Periodic Classification Of Elements

Dobereiner created some groups having three elements in each group. He called these groups triads. He stated that when the elements in a group are arranged in increasing order of their atomic masses then the atomic mass of the middle element is approximately the average of the atomic masses of two other elements.

Newland arranged the known elements in increasing order of atomic mass. He found that the every eighth element had properties similar to that of first one. He compared it to the octaves found in music and thus it is called Newland’s law of octaves.

Mendeleev classified the elements on the basis of their fundamental property, i.e. atomic mass and chemical properties. He reacted the elements with hydrogen and oxygen and formulae of the hydrides and oxides of the elements were considered as a basic property for their classification.

Mendeleev left some gaps in his periodic table because he predicted the existence of some elements which had not been discovered at that time.

Electronic configuration of hydrogen is similar to that of alkali metals. Just like alkali metals, it reacts with halogens, oxygen and sulphur to form compounds of similar formulae. For e.g.: H₂S and Na₂S. On the other hand, hydrogen reacts with metals and non-metals just like halogens. Due to this anomalous behaviour of hydrogen, Mendeleev was unable to fix a position for it.

Vertical columns in the periodic table are called “groups”. There are 18 such groups in modern periodic table. Horizontal rows are called “periods”. There are 7 such periods in modern periodic table.

The main difference between the Mendeleev’s and modern periodic law is that Mendeleev considered the properties of elements as periodic function of their atomic masses while modern periodic law considers the properties of elements as periodic function of their atomic numbers.

Elements present in the same group have a same number of electrons in their valence shells. On moving down a group, the number of shells increases but the number of electrons in the valence shell remains the same. That’s why the elements of the same group have similar properties.

(i) The electronic configuration is 2,8. It means that the given element contains 2 electrons in K shell and 8 electrons in L shell. Thus the element contains 2 shells. Thus it belongs to the second period. Secondly, there are 8 electrons in the valence shell (completely filled) which is true for the 18^th group.

(ii) The electronic configuration is 2,5. It means that the given element contains two shells, thus the period is 2^nd. Secondly, there are 5 electrons in the valence shell which is true for the 15^th group.

The atomic size increases on moving down the group. This is because of the increase in a number of shells due to which the distance between the outermost electron and the nucleus increases.

The atomic size decreases on moving along a period. This is because of the increase in the effective nuclear charge on moving from left to right which tends to pull the electrons towards the nucleus.

Since the Molecular formula of oxide of a metal (M) is M₂O₃ where valency of oxygen is 2 and the valency of the metal is 3. Since the metal has a valency of 3 and belongs to the third period, so the metal is Aluminium.

F does not fit in dobereiner’s triad.

Dobereiner stated that when the elements in a group are arranged in increasing order of their atomic masses then the atomic mass of the middle element is approximately the average of atomic masses of two other elements. Fluorine doesn’t fit into this rule while the atomic mass of Br is average of the atomic masses of Cl and I.

Elements were arranged in the increasing order of their atomic numbers. This leads to the modern periodic law that the properties of elements are the periodic function of their atomic numbers.

For group II elements, the group no. (2) is equal to the valency of the elements present in this group. So two such elements belonging to group II are Calcium and Magnesium.

Lithium is present in the 1^st group and the 2^nd period. On moving along the period from left to right, the atomic number increases. The atomic number of Li is 3 so the atomic number of next element will be 4. Thus the next element to Li is Beryllium.

Both Mg and Ca belong to the same group (group II) and are capable of losing electrons which is responsible for their reactivity. Mg lies above Ca in Group II. As we move down a group, the effective nuclear charge on valence shell electrons decreases due to which losing electrons become easier. Thus Ca is more reactive than Mg.

The electronic configuration of Carbon is: 2,4. We can see that there are 4 electrons in the valence shell of Carbon, thus it has a valency of 4.

The electronic configuration of Chlorine is: 2,8,7. Though there are 7 electrons in valence shell of Cl but instead of losing 7 electrons it tends to gain 1 electron. Thus its valency is 1.

Since the metal M belongs to the Group 13, so its valence shell will have 3 electrons. Thus the valency of the metal M belonging to 13^th group is 3.

The atomic number 20 indicates that the electronic configuration of element is 2,8,8,2. The number of electrons in valence shell is 2. Thus its valency is 2. So the formula of its chloride will be MCl₂. Moreover we know that calcium has an atomic number=20, therefore the exact molecular formula will be CaCl₂.

Ca, Mg and Sr all three belong to the same group and are present in the order Mg < Ca < Sr. The atomic size increases on moving down the group. This is because of the increase in number of shells due to which the distance between the outermost electron and the nucleus increases. Thus Sr will have atoms of biggest size.

Li, C and N all three belong to the same period. They are present in the order: Li < Be < B < C < N. On moving along a period from left to right, the metallic character decreases and non-metallic character increases. So Li has the most metallic character and N has the most non- metallic character. Since metal-oxides is basic in nature, therefore oxide of Li will be most basic and non-metal oxide is acidic in nature, therefore Oxide of Nitrogen will be most acidic.

(i) yes, Mendeleev’s suggestion was correct as it was found later on experimentally that beryllium’s atomic mass is 9.

(ii) He studied the formula of hydrides and oxides of beryllium and observed that they had similarity with those of magnesium oxides and hydrides. Thus he placed the beryllium above Mg and predicted its atomic mass as 9.

(i) Since the element belong to the 15^th group , so it must have 5 valence electrons in its atoms.

(ii) since the element is in 2^nd period and belongs to 15^th group, so it must be nitrogen. Nitrogen has a valency of 3.

(i) Both Na and K belongs to the same group (group I). Both contains same number of electrons in their valence shells i.e. 1

(ii) Both B and C belong to the same period (2nd period). Both contains same number of shells i.e. 2

The atomic size increases on moving down the group. This is because of the increase in number of shells due to which the distance between the outermost electron and the nucleus increases.

The atomic size decreases on moving along a period. This is because of the increase in the effective nuclear charge on moving from left to right which tends to pull the electrons towards the nucleus.

The electronic configuration tells about the number of electrons present in different shells. Thus total electrons can be found by adding the electrons present in all shells i.e. 2+8+6=16. The number of electrons and protons are equal. Therefore, atomic number is 16.

(ii) The element with atomic number(Z)=16 is sulphur. The other element which show chemical similarity with sulphur is oxygen.

(i) As the metallic character decreases, the tendency to loose valence electrons also decreases. In other words the tendency to gain electron increases, so their electronegative character increases:

Order: D<A<C<B

(ii) Since the element B has least metallic character, so it will have maximum non-metallic character in it. Non-metal oxides are acidic in nature, therefore oxide of B will be acidic in nature.

(i) Lithium and sodium contain only single electrons in their valence electrons which they can readily loose to react. Due to their high reactivity with water, they are considered as active metals.

(ii) Fluorine lies above chlorine in the 17^th group. This indicates that fluorine has much higher electronegativity than chlorine due to which it tends to gain electron more easily. That’s why fluorine is more reactive than chlorine.

(i) Li is bigger in size than beryllium. This is because the atomic size decreases on moving along a period as the effective nuclear charge increases on moving from left to right which tends to pull the electrons towards the nucleus.

(ii) Na is the most metallic element out of the given elements. This is because on moving along a period from left to right, the metallic character decreases and on moving down a group the metallic character increases. Thus Na is most metallic as it can easily loose its valence electron.

(i) C and D are the two elements which are chemically similar. This is because the electronic configuration of C is 2,8,2 and that of D is 2,2 , so both the elements contains same number of valence electrons.

(ii) Element B is an innert gas. This can be concluded by writing its electronic configuration as 2,8. Both the shells are completely filled. Thus it is an innert gas.

(iii) The element E belongs to the 3^rd period. Its electronic configuration is 2,8,9 from which we can see that there are 3 shells present in it. So it belongs to 3^rd period.

(iv) Element A is a non-metal. Its atomic number is 7, so its electronic configuration will be 2,5, we can see that it contains 5 valence electrons so it belongs to 15^th group and will lie on the right side of the second period. On moving from left to right in a period, non-metallic character increases. Thus A is a non-metal.

● The electrons present in the outermost shell of an electron are known as valence electrons.

● The number of valence electrons increases on moving from left to right in a period. This is because on moving from left to right the electrons are added in the same shell.

● The valency of elements first increases in the period and then decreases. For example the trend of valency can be illustrated as follows:

(i) ● The electronic configuration of element A is: 2,8,1. Number of valence electrons=1 So, it belongs to 1^st group.

● electronic configuration of element B is:2,8,4. It contains 4 valence electrons, so it belongs to 14th group.

● electronic configuration of element C is: 2,8,7. It contains 7 valence electrons, so it belongs to 17^th group.

(ii) The element B is Si and it has a valency of 4. The element C is Chlorine and it has a valency of 1. So the formula of compound when B reacts with C is SiCl4.

(a) From electronic configuration, we can observe that the number of shells is 4. Thus it belongs to the 4^th period.

(b) From electronic configuration, we can observe that the number of valence electrons is 2. Thus, it belongs to 2^nd group.

Valency also depends on the number of valence electrons. Given element X tends to loose two electrons thus it has a valency of 2.

(i) Cu, Mn, Ge, Sc all belong to same period. The order in which they appear in the period in left to right direction is: Sc → Mn → Cu → Ge. The atomic size decreases on moving along a period. This is because of the increase in the effective nuclear charge on moving from left to right which tends to pull the electrons towards the nucleus. Therefore, the increasing order of atomic sizes is: Ge < Cu< Mn< Sc.

(ii) Fe, Hs, Ru, Os all belong to the same group. On moving down a group, the number of shells increases. Thus the increasing order for number of shells is Fe<Ru<Os<Hs.

(iii) Number of valence electrons in given elements is as follows:

Hf → 4

Cs → 1

Ba → 2

Tl → 3

So the order for increasing valency is : Cs<Ba<Tl<Hf

(i) The electronic configuration of Q is :2,8,3. It has three valence electrons, so it has a valency of three.

(ii) We can classify an element as a metal or non-metal by looking at its position in periodic table. Those lying on left side of periodic table are metals and those lying on right side are non-metals.

(iii) The element P is most metallic because it can easily loose two electrons and lies on the left side of the periodic table. Metal oxides are basic in nature, therefore oxide of P will be most basic.

(i) Two elements that can show a valency of two with respect to oxygen are calcium and barium.

(ii) Two elements that can form oxides of type XO₃ are chromium (Cr) and sulphur (S). Formulae of their oxides are: CrO₃ and SO₃.

(iii) The innermost shell is k shell, followed by L shell and then comes the M shell. We can write an electronic configuration which can satisfy the given condition as: 2,8,4. We can see that the number of electrons present in L shell is double of the electrons present in M shell. This is possible for an element with atomic no.=14. Thus the element is Si.

(i) The element X has atomic number=20, it means that the element is Ca. It has electronic configuration as: 2,8,8,2. The number of valence electrons is 2 which it can lose easily. Thus element X (Ca) is a metal.

(ii) Mg has atomic number 12 and electronic configuration as: 2, 8, 2. Also, it has a lesser number of shell and thus its valence electrons lie closer to nucleus as compared to X(Ca) which has its valence electrons in 4^th shell, so Mg is less reactive than X(Ca). Thus X(Ca) is more reactive than Mg.

(iii) Since X contains 2 electrons in last shell and it can readily lose them so its valency is 2.

(i) As we can see it is given that X is surrounded by all other elements given above and there are numbers given 34 and 36. So, X must have atomic number 35. And this can be verified with periodic table where atomic number 35 corresponds to Bromine.

(ii) As atomic no. 36 corresponds to Neon (Ne) having noble gas configuration: 2, 8, 18, 8 which means 18^th group element so X (Br) with atomic no. 35 will lie in group just before this that is 17^th group. Also, X (Br) has electronic configuration: 2, 8, 18, 7 which means it has 4 shells or n=4 and lies in 4^th period. So X (Br) lies in 4^th period and 17^th group.

(iii) Electric configuration of elements with given atomic numbers,

Z=17: 2, 8, 7

Z=34: 2, 8, 18, 6

Z=36: 2, 8, 18, 8

Z=53: 2, 8, 18, 18, 7

It can be easily seen that elements with atomic number 34 and 36 lie in same period as have same number of shells n=4 and all other elements are from different periods.

Since, we know that elements in 16^th group have 6 electrons in their outer most shell i.e. the valence electrons are 6. Therefore, we can say that B has 6 valence electrons.

(Note: For this case ignore the period.)

(i) We know that element in lower period have smaller radius as compared other as they have less no. of shells. Also, going along period the size of atom decreases as the effective nuclear charge on outer electrons increases. So, B will have smallest size as it lies in 1^st period and 16^th group.

(ii) We know that, the metallic character increases going down the group and decreases going right along the period. Thus C will be most metallic of all.

(iii) Since C and A lie in 1^st and 2^nd group so these can readily lose electron and thus can form basic oxides.

(iv) We know that elements of 15^th group need only three electrons to complete their octet configuration and thus show valency 3. So, D will show valency 3.

(v) We know that tendency to gain electron or electron gain enthalpy increases going right along period and decreases going down the group. Thus, B will show highest tendency to gain electrons

(i) The largest non-metal constituent of air is nitrogen (N) and it is present as N₂ gas .It reacts with hydrogen in presence of Fe catalyst to give Ammonia (NH₃). Ammonia reacts with conc. H₂SO₄ to give ammonium sulphate (NH₄)₂SO₄.

Thus A= Nitrogen (N)

B=Ammonia (NH₃)

C= Ammonium sulphate (NH₄)₂SO₄

(ii) Element A is Nitrogen. It lies in the 2^nd period and 15^th group of the modern periodic table.

(iii) Carbon comes before nitrogen and oxygen comes after nitrogen in the 2^nd period.

(iv) The atomic number of nitrogen is 7. So, its electronic configuration is 2,5.

(i) The situation discussed above is similar to the “Newlands’s law of octaves”. Newland arranged the known elements in increasing order of atomic mass. He found that the every eighth element had properties similar to that of first one. He compared it to the octaves found in music.

(ii) The limitations of “Newland’s law of octaves” are as follows:

● This law was only applicable till calcium. After calcium, every eighth element didn’t possess properties similar to first one.

● Newland assumed that no more elements would be discovered in future but new elements were discovered later on which couldn’t fit in this law.

● Newland put two dissimilar elements in the same slot.

(iii) The value we can infer from above passage is that music is important in our lives.

(i) yes, we find the following similarities between rows of periodic table and that of assembly:

● In periodic table rows, elements are arranged in increasing order of atomic numbers just like in assembly rows students are made to stand in increasing order of their classes.

● Just like the height in assembly rows, the atomic size in rows also changes.

(ii) As the students of a same class have the same academic background similarly elements belonging to same group have same chemical properties.

(iii) The values of good morality and self-belief are inculcated during assembly of school.

A=sodium (Na), B=Calcium (Ca), C= Boron (B), D=Germanium(Ge)

The metallic character follows the order: Al>Si>P>S. The non-metallic character increases towards sulphur and oxides of non-metals are acidic nature. So the acidic nature increases as: (c) Al₂O₃< SiO₂ < P₂O₃ < SO₂

The element X is nitrogen which reacts with hydrogen to form NH₃. Nitrogen is a non-metal and a gas. It has 5 valence electrons (electronic configuration:2,5) and reacts with water to form ammonium hydroxide (NH₄OH) which is basic in nature.

I refers to the element Ge, II refers to the element S, III refers to the element Se, IV refers to the element Tl, V refers to the element F. Non-metallic character follows the order: Tl<Ge<Se<S<F

The element X is magnesium. Its valency is 2 because it loses 2 electrons readily to combine with other reactants. Only d) option has the valency 2 for element X.

The element with atomic number 17 is chlorine. It belongs to 17^th group. The element belonging to 5^th period of the same group is iodine which has an atomic number of 53.

The given table consists of group no. 8, 9 and 10 respectively. So the element p=Iron, q=Cobalt, r=Nickel and s=palladium.