classification of elements and periodicity in properties · classification of elements and...

Classification of Elements and Periodicity in Properties

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

Solution

The main idea of organising the elements in the periodic table is to obtain an orderly

arrangement that may enable us to study the properties of elements with ease.

2) Which important property did Mendeleev use to classify the elements in his periodic table

and did he stick to that?

Solution

Mendeleev used “atomic weight” as the fundamental property to classify the elements in the

periodic table.

Except for a few cases, he stuck to his principle of classification. For instance, the placement

of iodine and tellurium in his periodic tale. Even though the atomic weight of iodine is lower

than that of tellurium, he placed tellurium in the sixth group and iodine in the seventh group

along with the elements fluorine, chlorine and bromine because of similarities in their

properties.

3) What is the basic difference in approach between the Mendleev’s Periodic Law and the

Modern Periodic Law?

Solution

Mendeleev arranged the elements in his periodic table in increasing order of their atomic

weighs, whereas, in the Modern Periodic Table, the elements are arranged in increasing order

of their atomic numbers.

4) On the basis of quantum numbers, justify that the sixth period of the periodic table should

have 32 elements.

Solution

For the sixth period, the value of the principal quantum number(n) is 6.

In general, the available orbits for any nth period are ns, np, (n-1)d and (n-2)f. Therefore ,

when n=6, the available orbitals are 6s, 6p, 5d and 4f. We know that ‘s’ has one orbital, ‘p’

has 3 orbitals, d has 5 orbitals. and f has 7 orbitals. Hence, in total, 16(1+3+5+7= 16) orbitals

are available. As each orbital can accommodate 2 electrons, 16 orbitals can accommodate 32

electrons. Therefore, 32 elements are present in sixth period, and the electrons are filled in the

order of 6s, 4f, 5d and 6p.

5) In terms of period and group where would you locate the element with Z= 114?

Solution

In general, any element in the periodic table can be located y the quantum numbers of the last

filled orbital. The electron distribution in various orbitals of an element can be known through

its electronic configuration. Hence, writing the electronic configuration of the element with

atomic number 114 helps in identifying the last filled orbital in it. The electronic configuration

of the element with atomic number 114 can be written as [Rn] 7s25p146d107p2 (86

+2+14+10+2=114) . Thus, the last filled orbital in it is 7p2. Here, the principal quantum

number (n) indicates that it belongs to the seventh period. P2 indicates IV A group (ns2np2).

Thus, element with atomic number 114 belongs to the seventh period, and IV A group of the

periodic table.

6) Write the atomic number of the element present in the third period and seventeenth group

of the periodic table.

Solution

Given that the elements belongs to the third period, which indicates that the principal quantum

number (n)= 3, and the seventeenth group or VII A group, which indicates that the element

contains 7 valence electrons. Hence, as the element’s n value is 3 and it has seven valence

electrons, its valence configuration can be written as 3s23p5.The complete electronic

configuration of the element can e written as 1s22s22p63s22p5. The total number of electrons in

it= 2+2+6+2+5=17. Thus, the element present in the third period and the seventeenth group of

periodic table is chlorine, whose atomic number is 17.

7) Which element do you think would have been named by

i) Lawrence Berkeley Laboratory

ii) Seaborg’s group?

Solution

i) The element named by Lawrence Berkeley Laboratory are:

a) Lawrencium (Lr) with the atomic number 103.

b) Berkelium (Bk) with the atomic number 97.

ii) The element named by Seaborg’s group in Seaborgium (Sg) with the atomic number 106.

8) Why do elements in the same group have similar physical and chemical properties?

Solution

The physical and chemical properties of the elements depend on the number of valence

electrons or outer electronic configuration. As the elements in a group have similar outer or

valence electronic configuration, they have similar physical and chemical properties.

9) What does atomic radius and ionic radius really mean to you?

Solution

Atomic radius: The atomic radius of an atom is the distance between the centre of its nucleus

and the electron cloud of the outermost energy level. The electron cloud surrounding an atom

has no definite boundary; hence, the exact determination of the atomic size is not possible. To

get comparative values for atoms of different elements and also to explain a number of

chemical phenomena, an operational concept of atomic radius i needed.

There are three operational concepts of atomic radius based on the type of bond in which the

atom is involved. They are:

i) Covalent radius- When the bonding is covalent.

ii) Metallic radius – When the bonding is metallic.

iii) Vanderwaal’s radius- When two atoms are not bonded by a chemical bond(as in noble

gases).

Ionic radius: Ionic radius is the effective distance from the nucleus of an ion up to which it has

its influence on its electron cloud. A cation is formed y the loss of an electron by an atom, and

is always smaller than the corresponding atom. For example, the atomic radius of Na is 1.86Å,

while that of Na+ is 1.02 Å. An anion is formed by the gain of electron, and is always larger

than the corresponding atom. For example, the atomic radius of chlorine is 0.99 Å and that of

Cl- is 1.81 Å.

10) How does radius vary in a period and in a group? How do you explain the variation?

Solution

The atomic radius in a given period decreases from the left to the right. The decrease in size is

due to the fact that with the increase in atomic number in a period the nuclear charge

increases, but electrons enter into the same valence shell. As a consequence of which electrons

are attached closer to the nucleus, resulting in a decrease in size of the atom. In a given period,

alkali metal is the largest atom and halogen the smallest.

Atomic size increases down the group The increase in atomic size on moving down the group

is due to the addition of an extra shell, which results in an increase in the distance between

nucleus and the valence electrons. The shielding effect of the inner electrons outweighs the

effect of the increased nuclear charge, and consequently, the size of an atom increases down

the group.

11) What do you understand by isoelectronic species? Name a species that will be

isoelectronic with the atoms or ions of F-.

Solution

Species (atoms or ions) with the same number of electrons are known as isoelectronic species.

i) F-.

Fluorine atom: F(Z=9)

Fluorine ion (F-) = 9+1=10 electrons. Hence, any atom or ion having 10 electrons is

isoelectonic with a fluoride ion. They are:

Atom Atomic number (Z)

Ion Number of electrons

Oxygen 8 0-2 8+2=10 Neon 10 ---- 10 Sodium 11 Na+ 11-1=10 Magnesium 12 Mg+2 12-2=10 Aluminium 13 Al+3 13-3=10


ii)Ar

The atomic number(Z) of argon is 18. Hence, any species having 18 electrons would be

isoelectronic with argon.



Sulphur 16 S-2 16+2=18 Chlorine 17 Cl- 17+1=18 Potassium 19 K+ 19-1=18 Calcium 20 Ca+2 20-2=18


iii)Mg+2

The atomic number (Z) of magnesium is 12 . The number of electrons in Mg+2 = 12-2=10.

Hence, any species having 10 electrons would be isoelectronic with a magnesium ion.



Oxygen 8 O-2 8+2=10 Neon 10 ----- 10 Sodium 11 Na+ 11-1=10 Fluorine 9 F- 9+1=10 Aluminium 13 Al+3 13-3=10


iv) Rb+

The atomic number (Z) of rubidium is 37 . The number of electrons in rubidium= 37-1=36.

Hence, any species having 36electrons would be isoelectronic with a rubidium ion.



Bromine 35 Br- 35+1=36 Krpton 36 ----- 36 Strontium 38 Sr+2 38-2=36

12) Consider the following species:

N3- , O2- , F- , Na+ , Mg+2 and Al3+ .

a) What is common in them?

b) Arrange them in the order of increasing ionic radii.

Solution

The species N3- , O2- , F- , Na+ , Mg+2 and Al3+ have the same number of electrons, that is, 10.

Hence, they can be considered as isoelectronic species. The following illustrates that all the

given species are isoelectronic.

Ion Atomic number(Z) of the element

Total number of electrons

N3- 7 7+3=10 O2- 8 8+2=10

F- 9 9+1=10 Na+ 11 11-1=10 Mg+2 12 12-1=10 Al3+ 13 13-3=10

Increasing order of ionic radii.

In the case of isoelectronic species

(ions with same number of electrons, but with different charge),

the greater the nucleus charge, the greater is the attraction for the electrons and smaller is the

ionic radius.

Nuclear chargeα 1/ Ionic radius

Hence, the increasing order of ionic radii of the given isoelectronic species is :Increasing order

of nuclear charge: Al+3 < Mg+2 <Na+ <F- < O-2 < N-3

13) Explain why cations are smaller and anions larger in radii than their parent atoms?

Solution

When an atom loses one or more electrons, a positive ion or cation is formed. A cation is

always smaller than the corresponding atom, because when a cation is formed, the ratio of

nuclear charge to the number of electrons(Z/e) increase. As a result, the effective nuclear

charge increases and the electrons are pulled towards the nucleus resulting in the decrease of

atomic size. For example the atomic radius of sodium atom is 1.86Å while that of sodium ion

is 1.02 Å. When an atom gins one or more electrons, a negative ion or anion is formed. An

anion is always larger than the corresponding atom because, when an anion is formed, the

ration of nuclear charge to the number of electrons(Z/e) decreases. As a result, the effective

nuclear charge decreases, because of which the nuclear attraction on the valence electrons

decrease, and this results in an increase in ionic size. For example, the atomic radius of

chlorine is 0.99 Å and that of chloride ion is 1.81 Å.

14) What is the significance of the terms –‘isolated gaseous atom’ and ‘ground state’ while

defining the ionisation enthalpy and electron gain enthalpy?

Hint: Requirements for comparison purposes.

Solution

Ionisation enthalpy refers to the energy required to remove the outermost electron from an

isolated gaseous atom in its ground state.

Electron gain enthalpy is the amount of energy released when an electron is added to an

isolated gaseous neutral atom in its ground state.

An isolated gaseous atom means an atom in the gaseous state that is all by itself and is not

bound to any other atom. In the gaseous state, the inter atomic forces of attraction are at the

minimum, As far as the determination of ionisation enthalpy and electron gain enthalpy are

concerned, it is essential that the inter atomic forces of attraction between the atoms should be

at the minimum.

The f=ground state of an atom is its lowest possible energy or most stable state. The exact

determination of the ionisation enthalpy or electron gain enthalpy would e possible only when

the atom is in the ground state and influenced by any other external forces. This is because

when the atom is in an excited state, less amount of energy would be required to remove an

electron, and so the ionisation energy value will be less. On the other hand, when the atom is

not in the ground state, less amount of energy is released when an electron is added.

15) Energy of an electron in the ground state of the hydrogen atom is -2.18 x 10-18 J.

Calculate the ionisation enthalpy of atomic hydrogen in terms of Jmol-1.

Solution

Energy of an electron in ground state of hydrogen atom= -2.18 x 10-18 J

The negative sign indicates the attraction between the electron and the nucleus, and the same

amount of energy, which is 2.18 x 10-18 J, must be spent to break this attraction.

Therefore, ionisation enthalpy, that is, the energy required to remove the electron from the

ground state of the hydrogen atom is 2.18 x 10-18 J

To remove one mole or 6.023 x 1023 electrons, the amount of energy needed=?

= 6.023 x 1023 x 2.18 x 10-18 J = 1.31236 x 106 Jmol-1

Hence, the ionisation enthalpy of atomic hydrogen in terms of Jmol-1 is 1.31236 x 106 Jmol-1 .

16) Among the second period elements the actual ionization enthalpies are in the order of

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

Explain why

i) Be has higher ∆iH than B ii) O has lower∆iH than F?

Solution

i) In beryllium (Be-1s22s2), the paired electrons in the valence shell are present in the highly

penetrating (towards nucleus) s-orbital (due to its spherical shape). Hence, more energy is

required to remove the electron from this orbital. On the other hand, in boron (B-1s22s22p1),

the unpaired electron is present in the less penetrating p-orbital , which requires less energy

for removal. Hence, Be has higher ∆iH than B.

ii) In case of nitrogen (N-1s22s22p3). the p-orbital is half filled and has extra stability, so more

energy is required to remove an electron from it. On the other hand, in oxygen (O-1s22s22p4),

the fourth electron is present in the 2p subshell. Electrons being negatively charged repel each

other, resulting in an increase in the electron- electron repulsion .thus, it requires

comparatively less energy for the removal of the 4th 2p electron in oxygen. Therefore, oxygen

has lower ionisation enthalpy than nitrogen . On the other hand, fluorine, being smaller in size,

experiences more effective nuclear charge, and hence nuclear attraction . Thus, more energy is

needed to remove an electron from fluorine in comparison to oxygen.

17) How would you explain the fact that the first ionisation enthalpy of sodium is lower than

that of magnesium nut its second ionisation enthalpy is higher than that of magnesium?

Solution

The first ionisation enthalpy of sodium is lower than that of magnesium, because sodium

needs to lose only one electron to get the stable inert gas configuration of neon. This can be

clearly known by looking at the electronic configuration of sodium and sodium ions.

Na(Z=11)→Na+ + e− (first ∆iH = 5.138 eV)

(2,8,1) (2,8)

Magnesium has two electrons in its valence shell in the s – subshell. Because of some stability

gained in having a filled sublevel, more energy is needed to remove an electron from such an

orbital. Moreover, magnesium does not attain inert gas configuration after losing one electron.

A glance at the electronic configurations of magnesium and magnesium ion will help in

understanding this.

Mg(Z=12) → Mg+ + e− (First ∆iH = 7.6 eV)

(2,8,2) (2,8,1)

The second ionisation enthalpy of magnesium is lower than the second ionisation energy of

sodium. This can be explained on the fact that much more energy is needed to remove an

electron from the already attained stable inert gas configuration of a sodium ion.

Na+ → Na++ + e− (second ∆iH = 47.2 eV)

(2,8) (2,7)

Magnesium, on the other hand, tends to lose the second electron, because after losing the

second electron, it will attain the stable inert gas configuration of neon.

Mg+→ Mg++ + e−(second ∆iH = 15.0 eV).

Thus, the energy required to remove the second electron in magnesium is less than that in

sodium.

18) What are the various factors due to which the ionisation enthalpy of the main group

elements tends to decrease down a group?

Solution

The factors that influence the decrease in ionisation enthalpy down a group are :

a) Increase in atomic radii.

b) Increase in shielding effect or screening effect.

a) Increase in atomic radii:

As we descend down a group, with an increase in atomic radius, the distance between the

nucleus and the valence electrons increases, and hence, the nuclear attraction over the valence

electrons decreases. Thus, ionisation enthalpy or the energy required to remove the valence

electrons decrease down the group.

b) Increase in shielding effect:

In multi-electron atoms, the electrons present in the inner orbits screen the electrons present in

the outermost orbit from the attraction of the nucleus. The screening effect increases with an

increase in the number in the number of electrons in the inner orbits. When the screening

effect increases, the ionisation enthalpy decreases.

19) The first ionisation enthalpy values (in kJ mol-1) of group 13 elements are:

How would you explain this deviation from the general trend?

Solution

In general, as we descend down a group , the ionisation energy decreases with an increase in

atomic size. So, the expected order of ionisation enthalpy in the given IIIA group elements

would be B>Al>Ga>In>Tl. However, the order observed is B>Al<Ga>In<Tl. The anomaly

can be understood from their electronic configurations.

Element Atomic number

Electronic configuration

Ionisation enthalpy (kJmol-1)

B 5 [He]2s22p1 801

Al 13 [Ne]3s23p1 577

Ga 31 [Ar]3d104s24p1 579

In 49 [Kr]4d105s25p1 558

Tl 81 [Xe]4f145d106s26p1 589

B Al Ga In Tl

801 577 579 558 589

The following deductions can be drawn from the table.

1. As expected, the ionisation enthalpy decreases from boron to aluminium due to an increase

in the atomic size and the shielding effect.

2. Instead of decrease in ionisation enthalpy from aluminium to gallium, there is an increase.

The increase in ionisation enthalpy is because of the fact that in gallium (In indium and

thallium also) the poor shielding effect of ‘d’ electrons do not effectively screen the valence

electrons from the nuclear attraction. Hence, the valence electrons are held more firmly by

the nucleus and thus, more energy is required to remove the electrons. So, ionisation enthalpy

of gallium is higher than aluminium.

3. From gallium to indium, as expected , the ionisation energy decreases.

4. As we move from indium to thallium, instead of a decrease in the ionisation enthalpy, we

find an increases. The increase in the ionisation enthalpy is because of the poor shielding

effect of 5d10 and 4f14 electrons in thallium, which again do not effectively screen the valence

electrons from nuclear attraction, resulting in an increase in the ionisation enthalpy.

20) Which of the following pairs of elements would have more negative electron gain

enthalpy?

i) O or F ii) F or Cl

Solution

Electron gain enthalpy is considered more negative when more nergy is released

(exothermic) on the addition of an electron to an atom.

i) O or F

The electron gain enthalpy of fluorine is more negative than that of oxygen.

This is because:

1. The atomic size of fluorine atom (0.72 Å) is smaller than that of oxygen atom (0.72 Å).

Hence, the nucleus in the fluorine atom can attract the incoming electron more easily than the

oxygen atom. Thus, more energy is released in fluorine than in oxygen when an extra

electron is added.

2. Fluorine requires only one electron to attain the stable inert gas configuration of neon,

while oxygen needs two. Hence, the tendency to gain an electron is more in fluorine than in

oxygen.

ii) F or Cl

The electron gain enthalpy of chlorine (-348 KJ/mol) is m ore negative than that of fluorine

(-333kJ/mol). This is because the fluorine atom is smaller in size than the chlorine atom, and

has strong inter-electronic repulsions. Some amount of energy is used to overcome the inter-

electronic during the addition of an electron to the fluorine atom. Hence, the energy liberated

is less than that of chlorine atom.

21) Would you expect the second electron gain enthalpy of O as positive , more negative or

less negative than the first? Justify your answer.

Solution

The second electron gain enthalpy of oxygen is positive, because the uninegative ion that is

formed by accepting an electron prevents the entry of further electrons due to the repulsive

forces between the negative charges. Thus, energy is needed to overcome these repulsive

forces between the uninegative ion and the electron. Hence, the second electron gain

enthalpy of oxygen is positive.

22) What is the basic difference between the terms electron gain enthalpy and electro

negativity?

Solution

The basic difference between the electron gain enthalpy and electronegativity is that electron

gain enthalpy refers to the energy released (-ve) or absorbed (+ve) when an electron is added

to a neural gaseous atom, whereas, electronegativity refers to the tendency of an atom to

attract the shared electron pair towards itself in a diatomic molecule.

23) How would you react to the statement that the electronegativity of N on Pauling scale is

3.0 in all nitrogen compounds?

Solution

Pauling has determined the electronegativities of all the elements by taking hydrogen as the

reference element. The difference in the electronegativity between the two combined atoms

may change, but the electronegativity of an element remains the same. For example, the

electronegativity of nitrogen on the Pauling scale remains 3.0 even when it is bonded to

hydrogen in ammonia or bonded to oxygen in NO2 . However, the difference between the

electronegativities between the two elements in these two compounds would vary. Hence, the

electronegativity of nitrogen on the Pauling scale in all nitrogen compounds remains the same.

24) Describe the theory associated with the radius of an atom as it

a) Gains an electron

b) Loses an electron

Solution

a) When an atom gains one or more electrons, a negative ion or anion is formed. An anion is

always larger than the corresponding atom, because the ratio of nuclear charge also decreases.

As a result , the nuclear attraction on the valence electrons decreases, and this results in an

increase in ionic size. For example , the atomic radius of chlorine is 0.99 Å and that of

chloride ion is 1.81 Å.

b) When an atom loses one or more electrons, a positive ion or cation is formed. A cation is

always smaller than the corresponding atom, because the ratio of nuclear charge to the number

of electrons(Z/e) increases in a cation, and the effective nuclear charge also increase. As a

result, the electrons are pulled towards the nucleus, resulting in a decrease in atomic size. For

example, the atomic radius of sodium atom is 1.86 Å while that of sodium ion is 1.02 Å.

25) Would you expect the first ionisation enthalpies for two isotopes of the same element to be

the same or different ? Justify your answer.

Solution

Atoms of a given element with the same atomic number but with different mass numbers are

called isotopes. That is, isotopes have the same number of electrons and protons, but a

different number of its electrons and protons (nuclear charge), and not on the number of

neutrons, the ionisation enthalpies of two isotopes of the same element would remain the

same.

26) What are the major differences between metals and non-metals?

Solution

27) Use the periodic table to answer the following questions.

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

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

Metals Non-metals They exhibit metallic lustre when polished or freshly cut.

They are non- lustrous, except for graphite and iodine.

They are solids, except for mercury, which is a liquid.

They exist in all the three forms of matter.

They possess high electrical and thermal conductivities.

They are bad conductors of heat and electricity, except for graphite, which is a good conductor of electricity.

They are malleable and ductile. They are neither malleable nor ductile.

They have the tendency to lose electrons.

They have the tendency to gain electrons.

They have low ionisation enthalpies. They have high ionisation enthalpies.

They are good reducing agents. They are good oxidising agents. They have high melting and boiling points.

They have low melting and boiling points.

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.

Solution

a) Elements with five electrons in the outer subshell belong to the Vth group in the

periodic table. The elements in the Vth group are N, P, As, Sb, Bi. Hence, an element with

five electrons in the outer subshell can be any one of the Vth group members.

b) II A group or alkaline earth metals have the tendency to lose two electrons to attain the

nearest inert gas configuration. The elements of the IIA group or alkaline earth metals are

Be, Mg, Ca, Sr, Ba and Ra. Hence, an elements that would tend to lose two electrons could

e any of the II A group members.

c) Elements of the VIA group require two electrons to attain the nearest stable inert gas

configuration . The elements of the VI A group are O, S, Se, Te, Po. Hence, an element

that would tend to gain two electrons could be any one of the members of VIA group.

d) The VII A group has elements that exist in all the three states of matter at room

temperature, F, Cl, Br, I and At are the members of the VII A group. Among the members

of the VII A group, F and Cl are gases, Br is a liquid, and I and At are solids.

28) The increasing order of reactivity among group I elements is Li< Na< K< Rb< Cs

whereas that among group 17 elements is F>Cl>Br>I. Explain.

Solution

As we descend down the group of alkali metals, the atomic size increases and ionisation

enthalpy decreases. Due to this, the tendency to lose electrons from the valence shell, and

hence, the reactivity increase. Hence, the order of reactivity of IA group elements is Li <

Li< Na< K< Rb< Cs. On the other hand, when we descend down the group of non-metals,

that is , group 17 elements, the atomic size, and hence, the tendency to gain electrons,

increases. Hence, the reactivity decreases. Thus, the order of reactivity of group 17

elements is F>Cl>Br>I.

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

Solution

The general outer electronic configuration of s, p ,d, and f-block elements is:

Block General outer electronic configuration

s ns1-2 p ns2np1-6 d (n-1)d1-10ns1-2 f (n-2)f1-14(n-1)d0-1ns2

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)d1 ns2 for n=6, in the periodic table.

Solution

i) As n=3, the element belongs to the third period. Writing the outer electronic

configuration for the element with n= 3 gives 3s23p4. Sulphur is the element in the third

period with 3s23p4 valence configuration. Sulphur belongs to the third period and the VIA

or sixteenth group of the periodic table.

ii) As n= 4, the element belongs to the fourth period. Writing the outer electronic

configuration for the given element with n=4 gives 3d24s2. Titanium with atomic number

22 is the element that has 3d23s2 outer electronic configuration. Titanium belongs to the

fourth period and the IVB group of the transition series.

iii) As n=6, the element belongs to the sixth period. Writing the outer electronic

configuration for the element with n= 6 gives 4f75d16s2. Lanthanum with atomic number

57 is the element that has 4f75d16s2 outer electronic configuration. Lanthanum belongs to

the sixth period and the IIIB group of the periodic table.

31) The first (∆iH) 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 ∆iH2 ∆egH Elements ∆iH ∆iH2 ∆egH

I 520 7300 -60 IV 1008 1846 -295

II 419 3051 -48 V 2372 5251 +48

III 1681 374 -328 VI 738 1451 -40

Which of the above elements is likely to be:

a) The least reactive element. b) The most reactive metal.

Solution

a) Element V with the highest first ionisation enthalpy of 2372 indicates that is has the

least tendency to lose electrons, and a positive electron gain enthalpy of +48 indicate its

least tendency to gain electrons. Hence, element V with no tendency to gain or lose

electrons is the least reactive element.

b) Element II with the lowest first ionisation enthalpy of 419 indicates its greater metallic

nature or tendency to lose electrons, and a low negative value of electron gain enthalpy of

-48 indicates its least tendency to gain electrons. Hence, element II with a greater tendency

to lose electrons and least tendency to gain electrons is the most reactive metal.

c) Element III with the highest negative electron gain enthalpy of -328 indicates its greater

tendency to gain electrons or non-metallic nature . Element III with a higher value of first

ionisation enthalpy of 1681 also indicates its least tendency to lose electrons. Hence,

element III with a greater tendency to gain electrons is the most reactive non-metal.

d)In the above table, elements III and IV with high negative electron gain enthalpy values

indicate that they are non-metals . Among these two, the one with comparatively lower

negative electron gain enthalpy value of -295 indicates that is has less tendency to gain

electrons, and hence, less reactivity. Therefore, element IV is the reactive non-metal.

e) Element VI with the lowest values of first (-738) and second (1451) ionisation enthalpy

values indicates it can lose two electrons easily, this shows its metallic nature . A low

electron gain enthalpy of -40 also indicates its least tendency to gain electrons. Hence,

element VI with a valency of 2 indicates that it can form a stable binary halide of the

formula MX2 .

f) Element I with lower value of first ionisation enthalpy (520) and a very higher value of

second ionisation enthalpy (7300) indicates that it is a metal that can lose only one

electron easily but not two. Hence, element I with a valency of 1 can form a stable

covalent halide of the formula MX.

32) Predict the formula of the stale binary compounds that would be formed by the

combination of the following pairs of elements.

a) Lithium and oxygen b) Magnesium and nitrogen c) Aluminium and iodine

d) Silicon and oxygen e) Phosphorus ad fluorine f) Element 71 and fluorine

Solution

a) Li1+ O2-

LiO2------Lithium oxide

b)Mg2+N3-

Mg3N2------Magnesium nitride

c) Al3+I-

AlI3-------Aluminium iodide

d)Si4+O2-

SiO2------Silicon dioxide

e)P+3F1- P+5F1-

PF3----Phosphorus trifluoride PF5 ----Phosphorus pentafluoride

f) The element with atomic number 71 is lutetium. Its oxidation state is 3+.Lu3+F1-

LuF3------Lutetium trifluoride

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.

Solution

The correct option is ‘c’.

The period indicates the value of n(the principal quantum number) fir the outer most or

valence shell of the atom of an element.

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

orbitals in a p-shell.

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

orbitals in a d-subshell .

c) Each block contains a number of columns equal to the number of electrons that can

occupy that subshell.

d) A block indicates the value of the azimuthal quantum number(l) for the last subshell

that received electrons in building up the electronic configuration.

Solution

The incorrect statement is ‘b’.

The ‘d block ‘ has ten columns or groups as the five d-orbitals can accommodate a

maximum of ten electrons.

35) Anything that influences the valence electrons will affect the chemistry of the element.

Which one of the 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.

Solution


Nuclear mass has no effect on the valence electrons of an element.

36) The size of isoelectronic species-F-, Ne and Na+ is affected y

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.

Solution

The correct option ‘a’.

Ion/atom Atomic number(nuclear charge), Z

Number of electrons in the ion or atom, e−

Z/e ratio

Ionic or atomic radii

F- 9 10 0.9 1.36

Ne 10 10 1.0 1.12

Na+ 11 10 1.1 0.95

In the case of isoelectronic species, the greater the nuclear charge, the greater is the

attraction for electrons and smaller is the ionic radius. The variation in the radii of the

given isoelectronic species can be understood from the following table.

37) Which one of the following statements is incorrect in relation to ionisation enthalpy?

a) Ionisation enthalpy increase for each successive electron.

b) The greatest increase in ionisation enthalpy is experienced on the removal of an electron

from core noble gas configuration.

c) The end of valence electrons is marked by a big jump in ionisation enthalpy.

d) The removal of electrons from orbitals bearing lower n value is easier than from orbitals

having higher n value.

Solution

The correct option is ‘d’.

The higher the value of the principal quantum number (n), the greater is the distance

between the nucleus and the valence orbital, and lesser is the nuclear attraction on the

valence electrons. Hence, as the value of n increases, the nuclear attraction on the valence

electrons decreases, and the energy required to remove the valence electrons also

decreases.

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

Solution

The correct option is’d’.

‘K’ belongs to the IA group and Mg belongs to the IIA group. The elements ‘B’ and ‘Al’

belong to the same group of IIIA. Metallic character decreases from the left to the right in

a period, and increases down a group. Hence, the order of metallic character in the given

set of element is K>Mg>Al>B.

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

Solution


Elements B,C,N and F belong to the second period of the periodic table, and the elements

C and Si belong to the IV A group of the periodic table. Non-metallic character increases

across a period and decreases down a group.

Hence, based on the positions of these elements in the periodic table, and their

eletronegativity values.

Order of non-metallic character F> N> C> B> Si

Electronegativity values (4.0) (3.0) (2.5) (2.0) (1.8)

40) Considering the elements F, Cl, O,N , the correct order of their chemical reactivity in

terms of oxidising property is:

a) F > Cl >O > N b) F > O> Cl >N c) Cl > F >O> N d) O> F> N >Cl

Solution

The correct option is ‘b’.

The elements F, O, N belong to the second period of the periodic table. The elements F

and Cl belong to the VII A group of the periodic table. The oxidising property increases

from the left to the right in a period due to an increase in the electronegativity. Hence, the

oxidising property of F, O, N follows the order F> O> N. Chlorine, being less

electronegative than oxygen, has less oxidising ability than oxygen. Hence, the oxidising

ability of these elements follow the order F > O > Cl > N.

classification of elements and periodicity in properties · classification of elements and...

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