns and (n – l)d orbitals can participate in bonding. Hence the transition elements exhibit variable valency. The minimum oxidation state that can be exhibited by a transition metal is equal to the number of ns electrons. The maximum oxidation state that can be exhibited by a transition metal is equal to the sum of the ns and (n-1) d electrons. The common oxidation exhibited by transition element is + II.

The outer electronic configuration of manganese 3d5 4s2. Manganese exhibit + II, + III, + IV, + V, + VI and + VII oxidation states. When the ns electrons only participate in bonding manganese exhibit + II. But when (n – l)d electrons also participate in bonding along with ns electrons the other oxidation states are exhibited by manganese.

The atomic radiis’ along with 3d series decreases. Explain on the basis of their electron configurations. But the A.R’s increase from Sc to Y to La why? 

In the first transition series electrons are added to the 3d orbitals, all of which have the same radial distribution. Due to such addition the size of the atom does not alter. But the d – electrons have poor shielding capacity. Hence the increased nuclear charge attracts the electrons closer to the nucleus. So the orbits are contracted and brings aboilt slow contraction in the size.

In any given d – series the differentiating electron enters initially into different degenerate d – orbitals of the same orbit. When all the d – orbitals are partially filled, pairing starts. The repulsion between paired electrons reduce the attractive forces due to increased nuclear charge. This effect increases so much that the s electrons of the outer obit are shielded from nuclear attractions. As a result the atomic radii of the elements at the end of each series increases.

In the same group transition elements the number of orbits increase as the nuclear charge increases. Therefore atomic radii and ionic radii increase as we move from top to bottom. So the atomic radii of Sc, Y & La increases from Sc to Y to La because they are in the same group.

 

Explain the colour of [Ti H2O)6]3+will the colours of the metallic species be as predicted by absorption of visible radiations.         

White light consists of seven colours VIBGYOR. When light fells on a substance the following changes may occur.

  1. All the light radiations may be reflected. Then the substance will appear white in colour.
  2. All the light radiations may be absorbed. Then the substance will appear black in colour.
  3. If one colour of the visible radiations is absorbed and the remaining colours are reflected then the substance will appear in the complimentary colour of the absorbed colour.

The colour of the transition metal ions is due to the incompletely filled d – orbitals. On the, free, isolated gaseous metal ion all the d – orbitals of transition metal ions are degenerate. But in the compounds or hydrated ions or complexes the d – orbitals split up into two sets one set consists of dx²-y²anddz²  and another set consists of dxy, duz and dxz  One set will have lower energy while other will have higher energy but the average d-orbital energy remains same.

When white light falls on the metal ion, the electron in the lower energy set of orbital absorb energy and are excited to another set of d – orbitals having higher energy. Since the energy difference between the two sets of d – orbitals is less, transition of electrons is possible due to the absorption of visible radiations (  λ= 400 – 750nm). So the transition metal ion appears in the complimentary colour of the absorbed radiation.

For example, hydrated titanium *(III) ion [Ti (H2O)6] +3 has a lone 3d electron. It absorbs light of a wavelength of about 500 – 575 nm (green and yellow region) and transmits the light that is rich in its compliment any colour (a mixture of red and violet) about 400 – 450 nm. Hence the hydrated titanium (III) ion has a pink colour.

The observed colour of the metallic species may not be simply the predicted complimentary colour. The colour of the ions with incomplete d or f – subshells may arise due to entirely different reasons than those given in the above example.

How does a catalyst work in a chemical reaction? Discuss with a suitable example. Mention one important catalyst used in an industry.

Transition metals and their compounds act as important catalysis in industry and in biological systems. The catalytic activity of the compounds of transition metals is due to their variable valancy or to form coordination compounds.

Example : In contact process for the manufacture of H2SO4 either V2O5 or vanadate is used as a catalyst to oxidise SO2 to SO3.

It is believed the calysis takes place in two stages.

Stage 1: Oxidation of SO2 to SO3 in which V5 + is reduced to V4 +

Stage 2: V5+ is again regenerated from V4+ by oxygen.

In some cases the catalysts provide a suitable surface for the reactants. Catalysts bind the reactants to the surface and reduce the activation energy. Ea, thus the reaction between reactants may proceed and convert into products.

Example:  Polymerization of ethylene using titanium (IV) chloride as catalyst.

How do you classify magnetic substances. Give two examples for each type. Ans:-Magnetic substances can be classified into different types.

Diamagnetic substances ii) Paramagnetic substances (iii) Ferromagnetic substances. When a substance is placed in a strong magnetic field, the magnetic lines of force are effected in one or three ways.

  1. Diamagnetic substances: If the magnetic lines of force are repelled by the substance, it is called diamagnetic substance. The’ diamagnetic materials move from a stronger part of the magnetic field to the weaker part of the magnetic field. In diamagnetic substances all the electrons are paired Examples: KCl, Ti 4+, V5+, Cu+, Zn2+
  2. Paramagnetic substances: If the magnetic force of lines is drawn into the substance, it is called paramagnetic substance. The magnetic force of lines passes through the substance. A paramagnetic substance moves from the weaker part of the field to the stronger part of the field. Paramagnetic substances contain unpaired electrons. Example: K3 (Fe [CN)6], Sc2+, Cr3+ etc.
  3. Ferromagnetism: It is a special case of paramagnetism. When a substance is kept in an external magnetic field the magnetic property of the substance increases by million times it is called ferromagnetic substance. In these substances the magnetic moments of the individual atoms are aligned in the same direction.  Example: Fe; Co; Ni.

Among these three elements iron is strongest ferromagnetic substance. Ferromagnetism in crystalline substances disappears in the solution form of the substance.

What do you understand by non – stoichiometric compounds? Why is ZnO white when cold and yellow when hot? Explain. Do you classify steel as an interstitial compound?

Compounds formed by the occupation of small atoms .like H, B, C and N in the holes or interstitial voids of the metal crystal are called interstitial compounds. In these compounds component elements do not combine in definite ratios. So these substances are also called as non – stoichiometric substances. Interstitial compounds have metallic nature, hard and brittle, high melting and high boiling materials. The non metallic atoms occupy the holes and so the metal crystal structure is not altered. But due to the presence of non metal atoms the crystal lattice will expand. So the density of the interstitial compounds is less than that of the metal. Metals like Ti, V, Zr, Nb, Hf and Ta form such compounds.

Sometimes stoichiometric compounds may become non stoichiometric compounds at higher temperature and therefore become coloured.

Example: ZnO is white when cold but yellow when hot. On heating ZnO the oxide ion lose electron and leaves from the substance. The electrons occupy the position of oxygen so that the material is electrically neutral. Thus at high temperature ZnO becomes                non-stoichiometric compound. Due to the presence of unpaired electrons ZnO has yellow colour at high temperature. In cold condition again oxygen is taken inside forming stoichiometric ZnO. So it is white in colour.

Steel is a non-stoichiometric compound of carbon and iron. It is an interstitial compound.

What are alloys? How are they prepared?  

An intimate mixture having physical properties similar to that of the metal formed by a metal with other metals or metalloids or sometimes a non – metal is called as an alloy. Alloys can be prepared by the following methods.

i) Melting a mixture of metals or components: This is the most commonly used method. Different, metals that should present in an alloy are taken in proper proportion and heated until they melt. The molten mass is cooled to get the alloy. Example : Brass, a copper – zinc alloy is prepared by this method.

ii) Simultaneous electrolytic deposition of metals: Under suitable condition generally dilute solutions of complex compounds of the respective metal ions are used. When electrolysis is carried simultaneous electrolytic deposition of metals takes place at cathode giving an alloy. Ex: By the electrolysis of a mixture of Cu and Zn cyanides brass is prepared.

iii) Amalgamation process: Alloys of mercury with metals are known as amalgams. They are prepared by grinding the metals with mercury. Example: M – Hg, M = Zn, Na, Cu, Ag or Au.

iv) Reduction method: In this method a compound of one of the component elements is reduced in] the presence of the second component. Generally a metal oxide is reduced in this way. Example : Fe, Cr alloy by reduction of chromate ore can be prepared.

v) Compression method or powder metallurgy: In this method separately powdered metals are mixed and compressed at a temperature slightly below the melting points to get the alloy. In this method alloys having proper structures are obtained and raw materials are not wasted. Example: Alloys involving metals like Mo &W are prepared by this method.

vi) Quenching: In this method Fe – C alloy is produced. Red hot steel is suddenly immersed in oil 01 water. Steel from this method is tougher and its property depends on the percentage of C.

Mention any four alloys and give their uses.

i) Invar    ii) Nichrome iii) Aluminum bronze   iv) Brass.

i) Invar: It is a nickel steel. Its compositions 64% Fe, 35% Ni containing places of Mn &C. It has a low temperature coefficient. It is used for pendulum rods,

ii) Nichrome: It is nickel chromium steel. Its composition is 60% Ni, 25% Fe, 15% Cr. It is used i| heating elements of fire stoves and furnaces.

iii) Aluminum bronze :  It is an alloy of copper and aluminum. Its composition is 88 – 90% Cu 10 -12% Al. It appears like gold. So used in cheap ornaments, photo frames and currency coins,

iv) Brass: It is an alloy of copper and zinc. Its composition is 60 – 80% Cu, 20 – 40% Zn. It is used in machinery parts and in house hold utensils.

Disuss the colour of the transition metal compounds with suitable examples.

White light consists of seven colours VIBGYOR. When light falls on a substance the following changes may occur.

  1. All the light radiations may be reflected. Then the substance will appear white in colour.
  2. All the light radiations may be absorbed. Then the substance will appear black in colour.
  3. If one colour of the visible radiations is absorbed and the remaining colours are reflected then the substance will appear in the complimentary colour.

The colour of the transition metal ions is due to the incompletely filled d – orbitals. In the free, isolated gaseous metal ion all the d – orbitals of transition metal ions are degenerate. But in the compounds or hydrated ions or complexes the d – orbitals split up into two sets one set consists of  and  and another set consists of dxy, dyz and dxz.One set will have lower energy while other will have higher, energy but the average d – orbital energy remains same.

When white light falls on the metal ion, the electron in the lower energy set of orbital absorb energy and is excited to another set of d – orbitals having higher energy. Since the energy difference between the two sets of d – orbitals is less, transition of electrons is possible due to the absorption of visible radiations (  λ= 400 – 750 nm). So the transition metal ion appears in the complimentary colour of the absorbed radiation.

Explain the magnetic properties of first transition series of metal ions.

When a substance is placed in strong magnetic field, the following type magnetic characters are exhibited by the substances.

  • Some substances attract the magnetic force of line and allow them to pass through them. They will move from weaker part of field to the stronger part of the magnetic field. Such substances are said to be paramagnetic These substances contain unpaired electrons.
  • Some substances repel the external magnetic force. They are hard for the magnetic force of lines to pass through them. They move from a stronger part of magnetic field to weaker part of magnetic field. Such substances are said to be diamagnetic      These substances contain all paired electrons.
  • In some substances the magnetic moment increases by million times, when kept in external magnetic field. Such substances are called ferromagnetic Ex: Fe, Co, Ni.

The magnetic character of substances is due to the spin and orbital momentum of electrons. The magnetic moment of paired electrons having opposite spin will be cancelled.

The magnetic moment of a substance is the resultant of all the magnetic moments of unpaired electrons. This can be calculated by the following formula.

In the above formula S is the sum of the spin of all the unpaired electrons and L is the sum of the azimuthal quantum number values of all unpaired electrons in the substance.

In Several compounds of transition elements the angular momentum due to orbital motion of unpaired electrons is small and can be neglected. Then the above formula can be written as.

Where n is the number of unpaired electrons. This formula is called spin only formula. Spin only unpaired electrons are small and can be neglected. Then the above formula can be written as follows.

 

Metal ion 3d configuration No. of unpaired electrons Magnetic moment BM
Sc3+ 3d0 0 0
Ti3+ 3d1 1 1.73
Ti2+ 3d2 2 2.84
V2+ 3d3 3 3.87
Cr2+ or Mn3+ 3d4 4 4.90
Mn2+ or Fe3+ 3d5 5 5.92
Fe2+ 3d6 4 49
Co2+ 3d7 3 3.87
Ni2+ 3d8 2 2.84
Cu2+ 3d9 1 1.73
Zn2+ 3d10 0 0.0

 

Where n is the number of unpaired electrons. This formula is called spin only formula. Spin only magnetic moments for first transition series metal ions will be as follows.

 

Write the 4f elements with their outer electronic configurations?

ELECTRONIC CONFIGURATIONS OF LANTHANOIDS

Name At. No. Electronic Configuration
Lanthanum 57 [Xe] Sd1 6s2
Cerium 58 [Xe] 4f1 5d1 6s2
Praseodymium 59 [Xe] 4f2 5d1 6s2
Neodymium 60 [Xe] 4f3 5d1 6s2
Promethium 61 [Xe] 4f4 5d1 6s2
Samarium 62 [Xe] 4f5 5d1 6s2
Europium 63 [Xe]4f6 5d° 6s2
Gadolinium 64 [Xe] 4f7 5d1 6s2
Terbium 65 [Xe] 4f8 5d1 6s2
Dysprosium 66 [Xe] 4f9 5d1 6s2
Holmium 67 [Xe] 4f10 5d1 6s2
Erbium 68 [Xe] 4f11 5d1 6s2
Thulium 69 [Xe] 4f12 5d1 6s2
Ytterbium 70 [Xe]4f14 5d° 6s2
Lutetium 71 [Xe] 4f14 5d1 6s2

 

What are the oxidation states exhibited by the lanthanide elements ? Write a note with atleast one example each.

All the lanthanides have 6s2 electrons. So they are expected to show + II oxidation state. But the common oxidation state for all these elements is + III. They use one of the 4f electrons to exhibit + III oxidation state. Other oxidation states which are encountered less frequently are + II and + IV states.

The + IV oxidation state occurs in the elements whose ions have almost empty or slightly more than half – filled 4f level. Cerium and terbium exhibit + IV oxidation state because they contain stable completely vacant and exactly half filled 4f subshells respectively in their + IV oxidation states.

Europium and ytterbium exhibit + II oxidation state because they contain stable half filled and completely filled 4f subshell respectively in their + II oxidation states.

Except in Ce +4 and Eu+2 the oxidation states other than + III are generally unstable.

Write note on the characteristics of lanthanides.

 

  1. In the lanthanoid series the differentiating electron enters into 4f subshell.
  2. Lanthanoid series includes 15 elements i.e. La57 to Lu71.
  3. AH these elements contains same number of outer electrons Sd1 6s2 while they differ only in the number of f electrons which are deep – seated and cannot participate in reactions. Hence they show similar properties.

Oxidation states

4. The principal oxidation state exhibited by all the elements is + III (Ln+3) in aqueous solution and in their solid compounds.

5. La, Gd and Lu exhibit only +3 oxidation state because they contain stable completely vacant or exactly half – filled or completely filled 4f subshell respectively in their + III oxidation states.

6. Eu and Yb exhibit + II oxidation state also because they contain stable half- filled and completely filled 4f subshell respectively in their + II oxidation states. ;

7. Ce and Tb also exhibit + IV oxidation state because they contain stable completely vacant and completely filled 4f subshells respectively in their + IV oxidation states.

8. The atomic and ionic radii of lanthanides decrease from lanthanum to Lutetium. This decrease is more; pronounced in the Ln3+ ions and it is called lanthanide contraction. The lanthanide contraction is due to poor shielding capacity of 4f electrons from one another in the same sub shells.

9. Due to lanthanide contraction the hydroxides of the earlier lanthanides are ionic while the hydroxides of later lanthanides are covalent. The basic character of hydroxides of lanthanides decreases from lanthanum to Lutetium.

10. Due to lanthanide contraction the atomic sizes of second and third row transition elements-in a group II group  become equal,

11. Due to lanthanide contraction the atomic sizes of Zr and Hf become equal. So they have almost same chemistry and occur together in nature. Their separation is also very difficult

12. Many trivalent lanthanide ions are coloured both in solid state and in aqueous solution,due to the presence of incompletely filled f – sub shell.

13. Lanthanides and their ions are paramagnetic due to the presence of unpaired electrons.

What is meant by lanthanide contraction? Explain in detail.

In lanthanides and their trivalent ions the nuclear charge increases with atomic number. If the atomic number increases from one element to the next element the differentiating electron enters the inner 4f orbitals. The mutual screening effect of f electrons is very poor. Therefore effective nuclear charge causes the atomic size or ionic size to decrease. This decrease in size of the atom or an ion among the lanthanides is known as lanthanide contraction.

Consequences of lanthanide contraction:

  1. Because of lanthanide contraction the chemical properties of lanthanides vary little from one another as a result their separation is extremely difficult
  2. Due to lanthanide contraction the hydroxides of earlier elements are ionic while the hydroxides of later lanthanides are covalent (Fajan’s rule). So the basic character of hydroxides of lanthanides decreases from lanthanum to lutesium.
  3. Due to lanthanide contraction the atomic sizes of II nd and III rd row transition elements in a group become equal
  4. Due to lanthanide contraction the atomic sizes of Zr and Hf become equal. So they have almost same chemistry and occur together in nature. Their separation is also very difficult.

What are the consequences of lanthanide contraction?

  1. Because of lanthanide contraction the chemical properties of lanthanides vary little from one another, As a result, their separation is extremely difficult.
  2. Due to lanthanide contraction the hydroxides of earlier elements are ionic while the hydroxides of later lanthanides are covalent (Fajan’s rule). So the basic character of hydroxides of lanthanides decreases from lanthanum to lutesium.
  3. Due to lanthanide contraction the atomic sizes II nd and III rd row transition elements in a group become equal.

Due to lanthanide contraction the atomic sizes of Zr and Hf become equal. So they have almost same chemistry and occur together in nature. Their separation is also very difficult.

Giving examples explain Werner’s theory of complex compounds.  

Werner’s theory of complex compounds: Werner’s theory of complex compounds is based on the following postulates.

1. Every complex compound has a central metal atom or ion.

2. central metal exhibits two types of valencies

a) Primary valency and b) secondary valency.

a) Primary Valency:

  1. The number of primary valencies of metal is equal to its oxidation number.
  2. The primary valencies are non-directional, represented by dotted lines,
  3. Primary valencies are ionizable in water.
  4. Primary valencies can be satisfied only by negative ions.

In modern theory primary valencies are nothing but ionic bonds.

b) Secondary Valency:

  1. The central metal has a fixed number of secondary valencies called co-ordination number. The co-ordination number varies from 2-10.
  2. Secondary valencies can be satisfied by neutral molecules atoms or by negative ions. Rarely positive ions can also satisfy secondary valencies.
  3. Those which satisfy secondary valencies are called ligands and represented by solid lines.
  4. Secondary valencies are non-ionizable in water.
  5. Secondary valencies are directional in nature so the complex compounds exhibit stereo-isomerism.
  6. Some negative ligands may satisfy both primary and secondary valencies.These donot ionize in water.

The following examples will explain Werner’s theory.

1)    CoCl3 • 6NH3 :

In the above complex Co has co-ordination number 6. Six neutral NH3 molecules satisfy secondary valencies.  The Cl ions satisfy primary valencies and ionise in water giving 3Cl ions.

CoCl3 • 5NH3:

In this complx 5NH3 molecules and one Cl satisfy the secondary valencies. The Cl ion can also satisfy primary valency. So it will not ionise in water. Only two Cl ions satisfying primary valencies and ionise in water.

CoCl2 4NH3:

 

Two chloride ions satisfy primary as well as secondary valencies. They will not ionize in water. The remaining four secondary valencies are satisfied by NH3 one Cl ion which satisfy only primary valency ionize in water.

C0CI3 •3NH3:

Three Cl ions satisfy both primary and secondary valencies.   The remaining three secondary valencies are satisfied by NH3 molecules. No ions are formed when dissolved in water.

Why was Werner’s theory necessary? Discuss the theory.

Necessity for Werner’s theory:

  1. All transition elements form complex compounds.
  2. Complex compounds are formed by the addition of two stable independent compounds.
  3. The complex compound is a new chemical species different from the compounds from which it is formed.
  4. Complex compounds do not give tests for all the ions present in it.
  5. Complex compounds retain their identity in solution.
  6. The physical properties such as colour conductivity etc. of complex compounds are different from those substances from which they are formed.

The above facts of complex compounds could not be explained by the valence theories known. So Werner’s theory is necessary to explain the above facts.

Werner’s theory: Werner’s theory of complex compounds is based on the following postulates.

1.  Every complex compound has a central metal atom or ion.

2. The central metal exhibits two types of valencies a) primary valency and b) secondary valency.

Primary Valency:

  1. The number of primary valencies of metal is equal to its oxidation number.
  2. The primary valencies are non-directional, represented by dotted lines,
  3. Primary valencies are ionizable in water.
  4. Primary valencies can be satisfied only by negative ions.

In modern theory primary valencies are nothing but ionic bonds.

b) Secondary Valency:

  1. The central metal has a fixed number of secondary valencies called co-ordination number. The co-ordination number varies from 2-10.
  2. Secondary valencies can be satisfied by neutral molecules atoms or by negative ions. Rarely positive ions can also satisfy secondary valencies.
  3. Those which satisfy secondary valencies are called ligands and represented by solid lines.
  4. Secondary valencies are non-ionizable in water.
  5. Secondary valencies are directional in nature so the complex compounds exhibit stereo-isomerism.
  6. Some negative ligands may satisfy both primary and secondary valencies.These donot ionize in water.

Discuss the colour of the transition metal complexes with suitable examples.

White light consists of seven colours VIBGYOR. When light falls on a substance the following changes may occur.

  1. All the light radiations may be reflected. Then the substance will appear white in colour.
  2. All the light radiations may be absorbed. Then the substance will appear black in colour.
  3. If one colour of the visible radiations is absorbed and the remaining colours are reflected then the substance will appear in the complimentary colour.

The colour of the transition metal ions is due to the incompletely filled d – orbitals. On the free, isolated gaseous metal ion all the d 5 orbitals of transition metal ions are degenerate. But in the compounds or hydrated ions or complexes the d – orbitals split up into two sets one set consists of and and another set consists of dxv, duz and dxz.  One set will have lower energy while other will have higher energy but the average d – orbital energy remains same.

When white light falls on the metal ion, the electron in the lower energy set of orbital absorb energy and is excited to another set of d – orbitals having higher energy. Since the energy difference between the two sets of d – orbitals is less, transition of electrons is possible due to the absorption of visible radiations (  λ= 400 – 750 nm). So the transition metal ion appears in the complimentary colour of the absorbed radiation.

Write postulates of valence bond theory?

Valence bond theory (VBT) of complex compounds was proposed by Linus pauling. The formation of a complex between a metal and the ligands according to this theory may be explained by the following postulates

  1. The metal M first loses requisite number of electrons to form the ion. necessary 2. number of vacant orbitals are created.
  2. The number of electrons lost is numerically equal to the oxidation state of the metal
  3. The vacant orbitals of the metal ion undergo suitable hybridisation so that the desired shape is obtained by the complex.
  4. The hybrid orbitals are filled with electron pairs donated by the ligands.

 

Coordination number Type of hybridization Molecular geometry
2 sp linear
3 sp2 trigonal planar
4 sp3 tetrahedral
4 dsp2 square planar
5 Sp3d (or)

dsp3

trigonal bipyramid
6 sp3 d2 or d2sp3 octahedral
  1. The non bonding metal electrons occupy the inner d – orbitals and do not participate in the hybridization
  2. If the ligands are strong like CN, CO, NH3 the d electrons are rearranged vacating some d – orbitals which can participate in hybridization.
  3. If the ligands are weak like F, CI and H2O the d electrons are not rearranged.
  4. The d – orbitals involved in the hybridisation may be either (n-1) d orbitals or nd- orbitals.
  1. The complexes formed by the involvement of (n -l)d orbitals in hybridisation are called inner orbital complexes or low spin complexes,
  2. The complexes formed by the involvement of (n-1)d – orbitals in hybridization are called outer orbital complex or high spin complex.
  3. Each ligand contain a lone pair of electrons. A dative bond is formed by the overlap of a vacant hybrid orbital of metal ion and a filled orbital of ligand.
  4. The complex will be paramagnetic, if any unpaired electron present, otherwise diamagnetic.

Discuss the complex on the basis of valence bond theory.

The formation of complex CoCl3, 6NH3 on the basis of valence bond theory is discussed here.

1)   The metal atom first loses the requisite number of electrons to form the ion.

In the complex CoCl3. 6NH3 the atomic number of cobalt is 27. In this complex the oxidation number of cobalt is + III. So it loses three electrons and convert into Co3+ion.

NH3 is a strong ligand so the inner d electrons are rearranged vacating some d – orbitals which can participate in hybridization.

In this ‘x’ represents the electron from the donor atom of the ligand.

In hexammine cobalt (III) ion [Co (NH3)6]3+ the coordination number of Co+3 is 6.  So six vacant orbitals undergo hybridization i.e., d2sp3. Hence the complex ion has an octahedral structure.

Since the complex ion has no unpaired electrons the magnetic moment s= O. So the complex diamagnetic.

Explain the complex formation of the following compounds on the basis of VBT

 i)[Co(NH3)6]Cl3        ii) [Cu (NH3)] SO4

  1. Same as in the above question.
  2. [Cu (NH3)4]So4: In the complex [Cu(NH3)4] SO4 the atomic number of copper is 29. In this complex the oxidation number of copper is + II. So it loses two electrons and converts into Cu+2

In this the x represents the electron from the donor atom of the ligand.

The complex ion [Cu (NH3)4]+2 is having square planar structure. To explain this it was suggested that the 3d electron is promoted to 4p orbital vacating a d. – orbital. Now dsp2 hybridization takes place. The coordination number of copper in this complex is 4. The complex has square planar structure.

Since the complex has one unpaired electron (i.e. n = 1) the paramagnetic moment µs = 1.73. So the complex is paramagnetic.

Define EAN. Calculate the EAN of the following metals in their respective complexes, I) [Cu (NH3)4] (OH)2   II) [Co (H2O)6 (NO3)]3 III) K4 [Fe (CN)6] 

 Effective Atomic Number (EAN): The total number of electrons possessed by the central metal in a complex after coordination is known as Effective Atomic Number (EAN) of the metal in that complex.

The EAN of a metal in a complex is obtained by subtracting the number of electrons lost by the metal in its ion formation from the atomic number Z and then adding the number of electrons gained through coordination.

EAN of a metal in a given complex =

[Z – (no. of electrons lost by the metal) + (no. of electrons gained by the metal through coordination)]
  1. i) [Cu (NH3)4] (OH)2

This complex contains [Cu(NH3)4]+2 and 20H ions! Since the NH3 is neutral ligand copper is in + II oxidation state in [Cu (NH3)4]+2 ion.

Atomic number of copper is 29 coordination number of copper ion this complex is 4. EAN-29-2+ (2×4) = 35

  1. ii) [Co (H2O)6] (NO3)3

This complex contain[Co (H2 O)6]3+ and 3 NO3 ions. Since H2O is neutral ligand cobalt is in+111oxidation state in [Co (H2O)6]+3 ion.

Atomic number of cobalt is 27. Coordination number of cobalt in this complex is 6.

EAN = 27-3 + 2(6)=36

iii)  K4 [Fe (CN)6]

This complex contain 4K+ and one [Fe (CN)6]4- ions. CN is a negative ligand. There are 6 CN ions but the complex ion has only four negative charges. So two negative charges are neutralised by two positive charges in Fe+2. So the oxidation number of ion in this complex is + 11. Coordination number of iron in this complex is 6.

EAN = [26 – 2 + (2 x 6) ] = 36

What are the distinguishing features of complex compounds?

  1. i) Complexes are formed by chemical union of a apparently stable compounds

Ex : CuSO4 reacts with NH3 to form [Cu (NH3)4]SO4.

  1. ii) The complex compound is a hew chemical species different from the compounds from which it is formed

Ex : Potassium ferrocyanide contains the anion [Fe (CN)6l4- which is not present in either KCN or Fe (CN)2 from which it is formed

iii)  Complex molecule or ion retain its identity in solution.

iv) The physical properties such as colour conductivity etc. of the complex species are distinctly different from those of the substances from which it is formed.

Ex: CuSO4.4NH3 is dark blue while the compounds forming the complex possess different colours. NH3 is colourless and CuSO4 is pale blue in colour.

What do you understand by the term Iigand?

The ions or groups of atoms that surround the metal ion or atom in a coordination complex are called ligands.

The Iigand may be a molecule or ion bonded to the central metal through secondary valencies of the metal. Some times a Iigand may satisfy both the primary and secondary valencies of the central metal ion or atom. The legand is a donor of electrons and the metal ion is an acceptor. Sometimes the Iigand may possess acceptor orbitals in addition to donor orbitals.

Ligands may be classified according to the number of positions they occupy around the metal ion. As per the number of,electron pairs donated they are named as mono, bi, tridentat ligands.

Type of Iigand Name of Iigand No. of electron pairs donated Donor atom in the the Iigand
Monodentate

 

 

Bidentate

Ammonia

aqua / aquo

 

Oxalate ion

1

1

 

2

N in NH3

O in H2O

 

O in COO-

 

Glycinate ion

 

2

        COO-

H2 NCH2COO-

OandN

Polydentate Ethylene diamine- tetra acetic acid EDTA More than 2 4, 5 or 6

 

Another way of classification of ligands on the donor and acceptor properties of ligands. Example : i)   Ligands with one or mdre lone pairs of electrons.

  1. ii) Ligands without a lone pair of electrons but with the bonding electrons.

Third type of classification of ligands is based on the charge they possess

  1.  Neutral ligands:         NH3, H2O etc.
  2.  Negative ligands:       CN, Cl, NO2
  3.  Positive ligand:           NO+

what are the shapes of complex ions with coordination numbers 4 and 6?

The complex ion in which the coordination number of metal is atom or ion is 4 can have two geometries. When the central metal ion is involved in dsp2 hybridisation the complex will have square planar structure. If the central metal ion is involved in sp3 hybridisation the complex will have square planar structure. For example [Ni (CN) 4]2- has square planar structure while [NiCl4]2‑ has tetrahedral structure in [Ni(CN)4]-2hybridisation in nickel is dsp2 while in [NiCl4] the hybridisation in nickel is sp3.

Atomic number of nickel is 28. In both these complexes nickel is in +II oxidation state. But CN being strong ligand rearrangement of electrons takes place leading to dsp2 hybridisation. While in [NiCl4]2-Cl being weak ligand no rearrangement of electrons takes place leading to sp3 hybridisation outer electronic configuration

Here x is the electron from the donor atom of ligand.

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