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Introduction
One hundred and fifteen elements are known till date and many more may be known in future. All elements differ in their properties due to the presence of different kinds of atoms in them. Elements combine to form millions of compounds.
Important Terms and Concepts :
1. Need of Classification : It is difficult to study each and every element individually and to know its properties and uses.
Therefore, they have been classified into groups on the basis of their similarities in properties.
2. Classification : Classification means grouping of elements on the basis of similarities in properties, e.g., All of you belong to class X because you resemble in you properties, e.g., All of you have passed class IX, all of you are in age group of 13-15 years.
3. Basis of Classification : Classification is done on the basis of similarities in properties so that the systematic study could be made about them
Early attempts of classification
Lavoisier’s Classification :
Lavoisier classified elements into metals and nonmetals. This classification was based on certain distinctive physical properties such as hardness, malleability and luster. On the basis of these properties, sodium and lead were classed together as belonging to the group of metals.
Limitations
(1) Hardness, malleability and luster were found to be the only common properties of sodium and lead, otherwise the two elements were entirely different.
(2) In such a classification there was no place for elements with properties resembling those of metals as well as nonmetals.
There fore, Lavoisier’s classification was found to be inadequate.
Dobereiner’s Classification :
Law of triads In 1817, German chemist Johann Dobereginer classified elements having similar chemical properties into groups of three. These groups were called triads. He proposed a law known as Dobereiner’s law of triads. According to this law, when elements are arranged in the order of increasing atomic mass in a triad, the atomic mass of the middle element was found to be approximately equal to the arithmetic mean of the atomic masses of the other two elements.
The classification of elements into triads was very successful in predicting the atomic mass and properties of the middle element. Further, this classification showed that there exists some relationship between the properties of elements and their atomic masses. This paved the way for future attempts at classification of elements. Limitation : All the elements could not be grouped into triads.
Newlands’ Classification :
The classification of elements into triads was very successful in predicting the atomic mass and properties of the middle element. Further, this classification showed that there exists some relationship between the properties of elements and their atomic masses. This paved the way for future attempts at classification of elements. Limitation : All the elements could not be grouped into triads.
sa re ga ma pa dha ni sa,
where the first and the eighth note are same.A part of Newlands’ classification is given below where the figures under the symbols show the atomic masses
Starting from lithium (Li) the eight element is sodium (Na). The eight element starting from sodium is potassium. The properties of lithium, sodium and potassium are similar. The properties of beryllium, magnesium and calcium are similar too.
Limitation :
(i) This law worked well for lighter elements (up to calcium), but it could not be applied to heavier ones (elements of higher atomic masses) because starting from calcium every eight element was found to have properties different from those of the first element.
(ii) Newlands emphatically said that only 56 elements do exist in nature and no more element is likely to be discovered in future. But this concept was later on found to be untrue with the discovery of many new elements which defined the law of octaves.
(iii) In arranging elements in the form of a table, Newlands clubbed two elements together at the same place and in the same column. Not only this, he also placed some dissimilar elements in the same column. For example, cobalt (Co) and nickel (Ni) were clubbed together in the column of fluorine (F), chlorine (Cl) and bromine (Br) (under sa/do). We know that cobalt and nickel have properties entirely different from those of fluorine, chlorine and bromine. It is also known that cobalt and nickel have properties similar to those of iron. But iron (Fe) was placed in a column (under ni/ti) different from the column of cobalt and nickel. However, this law support to the idea that the properties of elements depend upon the atomic masses. It also showed that the properties of elements are repeated after a certain interval, i.e., the properties of elements are periodic in nature.
Mendeleev’s periodic law and periodic table
While working systematically on the physical and chemical properties of elements, Dmitri Invanovich Mendeleev noticed that properties of elements varied regularly with the atomic mass. He arranged the 63 elements then known in a table on the basis of similarities in properties. It was found that most of the elements occupied places in the table in order of their increasing atomic masses. In 1869, Mendeleeve formulated a law, now known as the periodic law. The law is stated as follows.
The properties of elements are periodic functions of their atomic masses. This means, if the elements are arranged in order of increasing atomic masses then those with similar properties are repeated at regular intervals.
On the basis of the periodic law, Mendeleev presented his classification in the form of a table, now known as Mendeleev’s periodic table. A simplified version of this periodic table is given below. In this table, copper, silver and gold find places in groups I as well as VIII.
This table consists of vertical columns called groups and horizontal rows called periods. There are only eight groups in the table. Mendeleev left some vacant places (shown by question marks) for the yet undiscovered elements. Noble gases were not discovered then. So, he did not provide any place for them in his periodic table.
Mandeleev’s idea was remarkable in that he used a fundamental atomic property (atomic mass) as the basis of classification. While classifying elements he laid special emphasis on tow factors.
1. Similar elements were grouped together.
2. Elements were arranged in order of increasing atomic masses.
Modified version of Mendeleev’s Periodic Table :
The elements which were undiscovered and for whom Mendeleev had left vacant places were discovered later. Some of these are scandium (Sc), gallium (Ga) and germanium (Ge). These elements were accommodated in their proper places in the table. The elements helium (He), neon (Ne), argon (Ar), Krypton (Kr), Xenon (Xe) and radon (Rn) became known only towards the end of the nineteenth century. These elements, called noble gases, were placed in the table as a separate group, called 0 group. The periodic table had to be modified then. The modified version of the table is shown below.
Features of the modified version of Mendeleev’s periodic table :
1. Groups into subgroups : Each group of this periodic table is further divided into two subgroups A and B. The properties of elements within a subgroup resemble more markedly but they differ from those of the elements of the other subgroups. For example., lithium (Li), sodium (Na), potassium (K), etc., of subgroups IA have close resemblance of properties but they have hardly any resemblance to the coinage metals (Cu, Ag and Au) of subgroup IB. Mendeleev allowed the subgroups to be represented within the same group.
2. Prediction of errors : This periodic table could predict errors in the atomic masses of some elements on the basis of their position in the periodic table. For example, when the periodic table was published, the experimental value of the atomic mass of beryllium (Be) we was supposed to be 13.65 and its valency, 3. So, the position of Be should have been somewhere else, but Mendeleev placed it at its appropriate position on the basis of its properties. He further suggested that the atomic mass of Be needed correction. Mendeleev predicted its atomic mass to be 9.1 and valency, 2. Latter investigations proved him right.
Similarly, the atomic mass of uranium was corrected from 120 to 240. Corrections were also made in the atomic masses of gold, platinum, etc.
3. Predictions of properties of higher to undiscovered element : We know that Mendeleev classified the elements in order of their increasing atomic masses. However, this order had to be ignored at some places to make sure that the elements with similar properties fell in the same group. In doing so, he left some vacant places in the table. These vacant places were kept reserved for elements not discovered till then. Mendeleev was confident that these elements would be discovered later and they would occupy these vacant places. Not only this, he also predicted the properties of these undiscovered elements on the basis of this study of his the properties of the neighboring elements. Amazingly, when the missing elements of Mendeleev’s periodic table were discovered subsequently, their properties were found to be very similar to those predicted by Mendeleev.
The elements scandium, gallium and germanium were not known in 1871 but their existence was predicted by mendelev. He named these elements as eka-boron, eka- Aluminium and eka silicon when these elements were discovered, they were named scandium, gallium and germanium respectively and their properties were found to be in good agreement with those predicted by Mendeleev. Properties of ka-aluminium (predicted by Mendeleev) and those of the gallium (discovered later) are given in table.
Considering its atomic mass, titanium (Ti) should have been placed below aluminium in the periodic table, but Mendeleev placed is below silicon (Si) because the properties of titanium were similar to those of silicon. Thus, a gap was left below aluminium in the periodic table. This gap was filled up by gallium which was discovered later. The properties of gallium (Ga) were found to be similar to those of boron and aluminium.
4. Basic features intact : All the basic features of Mendeleev’s periodic table are intact even today. Even when a new class of elements, i.e., noble gases, were discovered, they found place in a separate group called the zero group. The existing order of the periodic table was not at all disturbed.
Discrepancies in Mendeleev’s periodic table :
Mendeleev’s periodic table has the following defects.
1. Position of hydrogen : The position of hydrogen in the periodic table is anomalous.Hydrogen resembles alkali metals (Li, Na, K,etc). So it may be placed in the group of the halogens (VII A).
2. Position of lanthanides and actinides : The elements from atomic number 57 to 71 are collectively known as lanthanides. They do not have a proper place in the periodic table. They all have been placed at the same position in group III and period 6. Similarly, the actinides (atomic numbers 89-103) also have no proper place in the periodic table. These elements have also been placed in the same position, in group III and period 7.
3. Some similar elements are separated, while some dissimilar elements have been placed in the group : Some similar elements are separated in the periodic table. For example, copper (Cu) and mercury (Hg), silver (Ag) and thallium (Tl), and barium (Ba) and lead (Pb). On the other hand, some dissimilar elements have been placed together in the same group. For example, copper (Cu), silver (Ag) and gold (Au) have been placed in group I along with the alkali metals. Similarly, manganese (Mn) is placed in the group of the halogens.
4.Presence of some anomalous pairs of elements : In Mendeleev’s periodic table, elements are arranged in order of increasing atomic mass. In some places, this order has been ignored.
(a) The atomic mass of argon is 40 and that of potassium is 39. But argon is placed before potassium in the periodic table.
(b) The positions of cobalt and nickel are not in proper order. Cobalt (at. mass = 58.9) is placed before nickel (at. mass = 58.6).
(c) Tellurium (at. mass = 127.6) is placed before iodine (at. mass = 126.9).
(d) Thorium (at. mass = 232.12) is placed before protactinium (at. mass = 231)
5. Position of isotopes : The isotopes of an element have no place in the periodic table. The failure of Mendeleev’s periodic law to explain the wrong order of the atomic masses of some elements and the position of isotopes led scientists working in this field to conclude that atomic mass cannot be the basis for the classification of elements. There must be a more fundamental property of elements which can be the basis of classification.
Modern Periodic Table :
Henry Moseley, an English physicist found that the atomic number (Z) was the fundamental property of an elements and not the atomic mass for classification of elements.
Modern Periodic Law :
‘‘Properties of elements are periodic functions of their atomic numbers, i.e., the number of protons or electrons present in the neutral atom of an element.’’
Long form of Periodic Table :
Arranged in increasing order of their atomic numbers.
The prediction of properties elements and their compounds can be made with precision. All drawbacks of Mendeleev’s Periodic Table vanish when the elements are arranged on the basis of increasing atomic numbers.
Elements in a Group :
(1) They show similar chemical properties due to similar outer electronic configuration, i.e., same number of valence electrons.
(2) They have gradation in properties due to gradually varying attraction of the nucleus and the outer valence electrons as we go down the group.
Main Features of the Long Form of the Periodic Table :
(1) It shows arrangement of elements based on modern periodic law.
(2) There are 18 vertical columns known as groups.
(3) There are 7 horizontal rows known as periods.
(4) Elements having similar outer electronic configurations, i.e., having same valence electrons have been placed in same groups, e.g.,
(6) Each group in the table signifies identical outer shell electronic configuration i.e., same valence electrons, e.g., group 1 has 1 valence electron,group 2 has 2 valence electrons, group 13 has 3, group 14 has 4 valence electrons.
(7) Each period starts with filling of new shell, e.g.,
1st Period – K shell (1st shell) starts filling with Hydrogen and ends at Helium.
2nd Period – L shell (2nd shell) starts filling from Li (3) upto Ne (10)
3rd Period – M shell (3rd shell) start filling from Na (11) upto Ar (18)
4th Period – N shell (4th shell) starts filling from K (19) upto Kr (36) and so on.
(8) The periodic table is divided in four blocks :
(a) s-block elements : Group 1 and 2 elements are called s-block elements.
(b) p-block elements : Group 13 to 18 elements are called p-block elements
(c) d-block elements : Group 3 to group 12 are called d-block elements or transition elements (in between s- block and p-block elements)
(d) f-block elements : The elements placed at the bottom of the periodic table are known as f-block elements. The fourteen elements after La(57) (Lanthanum) are called Lanthanoides and 14 elements after Actinium Ac (89) are called Actinoides.
Naturally occurring Elements :
Elements upto atomic number 92 occur in nature except Technetium, Tc (Z = 43) and Promethium, Pm (Z = 61) which are formed from radioactive elements where ‘Z’ represents atomic number. Synthetic Elements : Elements beyond atomic number 92 are man-made elements. They are also called synthetic elements
Groups :
(1) Elements in a group have same number of valence electrons.
(2) The chemical properties of valence electrons, e.g., all the group 1 elements have 1 valence electron. They form positively charged ions by losing one electron, when required amount of energy is supplied to them i.e., Li+, Na+, K+. Group 1 elements are called alkali metals. Group 2 elements are called alkaline earth metals.
Group 2 elements when 2 valence electrons in the outermost shells. They can lose both the valence electrons to form dipositive cations, i.e., Be2+, Mg2+, Ca2+, etc. Positively charged ions are called cations. Group 13 elements belong to boron family, 14 to carbon family, 15 to Nitrogen family, 16 to Oxygen family.
Group 16 elements contain 6 valence electrons in their outermost shells, i.e., two electrons less than the maximum number of electrons that can be present in the outermost shell. They can gain 2 electrons more easily rather than lose 6 electrons.They change into dinegative ions such as O2–, S2–.Group 17 elements called Halogens contain 7 valence electrons. They can gain one electron to acquire stable electronic configuration, i.e., 8 electrons in the outermost shell and form uni-negative (single negative) ions such as F–, Cl–, Br–, I–.
Negatively charged ions are called anions.
Group 18 elements called noble gases, have their outermost, shell completely filled. The elements of this group have no tendency to lose or gain electrons. Thus, the elements of this group have zero valency and are almost unreactive. Hence they are called Noble gases. However, nowadays, compounds of Kr, Xe and Rn have been prepared.
In any particular group, the number of shells increase but the number of valence electrons remains the same.
Periods :
(1) The horizontal rows in the periodic table are called periods.
(2) There are 7 periods in the long form of periodic table
(3) The first period contains 2 elements, Hydrogen and Helium. They have only one shell.
(4) The second period contains 8 elements : Lithium Li(3), Beryllium Be(4), Boron B(5),Carbon C(6), Nitrogen N(7), Oxygen O (8),Fluorine F(9) and Neon Ne (10). The second period has 2 shells (K and L) and L shell is progressive filled.
*(v) The elements of 3rd period are :
In 3rd period, 3rd shell (M-shell) is being progressively filled and there are three shells.
4th period has 18 elements
5th period has 18 elements
6th period has 32 elements
7th period has 32 elements
In periods, the number of valence electrons increases from left to right in s and p-blocks
Periodicity in Properties :
The properties of elements depends upon the electronic configuration which changes along a period and down a group in periodic table.
There is periodicity in properties, i.e., repetition of properties after a regular interval due to similarity in electronic configuration.
Atomic Size (Atomic radii) :
Atomic size means radius of an atom. It is defined as distance between centre of nucleus and outermost shell (valence shell) of an isolated atoms.
Covalent Radii :
It is defined as half of the distance between the centres of nuclei of two atoms (bond length) bonded by a single covalent bond, e.g., Bond length in case of
H—H (Hydrogen molecule) is 74 pm.Covalent radius = 1/2 × 74 pm = 37 pm (picometre)[1 pm = 10–12 m]
It can be measured in case of diatomic molecules of non-metals.
Metallic Radii : If is defined as half of the internuclear distance between the two metal ions in a metallic crystal. It is measured in case of metals.
Variation of Atomic size in a Group :
Size generally increases from top to bottom in a group.
*Reason : It is due to addition of a new shell, i.e.,number of shells go one increasing, e.g., pm stands for picometre, i.e., 10–12 m.
Variation of Atomic size along a Period :
Atomic size goes on decreasing along a period from left to right
*Reason : It is due to increase in nuclear charge (number of protons in nucleus) which pulls the electrons towards it, i.e., force of attraction between nucleus and valence electrons increase, therefore atomic size decreases, e.g.,
Ionisation Energy and Electron Affinity :
Chemical nature and reactivity of an element depend upon the ability of its atoms to donate or accept electrons. This can be measured quantitatively with the help of ionisation energy and electron affinity of its atoms :
Ionisation energy : It is defined as the energy required to remove an electron completely from an isolated gaseous atom of an element. The energy required to remove the 1st electron is called first ionisation energy.
A(g) + I.E1 → A+(g) + e–
Second Ionisation Energy : he energy required to remove an electron from a unipositive ion is called the second ionisation energy.
A+(g) + I.E2 → A2+ (g) + e–
Te second ionisation energy is greater than the first ionisation energy due to increase in positive charge,hence increase in force of attraction between the nucleus and the valence electron.
Ist I.E. < 2nd I.E. < 3rd I.E.
Variation of Ionisation energy down a Group :
Ionisation energy goes on decreasing down a group.
Reason : It is due to the increase in the distance between the valence electrons and the nucleus as the atomic size increase down a group, the force of
attraction between the nucleus and the valence electrons decrease, therefore, the energy required to remove the electrons, i.e., the ioisation energy goes on decreasing
*Example :
Variation of Ionisation energy along a Period :
It goes on increasing generally along a period from left to right with decrease in atomic size. Reason : Due to decrease in atomic size, the force of attraction between the valence electrons and the nucleus increase and, therefore, more energy is required to remove electron.
*Example :
There is a decrease in ionisation energy from Be to B and from N to O, the reason of which you will study in higher classes. Group 18 elements (noble gases) have the highest ionisation energy in respective periods due to stable electronic configuration, i.e., 8 electrons in their valence shells except He which has 2 electrons.
Electron Affinity :
It is the energy change when an electron is gained by a neutral gaseous atom converting it into a negatively charged ion.
It is a measure of attraction or affinity of the gaseous atom for an extra electron to be added.
Cl(g) + e– → Cl–(g) + E.A.
Factors :
It depends upon atomic size as well as electronic configuration.
Variation down the Group :
Electron affinity goes on decreasing down the group in general.
Reason : Due to the increase in atomic size, the force of attraction between the nucleus and the electron to e added becomes less.
Variation along a Period :
Electron affinity increase from left to right in period.
Reason : It is due to decrease in atomic size which leads to an increase in the force of attraction between the nucleus and the electrons to be added.
*Example :
However, deviations to this rule are observed in variation of electron affinity.
Metallic and Non-metallic Character :
Group 1 to 12 are metals. Group 13 to 18 comprise non-metals, metalloids and metals.
Metalloids : Those elements which resemble both metals and non-metals are called metalloids. They are also called semi-metals, e.g., Boron, Silicon,Germinaium, Arsenic, Antimony, Tellurium and Polonium.
Properties of Metals :
(i) The are malleable.
(ii) They are ductile.
(iii) They are good conductors of heat and electricity.
(iv) They have generally 1 to 3 valence electrons.
(v) They have the same or less number of electrons in their outermost shell than the number of shells.
(vi) They are mostly solids.
Properties of Non-metals :
(i) They exist in solid, liquid or gaseous state.
(ii) Non-metals are generally brittle.
(iii) They are non conductors.
(iv) They have 4 to 8 valence electrons
Variation of Metallic Character :
Metallic character increases down a group due to decrease in ionisation energy. It decrease along a period due to increase in ionisation energy from left to right
Variation of Non-metallic Character :
Non-metallic character decreases down a group because of decrease in electron affinity which is due to increase in atomic size.
Along a period, non-metallic character increases from left to right due to increase in electron affinity which is due to decrease in atomic size
POINTS TO REMEMBER
- Dobereiner’s law of triads states that in a given set of three elements (triad) the atomic mass of the middle element is approximately equal to the average of the atomic masses of the other two elements.
- Newland’s law of octaves states that if the elements are arranged in order of increasing atomic mass, the eighth element starting from a given element shows a repetition of the properties of the first element.
- According to Mendeleev’s periodic law, the physical and chemical properties of elements are a periodic function of their atomic masses.
- Mendeleev’s periodic table (i) helped correct the wrongly assigned values of atomic masses of some elements, and (ii) predicted the properties of some undiscovered elements.
- In Mendeleev’s periodic table (i) isotopes had no place, and (ii) some similar elements are separated, while some dissimilar elements are placed in the same group.
- In Mendeleev’s periodic table there are eight groups (I to VIII). Each group is divided into subgroups A and B, except group VIII. In the modern periodic table there are 18 groups.
- The vertical columns of the periodic table are called groups.
- The horizontal rows of the periodic table are called periods.
- Modern periodic law. The physical and chemical properties of elements are a periodic function of their atomic numbers.
- The modern periodic table (long form) is based on atomic number.
- There are seven periods in the periodic table. The first six periods consists of 2, 8, 8, 18, 18 and 32 elements respectively. The seventh periods is still incomplete.
- Elements in a group have similar properties.
- The regular occurrence of similar chemical properties of elements with increasing atomic number is known as chemical periodicity.
- A series of transition elements starting from actinium are called actinides or actinoids.
- A family of elements headed by helium in the periodic table constitutes the noble gases. They are unrecactive.
- An element that is intermediate between metals and nonmetals is known as a metalloid.
- The elements of group 1 are called alkali metals.
- The elements of group 2 are called alkaline earth metals.
Passage Based Question Periodic Classification of Elements Class 10 Science
The following table shows a part of the periodic table in which the elements are arranged according to their atomic numbers. (The letters given here are not the chemical symbols of the elements):
(A) Which element has a bigger atom, a or f?
(B) Which element has a higher valency, k or o?
(C) Which element is more metallic, i or k?
(D) Select a letter which represents a nonmetal of valency 2.
Answer : (A) a (size decreases from left to right in a period).
(B) k (valency of k = 3; valency of o = 1).
(C) i (metallic character decreases from left to right in a period).
Analyse the following table given below and answer the questions that follow:
Given alongside is a part of the periodic table.
As we move vertically downward from Li to Fr:
| Li | Be |
| Na | |
| K | |
| Rb | |
| Cs | |
| Fr | Ra |
(A) Which among the following has larger size of atoms?
(a) Li
(b) Na
(c) Cs
(d) Fr
(B) Which one of the following does not increase while moving down in group of the periodic table?
(a) Atomic radius
(b) Metallic character
(c) Valence
(d) Number of shells in an element
(C) Name two properties of elements whose magnitude change while going from top to bottom in a group of the periodic table.
In what manner do they change?
(D) Rewrite the following statement after correction, if necessary:
“Groups have elements with consecutive atomic numbers”
Answer : (A) (d) Fr
Important Questions Periodic Classification of Elements Class 10 Science
Very short answer Type Questions :
Question. What is the number of valence electrons in the last element of the 3rd period?
Answer: 8
Question. 35 37 17 17 Cl and Cl are isotopes of chlorine, would you place them in different slots because their atomic masses are different? Or would you place them in the same position because their chemical properties are the same?
Answer: They will be placed in the same slot.
Question. Is it possible to have an element with atomic number 1.5 placed between hydrogen and helium?
Answer: No, atomic number cannot be in fractions.
Question. If an element ‘X’ is placed in group 14, what will be the formula and nature of bonding of its chloride?
Answer: XCl4, it has covalent bonding.
Question. An element ‘A’ has atomic number 17. To which group and period does it belong?
Answer: Its electronic configuration is 2, 8, 7. It belongs to group 17 and 3rd period.
Question. Find the atomic number of the element whose electronic configuration is 2, 8, 5.
Answer: Its atomic number is equal to 2 + 8 + 5 = 15.
Question. Name the scientist who first of all showed that atomic number of an element is a more fundamental property than its atomic mass.
Answer: Henry Moseley
Question. An element ‘X’ belongs to the second group of periodic table. What is the formula of its chloride?
Answer: XCl2
Question. An element ‘B’ belongs to the second period and Group 13. Give the formula of its oxide.
Answer: B2O3
Question. How many metals are present in second period of periodic table?
Answer: There are two metals Lithium (Li) and Beryllium (Be) in second period of periodic table.
Question. A metal ‘M’ belongs to 13th group in the modern periodic table. Write the valency of the metal.
Answer: 3
Question. Did Dobereiner’s triads also exist in the columns of Newlands’ octaves?
Answer: Yes, e.g. Li, Na, K is a part of Newlands’ law of octaves.
Question. What were the limitations of Dobereiner’s classification?
Answer: (i) All the elements could not be classified into triads.
(ii) Only three triads could be identified resembling each other.
Question. Write the name, symbol and electronic configuration of an element X whose atomic number is 11.
Answer: Sodium, Na: 2,8,1
Question. P(3), Q(12), R(13), S(20), which two elements have similar chemical properties and why?
Answer: P(3): 2, 1; Q(12): 2, 8, 2; R(13): 2, 8, 3; S(20): 2, 8, 8, 2
‘Q’ and ‘S’ because they have the same number of valence electrons.
Question. Lithium, sodium and potassium form a Dobereiner’s triad. The atomic masses of Li and K are 7 and 39, respectively. Predict the atomic mass of sodium.
Answer: Atomic mass of sodium = 7 + 39/2 = 23
Question. Write the number of valence electrons present in a nitrogen atom (147N).
Answer : Valence electrons present in a nitrogen atom(147N) = Number of electrons in outermost or valence shell
7—Atomic number
14—Mass number
Atomic number = Number of protons = Number of electrons
∴ Number of electrons in nitrogen 7
Electronic configuration k L 2 5
Hence, valence electrons present in nitrogen are.
Question. Give reason: The system of classification into triads was not found to be useful.
Answer : The system of classification into Dobereiner’s triads was not found to be useful as he failed to arrange all the then-known elements in the form of triads of elements having similar chemical properties. He could identify only three triads.
Question. Consider the following :
20Ca, 8O, 16S, 4Be
Which of the above elements would you expect to be in group 2 of the Modern Periodic table?
Answer : Consider the electronic configuration of elements:
Ca: 2, 8, 8, 2
O: 2, 6
S: 2, 8, 6
Be: 2, 2
As Ca and Be all have the same number of valence electrons (2), they belong to Group 2 of the Modern Periodic Table.
Question. How many metals are present in second period of periodic table?
Answer : Second period has 8 elements. There are only two metals in Second period. There are Lithium and Beryllium.
Question. State Modern Periodic Law of classification of elements.
Answer : The Modern Periodic Law states that ‘Properties of elements are a periodic function of their atomic number.’
Question. How does valency of an element vary across a period?
Answer : The valency of an element first increases and then decreases across a period.
Question. How does the metallic character vary across the period? Explain.
Answer : Metallic character decreases as we move across a period in the periodic table from left to right. This occurs because as we move across a period, the nuclear charge increases and the tendency of the element to lose electrons decreases.
Short Answer Type Questions
Question. How it can be proved that the basic structure of the Modern Periodic Table is based on the electronic configuration of atoms of different elements?
Answer: Position of element in periodic table is decided with the help of electronic configuration e.g. group number is decided on the basis of valence electrons e.g., elements having valence electrons 1,2,3,4,5,6,7,8, belong to Group 1, 2, 13, 14, 15, 16, 17 and 18 respectively.
Period is equal to number of shells e.g. 2,8,3 belong to third period.
Question. The electronic configuration of an element is 2,8,4. State its:
(a) Group and period in the Modern Periodic Table.
(b) Name and write its one physical property.
Answer: (a) It belongs to Group 14, third period.
(b) Silicon is the element. It is a metalloid, forms covalent bond. It is a semiconductor.
Question. An element ‘X’ has atomic number 13.
(a) Write its electronic configuration.
(b) State the group to which ‘X’ belong.
(c) Is ‘X’ a metal or non‑metal?
(d) Write the formula of its bromide.
Answer: (a) 2, 8, 3, (b) Group 13, (c) Metal, (d) AlBr3
Question. Elements have been arranged in the following sequence on the basis of their increasing atomic masses: F, Na, Mg, Al, Si, P, S, Cl, K
(a) Pick up two sets of elements which have similar properties.
(b) The above given sequence represents which law of classification of elements?
Answer: (a) (i) F, Cl (ii) Na, K (b) Newlands’ law of octaves
Question. Can the following group of elements be classified as Dobereiner’s triads? Explain by giving reasons.
(a) Na, Si, Cl (b) Be, Mg, Ca
Answer: (a) No, because atomic mass of Si = 58.5/2 = 29.25
which is nearly equal to 28 but these elements do not resemble with each other.
(b) Yes, atomic mass of Mg = 9 + 40 /2 = 24.5
which is nearly equal to 24 and these three elements resemble with each other.
Question. Write two drawbacks of Mendeleev’s periodic table.
Answer: (i) Isotopes challenged the basis of Mendeleev’s periodic table.
(ii) Increasing order of atomic masses could not be maintained in classifying all the elements.
Question. Predict the maximum number of valence electrons possible for the elements in the first period of periodic table.
Answer: 2 valence electrons are present in the last element ‘Helium’ of 1st period.
Question. (a) Predict the following which will form anions and which will form cations:
(i) Na (ii) Al (iii) Cl (iv) O
(b) Name two elements that are inert.
Answer: (a) Cl and O will form anions
Na and Al will form cations
(b) He, Ne are inert.
Question. List two anomalies of Mendeleev’s periodic table which were solved by modern periodic table law.
Be(4) and Mg(12) are first two elements of Group 2.
Answer: (i) Position of isotopes were not justified in Mendeleev’s periodic table but it is justified in the modern periodic table.
(ii) Increasing order of atomic masses could not be followed but increasing order of atomic numbers has been followed.
Question. (a) Among the following elements identify the one that would form anions:
K, O, Na, F, Ca, Cl, Hg
(b) Write the electronic configuration of the anions identified above.
Answer: (a) O, F, Cl will form anions.
(b) O2–(10) 2, 8
F–(10) 2, 8
Cl–(18) 2, 8, 8.
Question. From the elements 3919 A, 2814 B, 168 C and 4018 D identify:
(a) the most electropositive element.
(b) a noble gas.
(c) a metalloid.
(d) an element which will gain 2 electrons to attain nearest noble gas configuration.
(e) formula of compound formed between A and C.
(f) elements belonging to same period.
Answer:
Question. State the Modern Periodic Law. What is the number of groups and periods in the Modern Periodic Table?
Answer : The Modern Periodic Law states that ‘properties of elements are a periodic function of their atomic number.’ The number of groups in the Modern Periodic Table are 18 and the number of periods are 7.
Question. From the elements Li, K, Mg, C, Al, S identify the:
(a) elements belonging to the same group.
(b) element which has the tendency to lose two electrons.
(c) element which prefers sharing of electrons to complete its octet.
(d) most metallic element.
(e) element that forms acidic oxide.
(f ) element that belongs to group 13.
Answer : (a) Li, K
(b) Mg
(c) C
(d) K
(e) S (strongly acidic, C (weakly acidic)
(f ) Al
Question. The position of three elements A, B and C in the Modern periodic table is as follows:
(A) Write formula of compound formed between:
(i) B and A
(ii) B and C
(B) Is any of the three elements a metal? Give reason to justify your answer.
Answer :
(A) (i) Formula between B and A.
Or it can be written as
Question. Elements have been arranged in the following sequence on the basis of their increasing atomic masses:
F, Na, Mg, Al, Si, P, S, Cl, Ar, K
(A) Pick two sets of elements which have similar properties.
(B) The given sequence represents which law of classification of elements?
Answer : (A) In this sequence, the elements are arranged in increasing order of their atomic masses. According to Newlands’ law of octaves, every eighth element will show similar properties as that of the first element. Hence, F and Cl, Na and K will show similar properties. Therefore, set I is F and Cl and set II is Na and K. Note: Although Na and K have similar properties but they are not related as first and eighth element in the above sequence.
(B) The given sequence represents Newland’s law of octaves. In the given sequence, elements are arranged in increasing order of their atomic mass and each 8th element shows similar properties. So, it represents Newlands’ law of octaves. F Na Mg Al Si P S Cl Ar K.
Question. How it can be proved that the basic structure of the Modern Periodic Table is based on the electronic configuration of atoms of different element?
Answer : The Modern Periodic table consists of 18 vertical columns or ‘groups’ and 7 horizontal rows or ‘periods’. As we move from left to right along a period, we observe that the elements have the same number of valence shells but the number of valence electrons increases by one unit. Similarly, as we move down a group, we observe that elements have the same number of valence electrons but the number of shells increases by one. This proves that the Modern Periodic Table is based on electronic configuration of atoms of elements.
Question. Na, Mg and Al are the elements of the 3rd period of the Modern Periodic Table having group number 1, 2 and 13 respectively. Which one of these elements has the (a) highest valency, (b) largest atomic radius, and (c) maximum chemical reactivity? Justify your answer stating the reason for each.
Answer :
(a) Higher valency
K L M K L M K L M
Na 2 8 1 , Mg 2 8 1 Mg 2 8 3
Clearly highest valency is 3 i.e of aluminium as it can lose at 3 valence electrons to become ‘Al 3+’
(b) largest atomic radius – sodium (Na)
Question. Arrange the following elements in increasing order of their atomic radii: (A) Li, Be, F, N (B) Cl, At, Br, I
Answer : (A) F < N < Be < Li. Li, Be, F and N belong to the same period of the modern periodic table. As we move from left to right in the periodic table, the atomic radii of elements decreases due to high atomic charge and the same number of shells. Thus, the atomic radii of Li, Be, F and N increase in the order: F < N < Be < Li. (B) Cl < Br < I < At. Cl, At, Br and I belong to the same group of the modern periodic table that is group 17. As we move down in a group, the atomic radii of elements increases due to increase in the number of shells. Thus, the atomic radii of Cl, At, Br and I increase in the order:
Cl < Br < I < At.
Question. The electronic configuration of an element is 2, 8, 4. State its:
(A) group and period in the Modern Periodic Table.
(B) name and write its one physical property.
Answer : (A) As the number of valence electrons in the given element is 4, it belongs to Group 14 of the Modern Periodic Table.
As the number of occupied shells is 3 (since electrons are filled in K, L and M shells), the element belongs to third period.
(b) The given element is Silicon (atomic symbol: Si). It’s physical property is:
(i) Silicon is a solid
(ii) It is a semi-conductor
(iii) It exhibits allotropy
(iv) It has a metallic luster
Question. List the limitations of Newlands’ Law of Octaves?
Answer : Limitations of Newlands’ Law of Octaves:
(1) It was not applicable for all the elements.
It was applicable only till calcium, as after calcium every eighth element did not possess properties similar to that of the first.
(2) It was assumed by Newlands that only 56 elements existed in nature and no more elements would be discovered in the future.
But, later on, several new elements were discovered, whose properties did not fit into the Law of Octaves.
(3) Cobalt and nickel are placed in the group of halogens with the elements fluorine and chlorine (F, Cl) which have different properties and it was not explained for the same.
(4) Placing of iron far away from cobalt and nickel, which have similar properties as iron, could also not be explained.
Thus, Newlands’ Law of Octaves worked well with lighter elements only.
Question. Can the following groups of elements be classified as Dobereiner’s triad? (A) Na, Si, Cl (B) Be, Mg, Ca Atomic mass of Be 9; Na 23; Mg 24; Si 28; Cl 35; Ca 40. Explain by giving reason.
Answer : For a group to be Dobereiner’s triad, the atomic mass of the middle element must be average of the atomic masses of the first and the third elements.
(A) No, Na, Si, Cl cannot be classified as Dobereiner’s triad because all these elements do not have similar properties, although the atomic mass of silicon is the average of the atomic masses of sodium (Na) and chlorine (Cl).
Atomic mass of Si = 23+35/2
=58/2 = 29
(B) Yes, Be, Mg, Ca can be classified as Dobereiner’s triad because they have similar properties and the mass of magnesium (Mg) is roughly the average of the atomic mass of Be and Ca.
Atomic mass of Mg = 9+40/2
= 49/2 = 245
Question. An element ‘X’ (Atomic number = 19) burns in the presence of oxygen to form a basic oxide.
(A) Identify the element and write its electronic configuration.
(B) State its group number and period number in the modern periodic table.
(C) Write a balanced chemical equation for the reaction when this oxide is dissolved in water.
Answer : (A) The element ‘X’ is potassium (K) Its electronic configuration is 2, 8, 8, 1
(B) Group number 1 Since it has valence electron So it belongs to group Period number 4.
Question. Name two elements you would expect to show chemical reactions similar to magnesium. What is the basis for your choice?
Answer : Magnesium (Mg) belongs to group 2 known as alkaline earth family. The two other elements belonging to the same group are Calcium (Ca) and Strontium (Sr) which are expected to show chemical reactions similar to Magnesium (Mg).
The basis of choice is the electronic distribution in the valence shell of these elements. All of them have two valence electrons each.
Question. Write the formulae of chlorides of Eka-silicon and Eka-aluminium, the elements predicted by Mendeleev.
Answer : Eka-silicon is identified as germanium (Ge), which is placed in group 4 of Mendeleev’s periodic table. The valency of germanium is 4, so the chemical formula of its chloride must be GeCl4.
Eka-aluminium was later identified as gallium (Ga). It is placed in group 3 of Mendeleev’s periodic table. Hence, the valence of gallium is 3 and the formula of chloride would be GaCl3.
Long Answer Type Questions
Question. Define atomic size. Give its unit of measurement. In the modern periodic table what trend is observed in the atomic radius in a group and a period and why is it so?
Answer : Atomic Size: The distance between the centre of the nucleus and the outer most shell which contains electrons is known as atomic size.
Unit of Atomic Size: Atomic size is measured in angstroms (À) or in picometers (pm).
1À = 10 – 8 cm = 10–10 m
1 pm = 10–12 m
Variations of Atomic radii in a group : On moving down a group, the atomic radii of elements of increases gradually.
When we move from to P top bottom in a group in modern periodic table, the atomic radius increases as a new shell is added at each succeeding element.
Question. Mendeleev predicted the existence of certain elements not known at that time and named two of them as eka-silicon and eka-aluminium.
(A) Name the elements which have taken the place of these elements.
(B) Mention the group and the period of these elements in the modern periodic table.
(C) Classify these elements as metals, nonmetals or metalloids.
(D) How many valence electrons are present in each one of them?
Answer : (A) The two elements that have taken the place of eka-silicon and eka-aluminium are germanium (Ge) and gallium (Ga), respectively.
(B) Germanium is placed in group 14 and period 4 in the modern periodic table. Gallium is placed in group 13 and period 4 in the modern periodic table.
(C) Germanium (Ge) is a metalloid. gallium (Ga) is a metal.
(D) Valence electrons in germanium (Ge) are 4. Valence electrons in gallium (Ga) are 3.
Question. Atomic number of a few elements are given below:
10, 20, 7, 14
(A) Identify the elements.
(B) Identify the group number of these elements in the periodic table.
(C) Identify the periods of these elements in the periodic table.
(D) What would be the electronic configuration for each of these elements?
(E) Determine the valency of these elements.
Answer : (A) Elements — Neon (Ne), calcium (Ca), nitrogen (N), silicon (Si)
(B) Group — 18, 2, 15, 14
(C) Period — 2, 4 , 2, 3
(D) Electron configuration — (2, 8); (2, 8, 8, 2);(2, 5); (2, 8, 4)
(E) Valency — 0, 2, 3, 4
Question. Complete the following crossword puzzle (Figure).
Across:
(1) An element with atomic number 12.
(3) Metal used in making cans and member of group 14.
(4) A lustrous non-metal which has 7 electrons in its outermost shell.
Down:
(2) Highly reactive and soft metal which imparts yellow colour when subjected to flame and is kept in kerosene.
(5) The first element of the second period.
(6) An element which is used in making florescent bulbs and is the second member of group 18 in the modern periodic table.
(7) A radioactive element which is the last member of the halogen family.
(8) Metal which is an important constituent of steel and forms rust when exposed to moist air?
(9) The first metalloid in modern periodic table whose fibres are used in making bullet-proof vests.
Answer : Across:
(1) An element with atomic number 12 is magnesium
(3) Metal used in making cans and member of group 14 is tin
(4) A lustrous non-metal which has 7 electrons in its outermost shell is iodine.
Down:
(2) Highly reactive and soft metal which imparts yellow colour when subjected to flame and is kept in kerosene is sodium.
(5) The first element of the second period is lithium.
(6) An element which is used in making fluorescent bulbs and is the second member of group18 in the modern periodic table is neon.
(7) A radioactive element which is the last member of the halogen family is astatine.
(8) Metal which is an important constituent of steel and forms rust when exposed to moist air is iron.
(9) The first metalloid in modern periodic table whose fibres are used in making bulletproof vests is boron.
Question. ‘‘Atomic number of an element is considered to be a more appropriate parameter than its atomic mass for a chemist.’’ Take the example of the element X (atomic number 13) to justify this statement.
Answer : The electronic configuration of element X having atomic number 13 is: 2, 8, 3 It is a metal having valency = 3 (number of valence electrons). It is electropositive in nature and can form X3+ ions.
It is placed in 3rd period and 13th group of the Modern Periodic table.
Formula of its oxide will be X2O3.
Therefore, atomic number of an element is a more fundamental property and prediction of properties of elements could be made more accurately when elements were arranged on the basis of increasing atomic number.
Question. (A) In this ladder (figure), symbols of elements are jumbled up. Rearrange these symbols of elements in the increasing order of their atomic numbers in the periodic table.
(B) Arrange them in the order of their group also.
Answer : (A) The symbols of elements in increasing order of their atomic numbers are:
H, He, Li, Be, B, C, N, O, F, Ne, Na, Mg, Al, Si, P, S, Cl, Ar, K and Ca.
(B) The symbols of elements in increasing order of their group numbers are as follows:
Group 1 — H, Li, Na, K
Group 2 — Be, Mg, Ca
Group 13 — B, Al
Group 14 — C, Si
Group 15 — N, P
Group 16 — O, S
Group 17 — F, Cl
Group 18 — He, Ne, Ar
| Group→ Period↓ | 1 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
| 1 | H | He | ||||||
| 2 | Li | Be | B | C | N | O | F | Ne |
| 3 | Na | Mg | Al | Si | P | S | Cl | Ar |
| 4 | K | Ca |
Question. Give reasons for the following:
(A) K atom is bigger than Na atom even though both these elements belong to the same group.
(B) The metallic character of elements increases as we move down a group.
(C) Mg atom is smaller than Na even though these elements belong to the same period.
Answer : (A) On moving down in a group of the periodic table, the atomic radius of elements increases. When we go from top to bottom in a group, a new shell of electrons is added to the atoms regularly. In this way, the number of electron shells in the atoms increases gradually due to which the size of atoms also increases. For example, a sodium atom (Na) has three electron shells K, L and M in it and a potassium atom
(K) has four electron shells K, L, M and N, and hence it is bigger than a sodium atom.
(B) The metallic character of elements increases as we move down a group because one more electron shell is added to the atom. As a result, the size of the atom increases. The valence electrons move away from the nucleus at regular intervals. The holding capacity of the nucleus on valence electrons decreases. Due to this, the atom can lose valence electrons more easily to form positive ions, and hence the metallic character increases. (C) Mg atom is smaller than Na because on moving from left to right in a period of the periodic table, the atomic radius of elements decreases. This happens because the number of protons and electrons in the atoms increases. Due to a large positive charge on the nucleus, electrons are pulled in close to the nucleus and the size of atom decreases.
Question. An element X of group-15 exists as diatomic molecule and combines with hydrogen at 773 K in presence of the catalyst to form a compound, ammonia which has a characteristic pungent smell.
(A) Identify element X. How many valence electrons does it have?
(B) Draw the electron dot structure of the diatomic molecule of X. What type of bond is formed in it?
(C) Draw the electron dot structure for ammonia and what type of bond is formed in it?
Answer : (A) Element X which is placed in group 15 must be nitrogen (N). As it exists as diatomic molecule (N2) and combine with hydrogen to form ammonia.
The electronic configuration of nitrogen is 2, 5; hence, it has 5 valence electrons.
(B) Nitrogen has 5 valence electrons, so it requires three more electrons to complete its octet. Hence, it shares three of its electrons with three electrons of the other nitrogen atom to form a diatomic molecule of N2 gas. Thus, three covalent bonds are formed between two nitrogen atoms and each nitrogen atom is left with one lone pair of electrons. The electron dot structure of nitrogen molecule is given below:
(C) The chemical formula of ammonia is NH3.
Here, center nitrogen atom is bonded with three hydrogen atoms through three single covalent bonds. The electron dot structure for ammonia is as follows:
Question. Based on the group valency of element write the molecular formula of the following compounds giving justification for each:
(A) Oxide of first group element.
(B) Halide of the element of group thirteen, and (C) Compound formed when an element, , A of group 2 combines with element, B of group seventeen.
Answer : While writing the formula of compounds, we need to take the valencies of the combining elements into account.
(A) First group elements have a valency one, as they have one valence electron. Oxygen belongs to group 16 and has a valency of 2 as it has 2 valence electrons.
The molecular formula of compound formed by oxides of first group element will be X2O, where X is the first group element. Example of such a compound is Na2O.
(B) Elements of group 13 have a valency 3, as they have 3 valence electrons. Halogens belong to group 17 and have a valency of 1, as they have 3 valence electrons.
The molecular formula of halide of elements of group 13 will be YX3, where Y is the element of group 13 and X is a halogen.
Example of such a compound is AlCl3.
(C) All elements of group 2 have a valency 2, as they have 2 valence electrons. Group 17 elements are halogens, having a valency of 1. Therefore, valency of A is 2 and that of B is 1. The molecular formula of compound formed when A combines with B will be AB2. Example of such a compound is MgCl2.
Question. (A) How is the valency of an element determined if its electronic configuration is known? Determine the valency of an element of atomic no. 9.
(B) Given below are some elements of the Modern Periodic Table. Atomic numbers of the elements are given in parentheses: A(4), B(9), C(14), D(19), E(20)
(i) With the help of the electronic configuration, find out which one of the above elements will have one electron in its outermost shell.
(ii) Which two elements belong to the same group? Give reasons for your answer.
(iii) Which one of the above elements belonging to the fourth period has bigger atomic radius and why?
Answer : (A) Valency of an element is the number of electrons an element can donate or gain or share in order to complete its valence or outermost shell.
The electronic configuration of an element of atomic number 9 is (2, 7). As this element requires only 1 electron (8 – 7 = 1) to complete its valence shell, its valency is 1.
(B) The electronic configuration of the elements given is as under:
(i) Element D will have one electron in its outermost shell.
(ii) Elements A and E belong to the same group as they have the same number of valence electrons.
(iii) D has bigger atomic radius than E as E has more number of electrons and protons as compared to D. Due to the larger positive charge on its nucleus, electrons are pulled more strongly towards the nucleus in E than D.
Question. Two elements A and B have atomic numbers 11 and 19 respectively.
(A) State their position in the moden periodic table.
(B) Which element has a bigger atomic radius?
(C) What is the nature of their oxides?
Answer : (A) Electronic configuration of A: 2, 8, 1 Electronic configuration of B: 2, 8, 8, 1 From its electronic configuration, it is clear that element A belongs to Group 1 (since valence electron is 1) and Period 3 (since it has 3 electron shells).
Similarly, we can say that the element B belongs to Group 1 and Period 4.
(B) The atomic radius of element B is bigger as compared to element A. This is because the atomic radius increases (due to the presence of more shells) as we move down in a group.
(C) They form basic oxides.
Question. Explain giving justification the trends in the following properties of elements, on moving from left to right in a period, in the Modern periodic Table.
(A) Variation of valency.
(B) Change of atomic radius.
(C) Metallic to non-metallic character.
(D) Electronegative character.
(E) Nature of oxides.
Answer : The trends in the properties of elements, on moving from left to right in a period, in the Modern periodic Table are given below:
(A) Variation of valency: On moving from left to right, the valency increases from 1 to 4 and then decreases to 0 as the valency is determined by the number of valence electrons.
(B) Change of atomic radius: On moving from left to right in a period, the atomic radius decreases because the number of electrons and protons increases. Due to the large positive charge on the nucleus, electrons are pulled more strongly towards the nucleus.
(C) Metallic to non-metallic character: The metallic character decreases and nonmetallic character increases as we move from left to right in a period because the tendency to lose electrons decreases as the effective nuclear charge on the valence shell electrons increases.
(D) Electronegative character: The electronegative character, or the tendency to gain electrons, increases on moving from left to right along a period because of the increase in effective nuclear charge on the valence shell.
(E) Nature of oxides: The basic nature of oxides decreases and the acidic nature increases as we move from left to right since metals form basic oxides whereas non-metals form acidic oxides.
Question. (A) What was the basis of Mendeleev classification of elements?
(B) List two achievements of Mendeleev periodic tables.
(C) List any two observations which posed a challenge to Mendeleev periodic law.
Answer : (A) Atomic Mass of elements was the basis of Mendeleev’s classification of elements.
(B) Achievements of Mendeleev’s periodic table:
(1) Mendeleev’s periodic law predicted the existence of some elements that had not been discovered at that time. He therefore left some gaps in his Periodic Table
(2) Mendeleev’s periodic table could predict the properties of several elements on the basis of their positions in the periodic table.
(3) It could accommodate noble gases when these gases were discovered in a new group without disturbing the existing order. (Any 2 of 3 points can be written to get full marks)
(C) Two observations that posed a challenge to Mendeleev’s periodic law:
(1) The position of isotopes could not be explained since the elements are arranged according to their atomic masses.
(2) Wrong order of atomic masses of some elements could not be explained. For example, cobalt (atomic mass 58.9) appeared before nickel (atomic mass 58.7).
(3) A correct position could not be assigned to hydrogen in the periodic table.
(4) The atomic masses do not increase in a regular manner in going from one element to the next.
Question. The position of certain elements in the Modern Periodic Table are shown below.
Using the above table answer the following questions giving reasons in reach case:
(A) Which element will form only covalent compounds?
(B) Which element is a non-metal with valency 2?
(C) Which element is a metal with valency 2?
(D) Out of H2C and F which has largest atomic size?
(E) To which family does H, C and F belong?
Answer : (A) Element ‘E‘ will form only covalent compounds.
Question. Atoms of eight elements A, B, C, D, E, F, G and H have the same number of electronic shells but different number of electrons in their outermost shell. It was found that elements A and G combine to form an ionic compound. This compound is added in small amount to almost all vegetable dishes during cooking. The oxides of elements A and B are basic in nature while those of E and F are acidic. The oxide of D is almost neutral. Based on the above information answer the following questions:
(A) To which group or period of the periodic table, do the listed elements belong?
(B) What would be the nature of the compound formed by a combination of elements B and F?
(C) Which two of these elements could definitely be metals?
(D) Which one of the eight elements is most likely to be found in gaseous state at room temperature?
(E) If the number of electrons in the outermost shell of element C and G are 3 and 7 respectively, write the formula of the compound formed by the combination of C and G.
Answer : Element A is sodium (Na) and element G is chlorine (Cl). They form an ionic compound NaCl (also called common salt), which is added to dishes.
It is said in the question that the oxides of A and B are basic in nature. Since we now know that element A is sodium (Na), it combines with Na to form sodium oxide (Na2O), which is basic in nature.
Similarly, element B could be magnesium (Mg),which forms a basic oxide called magnesium oxide (MgO).
It is said that oxides of E and F are acidic in nature, which could probably be aluminium oxide and sulphur dioxide.
On the basis of this knowledge, let us answer the questions:
(A) A: 1st Group and 3rd Period
B: 2nd Group and 3rd Period
C: 13th Group and 3rd Period
D: 14th Group and 3rd Period
E: 15th Group and 3rd Period
F: 16th Group and 3rd Period
G: 17th Group and 3rd Period
H: 18th Group and 3rd Period
(B) Ionic compound: magnesium sulphate MgSO4.
(C) A and B (sodium and magnesium)
(D) H (argon)
(E) CG3
Question. The atomic number of an element is 20.
(A) Write its electronic configuration and determine its valency.
(B) Is it a metal or a non-metal?
(C) Write the formula of its chloride.
(D) Is it more reactive or less reactive than Mg (atomic number 12)? Give reason for your answer.
Answer : (A) The electronic configuration of element having atomic number 20 is 2, 8, 8, 2, i.e., electrons in K, L, M and N shells are 2, 8, 8 and 2 respectively. Valency of given element = 2 (B) It is a metal as it can lose electrons easily since it is electropositive.
(C) Let us denote the element by M. Since chlorine has valency 1, formula of the chloride of M is MCl2.
(D) This element, M, is more reactive than Mg as Mg lies in 2nd Group, 3rd period and M lies in 2nd Group but 4th period of the periodic table. M is more electropositive than Mg as the effective nuclear charge experienced by its valence electrons is less than that experienced by valence electrons of Mg because of increase in distance of valence electrons from the nucleus.
Question. Consider the following elements of the Modern Periodic Table:
Be(4); O(8); Cl(17); K(19); Ca(20)
(A) Select from these an element having only one valence electron.
Write the electronic configuration of this element.
(B) Select from these, two elements of the same group and state the reason for your answer.
(C) Select from these, two elements of the same period and state the reason for your answer.
Answer : The electronic configuration of the elements given is as under:
(A) The element having only one valence electron is Potassium (K).
Electronic configuration of K is (2, 8, 8, 1).
(B) Be (Atomic No. 4) and Ca (Atomic No. 20) belong to the same group 2 as they both have the same number of valence electrons 2.
(C) Be and O belong to the same period 2 as they both have the same number of occupied shells, which is 2.
K and Ca belong to the same period 4 as they both have the same number of occupied shells, which is 4.
Question. (A) Write the number of groups and periods in the Modern Periodic Table. How does the atomic size vary down a group in the periodic table? State its reason.
(B) The electronic configurations of four elements A, B, C and D are as follows:
A – (2, 8, 7); B – (2, 8, 1); C – (2, 8, 2); D – (2,8, 8, 2)
(i) Which amongst these elements will form acidic oxide and why?
(ii) Which amongst these elements has the smallest atomic radius and why?
(iii) Out of these select those two elements which have same valency and form compounds by ionic bonds.
Answer : (A) The number of groups in the Modern Periodic Table are 18 and the number of periods are 7. The atomic size increases as we go down in a group as a new shell is added to the atoms when going from top to bottom in a group which increases the distance between the valence electrons and the nucleus.
(B) (i) Element A (2, 8, 7) will form an acidic oxide as it is a non-metal and forms an oxide which on dissolution with water gives an acid.
(ii) A, B and C belong to 3rd period and A has the smallest atomic radius because as we move from left to right along a period, the nuclear charge increases which tends to pull the electrons closer to the nucleus and hence reduces the atomic size.
(iii) Elements A and B have same valency 1 and form compound by ionic bond. A is a non-metal whereas B is a metal. B transfers an electron to A thereby forming B+ ion and A forms A– ion by gaining an electron. The formula of the compound formed is AB.
Question. (A) Why do we classify elements?
(B) What were the two criteria used by Mendeleev in creating his periodic table?
(C) Why did Mendeleev leave some gaps in his periodic table?
(D) In Mendeleev’s periodic table, why was there no mention of noble gases like helium, neon and argon?
(E) Would you place the two isotopes of chlorine, Cl-35 and Cl-37 in different slots because of their different atomic masses or in the same slot because their chemical properties are the same? Justify your answer.
Answer : (A) Elements are classified to systematize their study and make the understanding of properties of elements and compounds simpler.
(B) The two main guiding factors for Mendeleev in the classification of the then known elements were:
(1) increasing atomic masses, and
(2) grouping together of elements having similar properties.
(C) These gaps were left for the elements which had not been discovered at that time. Mendeleev thought that these elements would be discovered later on in the future. The modern periodic table does not have any gaps because new elements were discovered later on, which were placed correctly in the gaps left by Mendeleev.
(D) Noble gases were not known at that time.
So, there was no group of noble gases in Mendeleev’s original periodic table.
(E) Both the isotopes of chlorine, Cl-35 and Cl-37, have the same atomic number of 17. Therefore, they have same chemical properties. Hence, both of them can be put at one place in the same group of the periodic table.
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