Question 1

Question 9State the contributions made by Mendeleeff towards the periodic table.

Accepted Answer

The contributions of Mendeleeff towards the periodic table are:Elements were arranged in increasing order of atomic weights in horizontal rows called 'periods' and vertical columns called 'groups'.Elements which are similar with respect to their chemical properties are grouped together and have atomic weights of nearly the same value.Elements in the same group had the same 'valency' and similar chemical properties.Based on the periodicity of properties a number of gaps were left in the table for undiscovered elements i.e., elements now discovered e.g., Scandium, Gallium and Germanium originally called eka-boron,eka-aluminium and eka-silicon respectively.The properties of undiscovered elements left in the vacant gaps was predicted.Incorrect atomic weights of some of the arranged elements were corrected with the knowledge of the atomic weights of the adjacent elements.

Question 2

Question 11How were elements arranged in the long form of the periodic table or the Modern Periodic Table. State the Modern Periodic Law and compare it with Mendeleeff's Periodic Law.

Accepted Answer

In the Modern Periodic Table elements are arranged in increasing order of atomic number.Modern Periodic Law states that the physical and chemical properties of elements are periodic function of their 'atomic numbers'.Mendeleeff's Periodic Law states that the physical and chemical properties of elements are periodic function of their 'atomic weights'.The basis of classification in Mendeleeff's Periodic Law was atomic weight of the element whereas in the Modern Periodic Law it is the atomic number of elements. This change in the Modern Periodic Law solves most of the defects in Mendeleeff's Periodic table.

Question 3

Question 16Write short notes on the following types of elements –(a) alkali metals(b) alkaline earth metals(c) halogens(d) noble gases(e) transition and inner transition elements(f) normal elements

Accepted Answer

(a) Alkali metals —The elements of group 1 [IA] from 3Li [Lithium] to 87Fr [Francium] are called alkali metals as they react with water forming strong alkalis.They are light metals and are univalent with one valence electron in outer shell. This makes them strong reducing agents as they readily donate the one electron to achieve stable electronic configuration.They are highly reactive electropositive metals, which are soft and can be cut with a knife.They are good conductors of heat & electricity.They form electrovalent compounds with non-metals. e.g. KCl, NaBr.(b) Alkaline earth metals —The elements of group 2 [IIA] from 4Be [Beryllium] to 88Ra [Radium] are called Alkaline earth metals.They are light metals and are divalent with two valence electrons in their outer shell. They are reducing agents as they can donate the two electrons to achieve stable electronic configuration.They have properties similar to alkali metals but are slightly less reactive than alkali metals.They are good conductors of heat & electricity.They form electrovalent compounds with non-metals. e.g. CaCl2, MgO.(c) Halogens —The elements of group 17 [VIIA] from 9F [Fluorine] to 53I [Iodine] are called Halogens. They are called Halogens as Halogens means salt producer and these elements react with metals to form salts.They are non-metals and are univalent with seven valence electrons in their outer shell. This makes them strong oxidising agents as they readily accept one electron to achieve stable electronic configuration.They are highly reactive, electronegative non-metals.They are bad conductors of heat & electricity.They form covalent compounds with non-metals. e.g. HCl, PCl3.(d) Noble gases —The elements of group 18 [0] from 9He [Helium] to 53Rn [Radon] are called Noble gases.They are non-metals and have eight valence elections in outer shell [except Helium (He) which has only two]. They have stable electronic configuration hence their valency or combining capacity is Zero.Noble gases do not react with other elements & are therefore inert or unreactive. They are all monatomic gases.(e) Transition and inner transition elements —Transition elements —The elements of group 3 to 12 [IB to VIIB & VIII] are called Transition elements.They are all metals.They lie at the centre of the periodic table between metals [highly electropositive] on the left and non-metals [highly electronegative] on the right.Inner transition elements —Lathanide [Rare earths] and Actinides series are called inner transition elements.They lie at the bottom of the periodic table in two horizontal rows [14 elements each].Lanthanide series : from Cerium 58Ce to Lutetium 71Lu : [Period - 6]Actinide series : from Thorium 90Th to Lawrencium 103Lr: [Period - 7]They are all metals and chemically very reactive [actinides are radioactive elements].(f) Normal elements —Elements of groups 1, 2, 13, 14, 15, 16, 17 are called normal elements.Atoms of these elements have all inner shells complete except outermost shell.They consist of metals, metalloids and non-metals.The number of electrons in their outermost shell signifies the group to which they belong.

Question 4

Question 18Name or state the following with reference to the elements of the first three periods of the periodic table.(a) The noble gas having duplet arrangement of electrons.(b) The noble gas having an electronic configuration 2, 8, 8.(c) A metalloid in period 2 and in period 3.(d) The number of electron shells in elements of period 1, period 2 and period 3.(e) The valency of elements in group 1 [IA].(f) The group whose elements have zero valency.(g) An alkaline earth metal in period 3.(h) The non-metallic element present in period 3 other than sulphur and chlorine.(i) A non-metal in period 2 having electronic configuration 2, 6.(j) An electrovalent compound formed between an alkali metal and a halogen.(k) A covalent compound formed between an element in period 1 and a halogen.(l) An alkali metal in period 3 which dissolves in water giving a strong alkali.(m) A metal in period 3 having valency 3.(n) The bridge elements of period 3 of group 1 [IA], 2 [IIA] and 13 [IIIA].(o) The periods which contain the inner transition elements.(p) The formula of the hydroxide of the element having electronic configuration 2, 8, 2.(q) The valency of the element in period 3 having atomic number 17.(r) A non-metal in period 2 which is tetravalent.

Accepted Answer

(a) The noble gas having duplet arrangement of electrons — Helium(b) The noble gas having an electronic configuration 2, 8, 8 — Argon(c) A metalloid in period 2 and in period 3 — Metalloid in period 2 is B (Boron) and in period 3 is Si (Silicon)(d) The number of electron shells in elements of period 1, period 2 and period 3 — Period 1 —1 shell, Period 2 — 2 shells, Period 3 — 3 shells(e) The valency of elements in group 1 [IA] — +1(f) The group whose elements have zero valency — Group 18 or 0 group(g) An alkaline earth metal in period 3 — Magnesium [Mg](h) The non-metallic element present in period 3 other than sulphur and chlorine — Phosphorus [P](i) A non-metal in period 2 having electronic configuration 2, 6 — Oxygen [O](j) An electrovalent compound formed between an alkali metal and a halogen — Sodium chloride [NaCl](k) A covalent compound formed between an element in period 1 and a halogen — Hydrogen chloride [HCl](l) An alkali metal in period 3 which dissolves in water giving a strong alkali — Alkali metal is Na and it dissolves in water to give strong alkali - NaOH(m) A metal in period 3 having valency 3 — Aluminium [Al](n) The bridge elements of period 3 of group 1 [IA], 2 [IIA] and 13 [IIIA] — Li & Mg, Be & Al(o) The periods which contain the inner transition elements — Period 6 and 7(p) The formula of the hydroxide of the element having electronic configuration 2, 8, 2 —Since valency of element X is 2+ and valency of OH is 1-Formula of the compound:X2+↗OH1−X22 ↘↙ OH1⇒X11 ↘↙ OH2\text{X}^{2+} \phantom{\nearrow} \text{OH}^{1-} \[0.5em] \overset{\phantom{2}{2}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{OH}} \Rightarrow \underset{\phantom{1}{1}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{OH}} \[0.5em]X2+↗OH1−X22 ↘↙ OH1⇒11X​ ↘↙ 2OH​So, we get the formula as X(OH)2\bold{X(OH)}\bold{_2}X(OH)2​As the element with electronic configuration 2,8,2 is Mg, Hence the compound is Mg(OH)2(q) The valency of the element in period 3 having atomic number 17 — -1(r) A non-metal in period 2 which is tetravalent — Carbon [C]

Question 5

Question 3(a) A, B and C are the elements of a Dobereiner's triad. If the atomic mass of A is 7 and that of C is 39, what should be the atomic mass of B?(b) Why were Dobereiner's triads discarded?

Accepted Answer

(a) When the atomic mass of A = 7, and that of C is = 39 then according to Dobereiner's triad, atomic mass of B should be mean value of atomic masses of A and C.7+392\dfrac{7 + 39}{2}27+39​ = 462\dfrac{46}{2}246​ = 23Hence, atomic mass of B = 23(b) Reasons for discarding Dobereiner's triads:Dobereiner failed to arrange all the then known elements in the form of triads.The law did not fully apply even within the same family.

Question 6

Question 4Explain 'Newland's law of Octaves'. Why was the law discarded?

Accepted Answer

According to Newland's law of octaves, when elements are arranged by increasing atomic mass, the properties of every eighth element starting from any element are a repetition of the properties of the starting element.Reasons for discarding the Newland's law of Octaves:This classification did not work with heavier elements i.e., those lying beyond Calcium. As more and more elements were discovered, they could not be fitted into Newland's Octaves.Newland adjusted two elements Cobalt (Co) and Nickel (Ni) in the same slot, and these were placed in the same column as fluorine, chlorine and bromine which have very different properties than these elements.Iron, which resembles cobalt and nickel in properties has been placed far away from these elements.

Question 7

Question 7(a) What was Mendeleev's basis for classification of elements?(b) Mendeleev's contributions to the concept of a periodic table laid the foundation for the Modern Periodic Table. Give reasons.

Accepted Answer

(a) Mendeleev classified elements according to Mendeleev's Periodic Law. Mendeleev's Periodic Law states that physical and chemical properties of elements are a periodic function of their atomic masses.(b) As many features of the Mendeleev's periodic table were later used when Modern Periodic Table was made so we can say that Mendeleev's contributions to the concept of a periodic table laid the foundation for the Modern Periodic Table. For example:Mendeleev's periodic table contains vertical column called groups and horizontal rows called periods.They were all in eight groups.All elements in a group had similar properties and same valency.In a period elements gradually changed from metallic to non-metallic character.Mendeleev left some gaps in his periodic table for subsequent inclusion of elements not known at that time.He predicted the properties of then unknown elements on the basis of the properties of elements lying adjacent to the vacant slots.He was able to correct the atomic mass of elements like gold and platinum by placing these elements strictly on the basis of similarities in their properties.

Question 8

Question 11State the merits of Mendeleev's classification of elements.

Accepted Answer

Merits of Mendeleev’s classification of elements:Grouping of elements: He generalized the study of the elements then known to a study of mere eight groups.Gaps of undiscovered elements: Mendeleev left some gaps in his periodic table for subsequent inclusion of elements not known at that time.Prediction of properties of undiscovered elements: He predicted the properties of then unknown elements on the basis of the properties of elements lying adjacent to the vacant slots.Incorrect atomic mass corrected: He was able to correct the atomic mass of elements like gold and platinum by placing these elements strictly on the basis of similarities in their properties.

Question 9

Question 12Why did Mendeleev leave some gaps in his periodic table of elements? Explain your answer with an example.

Accepted Answer

In order to make sure that elements having similar properties fell in the same vertical column or group, Mendeleev left some gaps in his periodic table. These gaps were left for subsequent inclusion of elements not known at that time.For example,(i) Eka-aluminium (means one place below aluminium in the group); its atomic mass and chemical properties are quite similar to those of the element gallium discovered later in 1876.(ii) Properties of eka-silicon are the properties of germanium.

Question 10

Question 18This question refers to elements of the periodic table with atomic numbers from 3 to 18. In the table below, some elements are shown by letters, even though the letters are not the usual symbols of the elements.345678910ABCDEFGH1112131415161718IJKLMNOP(a) Which of these is(i) a noble gas(ii) a halogen(iii) an alkali metal(iv) an element with valency 4(b) If A combines with F, what would be the formula of the resulting compound?(c) What is the electronic configuration of G?

Accepted Answer

(a) (i) H and P are noble gases, as they are in group zero and have their octet complete.(ii) G and O are halogens, as they are in group VIIA and have 7 valence electrons.(iii) A and I are alkali metals, as they are in group IA and have 1 valence electron.(iv) D and L have valency 4, as they are in group IVA and have 4 valence electrons.(b) Valency of A [2, 1] is +1, so it tries to donate an electron and attain a stable state.Valency of F [2, 6] is -2, so it tries to gain two electrons and attain a stable state. Hence, one F atom takes electron from two A atoms. Therefore, the formula of their compound will be A2F(c) Electronic configuration of G is [2, 7] as its atomic number is 9.

Question 11

Question 20Write the electronic configuration of the element 3517T.(a) What is the group number of T?(b) What is the period number of T?(c) How many valence electrons are there in an atom of T?(d) What is the valency of T?(e) Is it a metal or a non-metal?(f) State number of protons and neutrons in T.

Accepted Answer

The electronic configuration of 3517T is [2, 8, 7](a) Group number of T — VIIA [∵ it has 7 valence electrons](b) Period number of T — Third period [∵ it has 3 shells](c) Valence electrons — Seven [from electronic configuration](d) Valency of T is -1. [∵ it needs to gain one electron to complete its octet](e) T is a non-metal [∵ it has a tendency to accept electrons](f) Protons = 17, Neutrons = 18

Question 12

Question 5State the defects of the table proposed by Mendeleev.

Accepted Answer

Defects of table proposed by Mendeleev:Anomalous pairs — The following elements did not follow Mendeleev's principles:Argon with atomic mass 39.9 precedes potassium with atomic mass 39.1Cobalt with atomic mass 58.9 precedes nickel with atomic mass 58.6Tellurium with atomic mass 127.6 precedes iodine with atomic mass 126.9Position of isotopes — According to Mendeleev's periodic table isotopes should be given separate place in the table since they have different atomic mass. But they were not assigned separate places.Grouping of chemically dissimilar elements — Elements such as copper and silver bear no resemblance to alkali metals (lithium, sodium etc.) but they have been placed together in the first group.Separation of chemically similar elements — Elements that are chemically similar, such as gold and platinum have been placed in separate groups.Electron arrangement — It does not explain the electron arrangement of elements.Position of hydrogen — Hydrogen was not given a fixed position. It was considered in group IA as well as in group VIIA because it forms forms both a positive ion, viz. in HCl and a negative ion, viz. in NaH.

Question 13

Question 7Discuss the characteristics of halogens with respect to:(a) reactivity(b) intensity of colours(c) formation of ions.

Accepted Answer

(a) Halogens are the most reactive non-metals. Their reactivity decreases down the group. For example: fluorine is the most reactive and iodine is the least reactive.(b) The intensity of the colour of the element increases from pale to dark. Fluorine is a pale yellow gas, chlorine is a greenish-yellow gas, bromine is a reddish-brown liquid and iodine is a violet solid.(c) They all form negative ions carrying a single charge [Fluoride ions F-, Chloride ions Cl-, Bromide ions Br-, Iodide ions I-.]

Question 14

Question 2(ii)Assertion (A): Lithium, sodium and potassium form a Dobereiner's triad.Reason (R): The atomic mass of the middle element is the average of the masses of the other two elements.Both A and R are true and R is the correct explanation of A.Both A and R are true but R is not the correct explanation of A.A is true but R is false.A is false but R is true.

Accepted Answer

Both A and R are true and R is the correct explanation of A.Explanation — The atomic masses of Lithium (Li) is 7 , Sodium (Na) is 23 and Potassium (K) is 39. The mean of atomic masses of Lithium and Potassium is= 7+392\dfrac{7 + 39}{2}27+39​ = 462\dfrac{46}{2}246​ = 23Therefore, the atomic mass of middle element sodium (23) is the average of the masses of the other two elements. Hence both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Question 15

Question 2(iii)Assertion (A): Chlorine with atomic number 17 is placed in group 17 and period 3.Reason (R): The number of shells signifies a period, while the number of valence electrons indicates the group.Both A and R are true and R is the correct explanation of A.Both A and R are true but R is not the correct explanation of A.A is true but R is false.A is false but R is true.

Accepted Answer

Both A and R are true but R is not the correct explanation of A.Explanation — Chlorine, a halogen with atomic number 17 has valency -1 so, it is placed in group 17. Chlorine has total three shells- K, L and M. So, it is placed in period 3. Hence, the assertion (A) is true.The number of shell(s) in an atom signifies in which period it is placed and number of valence electrons indicates its group. Hence reason (R) is true.But, reason (R) does not explain why chlorine is placed in group 17 and period 3. Hence reason (R) is not the correct explanation of assertion (A).

Question 16

Question 2(iv)Assertion (A): 19Fluorine, 35Chlorine and 80Bromine do not constitute a Döbereiner's triad.Reason (R): Atomic mass of chlorine is not the average mass of Fluorine and Bromine.Both A and R are true and R is the correct explanation of A.Both A and R are true but R is not the correct explanation of A.A is true but R is false.A is false but R is true.

Accepted Answer

Both A and R are true and R is the correct explanation of A.Explanation — The atomic masses of Fluorine is 19, Chlorine is 35 and Bromine is 80.The mean of atomic masses of fluorine and bromine is= 19+802\dfrac{19 + 80}{2}219+80​ = 992\dfrac{99}{2}299​ = 49.5The obtained mean is not the atomic mass of chlorine. So, they do not constitute a Döbereiner's triad. Hence the assertion (A) is true.Atomic mass of chlorine is not the average mass of Fluorine and Bromine. Hence the reason (R) is true.Reason (R) justifies why florine, chlorine and bromine are not Döbereiner's triad. Hence reason (R) is the correct explanation of assertion (A).

Question 17

Question 6(a) Name the alkali metals. How many electrons(s) do they have in their outermost orbit.(b) Take any one alkali metal and write its reaction with(i) Oxygen(ii) Water(iii) Acid

Accepted Answer

(a) Lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs), and francium (Fr) are alkali metals. They have 1 electron in their outermost orbit.(b) (i) Alkali metal like sodium reacts rapidly with oxygen forming metal oxide4Na + O2 ⟶ 2Na2O(ii) Alkali metal like sodium reacts with water violently and produces its hydroxide and hydrogen2Na + 2H2O ⟶ 2NaOH + H2(iii) Alkali metal react violently with dil. HCl and dil. H2SO4 to produce hydrogen2Na + 2HCl ⟶ 2NaCl + H2

Question 18

Question 1Western music DoReMeFaSoLaTiIndian music SaReGaMaPaDhaNee1234567HLiBeBCNOFNaMgAlSiPSClKCaCrTiMnFe(a) Name the scientist who gave this table.(b) Which group of elements were not known at his time?(c) This system worked quite well for the .............. (heavier/lighter) elements.(d) The table shown above relates the properties of the elements to their .............. (atomic number / atomic mass)(e) Why was this table discarded?

Accepted Answer

(a) John Newland(b) 18th group of inert elements(c) This system worked quite well for the lighter elements.(d) The table shown above relates the properties of the elements to their atomic mass.(e) Reasons for discarding the law of octaves:This classification did not work for heavier elements i.e., those lying beyond calcium. As more and more of elements were discovered, they could not be fitted in this table.Newland adjusted two elements cobalt (Co) and nickel (Ni) in the same slot and these were placed in the same column as fluorine, chlorine and bromine which have very different properties than these elements.Iron, which resembles cobalt and nickle in properties, has been placed far away from these elements.

Question 19

Question 2(a) Complete the table.Atomic NoElementElectronic configurationSelect element of the same group11Sodium...............(Ca/N/K)15Phosphorus...............(Al/N/C)16Sulphur...............(F/Cl/O)9Fluorine...............(Ca/Cl/K)(b) Classify the elements given in the table as metals and non-metals.(c) To which group does sodium belong? Why is Hydrogen not considered while describing the elements of this group?

Accepted Answer

(a) The completed table is given below:Atomic NoElementElectronic configurationSelect element of the same group11Sodium2, 8, 1K15Phosphorus2, 8, 5N16Sulphur2, 8, 6O9Fluorine2, 7Cl(b)ElementMetal/Non-metalSodium (Na)MetalPhosphorous (P)Non-metalSulphur (S)Non-metalFluorine (F)Non-metal(c) Sodium belongs to group 1.Position of hydrogen is not satisfactory. Hydrogen atom is very small in size and therefore many of its properties are different from group 1 elements (alkali metals). It is a gas at room temperature, while other elements of group 1 are solid metals. Secondly, it does not readily form cations, in most compounds, whereas group 1 metals readily form cations and form only ionic bonds. Due to these reasons Hydrogen is not considered while describing the elements of group 1.

Question 20

Question 6Complete the table pertaining to the following elements given in Column 1.Column 1 ElementsAtomic numberElectronic configurationUnderline the element present in that group1. Hydrogen1.....Be/Na/Ca/2. Carbon6....., .....N/P/Si3. Nitrogen7....., .....C/P/S4. Oxygen8....., .....N/S/Cl5. Sulphur16....., ....., .....O/N/F6. Chlorine17....., ....., .....I/O/S

Accepted Answer

Column 1 ElementsAtomic numberElectronic configurationUnderline the element present in that group1. Hydrogen11Na2. Carbon62, 4Si3. Nitrogen72, 5P4. Oxygen82, 6S5. Sulphur162, 8, 6O6. Chlorine172, 8, 7I

The Periodic Table

Additional Questions

Exercise 5A

Exercise 5B

Exercise 5C Descriptive Type

Exercise 5C Multiple Choice Type

Exercise 5C Short Answer Type

Exercise 5C Structuredapplicationskill Type

Unit Test Paper 5 - Periodic Table