Question 1

Question 6The metals of Group 2 in the periodic table from top to bottom are — Be, Mg, Ca, Sr, and Ba.(i) Which one of these elements will form ions most readily and why?(ii) State the common feature in the electronic configuration of all these given elements.

Accepted Answer

(i) Ba - Elements at the bottom of a group are most metallic, have large atomic size and lowest ionisation potential. So, the outer electrons are loosely held and will form ions from metals most readily and thus are more reactive.(ii) As the elements belong to group 2 thus they all have 2 electrons in the valence shell.

Question 2

Question 7In the table below, H does not represent hydrogen. Some elements are given in their own symbol and position in the periodic table while others are shown with a letter.1A1IIA2IIIA13IVA14VA15VIA16VIIA17O18Li D  OJNeAMgESi HK BC FG  LAnswer the following questions.Identify the most electronegative element.Identify the most reactive element of group IA or 1.Identify the element from period 3 with the smallest atomic size.Identify the element with the highest ionization potential.How many valence electrons are present in G ?Which element from group 2 will have the least ionization energy ?Identify the noble gas of the fourth period.In the compound between A and H, what type of bond is formed ? Give its molecular formula.

Accepted Answer

J is the most electronegative element.B is the most reactive element of group IA or 1.K is the element from period 3 with smallest atomic size.Ne the highest ionization potential.5 valence electrons present in G.C is the element from group 2 which has least ionization energy.L is the noble gas of the fourth period.ionic bond is formed between A and H.The molecular formula is :2A + H ⟶ A2H

Question 3

Question 9Among the elements given below, the element with the least electronegativity is :LithiumCarbonBoronFluorine

Accepted Answer

LithiumReason — Lithium, Boron, Carbon, Fluorine is the order of given elements from left to right in period 2. As electronegativity increases from left to right in a period, hence, Lithium has the least electronegativity in the given elements.

Question 4

Question 11Parts (i) to (iv) refer to changes in the properties of elements on moving from left to right across a period of the Periodic Table. For each property, choose the correct answer.(i) The non-metallic character of the elements :decreases,increases,remains the same,depends on the period(ii) The electronegativity :depends on the number of valence electrons,remains the same,decreases,increases(iii) The ionization potential :goes up and downdecreasesincreasesremains the same(iv) The atomic size :decreases,increases,remains the same,sometimes increases and sometimes decreases

Accepted Answer

(i) increases.Reason — On moving across a period from left to right, nuclear pull increases due to increase in atomic number and thus atomic size decreases. Hence, elements cannot lose electrons easily. Therefore, non-metallic character increases.(ii) increases.Reason — On moving across a period from left to right, nuclear charge increases due to an increase in atomic number. Hence, electronegativity increases from left to right in a period.(iii) increases.Reason — On moving from left to right across a period, the atomic radius decreases due to increase in the nuclear charge, and thus, more energy is required to remove the electrons. Hence, ionization potential increases from left to right across a period.(iv) decreases.Reason — On moving across a period from left to right, nuclear pull increases due to increase in atomic number and thus atomic size decreases.

Question 5

Question 14Which of these statements gives the correct picture regarding halogens and alkali metals with respect to an increase in the atomic number ?Reactivity decreases in alkali metals but increases in halogens.Reactivity increases in both.Reactivity decreases in both.Reactivity increases in alkali metals but decreases in halogens.

Accepted Answer

Reactivity increases in alkali metals but decreases in halogens.Reason — Atomic number increases while going down the group in alkali metals and halogens. The tendency of losing electrons increases down the group. Reactivity of metals depends on tendency to lose electrons, thus reactivity of alkali metals increases while going down the group. Reactivity of non-metals depends on tendency to gain electrons, thus reactivity of halogens decreases while going down the group.

Question 6

Question 16An element 13X combines with 17Y to form a compound. which of the following is true ?P → X is a metal, Y is a metal.Q → X is a metal, Y is a non-metal.R → X losses electron(s), y gains electron(s).Only POnly QBoth P and QBoth Q and R

Accepted Answer

Both Q and RReason — Element X with an atomic number of 13 is Aluminium (Al), which is a metal. Element Y with an atomic number of 17 is Chlorine (Cl), which is a non-metal.X being a metal tends to lose electrons to achieve a stable electron configuration and Y being a non-metal tends to gain electrons to achieve stable electron configuration. Hence, statements Q and R are true whereas statement P is false.

Question 7

Question 17Assertion (A): Second period consists of 8 elements.Reason (R): Number of elements in each period is four times the number of atomic orbitals available in energy level that is being filled.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

A is true but R is false.Explanation — The second period includes the elements from lithium (Z=3) to neon (Z=10): Li, Be, B, C, N, O, F, Ne → 8 elements. Hence assertion is true. Reason given is incorrect. In the first period, electrons are being added to the K electronic configuration.The number of available electronic configuration or orbital for elements of first period is 1.The number of elements present in the first period is 2, which is not four times the number of atomic orbitals available. Hence the (R) statement is false.

Question 8

Question 20Assertion (A): Hydrogen is placed in group I.Reason (R): Hydrogen can gain an electron to achieve noble gas configuration.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 — Hydrogen is placed in Group I with alkali metals because, like them, it has one electron in its outermost shell and forms (H⁺) ions by losing this electron. Although hydrogen can also gain an electron to form the hydride ion (H⁻) similar to halogens. Hence, Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Question 9

Question 21Assertion (A): Atomic size increases along a period.Reason (R): Effective nuclear charge increases with atomic number.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

A is false but R is true.Explanation — Atomic size decreases across a period (from left to right) in the periodic table. This is because, although electrons are being added, they are added to the same shell, and the increasing effective nuclear charge pulls the electrons closer to the nucleus. Hence, Assertion (A) is false.As atomic number increases, the number of protons increases, which increases the effective nuclear charge. Hence, Reason (R) is true.

Question 10

Question 22Assertion (A): Elements in the same vertical column have similar properties.Reason (R): Properties depend upon the number of electrons in the valence shell.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 — Elements in the same vertical column i.e., a group in the periodic table, have the same number of valence electrons. Since chemical properties of elements depend upon the number of valence electrons, thus elements in the same group have similar properties. Hence, Reason (R) correctly explains the Assertion (A).

Question 11

Question 2Arrange the following in order of :(i) increasing radii.(a) Cl-, Cl(b) Mg2+, Mg, Mg+(c) N, O, P(ii) increasing ionisation energy.(a) P, Na, Cl(b) F, O, Ne(c) Ne, He, ArExplain your choice.

Accepted Answer

(i) In order of increasing radii:(a) Cl < Cl-.Reason — Anion is greater in size than its parent atom.(b) Mg2+ < Mg+ < Mg.Reason — Cation is smaller in size than its parent atom.(c) O < N < P.Reason — Between Oxygen (O) and Nitrogen (N), atomic size of Nitrogen (N) is more as it is to the left of Oxygen (O) in period 2. Between Nitrogen (N) and Phosphorus (P), Phosphorus (P) has larger atomic size as it comes after Nitrogen (N) in group 5A.(ii) In order of increasing ionisation energy:(a) Na < P < ClReason — Ionisation energy increases when moving across a period from left to right.(b) O < F < NeReason — Ionisation energy increases when moving across a period from left to right.(c) Ar < Ne < HeReason — Ionisation energy decreases when moving down the group.

Question 12

Question 4Give reasons for the following assertions :(a) The oxidising power of elements increases from left to right along a period.(b) Ionisation potential of elements increases across a period from left to right.(c) Alkali metals are good reducing agents.

Accepted Answer

(a) The oxidising power of elements increases from left to right along a period because electro-negativity and the non metallic character increases from left to right. As oxidising power depends on tendency to gain electrons and non-metals are good oxidising agents hence oxidising power of elements increases across a period.(b) The ionisation potential of element increases across a period because the atomic size decreases due to an increase in nuclear charge and electrons in the outermost shell are more strongly held because of which greater energy is required to remove the electron.(c) Alkali metals have one electron in their valence shell. In order to be stable, they easily lose this electron and get oxidised. Hence, they are good reducing agents.

Question 13

Question 1Fill in the blanks by selecting the correct word from the brackets :(a) The element below sodium in the same group would be expected to have a ............... (lower/higher) electro-negativity than sodium and the element above chlorine would be expected to have a (lower/higher) ionization potential than chlorine.(b) On moving down a group, the number of valence electrons ............... (remains the same/increases/ decreases).(c) Metals are good ............... (oxidising agent/ reducing agent) because they are electron ............... (acceptors/donors).(d) Down the group, electron affinity ............... [increases, decreases, remains same].(e) Electronegativity across the period ............... [increases/ decreases].(f) Non-metallic character down the group ............... [increases/ decreases].(g) In a period, increase in electron affinity increases ............... (oxidation/reduction).(h) On descending a group, ............... (increase/decrease) in ionisation potential as well as electron affinity ............... (increases/decreases) oxidising capacity.(i) If an element has a low ionization energy then it is likely to be ............... (metallic/non metallic).(j) If an element has seven electrons in its outermost shell then it is likely to have the ............... (largest/smallest) atomic size among all the elements in the same period.

Accepted Answer

(a) The element below sodium in the same group would be expected to have a lower (lower/higher) electro-negativity than sodium and the element above chlorine would be expected to have a higher (lower/higher) ionization potential than chlorine.(b) On moving down a group, the number of valence electrons remains the same (remains the same/increases/ decreases).(c) Metals are good reducing agent (oxidising agent/ reducing agent) because they are electron donors (acceptors/donors).(d) Down the group, electron affinity decreases [increases, decreases, remains same].(e) Electronegativity across the period increases [increases/ decreases].(f) Non-metallic character down the group decreases [increases/ decreases].(g) In a period, increase in electron affinity increases reduction (oxidation/reduction).(h) On descending a group, decrease (increase/decrease) in ionisation potential as well as electron affinity decreases (increases/decreases) oxidising capacity.(i) If an element has a low ionization energy then it is likely to be metallic (metallic/non metallic).(j) If an element has seven electrons in its outermost shell then it is likely to have the smallest (largest/smallest) atomic size among all the elements in the same period.

Question 14

Question 8What do you understand by ?(i) Periodicity(ii) Typical elements(iii) Orbits

Accepted Answer

(i) The properties that reappear at regular intervals, or in which there is gradual variation (i.e. increase or decrease) at regular intervals, are called 'periodic properties' and the phenomenon is known as the periodicity of elements.(ii) The third period elements are typical elements. They are Sodium (Na), Magnesium (Mg), Aluminium (Al), Silicon (Si), Phosphorus (P), Sulphur (S), Chlorine (Cl) .(iii) Electrons revolve around the nucleus in certain definite circular paths called orbits.

Question 15

Question 10Name the (i) metals (ii) metalloids and (iii) non-metals in the first twenty elements.

Accepted Answer

Metals — Lithium (Li), Beryllium (Be), Sodium (Na), Magnesium (Mg), Aluminium (Al), Potassium (K), Calcium (Ca).Metalloids — Boron (B), Silicon (Si).Non-Metals — Hydrogen (H), Helium (He), Carbon (C), Nitrogen (N), Oxygen (O), Fluorine (F), Neon (Ne), Phosphorus (P), Sulphur (S), Chlorine (Cl), Argon (Ar).

Question 16

Question 5(i)Which is greater in size ?(a) An atom or a cation(b) An atom or an anion(c) Fe2+ or Fe3+(d) Fluorine or oxygen

Accepted Answer

(a) An atom.Reason — An atom is greater in size than a cation because cation is formed by the loss of electron(s), hence proton(s) are more than electron(s) in a cation. So electrons are strongly attracted by the nucleus and are pulled inward. Hence, the size decreases.(b) An anion.Reason — An anion is greater in size than atom because anion is formed by the gain of electron(s). Thus, the number of electron(s) are more than proton(s). The effective positive charge in the nucleus is less, so less inward pull is experienced. Hence, the size increases.(c) Fe2+.Reason — Fe2+ is greater in size than Fe3+ because Fe2+ has one more electron than Fe3+. So, nuclear pull is more in Fe3+ than Fe2+, thereby bringing the outermost shell closer to the nucleus. Hence, size decreases.(d) Oxygen.Reason — Oxygen is greater in size than fluorine. The position of oxygen in periodic table is before fluorine so, as we move from left to right in a period the outermost shell with more number of electrons is brought closer to the nucleus, decreasing the size of the atom.

Question 17

Question 7State the trend in chemical reactivity :(i) Across the third period from left to right,(ii) Down the group.(a) in group IA (1) (b) in group VIIA (17)

Accepted Answer

(i) On moving from left to right in the third period, the chemical reactivity of elements first decreases until silicon as the tendency to lose electrons deceases and then starts to increase from phosphorus to chlorine, as the tendency to gain electrons increases.(ii) Chemical reactivity in metals increases on going down the group, but in non-metals it decreases on going down the group.(a) As group IA (1) contains metals, hence, chemical reactivity increases on going down the group IA (1).(b) As group VIIA (17) contains non-metals, hence, chemical reactivity decreases on going down the group VIIA (17).

Question 18

Question 12Which one has the largest size?BrII-Cl

Accepted Answer

I-Reason — Cl, Br and I belong to the same group (group 17) and between the three, I is the bottom most. As atomic size increases down the group hence, between Cl, Br and I, I has the largest atomic size. Between I and I-, atomic size of I- is larger as anion is larger than the parent atom. Hence, I- has the largest atomic size in the given elements.

Question 19

Question 14Write the number of protons, neutrons and electronic configuration of 1939K_{19}^{39}	ext{K}1939​K , 1531P_{15}^{31}	ext{P}1531​P . Also state their position in the periodic table.

Accepted Answer

For 1939K_{19}^{39}	ext{K}1939​K:No. of Protons = 19No. of Neutrons = 39 - 19 = 20Electronic Configuration = 2, 8, 8, 1Position in periodic table = 4th period (as it has 4 shells) & 1st group (as it has one valence electron).For 1531P_{15}^{31}	ext{P}1531​P:No. of Protons = 15No. of Neutrons = 31 - 15 = 16Electronic Configuration = 2, 8, 5Position in periodic table = 3rd period (as it has 3 shells) & 15th group (as it has five valence electrons).

Question 20

Question 18Explain why are the following statements not correct:(i) All groups contain metals and non-metals.(ii) Atoms of elements in the same group have the same number of electron(s).(iii) Non-metallic character decreases across a period with increase in atomic number.(iv) Reactivity increases with atomic number in a group as well as in a period.

Accepted Answer

(i) Incorrect — A few groups contain metalloids and inert gases also.(ii) Incorrect — as we move down the group the number of electrons increases.(iii) Incorrect — Non-metallic character increases across a period with increase in atomic number.(iv) Incorrect — Chemical reactivity in metals increases on going down the group, but in non-metals decreases on going down the group.

Periodic Table, Periodic Properties and Variations of Properties

Exercise 1 Long Answer Type

Exercise 1 Multiple Choice Type

Exercise 1 Short Answer Type

Exercise 1 Very Short Answer Type

Intext Questions 1

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Intext Questions 3