Periodic Table

INORGANIC CHEMISTRY

By-- P.K.SINHA

THE PERIODIC TABLE

DOBEREINER’S TRIADS:-

Dobereiner classified elements into groups of three, called triads. The properties of the middle element in a triad is the average of the properties of the first and the third elements. For example, (Li, Na, K) and (Cl, Br, I). However, only a limited number of such triads could be found.

NEWLAND’S LAW OF OCTAVES:-

John A. R. Newlands in 1865-66 developed his law of octaves. He found that when the elements were arranged in order of their increasing atomic weights, any given elements was similar to the either element that followed it. For example, K, the sixteenth element in Newland's list and properties similar to Li, the second element in the list.

MENDELEVE’S PERIODIC TABLE

In these tables, the elements were arranged in the order of their increasing atomic weights. Lother Meyer used the physical properties such as the atomic volume, melting point and boiling point to arrive at his table of elements. Mendellev’s system was more elaborate. He used a broader range of physical and chemical properties to classify the elements. In particular, Mendelleve relied on the similarities in the chemical formulae of the compounds formed by the elements. Mendelleve stated the Periodic Law.

The properties of the elements, as well as the formulae and properties of their compounds depend in a periodic manner on the atomic weight of the elements” There were a few positions in the table (e.g. Ar, K, and Te, I) where elements having the higher weight preceded the element with lower mass for proper periodicity in the properties. He therefore, ignored the order of atomic weights to group together elements which had similar chemical properties. His proposal was even backed by the predictions for the undiscovered elements. He had the courage and foresight to leave gaps in the table for elements which were not known at that time. He could predict the properties of those missing elements from a study of the properties of other elements in the same group. For example, germanium, gallium and scandium sere not discovered at that time when Mendeleev proposed his periodic table. He named these elements as eka-sillicon, eka – aluminum and eka-boron because he believed that they would be similar to silicon, aluminum and boron respectively.

Note: 1

2.

In true sense of words Mendelleve organized the elements according to a regular increase or decrease in valence (the capacity of an element to combine with another element). Properties were grouped together. Due to this systematic work and far – reaching ideas, Mendelleve is usually given the credit for the design of the periodic table as we know it today. The modern periodic table is essentially similar to that of Mendellev’s with a separate column added for noble gases which were not discovered until then

THE MODERN PERIODIC TABLE The modern version of the periodic law is stated as : “The physical and chemical properties of the elements are the periodic functions of their atomic masses”. LONG FORM OF THE PERIODIC TABLE There are many forms of the periodic tale. The long form of the periodic table is the most convenient and the most widely used and is presented here. The horizontal rows are called PERIODS. Elements having similar chemical and physical properties appear in vertical columns and are known as GROUPS or FAMILIES. Altogether there are seven periods and 18 groups.

PERIODS ST

1 Period IInd Period IIIrd Period

Contain only 2 elements namely 1H, 2He and is the shortest period. Contains 8 elements namely 3Li, 4Be, 5B, 6C, 7N, 8O, 9F and 10Ne is known as the short period. Contains 8 elements : 11Na, 12Mg, 13Al, 14Si, 15S, 16S, 17Cl, and 18Ar. The IIIrd period is also known as short period. Contact at [email protected]

INORGANIC CHEMISTRY IV Period

V Period VI Period

VII Period

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starts with potassium and contains 18 elements : 19K, 20Ca, 21Sc, 22Ti, 23V, 24Cr, 25Mn, 26Fe, 27Co, 28Ni, 29Cu, 30Zn, 31Ga, 32Ge, 33As, 34 Se, 35Br and 36Kr and is known as long period. Starts with rubidium and contains 18 elements : 37Rb, 38Sr, 39Y, 40Zr, 41Nb, 42Mo, 43Tc, 44Ru, 45Rh, 46 PD, 47Ag, 48 CD 49In, 50Sn, 51Sb, 52Te, 53I, 54Xe.The fifth Consists of 32 elements, starting from cesium (55Sc) and ending with radon. It is called as longest period. This period includes the 4f shell that includes Lanthanides (58Ce, …..71Lu). Like the sixth period would have a theoretical maximum of 32 elements. This period however is incomplete and at present contains 19 elements starting from 87Fr (francium) to 92U. All these elements are naturally occurring but rest are radioactive with very short half lives. These also include a part of inner transition elements, 90Th, …. 103Lr.

GROUPS The atom of the element in a single vertical column have the same or very similar electronic configurations in the highest occupied orbitals and are therefore said to belong to the same GROUP or FAMILY of elements. According to the new IUPAC recommendations the groups are numbered form 1 to 18. Bases on the electronic configuration, we can classify elements in to four types. 1. Noble Gases 2. Representative Elements 3. Transition Elements and 4. Inner Transition Elements

1.

Noble Gases The noble gases are found at the end of each period in group 18. With the exception of 2 6 helium, these elements have ns np electronic configuration in the outermost shell. Helium has 1s2 configuration. All the energy levels that are occupied by the electrons are completely filled and this stable arrangement of electrons cannot be easily altered. These elements have very low chemical reactivity.

2.

Group Group Group Group Group Group Group

Representative Elements (s and p block elements)

The elements of Group 1 (alkali metals), Group 2(alkaline earth metals) and Group 13 – 17 constitute the Representative Elements. For the representative elements the period in which the element is located equal the principal quantum number of the differentiating electron, i.e., 1-2 1-5 if an element is in nth period then the electronic configuration will be either ns or np . 1 1 1 1 1 1 1 Consists of 1H(1s ), 3Li(2s ), 11Na(3s ), 19K(4s ), 37Rb(5s ), 55Cs(6s ), 87Fr(7s1).The -1 common outermost electronic configuration is ns . Elements belonging to this group are known as Alkali Metals. 2 2 2 2 2 2 2 Contains 4Be(2s ), 12Mg(3s ), 20Ca(4s ), 38Sr(5s ), 56Ba(6s ), 88Ra(7s ). The elements belonging to this group are known as Alkaline Earth Metals. The common 2 outermost electronic configuration of the elements of this group is ns . 2 1 3 Starts with 5B(2s 2p ). The general electronic configuration of the elements of this 2 1 group is ns np . This group is also known as the Boron Family. 2 2 2 2 4 Starts with 6C(2s 2p ). The general electronic configuration is ns np . This group is also referred to as the carbon family. 2 3 5 Starts with 7N(2S 2P ).the general outermost electronic configuration of the elements 2 3 of this family ns np . This group is also referred to as the Nitrogen Family. 2 4 6 Starts with 8O(2S 2p ) and is known as Oxygen Family. The general outermost 2 4 electronic configuration of the elements of this family ns np . The elements of this family are also known as Chalcogens. 2 5 2 5 2 5 2 5 2 5 7 Contains 9F(2S 2p ), 17Cl(3s 3p ), 35Br(4s 4p ), 53I(5s 5p ) and 85At(6s 6p ) The elements of this group are commonly known as Halogens.

Typical Elements Elements of the third period are known as Typical elements, examples, 11Na, 12Mg, 13Al, 14Si, 15P, 16S and 17Cl Properties of all elements present in a particular group e.g. of group 1 resemble with the properties of 11Na and not with 3Li.

Bridge Elements Elements of second period are known as Bridge Elements. Properties of the bridge elements resemble with the properties of the diagonal elements of the third period. For example, Li resembles Mg; Be resemble Mg; Be resembles Al; B resembles Si etc.

Note:- Noble gases are also grouped with representative P – block elements as they come at the end of each period. The chemical and physical properties of the representative elements is determined by the Contact at [email protected]

INORGANIC CHEMISTRY

By-- P.K.SINHA

number of electrons in the outer most shell called the Valence Shell. The number of valence electrons for groups 1 and 2 is the same as the group number for group 13 – 17, this number is obtained by subtracting 10 from the group number.

3.

Transition Elements (d- block elements) These are the elements of Groups 3 to 12 in the center of the period table. The elements in which the last electron enters the d sub-shell of the penultimate energy level are called d 1 - 10 1 -2 block elements. The outer most configuration of these elements is (n-1)d ns They are metals. They form colored ions and exhibit variable valency. However, Zn, Cd and Hg which 1 – 10 2 too have (n – 1) d ns configuration in their outermost shell do not form colored ions and are not regarded as transition elements.

4.

The Inner Transition elements ( f – Block Elements)

The rows of elements at the bottom of the periodic table are called the Lanthanide and actinide series. These elements in which the last electron enters the f sub-shell of the antipenultimate (third to the outermost shell) shell are called f block elements Their outer electronic configuration is (n-2)f1 – 14 (n-1) d0-1 ns2. The differentiating electron is an felectron. They are all metals. Within each series the properties of the elements are quite similar .

PERIODIC TRENDS IN PROPERTIES 1. VALENCE:An important chemical property of the elements exhibiting periodic trends is their Valence. It is defined as combining capacity of an element. It can also be defined it terms of valence electrons (electrons in the outermost shells). The valency is equal to number of valence electrons (or equal to 8 minus the number of valence electrons. 1. the valence of representative elements is usually equal to the number of electrons in the outermost orbitals and /or equal to eight minus the number of outermost electrons. 2. Transition elements do not exhibit any general trend. The reason for this that those elements have variable valencies due to availability of vacant d- subshells in them. 3. Inner transition elements also do not exhibit any general trend in the valency.

2.

ATOMIC AND IONIC RADII It is impossible to define the size of atoms as we know that atoms have no shop boundaries due to the delocalized picture of electron cloud. An estimate of he atomic size can be made by knowing the distance between the atoms in the combined state. There are three operational concepts of atomic radius. a. If the bonding is covalent, the radius is called covalent radius. b. If the bonding is ionic, the radius is called ionic radius. c. If the two atoms are not bounded by a chemical bond (as in noble gases), the radius is called Vander Walls radius. a. Covalent Radius : It is half of the distance between the nuclei of two like atoms bounded together by a single bond. For Example, the bond distance in hydrogen molecule (H2) is 74 pm and half of this distance is taken as the atomic radius of hydrogen. This radius is known as the covalent radius. b Ionic Radius: It is the effective distance from the nucleus of an ion up to which it has its influence on its electron cloud. c van der Wall’s radius : It is one half of the distance between the nuclei of two adjacent atoms belonging to two neighboring molecules of an element in the solid state. The covalent radius is always smaller than the van der Wall’s radius because in the formation of chemical bond, the tow atoms have to come closer to each other. This is why the inert gases (where covalent radius is generally not possible) tend to have a larger size. 1. The size of atoms increases as we go down a column of periodic table. This increase is attributed to the increase in the number of shells around the nucleus. 2. The size of the atoms decrease as we go across the period from left to right except group 18 (Noble Gases). This decreases in the size is attributed to the increases in the nuclear charge and hence the attraction. 3. A positive ion is always smaller in size than the corresponding neutral atom. 4. A negative ion is always bigger in size than the corresponding neutral atom. 5. The size of ions increases as we go down a group provided that we are comparing ions of same charge. 6. Atoms or ions with the same electronic configurations are called as iso-electronic. If we consider a series of iso-electronic species (atoms or ions), the size decreases with the increasing atomic number. To illustrate the concept: consider the radius of the following iso-electronic species, all having 10 electrons. Contact at [email protected]

INORGANIC CHEMISTRY

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N3- > O2- > F- > Ne > Na+ > Mg2+ > Al3+ Note that the successive increase in the values of Z/e ratio decreases the values of ionic (atomic ) radii.

3.

IONIZATION ENERGY The chemical nature of an element depends on the ability of its atoms to accept of donate electrons. A quantitative measure of these tendencies is the Ionization Energy or the Electron or the Electron Affinity. The ionization Energy (IE) is defined as the energy required to remove an electron from an isolated gaseous atom (M) in its ground state. M(g) + IE M+(g) + e -1 The ionization energy is expressed in units of kJ mol or in ev / electron. The energy required to remove the second electron from the same element is known as the second ionization energy. The second ionization energy is higher than that required for the removal of the firs electron because it is mire difficult to remove an electron from a positively charged species than the second and so on. If the term ionization energy is not qualified (i.e., if the first, second and third is not motioned), it is taken as the first ionization energy. + +++ M (g) + second IE M (g) + e +++ M++(g) + third IE M (g) + e

FACTORS INFLUENCING IONIZATION ENERGY : The ionization energy depends upon the following factors :

Nuclear Charge Ionization energy increases with increase in the nuclear charge. With increase in the nuclear charge the force with which the electron is bound with the atom increases and hence it becomes more difficult to remove the outermost electron. For example the ionization energy of He is 567 kcal/mol, whereas that of H is only 314 kcal/mole. This increase in the ionization energy can be explained on the basis of the increase in the nuclear charge.

Atomic Size With the increase in the atomic size, the force with which the electron is bound to the atom decreases and hence with increase in the size, the ionization energy decreases. For example the ionization energy decreases as we don down the group.

Screening Effect The force of attraction between the valence electrons and the nucleus is greatly shielded by the presence of core electrons. With the increase in the electrons in the inner sub shells the force of attraction between the outermost electron and the nucleus decreases and hence the ionization energy decrease.

Penetration of Electrons The ionization energy also depends upon the penetrating power of the electrons. For example, the penetration of s-electrons. The f-electrons has the least penetration. The ionization energy increases with increasing penetration.

Stability of the Electronic Configuration The half and fully filled orbitals are most stable as compared to their neighbours and hence the ionization energy of the fully filled or half filled orbitals is higher as compared to their neighbors. For instance, he ionization energy for N is higher that for C and O.

PERIODIC TRENDS 1. It is observed that the ionization energy of an element strongly depends on its electronic

2.

3.

configuration and thus show periodic variations. The maxima are found at the noble gases which have completely filled electron shells. The high ionization energies of the noble gases can be connected with their extremely low chemical reactivity. Similarly, the high reactivity of alkali metals is reflected I their. In a group : First ionization energy decreases as we go down a group in the table. It measures the ease of removing an electron from the outer shell. As we go down a group, this shell is farther away from the nucleus. As a result nuclear attraction decreases. Though the positive charge of nucleus increases, its effect is weakened due to the shielding supplied by the inner shells to the outermost shell. In a period: As we go across a period from left to right, the atomic size decreases. As the number of shells in a particular period remain the same and the additional electrons are being continuously introduced in the same shell, the nuclear charge increases and hence the outer electrons are greatly attracted to the nucleus. Hence it becomes difficult to remove them and consequently ionization energy increases. The figure below shows the first ionization energy Contact at [email protected]

INORGANIC CHEMISTRY

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of elements of the second period as a function of the atomic number Z. a. IE (B) < IE (Be). Be has its 2s orbital fully filled whereas B has one unpaired 2p electron and it is easier to remove a lone electron rather than that from a paired orbit. Hence extra-stability of fully filled sub- shell is the cause of this irregularity. b. IE (O) < IE (N). nitrogen has an exactly half filled outermost electronic configuration and hence is extra stable. Thus extra-stability of half filled sub-shells is the cause of irregularity. 4. Whether all the outer shell electrons are removed, the next I.E. is much greater than the previous value of I.E. For the same element. Note that first I.E for the same is 72.64 eV.

Remarks: 1. 2. 3.

4.

Lower the ionization energy greater is the metallic character. Lower the ionization energy greater is the tendency to from ionic bond. Since higher oxidation states are associated with higher ionization energies, higher oxidation state ions from covalent compounds. 4. Lower the ionization energy, greater is the reducing power. ELECTRON AFFINITY The electron affinity is the amount of energy releases when an isolated gaseous atom accepts an electron to form a monovalent gaseous ion. -

-

X (g) + e X (g) + Energy Electron affinities can be positive or negative. When energy is released in the process of attachment of an electron to an atom, the electron affinity is taken as positive and if energy is absorbed, electron affinity is taken to be negative like in inert gases. Thus the magnitude of electron affinity measures the tightness with which the atom can hold the additional electron. The larger value of E.A reflects the greater tendency of an atom to accept the electron. Electron affinity values are influenced by a. size of the atom b. Nuclear charge c. Electronic configuration PERIODIC TRENDS 1 In a Period Electron affinity in general increases with atomic number across a period paralleling a decreasing in atomic size. It will be easier to add an electron to a smaller atom since the added electron on an average would be closer to the positively charged nucleus. Thus electron affinity of halogens is maximum, but the values are unexpectedly low in case of Be, N and Ne in the second row because of stable configurations. 2. In a Group Generally electron affinity decreases as we go down a group because the size of the atom increases and the electron added goes to the higher shells. It may be noted that the electron affinity of fluorine does not fall line the with the general trend. In this case, electron affinity of chlorine is unexpectedly higher than that of fluorine. This is, because of very small size of fluorine the inter electronic repulsions in relatively compact 2p sub-shell are much more then those in relatively larger 3p sub-shell of chlorine atom. Hence, of all elements, chlorine has the highest electron affinity. 3. The rare gases have no vacancies in their valence orbitals, and any electron added to them would have to be placed in an orbital of next higher quantum number. Because of the screening of the inner electrons, this added electron would feel very little net attraction of the atom, and consequently the electron affinities of the rare gases are essentially zero. The argument helps to show why the rare gases tend to be so inert. Since their electron affinities are so small, they never enter compounds as electron accepters. On the other hand, their ionization energies are so high that they are oxidized with great difficulty, and consequently they form only a limited number of compounds. 4. Let us now talk about the second electron affinity. On taking up an extra electron an atom becomes negatively charged (anion) and now a second electron is to be added to it. The anion will repel the incoming electron and an additional energy releases while the second E.A. si the energy required or the value of second e.A. is negative. O(g) + e  O (g) + 141 kJ O (g) + e- + 770 kJ  O2-(g) Remarks: 1. Electron affinity decides the oxidizing power : Greater the electron affinity greater is the oxidizing power. 2. Higher the electron affinity of an electron higher is its tendency to form ionic bond. Contact at [email protected]

INORGANIC CHEMISTRY 3.

By-- P.K.SINHA

Greater the electron affinity, more is the non metallic nature.

ELECTRONEGATIVITY Pauling introduced the concept of electro negativity and described it as a measure of the tendency of an atom in a molecule to attract the shared pair of electrons to itself. It is important to note that electron affinity and electronegativity both measure the electrons attracting power but the former refers to an isolated gaseous atom while the latter to an atom in a compound. Thus electron affinity is attraction for a single electron while electronegativity is for a pair of bonded electrons., Further , electron affinity is energy while electronegativity is a tendency. Electronegativity depends upon (a) the size of the atom and (b) electronic configuration. PERIODIC TRENDS 1. In a period, electronegativity increases from left to right. This is due to decrease in size and increase in nuclear charge. Thus the alkali metals possess the lowest value, while the halogens have the highest. Inert gases have zero electronegativity. 2. In a group, electronegativity decreases from top to bottom. This is due to increase in atomic size. 3. It follows the trends as shown by electron affinity or ionization energy. This means that higher I.E. and greater E.A., both results in high electronegativity. 4. The difference in electronegativity values of two atoms in a compound predicts the nature of chemical bond formed between two atoms. If the difference is 1.7, the bond formed between the two atoms will be 50% ionic in nature. If the difference is more than 1.7, bond is more than 50% ionic and vice versa. Since electronegativity is a relative property, it has no units. Electronegativity may be expressed on the following two scales: i. Mulliken’s scale: Mulliken regarded electronegativity as the average value of ionization potential and electron affinity of an atom. Electronegativity ii

=

Ionisation potential  electron affinity 2

Pauling scale : Pauling scale of electronegativity is most widely used. It is based on excess bond energies. He determined electronegativity difference between the two atoms and then by assigning arbitrary values to few elements (e.g. 4.00 to fluorine, 2.5 to carbon and 2.1 to hydrogen), he calculated the electronegativity of the other elements.

X A  X B  0.208

E

Where XA and XB are electronegativities are related as below x (Pauling) = 0.34 X (Mulliken) - 0.2 Some of the elements in their decreasing order of electronegativities are as follows: F(4.1) > 0(3.5)> N(3.1)>Cl(2.8)> S(2.4)> H(2.1) > metals

MAGNETIC PROPERTIES The magnetic properties of matter depend upon the electronic configuration of atoms contained in it. Any electron in an atom has two types of motions. One is its motion about the nucleus and other its spin about is own axis. A single electron spinning about its own axis possesses a magnetic dipole moment. For two electrons in an orbital, the spins are opposite and hence their magnetic dipole moment cancels. Consequently and atom which has an unpaired electron in its orbital possesses a dipole moment, while an orbital containing no unpaired electrons possesses no dipole moment. A substance that contains atom with unpaired electrons is termed as paramagnetic. The substance that contain such type of atoms are attracted by a magnetic field ad these substances show increase in weight. A substance that does not contain any unpaired electrons is known as diamagnetic. A diamagnetic. A diamagnetic substance when placed in a magnetic field, is weakly repelled by it and shows a decrease in weight. The strongest kind of magnetism of ferromagnetism. When a paramagnetic substance is placed in a magnetic field, the field will align the unpaired spins and magnetize the substance. If the substance keeps its magnetism after the fields is removed, it is called as ferromagnetism. It is much stronger than Paramagnetism. The most important examples of such type are Fe, Ni and Co. The magnetic moment is given by the formula

  n(n  2) BM

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By-- P.K.SINHA Where BM = Bohr Magneton N = no. of unpaired electrons.

If n = 1,  = 3 = 1.73 BM If n = 2,  = 8 = 2.83 BM Thus greater the number of unpaired electrons in substance, greater the magnetic moment.

MAIN FEATURES OF s-, p- d- AND F- BLOCK ELEMENTS s- BLOCK ELEMENTS 1.

2

3. 4 5

6 7 8.

They have low values of first ionization energies. Due to this, these elements can lose an electron quite easily and hence are strongly electropositive, good reducing agents, highly reactive. All of these elements when put in a flame, important colour to it. Due to low IE, the elements absorb some energy when brought near the flame and hence they jump to higher sates. Now when they fall back to ground state, emits radiations in visible region. Be an Mg do not give any colour to the flame because they are very small in size and hence possess high IE and do not absorb energy in visible region. The colours imparted by some important metals are Na (Yellow) , K (Violet), Li (Lilac), Ba (apple green), Ca (brick red). These are metals and hence posses typical metallic properties such as : lustre, ductility, malleability, conduction of heat and electricity. The cations of all of these elements are diamagnetic and colourless. The reason being, that there are no unpaired electrons in their cations. They form hydroxides, Which are very strong bases. These include NaOH, KOH, Mg (OH)2’ etc. The M – OH is highly polar and hence an easily give up OH- ion when dissolved in water. The basicity of the alkaline earth metals increases as we move down the group. the oxides of s-block elements are basic The mainly form ionic compounds with non – metals (NaCl, MgCl2 etc.). Some of these do form covalent compounds but have a very strong ionic character (BeCl2). Due to their low ionization energy , they can easily lose their valence electrons and hence are good reducing agents. Reducing character is related to their high oxidation potentials.

P - BLOCK ELEMENTS 1.

2.

3 4

5

Metallic character of these elements decreases as we move from left to right. (metal, rd metalloids, non metals) . In 3 period, Al is metal, Si is metalloid and P,s, Cl are non metals. On the other hand as one goes down a group metallic character increases. In group 14, c is non metal, Si & Ge are metalloids whereas Sn and Pb are metals. the reducing character of these elements decreases as we move from left to right; Al is reducing agent, Si is weakly reducing, P & S can act both as oxidizing and reducing whereas Cl is strongly oxidizing. On the other hand as one goes down a group reducing character increases while oxidizing character decreases due to increases in metallic nature. These generally form covalent compounds. However, non-metals (,Cl, O ect.) form both ionic as well as covalent compounds. Ionic character increases down a group, Elements of 3rd, 4th 5th periods show some characteristic properties due to presence of vacant d orbitals in them. Electrons from p sub-shells can be excited to vacant d sub-shell and hence exhibits variable valency in their compounds such as : PCl3, PCl5, SF4, SF6, SO2 , HclO4 etc. The formation of polyatomic molecules like S8, P4, etc. is the result of vacant d-sub shells. The oxides of p- block elements form mixed oxides. Non – metals belonging to P – block elements form acidic oxides (NO2, SO2, SO3, P2 O5 etc.).

d - BLOCK ELEMNTS 1. 2. 3.

4.

5. 6. 7.

All of these are metals and exhibits all typical properties. 0 with the exception of Zn, Cd, Hg, La and Ag all transition metals melt above 1000 C due to excessive covalent bonding apart from metallic bonding. d- block elements are paramagnetic due to presence of unpaired electrons in them. Many of the elements belonging to group 8 to 10 are ferromagnetic, such as Iron (Fe), Cobalt (Co) and Nickel(Ni). All transition elements expect first and the last member of each series exhibit variable oxidation states. The reason for this, is the presence of vacant d-sub shells, as these d orbitals enter into chemical bond formation with other atoms. They are good conductors of heat and electricity. They are electropositive but less than s block elements. Their ionization energies are greater than those of s block elements but lesser than those of p block elements. Contact at [email protected]

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8.

Compounds of these elements are usually coloured in solid state or in solution. The colour of metal ions is due to presence of unpaired electrons or incompletely filled d- orbitals. 9. Transition metals form complexes quite readily. A complex ion is formed when a metal ion forms coordinate bonds with neutral atoms capable of donating electrons to metal ions. Transition elements or one of their ions, due to presence of vacant d- orbitals can accept electrons and thus can form complexes. 10. Due to screening effect, there is very minute gradation observed in properties like ionization energies, atomic size and electron affinity. 11. Atomic volumes of d-block elements are low, since as the inner shells are filled, the increased nuclear charge pulls in the electron cloud. This results in an increase in density and hence decrease in atomic volume. 12. Screening or the Shielding Effect: In d – block elements the differentiating electrons enters rd the inner shell (penultimate shell) 3 after 4s, 4d after 5s and so on). This affects the nuclear attraction of the positive nucleus for the outer electrons. As the differentiating electron enters the inner shells, they tend to screen or shield the outer shell electrons from the nucleus and hence offset the nuclear attractive force. This is famous as screening or shielding effect. Due to this effect, the atomic size of transition elements remains nearly same or constant. The same is the case with ionization energy, electron affinity and other periodic properties when one moves across a period. F – BLOCK ELEMNTS 1. They are all highly dense metals. These form alloys very easily with iron. 2. All of the them very high melting points. 3. They are paramagnetic in nature due to unpaired electros in their f shell. 4. They form coloured compounds and complexes. Their common oxidation state is +3. 5. Many of these are radioactive, hence their nuclear properties are quite important. 6. Ionic and atomic radii decreases as atomic number increases. There is a steady decreases in size, which is rather surprising seeing the screening effect in case of transition elements. This is attributed to Lanthanide Contraction in f - block elements. “due to entering of electrons in 4f sub shell, screening effect of 4f should have existed like in 3d, 4d …..etc., but as 4f sub shell is too diffuse (i.e. spread out in larger space), so it fails to screen the nucleus effectively as more localized and concentrated d sub – shell.” Hence the attraction of the nucleus for the outermost electrons increases steadily with atomic number and thus a steady decreases in size.

PROPERTIES OF COMPOUNDS Periodic trends are also observed in the properties of many compounds within a group of the periodic table. 1. a. For the same alkali meal, the melting points of halides decreases in the order fluoride > chloride > bromide > iodide. b. For the same halide, the melting points of lithium halides are less than those of sodium halides and thereafter they decrease as we go from sodium down to cesium. 2. The solubilities of halides of alkali metals follow regular gradation. The solubilities of fluorides of alkali metals in water increase regularly from Li  Cs. 3. a. The hydroxides of alkali metals, are appreciably soluble in water and behaves as a strong bases. The basic strength increases form Li  Cs. b. The hydroxides of alkaline earth metals, the basicity of alkaline earth metal hydroxides in water increase as we go down the group. For example Be(OH)2 is amphoteric, Mg(OH)2 is a weak base, Ca(OH)2 and Sr(OH)2 are moderately strong bases and Ba(OH)2 4. a. The solubilities of alkali metal carbonates and hydrogen carbonates in water at 298 K increases as we go down the group from lithium to cesium. b. The solubilities of alkaline earth metal carbonates, they are almost insoluble in water and their solubilities further decrease in moving down the group. 5. The stability of carbonates of alkali and alkaline earth metals, alkali metal carbonates are quite stable towards heat. Except Li2CO3 all other carbonates do not decompose on heating. On the other hand, alkaline earth metal carbonates are less stable towards heat and decompose to give CO2 gas. However, their stability increases down the group. 7. The solubilities of sulphates of alkaline earth metals, decrease as we go down the group. MgSO4 is soluble in water, CaSO4 is slightly soluble in water and SrSO4 and BaSO4 are insoluble in water. Contact at [email protected]

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Exercises:

By-- P.K.SINHA

1. In each of the following pairs, circle the species with the higher first ionization energy:

(a) Li or Cs

(b) Cl- or Ar

(c) Ca or Br

(d) Na+ or Ne

(e) B or Be

(a) Mg or Ba

(b) S or S2-

(c) Cu+2 or Cu

(d) He or H-

(e) Na or Cl

2. In each of the following pairs, circle the species with the larger atomic radius: 3. Circle the best choice in each list: (a) highest first ionization energy: C, N, Si (b) largest radius: S2–, Cl–, Cl (c) highest electronegativity: As, Sn, S (d) smallest atom: Na, Li, Be (e) most paramagnetic: Fe, Co, Ni (f) lowest first ionization energy: K, Na, Ca (g) highest second ionization energy: Na, Mg, Al (h) lowest second ionization energy: Ar, K, Ca

Answers (be sure you can explain the reason for each answer!): 1. (a) Li; (b) Ar (isoelectronic pair); (c) Br; (d) Na+ (isoelectronic pair); (e) Be 2. (a) Ba; (b) S2-; (c) Cu; (d) H(isoelectronic pair); (e) Na. 3. (a) N; (b) S2- (S2- and Cl- are isoelectronic); (c) S; (d) Be; (e) Fe (f) K; (g) Na; (h) Ca.

PERIODIC TABLE

Explain the following Why the second ionization enthalpy of an element is higher that the first? First ionization enthalpy of nitrogen is higher than the first ionization enthalpy of Oxygen. Why noble gases have positive electron gain enthalpies. Why most of the compounds of the transition elements are colored. Why the atomic radii of 5d transition elements is same as 4d transition elements? Chlorine can be converted into chloride ion easily as compared to fluoride ion from fluorine. Why Be and Mg atoms do not impart color in flame? Why is potassium strongly electropositive element? Calculate the electro negativity of fluorine from the following data: -1 -1 -1 EH-H = 104.2 kcal mol , EF-F = 36.6 kcal mol , EH-F = 134.6 kcal mol ,XH = 2.1 10. Calculate the electro negativity of carbon from the following data: -1 -1 -1 EH-H =104.2 kcal mol , EC-C = 83.1 kcal mol , EC-H = 98.8 kcal mol , XH = 2.1 11. Ionisation enthalpy and electron gain enthalpy of fluorine are 17.42 and 3.45 eV respectively. Calculate the electronegativity of fluorine. 12. The electron gain enthalpy of chlorine is 3.7 eV. How much energy in kcal is released when 2 gm of chloride is completely converted to Cl ion in a gaseous state. 13. Calculate the electronegativity of silicon using Allred-Rochow method. Covalent radius of silicon is 1.175 A. 14. The first ionization enthalpy of Li is 5.4 eV and the electron gain enthalpy of Cl is 3.6 eV. -1 Calculate H is kcal mol for the reaction. 15. Calculate the electronegativity value of chlorine on Mulliken’s scale, given that IP=13.0 eV and EA=4.0 eV. 16. Find the electronegativity of lead with the help of the given values. Screening constant () of Pb=76.70,atomic number of lead = 82 and covalent radius of Pb= 5.3 A. 1 17. The ionisation enthalpies of atoms A and b are 400 kcal mol- respectively. The electron gain -1 enthalpies of these atoms are 80.0 and 85.0 kcal mol respectively. Prove that which of the atoms has higher electronegativity. 18. For the gaseous reaction, + K+F K + F- H was calculated to be 19 kcal under conditions where the cations and anions ere prevented by electrostatic separation from combining with each other. The ionisation enthalpy of K is 4.3 eV. What is the electron gain enthalpy of F? + 19. How many Cl Atoms can you ionize in the process ClCl + e by the energy liberated for the process Cl +e Cl for one Avogadro number of atoms? Given IE =13.0 eV and Eg =3.6 eV. 20. (A),(B) and (C) are elements in the third short period. Oxide of (A) is ionic, that of (B) is amphoteric and of (C) is giant molecule. (A),(B) and (C) have atomic number in the order: (a) (A)(B)(C) (b) (C)(B)(A) (c) (A)(C)(B) (d) (B)(A)(C) 21. Match the list I with list II and select the correct answer: 1. 2. 3. 4. 5. 6. 7. 8. 9.

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List I Most electronegative element The element having highest electron gain enthalpy Most abundant electron in the universe Most abundant gas in atmosphere A B C D 1 2 3 4 4 3 2 1 4 1 2 3 2 3 4 1

List II (1) Chlorine (2) Hydrogen (3) Nitrogen (4) Fluorine

Passage -1 The first iH1 and the second iH2 ionisation enthalpies in kJ mol and the electron gain -1 enthalpy in kJ mol of a few elements are given below: Elements iH1 iH2 egH (A) 520 7300 -60 (B) 419 3051 -48 (C) 1681 3374 -328 (D) 1008 1846 -295 (E) 2372 5251 +48 (F) 738 1451 -40 Answer the following questions. Only one alternative is correct. 1. Which on of the above elements is least reactive? (a) [C] (b) [D] (c) [E] (d) [F] 2. Which one of the above elements is most reactive metal? (a) [A] (b) [B] (c) [F] (d) [D] 3. Which one of the above elements is most reactive non-metal? (a) [C] (b) [D] (c) [E] (d) [F] 4. The metal which can form a stable binary halide of formula MX2 (X= halogen) (a) [A] (b) [B] (c) [F] (d) [D] 5. Which one of above elements is least reactive non-metal? (a) [C] (b) [D] (c) [E] (d) [F] 6. The metal which can form predominantly stable covalent halide of the formula MX (x= halogen) Table Matching Problems 1. Match List-I with List II: List I List II (Group/period) (Element) (a) The period (p) Transition elements (b) IIIB Group (q) Inner transition elements (c) IA (r) Hydrogen (d) VII A (s) Halogens 2. Match List-I with List II: List I List II (a) (P+1) N (p) Al (b) (P+1) = 7 (q) Pb (c) (P+1)N (r ) C (d) (P+1) = N (s) Si Here P = Number of period N = Number of electrons 3. Match the atomic number in List I with the block in List II and group in List III: List I List II List III (Atomic Number) (Block) (Group) (a) 55 (p) p VII (b) 45 (q) f IIIB (c) 81 (r ) s IIIA (d) 64 (s) d IA 4. Match List-I with List II: List I List II (a) Halogen (p) Oxygen (b) Chalcogen (q) Chlorine (c) Noble gas (r ) Sulphur (d) Representative element (s) Neon Contact at [email protected]

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6.

7.

8.

9.

10.

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Match List-I with List II: List I List II (Element) (Information) (a) Rutherfordium Z= 104 (b) Kurchatovium f-block (c) Thorium d-Block (d) Neptunium Transuranic element Match List-I with List II: List I List II (a) Dobereiner Law of octave (b) Alexander Newlands Davy Medal (c) A.E.B. de Chancourtois Law of triads (d) Dmitri Mendeleev Based on atomic weights Match List-I with List II: List I List II (a) Lanthanide contraction d-Block rd (b) Diagonal relationship Elements of 3 period (c) Typical elements f-Block (d) Inner transition elements Be-Al Match the electronic configuration (List I) with the ionisation energy (List II) List I List II (electronic configuration) (ionisation energy) 2 (a) ns 2100 2 1 (b) ns np 1400 2 3 (c) ns np 800 2 6 (d) ns np 900 Match the column I with column II and column III: Column I Column II Column III (Atomic Number) (Group) (Period) (a) 13 IIIA 6 (b) 35 VIIA 5 (c) 56 IIA 3 (d) 78 VIII 4 Match the elements in List-I with their properties in List-II: List I List II (a) Mercury Metal (b) Bromine Densest element (c) Osmium Element with highest melting point (d) Wolfram (Tungsten) Liquid at room temp

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SUBJECTIVE PROBLEMS

1.Explain why in aqueous solution, (a) Ti3+ is colored but Sc3+ is not. (b) Ti2+ is a reducing agent but Ca2+ is not. 1.Answer: (a) Ti3+ forms the octahedral complex, Ti(H2O)63+. The lone 3d electron is transferred between the split dûorbitals. Because the dûorbital splitting in most octahedral complexes corresponds to the energies of photons in the visible region, octahedral complex ions are usually colored. Sc3+ has no dûorbitals. (b) Ti2+ ion is a reducing agent because itÆs 3d2 electrons can be easily oxidized. The Ca2+ has hard to oxidize 3p electrons. 2.What is meant by the lanthanide contraction? Account for this phenomenon. Give two examples of its consequences. 2.Answer: In the lanthanide series, electrons are filling the 4f orbitals. Since the 4f orbitals are buried in the interior of these atoms, the additional electrons do not add to the atomic size. In fact, the increasing nuclear charge causes the radii to decrease significantly going from La to Lu. This contraction just offsets the normal increase in size due to going from one principal quantum level to another. Thus 5d elements are almost identical in size to 4d elements. This leads to great similarity in the chemistry of the 4d and 5d elements, such as Hf and Zr being remarkably similar in chemical properties. 3.There is a greater variation between the properties (both chemical and physical) of the first and second of a group or family in the periodic table than between the properties of the second and third members of the group. Consider as examples either the group containing nitrogen or the one containing oxygen. Select three properties and discuss the variation of these properties to illustrate the generalization expressed in the first sentence of the question. 3.Answer: Include discussion of small size and compact electron clouds of the first period atoms. Example properties: N P As O S Se atomic 0.70 1.10 1.21 0.66 1.04 1.17 radii (┼) specific 0.0012 1.82 5.73 0.0014 2.07 4.79 gravity 3 (g/cm ) melting 63 317 1090 55 386 490 point (K) 4.Consider the following melting points in degrees Celsius: Alkali metals Halogens Li 181_ F2 û119_ Na 98_ Cl2 û101_ K 63_ Br2 û7_ Rb 39_ I2 +104_ Cs 29_ (a) Account for the trend in the melting points of the alkali metals. (b) Account for the trend in the melting points of the halogens. 4..Answer: (a) Because of their large sizes and limited numbers of valence electrons, bonding between alkali metal atoms is not as strong as in most metals. Since the atoms increase in size down the family, those near the bottom (Rb and Cs) have the greatest internuclear distances. (b) Attractive forces between halogen molecules are rather weak, they are of the instantaneous dipoleûinduced dipole type (London forces), and increase in strength with increasing molecular weight (and increasing numbers of electrons). 5.Question Li Be

First ionization Energy (kilocalories/mole) Covalent Radii, ┼ 124 1.34 215 0.90 Contact at [email protected]

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B 191 0.82 C 260 0.77 N 336 0.75 O 314 0.73 F 402 0.72 The covalent radii decrease regularly from Li to F, whereas the first ionization energies do not. For the ionization energies, show how currently accepted theoretical concepts can be used to explain the general trend and the two discontinuities. 5.Answer:

450 400 350 300 250 200 150 100 50 0 Li Be

B

C

N

O

F

The trend in moving across a period is that the first ionization energy, I1, increases from group 1 (Li) to group 7 (F) because of an increase in effective nuclear charge, atoms get smaller (decrease covalent radii) and less metallic through the period. The I1 is less for B than Be because the electron to be ionized in B is in a higher energy orbital (2p) than is the electron (2s) to be ionized in Be. The I1 is less for O than N because the electron to be ionized in O is a paired electron in the 2p orbitals. At N, the outer sublevel of its atom is halfûfilled, resulting in a symmetrical spherical electron cloud. The extra electron in O reduces this symmetry and so less energy is required to remove this electron. 6. M(s) + Cl2(g) ----- MCl2(s) The reaction of a metal with chlorine proceeds as indicated above. Indicate, with reasons for your answers, the effect of the following factors on the heat of reaction for this reaction. (a) A large radius versus a small radius for M2+ (b) A high ionization energy versus a low ionization energy for M. 6.Answer: (a) As radius increases the heat of reaction decreases (less exothermic). Less energy released by ionic attraction (lattice energy inversely proportional to distance). (b) As ionization energy increases the heat of reaction decreases (less exothermic), More energy required to form M2+ while other factors remain unchanged. 7.The electron affinities of five elements are given below. 12 kcal/mole 13 Al 32 kcal/mole 14Si 17 kcal/mole 15P 48 kcal/mole 16S 87 kcal/mole 17Cl Define the term electron affinity of an atom. For the elements listed above, explain the observed trend with the increase in atomic number. Account for the discontinuity that occurs at phosphorus. 7.Answer: Electron affinity û the energy released when a gaseous atom gains an electron to form an ion. As an electron is added to the same valence shell of an atom, when Z increases, the atomic radius decreases. Therefore, the added electron in going from Al to Si and from P to S to Cl is closer to the nucleus and more energy is released (electron affinity greater). Also as the atomic radius Contact at [email protected]

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decreases, the shielding of the nucleus by the surrounding electrons is less effective, and the attraction for the added electron is greater. At P, the outer sublevel of its atom is halfûfilled, resulting in a symmetrical spherical electron cloud. The extra electron reduces this symmetry and so less energy is released when it enters the û atom to form P . 8.Discuss some differences in physical and chemical properties of metals and nonmetals. What characteristic of the electronic configuration of atoms distinguishes metals from nonmetals. On the basis of this characteristic explain why there are many more metals than nonmetals. 8.Answer: metals nonûmetals Physical properties: melting points rel. high rel. low elec. conduct. good insulators luster high little or none physical state most solids gases, liq. or solids [etc.] Chemical properties: redox agents reducing oxid. or reducing electropositive electronegative oxides basic or amphoteric acidic react with nonmetals metals & nonûmetals [etc.] Electron configurations: Metals: Valence electrons in s or d sublevels of their atoms. (A few heavy elements have atoms with one or two electrons in p sublevels.) Nonmetals: Valence electrons in the s and p sublevels of their atoms. There are more metals than nonmetals because filling d orbitals in a given energy level involves the atoms of ten elements and filling the f orbitals involves the atoms of 14 elements. In the same energy levels, the maximum number of elements with atoms receiving p electrons is six. 9. Properties of the chemical elements often show regular variation with respect to their positions in the periodic table. (a) Describe the general trend in acidûbase character of the oxides of the elements in the third period (Na to Ar). Give examples of one acidic oxide and one basic oxide and show with equations how these oxides react with water. (b) How does the oxidizing strength of the halogen elements vary down the group? Account for this trend. c) How does the reducing strength of the alkali metals vary down the group? Account for this trend. 9. Answer: (a) Oxides at left are basic and become less basic / more acidic as one moves to the right. Basic oxide: Na2O + 2 H2O - 2 Na+ + 2 OHû or: MgO + H2O - Mg(OH)2 Acidic oxide: any one of the oxides of Cl, S, or P SO2 + H2O --H2SO3 (or equivalent for another oxide) (b) Oxidizing strengths of halogen elements decrease down the group. Since atoms get larger down the group, the attraction for electrons decreases and oxidizing strength decreases. (c) Reducing strengths of alkali metals increases down the group. Since atoms get larger down the group, loss of outer electrons is easier and reducing strength increases.

Q. 10. Consider HF as an example H-F bond energy = 560 kJ/mol H-H bond energy = 430 kJ/mol F-F bond energy = 160 kJ/mol Find electronegativity of H & F. Answer The electronegativity difference is given by the equation (no derivation is required): ½ A -  B = 0.102  where A is electronegativity of A, and  is the difference between the heteronuclear bond energy and the geometric mean of the homonuclear bond energies. Contact at [email protected]

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Since F is the more electronegative of the two elements, we write: ½ F -  H = 0.102  The element having the larger electronegativity value is always placed first in the equation. ½ The geometric mean of the homonuclear bond energies = (160 x 430) = 262 kJ/mol Therefore,  = (560 - 262) = 298 kJ/mol. So to find the electronegativity of hydrogen,  H , given that the electronegativity of fluorine, F = 4, we rearrange the equation and solve for  H as shown: ½  4 -  H = 0.102 (298) ½   H = 4 - 0.102 (298)   H = 2.2 Q-11 Find Z effective of Ne . Answer Consider the Ne atom, which has electron configuration: 1s2 2s2 2p6 . There are only two shells containing electrons; those with n = 1 and those with n = 2. 1. Electrons in the same principal quantum number level contribute 0.35 to  (all electrons with n = 2 , except the one under consideration) so this is 7 x 0.35 = 2.45 This is because an electron cannot screen itself from the nuclear charge. So in this case, 7 electrons in the n = 2 shell, screen the 8 th electron, and this 8 th electron does not feature in the calculations. 2 2. Electrons in the (n-1) shell, i.e. 1s , contribute 0.85 to : so, 2 x 0.85 = 1.7 The total screening,  = 1.7 + 2.45 = 4.15 Therefore, Zeff = 10 - 4.15 = 5.85 Q12. Calculate Zeffective of discriminating electron of Magnesium. 2 2 6 2 Answer Consider Mg with electron configuration: 1s 2s 2p 3p 1. 1 electron in the valence shell (3p) will screen the other, so this electron will experience a screening of 0.35 2. The 8 electrons in the next shell (2p) will contribute 8 x 0.85 = 6.80 ( to  ) 3. The 2 electrons in the innermost shell will contribute 2 x 1 = 2 ( to ) So overall,  = 0.35 + 6.80 + 2 = 9.15 Hence, Zeff = 12 - 9.15 = 2.85

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