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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

3.1 Why do we Need to Classify Elements :

Q. What is the need to classify element ?

Solution : At present 114 elements are known. Of them, the recently discovered elements are man-made.Efforts to synthesise new elements are continuing. With such a large number of elements it is very difficultto study individually the chemistry of all these elements and their innumerable compounds individually. Toease out this problem, scientists searched for a systematic way to organise their knowledge by classifyingthe elements. Not only that it would rationalize known chemical facts about elements, but even predict newones for undertaking further study.

3.2 Genesis of Periodic Classification :

Q. Who has given group of Triads, explain it ?

Solution : 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 centralelement of the Triad was approximately the mean of the atomic weights of the other two members. Theproperties of the middle element were in between of those end members. e.g. Li, Na, K; Ca, Sr, Ba or Cl, Br,I etc.

Q. Explain the law of octaves.

Solution : John Newland had developed a law of octaves. He observed that similar elements are repeatedat 8th place like the 8th note of music. The elements are arranged in the order of increase of atomic weights.Similar element means that physical & chemical properties of element will be same. e.g. Li has the sameproperty as of Na.

Q. Explain the Lother Meyer arrangement of elements graphically.

Solution : He studied the physical properties such as atomic volume, melting point and boiling points ofvarious 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.

Q. What was Mandeleev’s Periodic Law ?

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

Q. Write the name of elements whose space Mendeleev’s left in his periodic table.

Solution : He predicted the properties of the missing elements from the known properties of the otherelements in the same group. e.g. gallium and germanium were not discovered at that time. He named theseelements as Eka-Aluminium and Eka-silicon.

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Q. Explain the arrangement of elements in Mendeleev’s Periodic table.

Solution : Mendeleev arranged all the elements in the order of increase of atomic 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) Mendeleev’s periodic table conists of seven horizontal rows known as periods and nine verticalcolumn known as groups.

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

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

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

(vii) The group number represents the valency.

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

Q. Write the defects in Mendeleev’s Periodic Table.

Solution : (i) Hydrogen resembles both the alkali metal and halogen. Hence its position in periodic table isundecided.

(ii) According to Mendeleev’s periodic table, isotopes should occupy different positions in the periodictable, but this is not so.

(iii) He predicted the properties of the missing elements from the known properties of the other elements inthe same group. e.g. gallium and germanium were not discovered at that time. He named these elements asEka-Aluminium and Eka-silicon.Nine elements in the VIII group do not fit into the system.

(iv) Their position was not justified according to the periodic law and cannot be arranged in the order oftheir increasing atomic weight.

(v) Alkali metals (Li, Na, K etc.) are placed with coinage metals (Cu, Ag, Au).

(vi) Chemically similar elements like Cu and Hg, Ag and Ti, Au and Pt have been placed in differentgroups.

(vii) Some elements of higher atomic weight have been placed before the elements 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) isplaced before nickel (At. wt. = 58.69); Tellurium (127.5) has been placed before iodine (126.9).

3.3 Modern Periodic Law and the Present Form of the Periodic Table :

Q. Write the modern periodic law.

Solution : As a result of modern researches it is estabished that atomic number is a fundamental propertynot the atomic weight.

Q. Who led the change in basis of classification of elements ? Explain.

Solution : Moseley change the basis of calssification of elements from atomic weight to atomic number.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

Q. Why elements with similar properties reoccurs after certain regular intervals ?

Solution : The cause of periodicity in properties is the repeatition of similar outer electronic configurationat certain regular intervals which indeed determines the physical and chemical properties of the elementsand their compounds.

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Q. Explain the arrangement of elements in modern periodic table.

Solution : (i) A modern version of table contains horizontal rows known as periods (which Mendeleevcalled series). Elements having similar outer electronic configuration in their atoms are grouped in verticalcolumns; these are referred as Groups or Families.

(ii) According to IUPAC (International Union of pure and Applied Chemistry), the groups are numberedfrom 1 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.

3.4 Nomenclature of Elements with Atomic Numbers > 100 :

Q. Write the naming of elements starting from atomic numbers 100 to 118.

Solution :

Atomic Name Symbol IUPAC IUPACNumber Official Name Symbol

101 Unnilunium Unu Mendelevium Md

102 Unnilbium Unb Nobelium No

103 Unniltrium Unt Lawrencium Lr

104 Unnilquandium Unq Rutherfordium Rf

105 Unnilpentium Unp Dubnium Db

106 Unnilhexium Unh Seaborgium Sg

107 Unnilseptium Uns Bohrium Bh

108 Unniloctium Uno Hassnium Hs

109 Unnilennium Une Meitnerium Mt

110 Ununnillium Uun Darmstadtium Ds

111 Unununnium Uuu Rontgenium* Rg*

112 Ununbium Uub * *

113 Ununtrium Uut +

114 Ununquadium Uuq * *

115 Ununpentium Uup +

116 Ununhexium Uuh * *

117 Ununseptium Uus +

118 Ununoctium Uno +

Q. What would be the IUPAC name and symbol for the element with atomic number 120 ?[NCERT Solved Example 3.1]

Solution : Hence, the symbol and the name respectively are Ubn and unbinilium.

3.5 Electronic Configurations of Elements and the Periodic Table :

Q. Why the first period contains only two elements ?

Solution : Each successive period in the periodic table is associated with the filling of next higher principalenergy shell i.e. n = 1, n = 2 etc. The first period starts with the filling of the lowest energy level (1s) and hasthe two elements – hydrogen (1s1) and helium (1s2).

Q. Why third period contains eight elements whereas fourth and fifth period contains eighteenelements ?

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

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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.

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

Q. Why sixth and seventh periods contains 32 elements ?

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

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.

Q. What are the Lanthanoide and Actinoid Series ?

Solution : Filling up of the 4f orbital begins with cerium (z = 58) and ends at lutetium (z = 71) to give the4f-inner transition series called the Lanthanoide Series. They are placed at the bottom of the periodic table.

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.

Q. How would you justify the presence of 18 elements in the 5th period of the Periodic Table ?[NCERT Solved Example 3.2]

Solution : When n = 5, l = 0, 1, 2, 3. The order in which the energy of the available orbitals 4d, 5s and 5pincreases is 5s < 4d < 5p. The total number of orbitals available are 9. The maximum number of electronsthat can be accomodated is 18; and therefore 18 elements are there in the 5th period.

Q. Explain group wise electronic configuration.

Solution : Elements in the same vertical column or group have similar electronic configuration, have samenumber of electrons in the outer orbitals and have similar properties.

3.6 Electronic Configurations and Types of Elements : s, p, d, f-Blocks :

Q. In how many blocks elements can be classify in periodic table ?

Solution : We can classify the elements in four blocks i.e., s-, p-, d-, f- blocks.

Q. Helium should belongs to s-block but its positioning is in p-block. Justify ?

Solution : Helium belongs to the s-block but its positioning in the p-block along with other group 18elements is justified because it has a completely filled valence shell (1s2) and as a result, exhibits propertiescharacteristic of other noble gases.

Q. Why the position of hydrogen is not fixed ?

Solution : It has a lone s-electron and hence can be placed in group 1 (alkali metals). It can also gain anelectron to achieve a noble gas arrangement and henct it can behave similar to a group 17 (halogen family)elements. Because it is a special case, we shall place hydrogen separately at the top of the Periodic Table.

Q. What are the s-Block elements and write its properties ?

Solution : These elements contain 1 or 2 electrons in the s-orbital of their respective outer most shell.

The elements of group 1 having outer-most electronic configuration ns1 are called as alkali metals. Theelements of group 2 having outer-most electronic configuration ns2 are are called alkaline earth metals.

Properties of s- block elements :

(i) They are reactive metals with low ionization energy.

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

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

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

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

Q. What are p-Block elements and write its properties ?

Solution : These elements contain 1-6 electrons in the P-orbital of their respective outermost shells.General electronic configuration of outermost shell is ns2np1-6, where n 2-7. These include elementsbelonging to group of 13, 14, 15, 16, 17 and 18 excluding helium.

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Properties of P-Block Elements :

(i) All the orbitals in the valence shell of the noble gases are completely filled by electrons. Thus it isdifficult to 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 togain electrons.

(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.

Q. What are d-Block elements and write its properties ?

Solution : These elements characterise by filling of inner d-orbital by electrons and therefore referred to d-Block elements. General electronic configuration of d-Block elements is (n – 1) d1-10 ns0-2. They are theelements 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.

Q. Why zinc, mercury and cadmium are transition elements although they have completely filledd-subshell ?

Solution : 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.

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

Q. What are f-Block elements and write its properties ?

Solution : The two rows at the bottom of periodic table called Lanthanoids and Antinoids have outer mostelectronic configuration as : (n – 2)f1-14(n – 1)d0-1ns2. Last electron enters the f-subshell. Thus two serieselements 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 calledas transuranium elements.

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

Q. What are the properties of Metals and Non-metals ?

Solution : Properties of Metals

(i) They are good conductors of heat and electricity

(ii) They are usually solids at room temperature

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

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.

Q. Write the variation of metallic character along the period and group.

Solution : Metallic character increases down the group and decreases along the period.

Q. Give the examples of metalloids.

Solution : The elements on the border line are semi-metals or metalloids e.g. (Si, As, Sb, Te).

Q. The elements Z = 117 and 120 have not yet been discovered. In which family / group would youplace these elements and also give the electronic configuration in each case. [NCERT SolvedExample 3.3]

Solution : The element with Z = 117, would belong to the halogen family (Group 17) and the electronicconfiguration would be [Rn] 5f146d107s27p5. The element with Z = 120, will be placed in Group 2 (alkalineearth metals), and will have the electronic configuration [Uuo]8s2.

Q. Considering the atomic number and position in the periodic table, arrange the following elementsin the increasing order of metallic character : Si, Be, Mg, Na, P. [NCERT Solved Example 3.4]

Solution : Metallic character increases down a group and decreases along a period as we move from left toright. Hence the order of increasing metallic character is : P < Si < Be < Mg < Na.

3.7 Periodic Trends in Properties of Elements :

Q. How the prediction of period, group and block of given element take place ?

Solution : (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 ofelectrons in valence shell (nth shell)

Q. What are the different types of Atomic Radii and define them ?

Solution : Atomic radii : is the distance from the centre of nuclei to the point upto which the density ofelectron cloud is 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]

Q. Write the variation of atomic radii in periodic table.

Solution : 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 effective nuclear charge increases while the number of shellsremain the same thus along the period size decreases.

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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.

Q. Compare the radii of magnesium atom and magnesium ion (Mg2+).

Solution : 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.

Q. Arrange the ionic radii of O–2, I–, Ne, Na+, Mg2+ ions in decreasing order.

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

O–2 > I– > Ne > Na+ > Mg2+

Q. Define ionisation energy.

Solution : Amount of energy required to remove a single electron from the outer shell of a neutral gaseousatom is known as first ionisation energy.

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.

Q. Write the trends of first I.E. along the period and down the group.

Solution : (a) First I.E. decreases as we move down the group. Reason is that as we move down the groupno. of shell increases thus nuclear attraction decreases. Although nuclear charge also increases but its effectis weakend by 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 ofsize of atom along the period. As the number of electrons are added to same shell along the period thusnuclear attraction for the outer electrons increases with increase of nuclear charge.

Q. Among Berilium and Boron, which has higher ionisation energy and why ?

Solution : I.E. (B) < I.E. (Be)

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

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Q. Among Nitrogen and Oxygen which has higher ionisation energy and why ?

Solution : 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.

Q. Show graphically variation of ionization enthalpy of elements of first group along the period andfor second group down the group.

Solution :

Q. Which has maximum ionisation energy among s, p, d, f orbitals electrons.

Solution : 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 orders > p > d > f orbitals.

Q. Define electron affinity or electron gain enthalpy, it is exothermic or endothermic process ?

Solution : It is defined as energy given out when an extra electron is taken up by a neutral gaseous atom, theenthalpy change accompanying the process is defined as electron gain enthalpy (

egH).

X(g) + e– X–(g)

Electron affinity measures the tightness with which an atom binds an extra electron to it. Electron affinitydecreases down the group, because an atom gets larger, the attractions of positive nucleus for an outsideelectron decreases. Depending on the element, the process of adding an electron to the atom can be eitherendothermic or exothermic. For many elements energy is released when an electron is added to the atomand the electron gain enthalpy is negative. For example, group 17 elements (the halogens) have very highnegative electron gain enthalpies because they can attain stable noble gas electronic configurations bypicking up an electron. On the other hand, noble gases have large positive electron gain enthalpies becausethe electron has to enter the next higher principal quantum level leading to a very unstable electronicconfiguration.

Q. Among fluorine and chlorine in which more energy is released when an extra electron is added totheir valence shell ?

Solution : E.A. of F < E.A. of Cl. It is due to extremely small size of F atom as compare to Cl atom. Extraelectron create strong electron-electron repulsion among all the electrons.

Q. Write the variation of (eg

H) along the period and down the group.

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

(i) (eg

H) of noble gases are very high.

(ii) (eg

H) 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 electronis to be added to it. The anion will repel the incoming of an electron and an additional energy will berequired to add it to the anion.

First (eg

H) is negative where as second (eg

H) is positive.

Down the group : We should also expect electron gain enthalpy to become less negative as we go down agroup because the size of the atom increases and the added electron would be farther from the nucleus. Thisis generally the case. However, electron gain enthalpy of oxygen and fluorine is less negative than that ofthe succedding element.

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Q. Why the (eg

H) of sulphur is more negative then oxygen ?

Solution : This is because when an electron is added to oxygen, the added electron goes to the smallern = 2 quantum level and suffers significant repulsion from the other electrons present in this level. For then = 3 quantum level (S), the added electron-electron repulsion is much less.

Q. Define Electronegativity. Write its trends along the period and down the group.

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

Electronegativity along the period increases & down the group decreases.Non-metals have high value ofelectronegativity than metals. F, O, N & Cl are highly electronegative than [K, Rb, Cs] (metals) which areelectropositive in nature. It is measured with the help of pauling scale & mulliken scale.

Q. Define screening (sheilding) effect.

Solution : In d-block elements (transition element) while writing electronic configuration of elements, it isseen that new electrons are added to inner shells i.e. penultimate shells. Thus nuclear attraction for outelectons gets affected. As the new electrons enter the inner shells they tend to sheild or screen the outershell electrons from nucleus and thus decreases the nuclear attractive force. This is called as screeningeffect. Due to this effect, the atomic size of transition elements remain same as we move along the periodwhich should decrease as we move from left to right. Thus ionisation energy, electron affinity and otherproperties remains nearly same as we move along the period.

Q. Which of the following species will have the largest and the smallest size ? Mg, Mg2+, Al, Al3+

[NCERT Solved Example 3.5]

Solution : Atomic radii increases across a period. Cations are smaller than their parent atoms. Amongisoelectronic species, the one with the larger positive nuclear charge will have a smaller radius.

Hence the largest species is Mg; the smallest one is Al3+.

Q. The first ionization enthalpy (iH) values of the third period elements, Na, Mg and Si are

respectively 496, 737 and 786 kJ mol–1. Predict whether the first iH value for Al will be more close

to 575 or 760 kJ mol–1 ? Justify your answer. [NCERT Solved Example 3.6]

Solution : It will be more close to 575 kJ mol–1. The value for Al should be lower than that of Mg becauseof effective shielding of 3p electrons from the nucleus by 3s-electrons.

Q. Which of the following will have the most negative electron gain enthalpy and which the leastnegative ? P, S, Cl, F. Explain your answer [NCERT Solved Example 3.7]

Solution : Electron gain enthalpy generally becomes more negative across a period as we move from left toright. Within a group, electron gain enthalpy becomes less negative down a group. However, adding anelectron to the 2p-orbital leads to greater repulsion than adding an electron to the larger 3p-orbital. Hencethe element with most negative electron gain enthalpy is chlorine; the one with the least negative electrongain enthalpy is phosphorus.

Q. Using the Periodic Table, predict the formulas of compounds which might be formed by thefollowing pairs of elements; (a) silicon and bromine (b) aluminium and sulphur. [NCERT SolvedExample 3.8]

Solution : (a) Silicon is group 14 element with a valence of 4; bromine belongs to the halogen family witha valence of 1. Hence the formula of the compound formed would be SiBr

4.

(b) Aluminium belongs to group 13 with a valence of 3; sulphur belongs to group 16 elements with avalence of 2. Hence, the formula of the compound formed would be Al

2S

3.

Q. Are the oxidation, state and covalency of Al in [AlCl(H2O)

5]2+ same ? [NCERT Solved

Example 3.9]

Solution : No. The oxidation state of Al is +3 and the covalency is 6.

Q. Show by a chemical reaction with water that Na2O is a basic oxide and Cl

2O

7 is an acidic oxide.

[NCERT Solved Example 3.10]

Solution : Na2O with water forms a strong base whereas Cl

2O

7 forms strong acid.

Na2O + H

2O 2NaOH

Cl2O

7 + H

2O 2HClO

4

Their basic or acidic nature can be qualitatively tested with litmus paper.

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Q. Define hydration and hydration energy.

Solution : Hydration energy is the enthalpy change that accompanies the dissolving of one mole of gaseousions in water.

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

Q. How the acid-base character of oxides vary along the period and down the group ?

Solution : (i) On moving across a period, the basic character of the oxides gradually changes first intoamphoteric and finally into acidic character.

(ii) 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–

(iii) Stability of oxides decreases across a period.

(Min. Max)

Q. What are the amphoteric substances ?

Solution : Which can behave both as an acid and as base are known as amphoteric substance.

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.

Q. Identify the group and valency of the element having atomic number 119. Also predict theoutermost electronic configuration and write the general formula of its oxide.

Solution : The present set up of the Long Form of the periodic table can accommodate at the maximum 118elements. Thereafter, in accordance with aufbau principle, the filling of 8s-orbital will occur. In otherwords, 119th electron will enter 8s-orbital. As such, its outermost electronic configuration will be 8s1.Since it has only one electron in the valence shell, i.e., 8s, therefore, its valency will be 1 and it will lie ingroup 1 along with alkali metals and the formula of its oxide will be M

2O where M represents the element.

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Q. First member of each group of representative elements (i.e., s and p-block elements) showsanomalous behaviour. Illustrate with two examples.

Solution : The first member of each group of the representative element (i.e., s- and p- block elements)shows anomalous behaviour from rest of the members of the same group because of the followingreasons : (i) small small, (ii) high ionization enthalpy, (iii) high electronegativity and (iv) absence ofd-orbitals.

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

Solution : The E.C. of Na is 1s22s22p63s1 and E.C. of Mg is 1s22s22p63s2. In both the cases, first electron isto be removed from the 3s orbital but nuclear charge of Mg (+12) is higher than that of Na (+11). Furthermore 3s-orbital in Mg is completely filled (more stable) while in Na it is only half-filled (less stable).Therefore,

iH

1 of Na is lower than that of Mg.

)p2s2s1(Na)p2s2s1(Na)s3p2s2s1(Na 5222H

)stablemore(ionconfiguratgasNeon

622H1622 2i1i

)stablemore(ionconfiguratgasNeon

6222H1622H2622 )p2s2s1(Mg)s3p2s2s1(Mg)s3p2s2s1(Mg 2i1i

After the removal of the first electron, Na+ acquires the more stable neon gas configuration. Thus, thesecond electron from Na is to be removed from its more stable noble (i.e., neon) gas configuration but theloss of second electron from Mg gives the more stable neon gas configuration. Thus,

iH

2 of Na is more

than i H2 of Mg.

Q. Arrange the elements N, P, O and S in the order of – (i) increasing first ionisation enthalpy.(ii) increasing non metallic character. Give reason for the arrangement assigned.

Solution : Arrange the elements N, P, O and S into different groups and periods in order of their increasingatomic numbers, we have,

Group 15 16

Period 2 N O

Period 3 P S

(i) The electron configuration of N(1s22s22p1x2p1

y2p1

z) in which 2p-orbitals are exactly half-filled is more

stable than the electronic configuration of O(1s22s22p2x2p1

y2p1

z) in which 2p-orbitals are neither half-filled

nor completely filled. Therefore, it is difficult to remove an electron from N than from O in spite of the factthat O (+8) has higher nuclear charge than N (+7). As a result,

iH

1 of N is higher than that of O. Similarly,

the electronic configuration of P (1s22s22p63s23p1x3p1

y3p1

z) in which 3p-orbitals are neither half-filled nor

completely filled. Therefore, it is difficult to remove on electron from P than from S in spite of the fact thatS (+16) has higher nuclear charge than P (+15). Thus,

iH

1 of P is higher than that S.

Further since iH

1 decreases down a group, therefore, H

i of N is greater than that of P and that of O is

greater than S. Combining the two results together, the first ionisation enthalpies of N, P, O and S increasein the order : S < P < O < N.

(ii) Since non-metallic character increases along a period and decreases down a group, therefore,non-metallic character increases in the order : P < S < N < O.

Q. How does the metallic and non metallic character vary on moving from left to right in a period ?

Solution : On moving from left to right in a period, the number of valence electrons increases by one ateach succeeding element but the number of shells remains the same. As a result, the nuclear chargeincreases and the tendency of the element to lose electron decreases and hence the metallic characterdecreases as we move from left to right in a period. Conversely, as the nuclear charge increases, thetendency of the element to gain electrons increases and hence the non-metallic increases from left to rightin a period.

Alternatively, metallic character decreases and non-metallic character increases as we move from left toright in a period. It is due to increase in ionization and electron gain enthalpy.

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Q. Among alkali metals which element do you expect to be least electronegative and why ?

Solution : Electronegativity decreases as the size of the atom increases. Since Fr has the largest size,therefore, it has the least electronegativity.

Q. Which one among the following elements has the lowest first ionisation enthalpy and which onehas the highest first ionisation enthalpy ? Li, K, Ca, S and Kr.

Solution : K has the lowest ionisation enthalpy and Kr has the highest ionisation enthalpy.

Q. Lanthanoids and actinoids are placed in separate rows at the bottom of the periodic table.Explain the reason for this arrangement.

Solution : These have been placed separately at the bottom of the periodic table for convenience. If theyare placed within the body of the periodic table, the periodic table will become extremely long andcumbersome.

Q. Which of the following elements has most positive electron gain enthalpy ? Fluorine, nitrogen,neon.

Solution : Both nitrogen and neon have positive electron gain enthalpies but neon has much higher positiveelectron gain enthalpy because of its much more stable inert gas configration than the less stable exactlyhalf-filled electronic configuration of nitrogen.

Alternatively, in neon, the new electron has to be placed in a much higher energy 3s-orbital of the new shellwhile in case of N, the new electron is to placed in the 2p-orbital of the same shell.

Q. From amongst Be, B and C, choose the element with highest first ionization enthalpy.

Solution : Carbon because of its higher nuclear charge and smaller size has the highest i H

1 amongst the

given elements.

Q. Name the elements in the periodic table which has the highest and lowest first ionizationenthalpy.

Solution : Highest : Helium and lowest : Francium.

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NCERT EXERCISE

3.1 What is the basic theme of organisation in the periodic table ?

A. To simplify and systematize the study of elements and their compounds

3.2 Which important property did Mendeleev use to classify the elements in his periodic table and did hestick to that ?

A. Mendeleev used atomic weight as the basis for the classification of elements in the periodic table andhe did stick to that. For example, he put gallium after aluminium and germanium as eka-aluminiumand eka-silicon

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

A. Mendeleev Periodic law states that the physical and chemical properties of elements are in periodicfunctions of their atomic weights, while Modern Periodic Law states that the physical and chemicalproperties of elements are in periodic functions of their atomic numbers.

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

A. In 6th period, electrons can be filled in only 6s, 4f, 5d and 6p-subshells whose energies increase in theorder as 6s < 4f < 5d < 6p.

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

A. 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 Aufbau principle inincreasing 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-block elements, thenext ten i.e., Z = 104 – 112 are d-block elements and last six i.e. Z = 113 – 118 are p-block elements.From this, it is concluded that Z = 114 is the second p-block element i.e., group 14 of the 7th period

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

A. In the third period, the filling up of only 3s- and 3p-orbital occurs, the element which will lie inseventeenth group will have Z = 12 + 5 = 17

3.7 Which element do you think would have been named by

(i) Lawrence Berkeley Laboratory

(ii) Seaborg’s group ?

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

A. Elements of the same group have similar electronic configuration.

3.9 What does atomic radius and ionic radius really mean to you ?

A. Atomic radius is one half of the distance between the nuclei of two identical atoms in a moleculebonded by a single bond. Ionic radius is the distance between the cations and anions in the ioniccrystals

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

A. The atomic radii of elements decreases in a period from left to right with an increase in atomicnumber. In a period electrons, enter one by one in the same energy level. On the addition of eachelectron, the nuclear charge increases by one unit. The result is that the electrons are attracted moreand more strongly towards the nucleus. This result in decrease of atomic radii. The atomic radiiincreases in a group while moving from top to bottom. In a group from top to bottom, new shells areadded with increasing atomic number. Nuclear size increases and hence atomic size should decreasewith increase in nuclear charge, but effect is dominated by increase in number of shells and henceover all size increases in moving from top to bottom

3.11 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+

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A. Ions of different elements having the same number of electrons but different magnitude of the nuclearcharge are termed as isoelectronic ions. (i) F– has (9 + 1) = 10 electrons. The other isoelectronicspecies with 10 electrons are : N3– (7 + 3); O2– (8 + 2); Na+(11 – 1); Mg2+(12 – 2) and Al3+(13 – 3). (ii) Arhas 18 electrons. The other 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 10electrons are : N3–, O2–, Na+, F–, Mg2+, Al3+. (iv) Rb+ has (37 – 1) = 36 electrons. The other isoelectronicspecies with 36 electrons are : Br(35 + 1), Kr(36), Sr2+ (38 – 2)

3.12 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.

A. (i) Each one of these have 10 electrons and hence all are isoelectronic

(ii) N3– > O2– > F– > Na+ > Mg2+ > Al3+

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

A. in the formation of cations their is a loss of one or more electrons which increases the effectivenuclear charge. As a result, the force of attraction of the nucleus for the electrons increases andhence the ionic radii decreases. The ionic radii of an anion is always larger than its parent atombecause the addition of one or more electrons decreases the effective nuclear charge. As a result, theforce of attraction of the nucleus for the electrons decreases and hence the ionic radii increases.

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

3.15 Energy of an electron in the ground state of the hydrogen atom is –2.18 × 10–18 J. Calculate theionization enthalpy of atomic hydrogen in terms of J mol–1.

3.16 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 ?

A. (i) The ionization enthalpy, beside other things, depends upon the type of electron to be removedfrom the same principal shell. In Be (electronic configuration; 1s2, 2s2), the outermost electron ispresent in 2s-orbital while in B (electronic configuration : 1s2, 2s22p1) it is present in 2p-orbital. Since2s-electrons are more strongly attracted to the nucleus as compared to 2p-electrons, therefore, lesserenergy is required to remove out a 2p-electron, as compared to 2s-electron. Because of this, H of Beis higher than that of

iH of B. (ii) Nitrogen has the electronic configuration : 1s2, 2s2 2p

x1 2p

y1 2p

z1 in

which 2p-orbital are half filled is more stable than the oxygen which has the electronic configuration1s2, 2s2 2p

x2 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 ofincreased nuclear charge outweights the effect of stability due to exactly half-filled orbitals. Becauseof 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 completely filled have higher ionizationenthalpy as compared to one having neither half-filled nor completely filled orbital. Also, ionizationenthalpy of half-filled orbital is < that of ionisation enthalpy of completely filled orbital)

3.17 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 ?

A. The sodium has electronic configuration 1s2, 2s22p6,3s1 and magnesium has electronic configuration1s2, 2s22p6, 3s2. Here, in both the cases first electron is to be removed from 3s-orbital, but the nuclearcharge of Na (+11) is lower than that of Mg (+12). Therefore, the first ionization enthalpy of sodiumis lower than magnesium. After the loss of one electron from sodium, the electronic configuration ofsodium becomes completely filled i.e., 1s2, 2s22p6 and after the 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

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

A. (i) Atomic size : On moving down the group, the atomic size gradually increases with the addition ofone energy shell to each succeeding element. As a result, the distance of valence electrons from thenucleus increases. Consequently, the force of attraction of the nucleus from the valence electronsdecreases and hence the ionization enthalpy decreases on moving down the group. (ii) Screeningeffect : On moving down the group, with the addition of new shells, the number of inner electronshells which shield the valence electrons increases regularly, thereby increasing the shielding effector screening effect. This means, that the force of attraction of the nucleus for the valence electronsfurther decreases and hence the ionization enthalpy further decreases

3.19 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 ?

A. On moving down the group from Boron to Thalium via aluminium, indium and thallium theionization enthalpy decreases (although not regular). This can be explained on the basis ofincreasing atomic size and screening effect. The abnormal behaviour of Al, Ga, In and Tl can beexplained as follows : Al follows immediately after s-block elements while Ga and In follow afterd-block elements, whereas Tl after d- and f-block elements. These d- and f- electrons do not shieldthe outer-shell electrons from the nucleus effectively. The result is that, the valence electron remainmore tightly held by nucleus and thus larger amount of energy is needed for their removal. Thisexplains why Ga has higher ionization enthalpy than Al. Further moving down the group from Ga toIn, the shielding effect increases due to the presence of additional 4d-electrons, outweighs the effectof 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-electrons andthus the

iH of Tl is higher than that of In

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

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

A. (i) Both oxygen and fluorine belongs to 2nd period. As we move from left to right i.e., from O to F, theatomic size decreases with the increase of nuclear charge. Both these factors are responsible toincrease the attraction of the nucleus for the incoming electron and thus electron gain enthalpybecomes more negative. Further, gain of one electron by O gives O– ion does not have stable inert gasconfiguration, whereas, gain of one electron by F gives F– ion which has stable inert gasconfiguration. Because of this, the energy released is much higher in going from F to F– than in goingfrom O to O–, i.e., the electron gain enthalpy of F(–328 kJ mol–1) is much more negative than that ofoxygen (–141 kJ mol–1). (ii) In general, the electron gain enthalpy becomes less negative on movingdown the group, but the electron gain enthalpy of chlorine (–349 kJ mol–1) is little more negativethan that of fluorine (–328 kJ mol–1). This is because fluorine has a small atomic size (only two shells)and hence the incoming electron is not accepted with the same ease as in the case of large chlorineatom. Because of this electron gain enthalpy of chlorine is more negative than that of fluorine

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

A. The second electron gain enthalpy of oxygen is positive which can be explained as given : When anelectron 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 anotherelectron is added to O– to form O2– ion, energy is released to overcome the strong electrostaticrepulsion between the negatively charged O– ion and the 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

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

A. The electron gain enthalpy refers to the tendency of an isolated gaseous atom to accept an additionalelectron to form a negative ion, whereas electronegativity refers to the tendency of the atom of anelement to attract the shared pair of electrons towards it in the formation of a covalent bond

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

A. The statement seems to be absurd. In fact, the electronegativity of any given atom is not constant. Itincreases as the percentage of s-character of a hybrid orbital increase or the oxidation state of theelement increase. For example, the electronegativity of N increases as we move from sp3-sp2-sphybrid orbitals

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

A. (a) When a neutral atom gains an electron to form an anion, the number of electrons in the anionincreases whereas its nuclear charge remains the same as that of the parent atom. Since the samenuclear charge now attracts larger numbers of electrons, therefore, the force of attraction of thenucleus on the electrons of all shells decreases. This means that the effective nuclear chargedecreases and the expansion of the electron cloud occurs. Because of this, the distance between thecentre of the nucleus and the last shell increases thereby increasing the ionic radius. (b) When aneutral atom loses an electron to form a cation, the number of electrons in the cation decreaseswhereas its nuclear charge remains the same as that of the parent atom. Since the same nuclearcharge now attracts lesser numbers of electrons, therefore, the force of attraction of the nucleus onthe electrons of all shells increases. This means that the effective nuclear charge increases and thecontraction of the electron cloud occurs. Because of this distance between the centre of the nucleusand the last shell decreases thereby decreasing the ionic radius

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

A. Two isotopes of the same element have the same atomic number; which means that they have thesame number of electrons and hence they have the same ionisation enthalpy

3.26 What are the major differences between metals and non-metals ?

3.27 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.

(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.

A. (a) The element with five electrons in the outermost sub-shell belongs to seventeenth group i.e.fluorine, chlorine, bromine, etc. (b) The element that would tend to loose two electrons belongs toalkaline earth metals i.e., magnesium, calcium, etc. (c) The element that would tend to gain twoelectrons belongs to sixteenth group i.e., oxygen and sulphur. (d) A metal which is liquid at roomtemperature is bromine belongs to fifteenth group

3.28 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.

A. Reactivity of an element depends upon the ease with which it can lose the outermost electron. Thetendency to lose electron, in turn, depends upon the ionization enthalpy. The ionization enthlapydecreases down the group, therefore, the reactivities of group 1 element increase in the same orderi.e., Li < Na < K < Rb < Cs. In contrast group 17 elements, have seven electrons in their respectivevalence shells and thus have a strong tendency to accept one electron to get the nearest inert-gasconfiguration. The tendency to accept electrons, in turn, depends upon their electrode potentials.The electrode potential of seventeenth group decreases in order i.e., F > Cl > Br > I

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

A. s-block elements : ns1 – 2, n = 2 to 7, where n represents period in all the cases, p-block elements :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

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3.30 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.

A. (i) The group number of the element = 10 + number of electrons in the valence shell = 10 + 6. Theelements with atomic number 16 is sulphur and the electronic configuration is 1s2, 2s22p53s23p4 (ii)the group number of the element = Number of d-electrons + number of s-electrons = 2 + 2 = 4. Thus,the element belongs to group 4 and 4th period i.e, Titanium with atomic number 22. The electronicconfiguration is 1s2, 2s2, 2p6, 3s23p63d2, 4s2 (iii) The complete electronic configuration of the elementis [Xe]4f7, 5d1, 6s2. Thus, the atomic number of the element = 54 + 7 + 1 + 2 = 64. The element isgadolinium with atomic number 64.

3.31 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) ?

A. (a) The element V has highest first ionization enthalpy (iH

1) and positive electron gain enthalpy

(eg

H) and hence it is the least reactive element. Element V must be inert gas because inert gaseshave positive

egH. 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 (

egH) is the

most reactive metal. The values of iH

1,

iH

2 and

egH match that of potassium. (c) The element III

which has high first ionization enthalpy (iH

1) and very high negative electron gain enthalpy (

egH)

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 earth metal and hence will form binary halide of the formulaMX

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 theformula MX (X = halogen), therefore, the alkali metal must be least reactive. The values of

iH

1,

iH

2

and eg

H match that of Lithium

3.32 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

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A. (a) Li2O (Lithium oxide) (b) Mg

3N

2 (Magnesium nitride) (c) AlI

3 (Aluminium iodide) (d) SiO

2

(Silicon dioxide) (e) PF3 (Phosporus tri-fluoride) or PF

5 (Phosphorus penta-fluoride) (f) LuF

3

(Lutetium fluoride)

3.33 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.

A. In the modern periodic table, each period begins with the filling of a new shell. Therefore, the periodindicates the value of principal quantum number. Thus, option (c) is correct.

3.34 Which of the following statements related to the modern periodic table is incorrect ?

(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitalsin a p-shell.

(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitalsin a d-subshell.

(c) Each block contains a number of columns equal to the number of electrons that canoccupy that subshell.

(d) The block indicates value of azimuthal quantum number (l) for the last subshell thatreceived electrons in building up the electronic configuration.

A. Statement (b) is incorrect while other statements are correct. The correct statement (b) is : thed-block has 10 columns, because a maximum of 10 electrons can occupy all the orbitals in ad-subshell.

3.35 Anything that influences the valence electrons will affect the chemistry of the element. Which one ofthe following factors does not affect the valence shell ?

(a) Valence principal quantum number (n)

(b) Nuclear charge (Z)

(c) Nuclear mass

(d) Number of core electrons

A. Nuclear mass does not affect the valence shell because nucleus consists of protons and neutrons.Whereas proton, i.e., nuclear charge affects the valence shell but neutrons do not. Thus, option (c) iswrong.

3.36 The size of isoelectronic species – F–, Ne and Na+ is affected by

(a) nuclear charge (Z)

(b) valence principal quantum number (n)

(c) electron-electron interaction in the outer orbitals

(d) none of the factors because their size is the same

A. The size of the isoelectric ions depends upon the nuclear charge (Z). As the nuclear charge increasesthe size decreases. For example, F– (+9) > Ne (+10) > Na+ (+11). Therefore, statement (a) is correctwhile all other statements are wrong.

3.37 Which one of the following statements is incorrect in relation to ionization enthalpy ?

(a) Ionization enthalpy increases for each successive electron.

(b) The greatest increase in ionization enthalpy is experienced on removal of electron fromcore noble gas configuration.

(c) End of valence electrons is marked by a big jump in ionization enthalpy

(d) Removal of electron from orbitals bearing lower n value is easier than from orbital havinghigher n value.

A. Statement (d) is incorrect. The correct statement is : Removal of electron from orbitals bearinglower n value is difficult than from orbital having higher n value. All other statements are correct.

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3.38 Considering the elements B, Al, Mg and K, the correct order of their metallic character is :

(a) B > Al > Mg > K (b) Al > Mg > B > K

(c) Mg > Al > K > B (d) K > Mg > Al > B

A. In a period, metallic character decreases as we move from left to right. Therefore, metallic characterof K, Mg and Al decreases in the order : K > Mg > Al. However, within a group, the metalliccharacter, increases from top to bottom. Thus, Al is more metallic the B. Therefore, the correctsequence of decreasing metallic character is : K > Mg > Al > B, i.e., option (d) is correct.

3.39 Considering the elements B, C, N, F and Si, the correct order of their non-metallic character is :

(a) B > C > Si > N > F (b) Si > C > B > N > F

(c) F > N > C > B > Si (d) F > N > C > Si > B

A. In a period, the non-metallic character increases from left to right. Thus, among B, C, N and F,non-metallic character decreases in the order : F > N > C > B. However, within a group, non-metalliccharacter decreases from top to bottom. Thus, C is more non-metallic than Si. Therefore, the correctsequence of decreasing non-metallic character is : F > N > C > B > Si, i.e., option (c) is correct.

3.40 Considering the element F, Cl, O and N, the correct order of their chemical reactivity in terms ofoxidizing property is

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

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

A. Within a period, the oxidising character increases from left to right. Therefore, among F, O and N,oxidising power decreases in the order : F > O > N. However, within a group, oxidising powerdecreases from top to bottom. Thus, F is a stronger oxidising agent than Cl. Further because O ismore electronegative than Cl, therefore, O is a stronger oxidising agent than Cl. Thus, overalldecreasing order of oxidising power is : F > O > Cl > N, i.e., option (b) is correct.

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