classification of elements and periodicity in … · the modern periodic law given by moseley is :...

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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

C1(a) Dabereiner : grouped elements in Traids. He pointed out that there were sets of three elements (Triads)which showed similar chemical properties. He also noted that the atomic weight of the central element ofthe Triad was approximately the mean of the atomic weights of the other two members. The properties ofthe middle element were in between of those end members. e.g. Li, Na, K; Ca, Sr, Ba or Cl, Br, I etc.

(b) John Newland : He developed a law of octaves.

He observed that similar elements are repeated at 8th place like the 8th note of music. The elements arearranged in the order of increase of atomic weights. Similar element means that physical & chemicalproperties of element will be same. e.g. Li has the same property as of Na.

(c) Lother Meyer arrangement : He studied the physical properties such as atomic volume, melting pointand boiling points of various elements. On this basis he plotted a graph between Atomic volume (cm3) Vs.Atomic weights.

Observation :

(i) The most electropositive alkali metals (Li, Na, K, Rb and Cs) occupy the peaks on the curve.

(ii) The less strongly electropositive elements i.e. alkaline earth metals (Be, Mg, Ca, Sr and Ba) occupy thedescending positions on the curve.

(iii) The most electronegative elements i.e., halogens (F, Cl, Br and I) occupy the ascending positions on thecurve.

On the basis of these observations, Lother Meyer proposed that the physical properties of the elements area periodic functions of their atomic weights.

(d) Mendeleev’s Periodic law and table : Mendeleev arranged all the elements in the order of increase ofatomic weight.

(i) A table formed with the help of classification of elements is called periodic table.

(ii) The method of arranging similar elements in one group and seprating them from dissimilar elements iscalled classification of elements.

(iii) He prepared the periodic table on the basis of periodic law i.e.,

“The properties of elements are periodic function of their atomic weight”.

(iv) Mendeleev’s periodic table conists of seven horizontal rows known as periods and nine vertical columnknown as groups.

(v) Periods : Out of seven periods, first three periods are short periods while the fourth, fifth and sixth periodsare called long periods.

(vi) There are nine groups in all including 8th group of transition elements and zero group of inert gases.

(vii) All the group from I to VII (except zero and VIII) were divided into sub-groups.

(viii) The group number represents the valency.

(ix) The elements of same sub group resemble to each other more closely and differ from other sub groups.

(x) He predicted the properties of the missing elements from the known properties of the other elements in thesame group. e.g. gallium and germanium were not discovered at that time. He named these elements asEka-Aluminium and Eka-silicon.

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Defects in Mendeleev’s Periodic Table :

1. Position of Hydrogen : Hydrogen resembles both the alkali metal and halogen. Hence its position inperiodic table is undecided.

2. Position of Isotopes : According to Mendeleev’s periodic table, isotopes should occupy differentpositions in the periodic table, but this is not so.

3. Position of VIII group elements : Nine elements in the VIII group do not fit into the system.

4. Positions of Lanthanides and Actinides : Their position was not justified according to the periodic lawand cannot be arranged in the order of their increasing atomic weight.

5. Dissimilar elements placed in the same group : Alkali metals (Li, Na, K etc.) are placed with coinagemetals (Cu, Ag, Au).

6. Similar elements placed Apart : Chemically similar elements like Cu and Hg, Ag and Ti, Au and Pt havebeen placed in different groups.

7. Anomalous pair of elements : Some elements of higher atomic weight have been placed before theelements of lower atomic weight. e.g. Argon (At. wt. = 39.9) has been placed before potassium(At. wt. = 39.1); cobalt (At. wt. = 58.94) is placed before nickel (At. wt. = 58.69); Tellurium (127.5) hasbeen placed before iodine (126.9).

(e) Modern Periodic law and the present form of periodic table : As a result of modern researches it isestabished that atomic number is a fundamental property not the atomic weight.

Thus this led Moseley to change the basis of calssification of elements from atomic weight to atomicnumber. The modern periodic law given by Moseley is :

“The properties of elements are periodic functions of their atomic numbers, i.e., if elements are arranged inthe order of their atomic numbers. Similar elements are repeated after regular intervals”. He also gave thefollowing formulae

i.e., = a(z – b) ; = frequency of X-rays

a, b = constant for all lines in a given series of X-rays

z = atomic number

Now we will discuss why elements with similar properties reoccurs after certain regular intervals.

Cause of Periodicity : The cause of periodicity in properties is the repeatition of similar outer electronicconfiguration at certain regular intervals which indeed determines the physical and chemical properties ofthe elements and their compounds.

(i) A modern version of table contains horizontal rows known as periods (which Mendeleev called series).Elements having similar outer electronic configuration in their atoms are grouped in vertical columns; theseare referred as Groups or Families.

(ii) According to IUPAC (International Union of pure and Applied Chemistry), the groups are numbered from1 to 18 replacing the older notation of groups O, IA, II A .......... VIII B.

(iii) There are seven periods (three short periods and four long ones).

(iv) The first period contains 2 elements. The subsequent periods consists of 8, 8, 18, 18 and 32 elementsrespectively.

(v) The 7th period is incomplete.

(vi) Till now elements upto 112 and 114 have been discovered.

(vii) Elements with z = 113, 115 and beyond are not known.

Practice Problems :

1. What is the basic theme of organisation in the periodic table ?

2. Which important property did Mendeleev use to classify the elements in his periodic table and didhe stick to that ?

3. What is the basic difference in approach between the Mendeleev’s Periodic Law and the ModernPeriodic Law ?

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4. Why do element in the same group have similar physical and chemical properties ?

5. In the modern periodic table, the period indicates the value of

(a) atomic number

(b) atomic mass

(c) principal quantum number

(d) azimuthal quantum number.

[Answers : (1) to simplify and systematize the study of elements and their compounds (2) Mendeleevused atomic weight as the basis for the classification of elements in the periodic table and he did stickto that. For example, he put gallium after aluminium and germanium as eka-aluminium andeka-silicon (3) Mendeleev Periodic law states that the physical and chemical properties of elementsare in periodic functions of their atomic weights, while Modern Periodic Law states that the physicaland chemical properties of elements are in periodic functions of their atomic numbers (4) Elementsof the same group have similar electronic configuration (5) The option (c) is correct, other’s arewrong]

C2 Arrangement of Elements in Periodic Table is according to Electronic Configuration of theElements :

Electronic configuration in Periods :

(i) Each successive period in the periodic table is associated with the filling of next higher principal energyshell i.e. n = 1, n = 2 etc.

(ii) Number of atomic orbitals in each period is twice the number of atomic orbitals available in the energylevel is being filled.

(iii) The first period starts with the filling of the lowest energy level (1s) and has the two elements – hydrogen(1s1) and helium (1s2).

(iv) Second period starts with lithium.

(v) The third period begins with sodium and added electrons enters a 3s orbital. This shell has nine orbitals(one 3s, three 3p and five 3d) but 3d orbital are of higher energy than 4s according to (n + l) rule. Therefore3d orbitals are only filled after filling the 4s-orbitals. Thus it contains only eight elements.

(vi) In the fourth period, the filling of electrons is in the fourth energy level i.e., n = 4. It starts with 4s. But inthis filling of 4d and 4f orbital does not takes place. After filling 4s orbital, filling of 3d orbital and then 4porbitals takes place. 4d and 4f are of higher energy than 5s. Thus it contains 18 elements.

(vii) In the fifth period the filling of electrons starts with 5s orbital and then 4d orbitals are filled and then three5p orbitals and filled.

Sixth Period : Corresponds to filling of sixth energy level i.e. n = 6. In this period filling takes place in(one 6s, seven 4f, five 5d and three 6p) orbitals.There are 16 orbitals are available, thus 32 elements can bethere.

(viii) Filling up of the 4f orbital begins with cerium (z = 58) and ends at lutetium (z = 71) to give the 4f-innertransition series called the Lanthanoide Series. They are placed at the bottom of the periodic table.

(ix) Similar to the sixth period, the seventh period corresponds to the filling of seventh energy shell i.e. n = 7. Itis also expected to contain thirty two elements corresponding to filling of sixteen orbitals i.e. one 7s,seven 5f, five 6d and three 7p.

(x) Filling up of 5f orbitals after actinium (z = 89) gives the 5f inner tansition series known as Actinoid Series.

Thus 4f, 5f transition series of elements are placed seprately in the periodic table to maintain its structure.

Group wise electronic configuration

(xi) Elements in the same vertical column or group have similar electronic configuration, have same number ofelectrons in the outer orbitals and have similar properties.

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Practice Problems :

1. On the basis of quantum numbers, justify that the sixth period of the periodic table should have32 elements.

[Answers : (1) in 6th period, electrons can be filled in only 6s, 4f, 5d and 6p-subshells whose energiesincrease in the order as 6s < 4f < 5d < 6p]

C3 Types of elements : s-, p-, d-, f- blocks

-s Block elements : These elements contain 1 or 2 electrons in the s-orbital of their respective outer mostshell.

(i) The elements of group 1 having outer-most electronic configuration ns1 are called as alkali metals.

(ii) The elements of group 2 having outer-most electronic configuration ns2 are are called alkaline earth metals.

(iii) Properties of s- block elements : They are reactive metals with low ionization energy.

(iv) They lose the outermost electron readily to form +1 oxidation state in case of alkali metals.

(v) Alkaline earth metals can loose two electrons to aquire +2 oxidation state very easily.

(vi) Metallic character and reactivity increases as we move down the group.

(vii) The compounds of s-block are predominantly ionic with exception of Be (beryllium)

P-block Elements : These elements contain 1-6 electrons in the P-orbital of their respective outermostshells.

(i) General electronic configuration of outermost shell is ns2np1-6, where n 2-7.

(ii) These include elements belonging to group of 13, 14, 15, 16, 17 and 18 excluding helium.

Properties of P-Block Elements :

(i) All the orbitals in the valence shell of the noble gases are completely filled by electrons. Thus it is difficultto alter this stable arrangement by addition or removal of electrons.

(ii) Thus noble gases exhibit low chemical reactivity.

(iii) Group - 17 elements are known as halogens and then have high negative electron gain enthalpy.

(iv) Similarly group - 16 elements i.e., oxygen family are also known as chalcogens, have high tendency to gainelectrons.

(v) The non-metallic character increases as we move from left to right in a period.

(vi) Metallic character increases as we move down the group.

(vii) Their ionization energies are higher than s-block elements.

(viii) They mostly form covalent compounds.

(ix) Some of them show more than one oxidation states in their compounds.

d-Block Elements : These elements characterise by filling of inner d-orbital by electrons and thereforereferred to d-Block elements.

(i) General electronic configuration of d-Block elements is (n – 1) d1-10 ns0-2.

(ii) They are the elements belonging to 3 to 12 groups.

Properties of d-Block Elements :

(i) There ionization energies are between s and p-block elements.

(ii) They show variable oxidation states.

(iii) They form both ionic and covalent compounds

(iv) Their compounds are generally coloured and paramagnetic.

(v) Most of transition metals form alloys.

(vi) Most of transition elements are used as catalyst i.e., V, Cr, Mn, Fe, Co, Ni, Cu etc.

Exception : Zn, Hg, Cd have (n – 1)d10ns2 electronic configuration, do not show most of properties oftransition elements as they have complete d subshell and in them last electron enters the s-subshell not thed-subshell.

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Similarly of Zn, Cd, Hg with transition elements are :

(i) they form complexes like d-block elements.

(ii) they form covalent compounds.

(iii) first ionization energies are much higher

f-Block Elements : The two rows at the bottom of periodic table called Lanthanoids and Antinoids haveouter most electronic configuration as : (n – 2)f1-14(n – 1)d0-1ns2

(i) Last electron enters the f-subshell.

(ii) Thus two series elements are called f-Block elements or inner-transition elements.

Properties of f-block elements :

(i) They are all metals.

(ii) They show variable oxidation states.

(iii) There compounds are generally coloured.

(iv) Most of the elements of actinide series are radioactive. In this series after uranium elements are called astransuranium elements.

(v) Classification of periodic table can be the basis of metals and non-metals.

Practice Problems :

1. Write the general outer electronic configuration of s-, p-, d-, and f-block elements.

2. Assign the position of the element having outer electronic configuration.

(i) ns2np4 for n = 3, (ii) (n – 1) d2ns2 for n = 4, and (iii) (n – 2) f7 (n – 1) d1ns2 for n = 6, in the periodictable.

[Answers : (1) s-block elements : ns1 – 2, n = 2 to 7, where n represents period in all the cases, p-blockelements : ns2np1 – 6, n = 2 to 6, d-block elements : (n – 1)d1 – 10 ns0 – 2, n = 4 to 7, f-block elements :(n – 2)f1 – 14(n – 1) d0 – 1 ns2; n = 6 to 7 (2) (i) The group number of the element = 10 + number ofelectrons in the valence shell = 10 + 6. The elements with atomic number 16 is sulphur and theelectronic configuration is 1s2, 2s22p53s23p4 (ii) the group number of the element = Number ofd-electrons + number of s-electrons = 2 + 2 = 4. Thus, the element belongs to group 4 and 4th periodi.e, Titanium with atomic number 22. The electronic configuration is 1s2, 2s2, 2p6, 3s23p63d2, 4s2

(iii) The complete electronic configuration of the element is [Xe]4f7, 5d1, 6s2. Thus, the atomicnumber of the element = 54 + 7 + 1 + 2 = 64. The element is gadolinium with atomic number 64]

C4 Properties of Metals

(i) They are good conductors of heat and electricity

(ii) They are usually solids at room temperature

(iii) They are malleable (can be flattened into thin sheets by hammering) and ductile (can be drawn into wires)

Properties of Non-Metals

(i) They are bad conductor of heat and electricity

(ii) They are usually solids or gases at room temperature

(iii) They cannot be drawn into thin sheets or wires.

* Metallic character increases down the group.

* Metallic character decreases along the period.

* The change from metals to non-metals in periodic table is not abrupt. The elements on the border line aresemi-metals or metalloids e.g. (Si, As, Sb, Te).

Practice Problems :

1. What are the major differences between metals and non-metals ?

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C5 Prediction of period, group and block of given elements :

(a) Period of an element corresponds to principal quantum number of the valence shell.

(b) The block of an element is type of orbital which receives the last electron.

(c) Group is predicted as follows :

(i) for s-block elements : group number is equal to the number of valence electrons.

(ii) for p-block elements : group number is equal to 10 + number of valence electrons.

(iii) for d-block elements : group number is equal to number of electrons in (n – 1) d subshell +number of electrons in valence shell (nth shell)

Practice Problems :

1. In terms of period and group, where would you locate the element with Z = 114.

2. Write the atomic number of the element present in the third period and seventeenth group of theperiodic table.

[Answers : (1) The filling of the 6th period ends at 86

Rn (1s2, 2s2 2p6, 3s2 3p6 3d10, 4s2 4p6 4d10 4f14, 5s2

5p6 5d10, 6s2 6p6). Thereafter the filling of 7th period starts and is filling according to the Aufbauprinciple in increasing order of energies as : 7s < 5f < 6d < 7p. Therefore, after

86Rn, the next two

elements with Z = 87 and Z = 88 are s-block elements, the next fourteen i.e., z = 90 – 103 are f-blockelements, the next ten i.e., Z = 104 – 112 are d-block elements and last six i.e. Z = 113 – 118 are p-blockelements. From this, it is concluded that Z = 114 is the second p-block element i.e., group 14 of the 7thperiod (2) In the third period, the filling up of only 3s- and 3p-orbital occurs, the element which willlie in seventeenth group will have Z = 12 + 5 = 17]

C6 Periodic trends in properties of elements :

1. Atomic radii : is the distance from the centre of nuclei to the point upto which the density of electron cloudis maximum.

It is of four types :

(i) covalent radii

(ii) vander wall radii

(iii) metallic radii

(iv) Ionic radii

Covalent radii : rcovalent

= ½ [Internuclear distance between two covalently bonded atoms ofsame molecule]

= ½ [bond length]

Vander wall radii : rvander wall

= ½ [Internuclear distance between two non-bonded atoms] ofdifferent or neighbouring molecules

Metallic radii : = ½ [internuclear distance between two adjacent atoms in the metalliclattice]

Variation of Atomic radii in periodic table :

1. Along the period it decreases as nuclear charge increases.

Exception :

The size of atoms of inert gases are however larger than the halogen elements.

Reason :

As we move along the period charge increases while the number of shells remain the same thus along theperiod size decreases.

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2. Along the group the atomic radii of elements increases with increase in atomic number as we move fromtop to bottom of group.

Reason : As we move down the number of shell increases thus distance between the outer electrons andnucleus increases. Also with increase of atomic number the nuclear charge down the group increases. Thusatomic radii should decreases, but effect of increased nuclear charge is reduced due to screening or sheildingeffect on the valence electrons by the electrons present in the inner shells.

– Comparison of the ionic radii with atomic radii of same atom – Mg > Mg2+

Reason : In Mg np = 12, 12n

e

Mg2+ np = 12, 10n

e

Thus in Mg 12 protons are attracting by the 12 electrons outer nuclear part. Whereas in Mg2+ the 12 protonsare attracting 10 e– outside the nucleus thus 12 protons will attract 10 e– electrons more strongly thus sizeof Mg2+ decreases as compared to Mg.

* Atoms or ions with same electronic configuration are called isoelectronic. If we consider a series ofisoelectronic species (atoms or ions) then size decreases with increase of atomic number.

e.g. O–2, I–, Ne, Na+, Mg2+

(1.40) (1.36) (1.31) (0.95) (0.65)

Atomic number 8 9 10 11 12

* The size of same cation decreases with increase of magnitude of positive charge i.e. Fe3+ is smaller thanFe2+, Cu2+ is smaller than Cu+ so on.....

Ionisation Energy

First I.E. : Amount of energy required to remove a single electron from the outer shell of a neutral gaseousatom.

M(g) M+(g) + e– .... IE1

M+ M2+ + e–.....IE2

(IE2) Second Ionisation Energy : Amount of energy required to remove the second electron from an atom

who has already lost one electron.

IE3 > IE

2 > IE

1

As the number of electrons in outer shell decreases so attraction of nucleus for remaining electronsincreases thus I.E. increases.

Trends of first I.E. along the period and down the group is as follows :

(a) First I.E. decreases as we move down the group. Reason is that as we move down the group no. of shellincreases thus nuclear attraction decreases. Although nuclear charge also increases but its effect is weakendby sheilding supplied by the inner shells to the outer most shell.

(b) As we move along the period ionisation energy increases. The reason for this is because of decrease of sizeof atom along the period. As the number of electrons are added to same shell along the period thus nuclearattraction for the outer electrons increases with increase of nuclear charge.

Exception :

(1) I.E. (B) < I.E. (Be)

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Electronic Configuration : B = 1s22s22p1, Electronic configuration Be = 1s22s2

As in Be e– is to be removed from completely filled shell thus its I.E. is more than B. Here is extra stabilityof subshell is the cause of this irregularity

(2) I.E. of O < I.E. N

Electronic configuration

N = 1s2 2s2 2p3

O = 1s2 2s2 2p4

The extra stability of half-shell is cause of irregularity in IE pattern.

(3) I.E. of Al < I.E. of Mg

Electronic configuration

Al = 1s2 2s2 2p6 3s2 3p1

Mg = 1s2 2s2 2p6 3s2

* The electrons in different orbitals (s, p, d, f) beloging to the same energy level experience different pull ofthe nucleus.

The I.E. for pulling out an s-electron is maximum and it decreases in pulling out p-electron.

Hence we can say that I.E. for pulling out an electron from a given energy level decreases in the order s >p > d > f orbitals.

Electron Affinity : It is defined as energy given out when an extra electron is taken up by a neutral gaseousatom.

X(g) + e– X–(g)

E. A measures the tightness with which an atom binds an extra electron to it.

E.A. decreases down the group, because an atom gets larger, the attractions of positive nucleus for anoutside electron decreases.

Exception : E.A. of F < E.A. of Cl

Reason : It is due to extremely small size of F atom as compare to Cl atom.

Extra electron create strong electron-electron repulsion among all the electrons.

Along the period : Along the period electron affinity increases as we move from left to right.

(i) E.A. of noble gases are zero.

(ii) E.A of N and Be atoms are quite low due to extra stability of half filled orbitals (p3 in N & s2 in Be)

(iii) After taking up an extra electron an atom becomes negatively charged (anion) and now second electron isto be added to it. The anion will repel the incoming of an electron and an additional energy will be requiredto add it to the anion.

First E.A. negative (energy released)

Second E.A. positive (energy released)

Electronegativity : It is the measure of the ability of an atom in a combined state (i.e. in a molecule) toattract itself to the electrons within a chemical bond or tendency to attract bond pair towards itself.

* Non-metals have high value of electronegativity than metals.

F, O, N & Cl are highly electronegative than [K, Rb, Cs] (metals) which are electropositive in nature.

* Electronegativity along the period increases & down the group decreases.

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* It is measured with the help of pauling scale & mulliken scale.

Screening (sheilding) effect :

In d-block elements (transition element) while writing electronic configuration of elements, it is seen thatnew electrons are added to inner shells i.e. penultimate shells. Thus nuclear attraction for out electons getsaffected. As the new electrons enter the inner shells they tend to sheild or screen the outer shell electronsfrom nucleus and thus decreases the nuclear attractive force. This is called as screening effect. Due to thiseffect, the atomic size of transition elements remain same as we move along the period which shoulddecrease as we move from left to right. Thus ionisation energy, electron affinity and other propertiesremains nearly same as we move along the period.

Practice Problems :

1. What does atomic radius and ionic radius really mean to you ?

2. How do atomic radius vary in a period and in a group ? How do you explain the variation ?

3. What do you understand by isoelectronic species ? Name a species that will be isoelectronic witheach of the following atoms or ions.

(i) F— (ii) Ar (iii) Mg2+ (iv) Rb+

4. Consider the following species : N3–, O2–, F–, Na+, Mg2+ and Al3+.

(i) What is common in them ?

(ii) Arrange them in the order of increasing ionic radii.

5. Explain, why cations are smaller and anions are larger in radii than their parent atoms ?

6. What is the significance of the terms – ‘isolated gaseous atom’ and ‘ground state’ while defining theionization enthalpy and electron gain enthalpy ?

7. Among the second period elements the actual ionization enthalpies are in the order :

Li < B < Be < C < O < N < F < Ne

Explain why ?

(i) Be has higher iH than B

(ii) O has lower iH and N and F ?

8. How would you explain the fact that the first ionization enthalpy of sodium is lower than that ofmagnesium, but its second ionization enthalpy is higher than that of magnesium ?

9. What are the various factors due to which the ionization enthalpy of the main group elements tendsto decrease down a group ?

10. The first ionization enthalpy values (in kJ mol–1) of group 13 elements are :

B Al Ga In Tl

801 577 579 558 589

How would you explain this deviation from the general trend ?

11. Which of the following pairs of elements have a more negative electron gain enthalpy ?

(i) O or F (ii) F or Cl

12. Would you expect the second electron gain enthalpy of O as positive, more negative or less negativethan the first ? Justify your answer.

13. What is the basic difference between the terms electron gain enthalpy and electronegativity ?

14. How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all thenitrogen compounds ?

15. Describe the theory associated with the radius of an atom as it (a) gains an electron (b) loses anelectron.

16. Would you expect the first ionization enthalpies of two isotopes of the same element to be the same ordifferent ? Justify your answer.

17. Use the periodic table to answer the following questions.

(a) Identify an element with five electrons in the outermost subshell.

(b) Identify an element that would tend to lose two electrons.

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(c) Identify an element that would tend to gain two electrons.

(d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.

18. The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs; whereas thatamong group 17 elements is F > Cl > Br > I. Explain.

19. The first (iH

1) and the second (

iH

2) ionisation enthalpies (in kJ mol–1) and the (

egH) electron gain

enthalpy in (kJ mol–1) of a few elements are given below :

Elements iH

1

iH

2

rgH

I 520 7300 –60

II 419 3051 –48

III 1681 3374 –328

IV 1008 1846 –295

V 2372 5251 +48

VI 738 1451 –40

Which of the above element is likely to be :

(a) the least reactive element

(b) the most reactive metal

(c) the most reactive non-metal

(d) the least reactive non-metal

(e) the metal which can form a stable binary halide of the formula MX2 (X = halogen).

(f) the metal which can form a predominantly stable covalent halide of the formulaMX (X = halogen) ?

20. Predict the formulas of the stable binary compounds that would be formed by the combination ofthe following pairs of elements.

(a) Lithium and oxygen (b) Magnesium and nitrogen

(c) Aluminium and iodine (d) Silicon and oxygen

(e) Phosphorus and fluorine (f) Element 71 and fluorine

[Answers : (1) Atomic radius is one half of the distance between the nuclei of two identical atoms ina molecule bonded by a single bond. Ionic radius is the distance between the cations and anions inthe ionic crystals (2) The atomic radii of elements decreases in a period from left to right with anincrease in atomic number. In a period electrons, enter one by one in the same energy level. On theaddition of each electron, the nuclear charge increases by one unit. The result is that the electronsare attracted more and more strongly towards the nucleus. This result in decrease of atomic radii.The atomic radii increases in a group while moving from top to bottom. In a group from top tobottom, new shells are added with increasing atomic number. Nuclear size increases and hence atomicsize should decrease with increase in nuclear charge, but effect is dominated by increase in numberof shells and hence over all size increases in moving from top to bottom (3) Ions of different elementshaving the same number of electrons but different magnitude of the nuclear charge are termed asisoelectronic ions. (i) F– has (9 + 1) = 10 electrons. The other isoelectronic species with 10 electronsare : N3– (7 + 3); O2– (8 + 2); Na+(11 – 1); Mg2+(12 – 2) and Al3+(13 – 3). (ii) Ar has 18 electrons. Theother isoelectronic species with 18 electrons are : S2–(16 + 2), Cl–(17 + 1), K+(19 – 1), Ca2+(20 – 2). (iii)Mg2+ has (12 – 2) = 10 electrons. The other isoelectronic species with 10 electrons are : N3–, O2–, Na+,F–, Mg2+, Al3+. (iv) Rb+ has (37 – 1) = 36 electrons. The other isoelectronic species with 36 electrons are: Br(35 + 1), Kr(36), Sr2+ (38 – 2) (4) (i) Each one of these have 10 electrons and hence all areisoelectronic (ii) N3– > O2– > F– > Na+ > Mg2+ > Al3+ (5) in the formation of cations their is a loss of oneor more electrons which increases the effective nuclear charge. As a result, the force of attraction ofthe nucleus for the electrons increases and hence the ionic radii decreases. The ionic radii of an anionis always larger than its parent atom because the addition of one or more electrons decreases theeffective nuclear charge. As a result, the force of attraction of the nucleus for the electrons decreasesand hence the ionic radii increases (7) (i) The ionization enthalpy, beside other things, depends uponthe type of electron to be removed from the same principal shell. In Be (electronic configuration; 1s2,2s2), the outermost electron is present in 2s-orbital while in B (electronic configuration : 1s2, 2s22p1)

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it is present in 2p-orbital. Since 2s-electrons are more strongly attracted to the nucleus as comparedto 2p-electrons, therefore, lesser energy is required to remove out a 2p-electron, as compared to2s-electron. Because of this, H of Be is higher than that of

iH of B. (ii) Nitrogen has the electronic

configuration : 1s2, 2s2 2px1 2p

y1 2p

z1 in which 2p-orbital are half filled is more stable than the oxygen

which has the electronic configuration 1s2, 2s2 2px2 2p

y1 2p

z1 in which the 2p-orbitals are neither half

filled nor completely filled. Therefore, it is difficult to remove electron from nitrogen than oxygen.Because of this

iH of nitrogen is higher than that of oxygen. Further the electronic configuration of

F is 1s2, 2s2 2px2 2p

y2 2p

z1. Because of high nuclear charge, that the first ionization enthalpy of F is

higher than of O. Also, the effects of increased nuclear charge outweights the effect of stability due toexactly half-filled orbitals. Because of this

iH of fluorine is higher than that of oxygen. (As a general

rule if the electron is to be removed from same orbital then the one having half-filled or completelyfilled have higher ionization enthalpy as compared to one having neither half-filled nor completelyfilled orbital. Also, ionization enthalpy of half-filled orbital is < that of ionisation enthalpy ofcompletely filled orbital) (8) The sodium has electronic configuration 1s2, 2s22p6,3s1 and magnesiumhas electronic configuration 1s2, 2s22p6, 3s2. Here, in both the cases first electron is to be removedfrom 3s-orbital, but the nuclear charge of Na (+11) is lower than that of Mg (+12). Therefore, thefirst ionization enthalpy of sodium is lower than magnesium. After the loss of one electron fromsodium, the electronic configuration of sodium becomes completely filled i.e., 1s2, 2s22p6 and afterthe loss of one electron from magnesium, the electronic configuration of magnesium becomes 1s2,2s22p6, 3s1 i.e., half-filled. Hence

iH of Na >

iH of Mg (9) (i) Atomic size : On moving down the

group, the atomic size gradually increases with the addition of one energy shell to each succeedingelement. As a result, the distance of valence electrons from the nucleus increases. Consequently, theforce of attraction of the nucleus from the valence electrons decreases and hence the ionizationenthalpy decreases on moving down the group. (ii) Screening effect : On moving down the group,with the addition of new shells, the number of inner electron shells which shield the valence electronsincreases regularly, thereby increasing the shielding effect or screening effect. This means, that theforce of attraction of the nucleus for the valence electrons further decreases and hence the ionizationenthalpy further decreases (10) On moving down the group from Boron to Thalium via aluminium,indium and thallium the ionization enthalpy decreases (although not regular). This can be explainedon the basis of increasing atomic size and screening effect. The abnormal behaviour of Al, Ga, In andTl can be explained as follows : Al follows immediately after s-block elements while Ga and In followafter d-block elements, whereas Tl after d- and f-block elements. These d- and f- electrons do notshield the outer-shell electrons from the nucleus effectively. The result is that, the valence electronremain more tightly held by nucleus and thus larger amount of energy is needed for their removal.This explains why Ga has higher ionization enthalpy than Al. Further moving down the group fromGa to In, the shielding effect increases due to the presence of additional 4d-electrons, outweighs theeffect of increased nuclear charge, and hence

iH of In is lower than that of Ga. Thereafter, the effect

of increased nuclear charge outweight the shielding effect due to the additional 4f and 5d-electronsand thus the

iH of Tl is higher than that of In (11) (i) Both oxygen and fluorine belongs to 2nd

period. As we move from left to right i.e., from O to F, the atomic size decreases with the increase ofnuclear charge. Both these factors are responsible to increase the attraction of the nucleus for theincoming electron and thus electron gain enthalpy becomes more negative. Further, gain of oneelectron by O gives O– ion does not have stable inert gas configuration, whereas, gain of one electronby F gives F– ion which has stable inert gas configuration. Because of this, the energy released ismuch higher in going from F to F– than in going from O to O–, i.e., the electron gain enthalpy ofF(–328 kJ mol–1) is much more negative than that of oxygen (–141 kJ mol–1). (ii) In general, theelectron gain enthalpy becomes less negative on moving down the group, but the electron gainenthalpy of chlorine (–349 kJ mol–1) is little more negative than that of fluorine (–328 kJ mol–1). Thisis because fluorine has a small atomic size (only two shells) and hence the incoming electron is notaccepted with the same ease as in the case of large chlorine atom. Because of this electron gainenthalpy of chlorine is more negative than that of fluorine (12) The second electron gain enthalpy ofoxygen is positive which can be explained as given : When an electron is added to O atom to form O–

ion, energy is released. O(g) + e–(g) O–(g), iH = –141 kJ mol–1. Because of this, first electron gain

enthalpy of oxygen atom is positive. But when another electron is added to O– to form O2– ion, energyis released to overcome the strong electrostatic repulsion between the negatively charged O– ion andthe second electron which is being added. O–(g) + e–(g) O2– (g);

iH = +780 kJ mol–1. Because of

this, the second electron gain enthalpy is positive (13) The electron gain enthalpy refers to thetendency of an isolated gaseous atom to accept an additional electron to form a negative ion, whereas

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electronegativity refers to the tendency of the atom of an element to attract the shared pair ofelectrons towards it in the formation of a covalent bond (14) The statement seems to be absurd. Infact, the electronegativity of any given atom is not constant. It increases as the percentage ofs-character of a hybrid orbital increase or the oxidation state of the element increase. For example,the electronegativity of N increases as we move from sp3-sp2-sp hybrid orbitals (15) (a) When aneutral atom gains an electron to form an anion, the number of electrons in the anion increaseswhereas its nuclear charge remains the same as that of the parent atom. Since the same nuclearcharge now attracts larger numbers of electrons, therefore, the force of attraction of the nucleus onthe electrons of all shells decreases. This means that the effective nuclear charge decreases and theexpansion of the electron cloud occurs. Because of this, the distance between the centre of the nucleusand the last shell increases thereby increasing the ionic radius. (b) When a neutral atom loses anelectron to form a cation, the number of electrons in the cation decreases whereas its nuclear chargeremains the same as that of the parent atom. Since the same nuclear charge now attracts lessernumbers of electrons, therefore, the force of attraction of the nucleus on the electrons of all shellsincreases. This means that the effective nuclear charge increases and the contraction of the electroncloud occurs. Because of this distance between the centre of the nucleus and the last shell decreasesthereby decreasing the ionic radius (16) Two isotopes of the same element have the same atomicnumber; which means that they have the same number of electrons and hence they have the sameionisation enthalpy (17) (a) The element with five electrons in the outermost sub-shell belongs toseventeenth group i.e. fluorine, chlorine, bromine, etc. (b) The element that would tend to loose twoelectrons belongs to alkaline earth metals i.e., magnesium, calcium, etc. (c) The element that wouldtend to gain two electrons belongs to sixteenth group i.e., oxygen and sulphur. (d) A metal which isliquid at room temperature is bromine belongs to fifteenth group (18) Reactivity of an elementdepends upon the ease with which it can lose the outermost electron. The tendency to lose electron, inturn, depends upon the ionization enthalpy. The ionization enthlapy decreases down the group,therefore, the reactivities of group 1 element increase in the same order i.e., Li < Na < K < Rb < Cs.In contrast group 17 elements, have seven electrons in their respective valence shells and thus havea strong tendency to accept one electron to get the nearest inert-gas configuration. The tendency toaccept electrons, in turn, depends upon their electrode potentials. The electrode potential ofseventeenth group decreases in order i.e., F > Cl > Br > I (19) (a) The element V has highest firstionization enthalpy (

iH

1) and positive electron gain enthalpy (

egH) and hence it is the least reactive

element. Element V must be inert gas because inert gases have positive eg

H. The values of iH

1,

iH

2

and egH match that of Helium. (b) The element II which has the least first ionization enthalpy (

iH

1)

and a low negative electron gain enthalpy (eg

H) is the most reactive metal. The values of iH

1,

iH

2

and eg

H match that of potassium. (c) The element III which has high first ionization enthalpy (iH

1)

and very high negative electron gain enthalpy (eg

H) is most reactive non-metal. The values of iH

1,

iH

2 and

egH match that of fluorine. (d) The element IV has high negative electron gain enthalpy

(eg

H) but not so high first ionization enthalpy (iH

1) is the least reactive non-metal. The values of

iH

1,

iH

2 and

egH match that of Iodine. (e) The element VI has low first ionization enthalpy (

iH

1)

but higher than that of alkali metals. Therefore, it appears that the element is an alkaline earthmetal and hence will form binary halide of the formula MX

2 (where X = halogen). The values of

iH

1,

iH

2 and

egH match that of Magnesium. (f) The element 1 has low first ionization (

iH

1) but a very

high second ionization enthalpy (iH

2), therefore, it must be an alkali metal. Since the metal form a

predominantly stable covalent halide of the formula MX (X = halogen), therefore, the alkali metalmust be least reactive. The values of

iH

1,

iH

2 and

egH match that of Lithium (20) (a) Li

2O (Lithium

oxide) (b) Mg3N

2 (Magnesium nitride) (c) AlI

3 (Aluminium iodide) (d) SiO

2 (Silicon dioxide) (e) PF

3

(Phosporus tri-fluoride) or PF5 (Phosphorus penta-fluoride) (f) LuF

3 (Lutetium fluoride)]

C7 Hydration and Hydration Energy :

Hydration energy is the enthalpy change that accompanies the dissolving of one mole of gaseous ions inwater.

Li+(g) + H2O [Li(H

2O)]+, H = –806 kJ mol–1

(i) Size of the ion and its charge determines extent of hydration.

(ii) Greater the charge, smaller the size of the ion, greater the attraction for the lone pair of O of H2O, hence

greater the extent of hydration and hence greater the hydration energy.

– size of the hydrated ions increases,

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– ionic mobility decreases [heavier (hydrated) ions moves slower]

Practice Problems :

1. Arrange the following ions in increasing Na+, Mg2+, Al3+

(i) extent of hydration (ii) hydration energy

(iii) size of the hydrated ions (iv) ionic mobility

[Answers : (1) (i) Na+ < Mg2+ < Al3+ (ii) Na+ < Mg2+ < Al3+ (iii) Na+ < Mg2+ < Al3+ (iv) Al3+ < Mg2+ < Na+]

C8 Acid-Base character of oxides :

On moving across a period, the basic character of the oxides gradually changes first into amphoteric andfinally into acidic character.

(i) On moving down the group, reverse behaviour is observed, i.e., from more acidic to more basic.

Oxides of the element M in H2O produce MOH

– If electronegativities difference of M and O is greater than that of H and O in H2O then MOH is acidic due

to formation of H3O+

M — O — H + H2O H

3O+ + MO–

– If electronegativities difference of M and O is less than that of H and O in H2O then MOH is basic due to

formation of OH–

M — O — H + H — O — H [MOH2]+ + OH–

(ii) Stability of oxides decreases across a period.

(Min. Max)

(iii) Oxides of the following elements are amphoteric

H, Be, Al, Ga, In, Tl, Sn, Pb, Sb, Bi, Po

– H2O is amphoteric (also called amphiprotic)

H2O + H

2O H

3O+ + OH–

Since it is H+ acceptor (base) as well as H+ donor (acid).

– BeO, Al2O

3, SnO

2, PbO

2,.... are amphoteric since they form salts with acid as well as with base

])OH(Al[Na2OH3NaOH2OAl

OH3AlCl2HCl6OAl

42baseacid

32

23base

32

– oxide is acidic if it reacts with a base.

– oxide is basic if it reacts with an acid.

Practice Problems :

1. Arrange the following oxides in order of increasing molecular (acidic) character :

SO3, Cl

2O

7, CaO and PbO

2

2. Arrange following oxides in increasing acidic nature Li2O, BeO, B

2O

3.

3. Which oxide is more basic, MgO or BaO ? Why ?

[Answers : (1) CaO < PbO2 < SO

3 < Cl

2O

7 (2)

acidic32

amphotericbasic2 OBBeOOLi (3) BaO]

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1. The following acids have been arranged in theorder of decreasing acid strength. Identify thecorrect order

ClOH(I) BrOH(II) IOH(III)

(a) I > II > III (b) II > I > III

(c) III > II > I (d) I > III > II

2. The statement is not true for the long form of theperiodic table

(a) it reflects the sequence of filling theelectrons in the order of the sub-energyshells s, p, d and f

(b) it helps to predict the stable valency statesof the elements

(c) it reflects trends in physical andchemical properties of the elements.

(d) it helps to predict the relative inicity ofthe bond between any two elements.

3. The electronic configuration

1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p6, 4d10, 5s2 is for

(a) f-block element

(b) d-block element

(c) p-block element

(d) s-block element

4. Which of the following statements is false :

(a) elements of I-B and II-B groups aretransition elements

(b) elements of V-B group do not containmetalloids

(c) elements of I-A and II-A groups arenormal elements

(d) elements of IV-B group are neitherstrongly electronegative nor stronglyelectropositive

5. Transition metals are characterised by theproperties :

(a) variable valency

(b) coloured compounds

(c) high melting and boiling points

(d) all of the above

6. The atomic number of an element is 16. Thiselement in Periodic Table belongs to

(a) group VI and period II

(b) period III and group VI

(c) group IV and period II

(d) period III and IV

SINGLE CORRECT CHOICE TYPE

7. Ionic radii of

(a) Ti4+ > Mn7+ (b) 35Cl— < 37Cl—

(c) K+ > Cl— (d) P3+ < P5+

8. Which of the following ions has the smallestradius ?

(a) Ti2+ (b) Pt2+

(c) Ni2+ (d) Zr2+

9. When the following five anions are arranged inorder of decreasing ionic radius, the correctsequence is

(a) Se2—, I—, Br—, O2—, F—

(b) I—, Se2—, O2—, Br—, F—

(c) Se2—, I—, Br—, F—, O2—

(d) I—, Se2—, Br—, O2—, F—

10. Which of the following ions has the largest heat ofhydration ?

(a) Na+ (b) Al3+

(c) F— (d) Sr2+

11. Which of the following anions is most easilypolarized ?

(a) Cl— (b) Se2—

(c) Br— (d) Te2—

12. Of the four H values needed to calculate a latticeenergy using the Born-Haber cycle, the one that ismost difficult to measure is

(a) the heat of sublimation of the metal

(b) the heat of formation of gaseous atomsof the non-metal

(c) the ionization energy of the metal

(d) the electron affinity of the non-metal

13. The melting point of RbBr is 6820C while that ofNaF is 9880C. The principal reason that themelting point of NaF is much higher than that ofRbBr is that

(a) the molar mass of NaF is smaller thanthat of RbBr.

(b) the bond in RbBr has more covalentcharacter than the bond in NaF

(c) the difference in electronegativitybetween Rb and Br is smaller than thedifference between Na and F.

(d) the internuclear distance, rc + r

a is greater

for RbBr than for NaF.

14. Going down in a group from F to I, which of thefollowing properties does not decreases

(a) ionic radius

(b) ionisation energy

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(c) oxidising power

(d) electronegativity

15. Which of the following will have maximumelectron affinity ?

(a) 1s22s22p5 (b) 1s22s22p6

(c) 1s22s22p63s23p5 (d) 1s22s22p63s23p6

16. Size of cation is smaller than that of the atombecause of

(a) the whole of the outer shell of electronsremoved

(b) increase in effective nuclear charge

(c) due to gain of electrons

(d) statement, that cation is smaller thanatom, is wrong

17. The sizes of the second and third row transitionelements are almost the same. This is due to

(a) d- and f-orbitals do not shield the nucleicharge very effectively

(b) lanthanide contraction

(c) both true

(d) none is true

18. The factors that influence the ionisation energiesare

(a) the size of the atom

(b) the charge on the nucleus

(c) how effectively the inner electron shellscreen the nuclear charge

(d) all the above

19. Following are the values of the electron affinity ofthe formation of O— and O2— from O

(a) –142, –702 (b) –142, 702

(c) 142, 702 (d) –142, –142

20. Covalency is favoured in the following cases

(a) a smaller cation

(b) a larger anion

(c) large charges on catio or anion

(d) all the above

21. A molecule H—X will be 50% ionicelectronegativity difference of H and X is

(a) 1.2 eV (b) 1.4 eV

(c) 1.5 eV (d) 1.7 eV

22. Representative elements belong to

(a) s- and p-block (b) d-block

(c) d- and f-block (d) f-block

23. Inert pair effect is shown by

(a) s-block (b) p-block

(c) d-block (d) f-block

24. Among the following stability of ions of Ge, Sn andPb will be in order except

(a) Ge2+ < Sn2+ < Pb2+

(b) Ge4+ > Sn4+ > Pb4+

(c) Sn4+ > Sn2+

(d) Pb2+ < Pb4+

25. Which is true statement

(a) Tl3+ salts are better oxidising agents

(b) Ga+ salts are better reducing agents

(c) Pb4+ salts are better oxidising agents

(d) all of above

26. The first element of a group in many ways differsfrom the other heavier members of the group. Thisis due to

(a) the small size

(b) the high electronegativity

(c) the unavailability of d-orbitals

(d) all of above

27. Catenation properties of C, Si, Ge, Sn, Pb in order

(a) C >> Si > Ge Sn >> Pb

(b) C < Si < Ge < Sn < Pb

(c) C > Si > Sn > Ge > Pb

(d) none is correct

28. Melting points of NaCl, NaBr, NaI and NaF will bein order

(a) NaI < NaBr < NaCl < NaF

(b) NaF < NaCl < NaBr < NaI

(c) NaBr < NaF < NaCl < NaI

(d) NaCl < NaI < NaF < NaBr

29. Which is amphoteric oxides

(a) BeO (b) SnO

(c) ZnO (d) All

30. Atomic number 64 will have electronicconfiguration

(a) [Xe]54

6s24f8

(b) [Xe]54

6s2 4f7 5d1

(c) [Xe]54

4f10

(d) [Xe]54

6s2 4f7 5p1

31. The correct order of second ionisation potential ofcarbon, nitrogen, oxygen and fluorine is

(a) C > N > O > F (b) O > N > F > C

(c) O > F > N > C (d) F > O > N > C

32. The element with the highest first ionisationpotential is

(a) boron (b) carbon

(c) nitrogen (d) oxygen

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ANSWERS (SINGLE CORRECTCHOICE TYPE)

1. a

2. d

3. b

4. d

5. d

6. b

7. a

8. c

9. d

10. b

11. d

12. d

13. d

14. a

15. c

16. b

17. a

18. d

19. b

20. d

21. d

22. a

23. b

24. d

25. d

26. d

27. a

28. a

29. d

30. b

31. c

32. c

33. b

34. a

35. a

36. c

37. a

38. d

39. b

40. d

41. b

42. d

43. b

44. b

45. b

33. The hydration energy of Mg2+ is larger than that of

(a) Al3+ (b) Na+

(c) Be2+ (d) Mg3+

34. The first ionisation potential in electron volts ofnitrogen and oxygen atoms are respectively givenby

(a) 14.6, 13.6 (b) 13.6, 14.6

(c) 13.6, 13.6 (d) 14.6, 14.6

35. Atomic radii of fluorine and neon in Angstrom unitsare respectively given by

(a) 0.72, 1.60 (b) 1.60, 1.60

(c) 0.72, 0.72 (d) None of these

36. The electronegativity of the following elementsincreases in the order

(a) C, N, Si, P (b) N, Si, C, P

(c) Si, P, C, N (d) P, Si, N, C

37. The first ionisation potential of Na, Mg, Al and Siare in the order

(a) Na < Mg < Al < Si

(b) Na > Mg > Al > Si

(c) Na < Mg < Al > Si

(d) Na > Mg > Al < Si

38. Which one of the following is the smallest in size ?

(a) N3– (b) O2–

(c) F– (d) Na+

39. Amongst the following elements (whose electronicconfigurations are given below), the one having thehighest ionisation energy is

(a) [Ne] 3s2 3p1

(b) [Ne] 3s2 3p3

(c) [Ne] 3s2 3p2

(d) [Ar] 3d10 4s2 4p3

40. The statement that is not correct for the periodicclassification of elements, is

(a) The properties of elements are theperiodic functions of their atomicnumbers

(b) Non-metallic elements are lesser innumber than metallic elements.

(c) The first ionisation energies of elementsalong a period do not vary in a regularmanner with increase in atomic number.

(d) For transition elements the d-subshellsare filled with electrons monotonicallywith increase in atomic number.

41. Which has most stable +2 oxidation state ?

(a) Sn (b) Pb

(c) Fe (d) Ag

42. Which of the following has the maximum numberof unpaired electron ?

(a) Mg2+ (b) Ti3+

(c) V3+ (d) Fe2+

43. The incorrect statement among the following is

(a) The first ionisation potential of Al is lessthan the first ionisation potential of Mg

(b) The second ionisation potential of Mg isgreater than the second ionisationpotential of Na

(c) The first ionisation potential of Na is lessthan the first ionisation potential of Mg

(d) The third ionisation of Mg is greater thanthird ionisation potential of Mg

44. Which of the following compounds is expectedcoloured ?

(a) Ag2SO

4(b) CuF

2

(c) MgF2

(d) CuCl

45. The correct order of radii is

(a) N < Be < B

(b) F– < O2– < N3–

(c) Na < Li < K

(d) Fe3+ < Fe2+ < Fe4+

classification of elements and periodicity in … · the modern periodic law given by moseley is :...

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