Chemistry Objective Questions and Answers: Chapter 03 Classification of Elements and Periodicity in Properties
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These assertion–reason and case based questions for Class 11 Chemistry Chapter 3, Classification of Elements and Periodicity in Properties, test your understanding of periodic trends, the modern periodic table and the anomalous behaviour of elements. Answers are given with complete explanations and a teacher’s note for each question.
Assertion–Reason Questions
Directions: In each question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.
Assertion (A): Noble gases are highly reactive.
Reason (R): Noble gases have stable outer electronic configuration.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. Noble gases are very unreactive because of their stable outer electronic configuration, ns2np6 (or 1s2 for He).
2. So the Assertion is false, but the Reason is true.
Teacher's Note:
1. Stable ns2np6 configuration is why noble gases are called inert gases.
2. Only a few compounds of Xe and Kr with F and O are known.
Assertion (A): Elements in the same vertical column have similar properties.
Reason (R): Elements have periodic dependence upon the atomic number.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Elements in the same group have the same number of electrons in their outer orbitals and similar outer electronic configuration.
2. Since properties depend periodically on atomic number (i.e. on electronic configuration), they show similar properties.
3. Both A and R are true, and R is the correct explanation of A.
Teacher's Note:
1. The modern periodic law links properties to atomic number.
2. Similar outer configuration in a group is the cause of similar chemical properties.
Assertion (A): The elements having 1s2, 2s2, 2p6, 3s2 and 1s2, 2s2 electronic configuration belong to same period.
Reason (R): Both have same outermost electronic configuration.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. 1s2 2s2 2p6 3s2 (Mg) belongs to the 3rd period; 1s2 2s2 (Be) belongs to the 2nd period.
2. Both have the outer configuration ns2, so they belong to the same group (group 2) but to different periods.
3. So the Assertion is false, but the Reason is true.
Teacher's Note:
1. Same outer configuration means same group, not same period.
2. The period is decided by the highest value of n.
Assertion (A): Atomic number of the element copernicium is 112.
Reason (R): IUPAC name of this element is ununbium in which un- and bi- are used for 1 and 2 respectively in Latin words.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. Copernicium (Cn) has atomic number 112.
2. Its temporary IUPAC name was ununbium: ‘un’ for 1 and ‘bi’ for 2.
3. Both A and R are true, but R (about the naming system) does not explain why its atomic number is 112.
Teacher's Note:
1. The temporary name only encodes the atomic number; it does not decide it.
2. Permanent name: copernicium (Cn), after Nicolaus Copernicus.
Assertion (A): The atomic radii of the elements of the oxygen family are smaller than the atomic radii of the corresponding elements of the nitrogen family.
Reason (R): The members of the oxygen family are more electronegative and thus, have lower values of nuclear charge than those of the nitrogen family.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. Atomic radii of oxygen-family elements are smaller than those of the corresponding nitrogen-family elements because of the higher effective nuclear charge, which pulls the electrons closer.
2. Oxygen-family elements have a higher (not lower) nuclear charge than the corresponding nitrogen-family elements.
3. So the Assertion is true, but the Reason is false.
Teacher's Note:
1. Moving one place to the right in a period adds one proton, increasing nuclear charge.
2. Higher nuclear charge gives a smaller radius and a higher electronegativity.
Assertion (A): Isoelectronic species have same radii.
Reason (R): They have different nuclear charges.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. Isoelectronic species have the same number of electrons but different nuclear charges.
2. The greater the nuclear charge, the more strongly the electrons are pulled in, so isoelectronic species have different radii (e.g. O2− > F− > Na+ > Mg2+).
3. So the Assertion is false, but the Reason is true.
Teacher's Note:
1. Isoelectronic means same number of electrons, not same size.
2. For isoelectronic ions, more protons means a smaller ion.
Assertion (A): The atomic and ionic radii generally decrease towards right in a period.
Reason (R): The ionisation enthalpy decreases on moving towards left in a period.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. The atomic and ionic radii decrease from left to right in a period because the effective nuclear charge increases.
2. Ionisation enthalpy increases from left to right in a period, i.e. it decreases on moving towards the left, so the Reason is also a true statement.
3. But the trend of ionisation enthalpy does not explain the decrease in radii; both are consequences of increasing effective nuclear charge.
4. So both A and R are true, but R is not the correct explanation of A.
Teacher's Note:
1. Read the direction words (‘towards left’) carefully before judging a statement false.
2. Increasing effective nuclear charge is the common cause of both trends.
Assertion (A): Boron has a smaller first ionisation enthalpy than beryllium.
Reason (R): The penetration of 2s electron to the nucleus is more than the 2p electron. Hence, 2p electron is more shielded by the inner core of electrons than the 2s electrons. (NCERT Exemplar)
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Boron has a smaller first ionisation enthalpy than beryllium because the 2s electron penetrates closer to the nucleus than the 2p electron.
2. So the 2p electron of B is more shielded by the inner core electrons and is removed more easily than a 2s electron of Be.
3. Both A and R are true, and R is the correct explanation of A.
Teacher's Note:
1. Penetration order: s > p > d > f.
2. The same reason explains Mg > Al in period 3.
Assertion (A): Ionisation enthalpy is the energy released to remove an electron from an isolated gaseous atom in its ground state.
Reason (R): Energy has to be supplied to remove an electron from an isolated gaseous atom, so ionisation enthalpy is always positive.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. Ionisation enthalpy is the energy required (absorbed), not released, to remove an electron from an isolated gaseous atom in its ground state.
2. Removing an electron always needs energy to overcome the attraction of the nucleus, so ΔiH is always positive.
3. So the Assertion is false, but the Reason is true.
Teacher's Note:
1. Ionisation is always endothermic.
2. Electron gain can release energy, but ionisation never does.
Assertion (A): Generally, ionisation enthalpy increases from left to right in a period.
Reason (R): When successive electrons are added to the orbitals in the same principal quantum level, the shielding effect of inner core of electrons does not increase very much to compensate for the increased attraction of the electron to the nucleus.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Across a period, nuclear charge increases while the electrons are added to the same shell, so the shielding by inner-core electrons hardly increases.
2. The effective nuclear charge therefore increases and atomic size decreases, so ionisation enthalpy increases.
3. Both A and R are true, and R is the correct explanation of A.
Teacher's Note:
1. Electrons in the same shell shield each other poorly.
2. Rising effective nuclear charge explains most trends across a period.
Assertion (A): Alkali metals have least value of ionisation energy within a period.
Reason (R): They precede alkaline earth metals in periodic table.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. Alkali metals (group 1) have the largest atomic size in their period and a single ns1 electron, so they have the lowest ionisation energy.
2. They do precede the alkaline earth metals in the periodic table, but this position is not the reason for their low ionisation energy.
3. Both A and R are true, but R is not the correct explanation of A.
Teacher's Note:
1. Low IE of alkali metals is due to large size and one loosely held outer electron.
2. A true statement about position is not automatically an explanation.
Assertion (A): Electron gain enthalpy becomes less negative as we go down a group.
Reason (R): Size of the atom increases on going down the group and the added electron would be farther from the nucleus.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Down a group, the atomic size and screening effect increase, and the added electron is farther from the nucleus.
2. So the atom attracts the extra electron less strongly and electron gain enthalpy becomes less negative.
3. Both A and R are true, and R is the correct explanation of A.
Teacher's Note:
1. The general trend has an exception at the top: Cl is more negative than F.
2. Larger atoms hold an incoming electron less firmly.
Assertion (A): Caesium and fluorine both react violently.
Reason (R): Caesium is most electropositive and fluorine is most electronegative.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Caesium is the most electropositive (stable) element, so it loses its electron very readily.
2. Fluorine is the most electronegative element, so it gains an electron very readily.
3. Hence both react violently. Both A and R are true, and R is the correct explanation of A.
Teacher's Note:
1. Reactivity of metals rises down group 1; reactivity of halogens falls down group 17.
2. Cs and F lie at opposite corners of the periodic table.
Assertion (A): Fluorine has a less negative electron affinity than chlorine.
Reason (R): There is relatively greater effectiveness of 2p electrons in the small fluorine atom to repel the additional electron entering the atom than to 3p electrons in the larger Cl atom.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Fluorine has a less negative electron affinity (−328 kJ/mol) than chlorine (−349 kJ/mol).
2. In the small F atom, the added electron enters the compact 2p subshell and is repelled strongly by the electrons already present; in the larger Cl atom, repulsion in the 3p subshell is less.
3. Both A and R are true, and R is the correct explanation of A.
Teacher's Note:
1. The same reason makes O less negative than S.
2. Electron–electron repulsion in a small shell outweighs the closer nucleus.
Case Based Questions
Case Study 1: Read the passage given below and answer the questions that follow.
When an electron is added to a gaseous atom in its ground state to convert it into a negative ion, the enthalpy change accompanying the process is called the electron gain enthalpy (ΔegH). It is a direct measure of the ease with which an atom attracts an electron to form an anion.
X(g) + e− → X−(g); ΔH = ΔegH
The most stable state of an atom is the ground state. If an isolated gaseous atom is in an excited state, comparatively less energy will be released on adding an electron. So, electron gain enthalpies of gaseous atoms must be determined in their ground states. Therefore, the terms ground state and isolated gaseous atom have also been included in the definition of electron gain enthalpy. Like ionisation enthalpy, electron gain enthalpy is measured either in electron volts per atom or kJ per mole.
(a) stable configuration
(b) large size
(c) high reactivity
(d) unstable configuration
Show Answer & Explanation
Answer: (a) stable configuration
Explanation:
1. Noble gases have a stable ns2np6 configuration, so an added electron must enter the next higher shell.
2. This is energetically unfavourable, so noble gases do not accept electrons readily and have positive electron gain enthalpy.
Teacher's Note:
1. Positive ΔegH means energy must be supplied to add an electron.
2. Noble gases have large positive values (e.g. Ne +116 kJ/mol).
(a) small size
(b) less repulsion
(c) large size
(d) high electronegativity
Show Answer & Explanation
Answer: (a) small size
Explanation:
1. When an electron is added to O or F, it enters the small n = 2 shell and suffers more repulsion from the electrons already present.
2. In S or Cl, the electron enters the larger n = 3 shell, where repulsion is less. So the small size of O and F lowers their electron gain enthalpy.
Teacher's Note:
1. Small size means more electron–electron repulsion, not less.
2. This is why Cl, not F, has the most negative ΔegH.
(a) −333, −325, −349 and −296
(b) −296, −325, −333 and −349
(c) −333, −349, −325 and −296
(d) −349, −333, −325 and −296
Show Answer & Explanation
Answer: (c) −333, −349, −325 and −296
Explanation:
1. All halogens have negative (exothermic) electron gain enthalpies, and ΔegH generally becomes less negative down the group.
2. However, F is an exception: inter-electron repulsion in its very small atom makes its ΔegH less negative than that of Cl.
3. So the values are F (−333), Cl (−349), Br (−325) and I (−296) kJ/mol, i.e. Cl > F > Br > I in magnitude.
Teacher's Note:
1. Cl has the most negative value; after Cl, the values fall steadily down the group.
2. The NCERT value for F is −328 kJ/mol; some books give about −333 kJ/mol.
(a) More penetration of s-electron
(b) More penetration of p-electron
(c) Large size
(d) Small size
Show Answer & Explanation
Answer: (a) More penetration of s-electron
Explanation:
1. In Be the electron is removed from 2s, while in B it is removed from 2p.
2. A 2s electron penetrates closer to the nucleus than a 2p electron, so it is held more tightly. Hence it is easier to remove the 2p electron of B, and Be has the higher ionisation enthalpy.
Teacher's Note:
1. Penetration of s electrons is greater than that of p electrons in the same shell.
2. Be also has a stable, fully filled 2s2 subshell.
(a) atomic size
(b) number of electrons
(c) number of neutron
(d) None of the above
Show Answer & Explanation
Answer: (a) atomic size
Explanation:
1. Electron gain enthalpy depends on atomic size (inversely), nuclear charge (directly) and electronic configuration.
2. Of the given options, atomic size is a factor; the number of neutrons has no effect.
Teacher's Note:
1. Smaller atoms with higher nuclear charge generally have more negative ΔegH.
2. Neutrons carry no charge, so they do not affect electron attraction.
Case Study 2: Read the passage given below and answer the questions that follow.
In the modern periodic table, elements are arranged in order of increasing atomic numbers, which is related to the electronic configuration.
Depending upon the type of orbitals receiving the last electron, the elements in the periodic table have been divided into four blocks, viz., s, p, d and f.
The modern periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Aufbau principle, the seven periods (1 to 7) have 2, 8, 8, 18, 18, 32 and 32 elements respectively.
The seventh period has been completed with the discovery of element 118. To avoid the periodic table being too long, the two series of f-block elements, called lanthanoids and actinoids, are placed at the bottom of the main body of the periodic table.
(a) s-block
(b) p-block
(c) d-block
(d) f-block
Show Answer & Explanation
Answer: (c) d-block
Explanation:
1. After barium (Z = 56), the 57th electron would be expected in a 4f orbital by the aufbau principle, but it enters the 5d orbital, which lies outside the xenon core and is only slightly higher in energy.
2. Thus, the outer electronic configuration of La (Z = 57) is 5d1 6s2 rather than 4f1 6s2, and La is placed in the d-block (group 3).
Teacher's Note:
1. La starts the 5d series; the 4f filling begins from Ce (Z = 58).
2. Lanthanoids are counted from Z = 58 to 71.
(a) 7s2 7p6
(b) 5f14 6d10 7s2 7p0
(c) 4f14 5d10 6s2 6p6
(d) 4f14 5d10 6s2 6p4
Show Answer & Explanation
Answer: (c) 4f14 5d10 6s2 6p6
Explanation:
1. The 6th period starts with the filling of the 6s orbital and ends when the 6p orbitals are completely filled; in between, the 4f and 5d orbitals are filled.
2. So the last p-block element of the 6th period (radon) has the outer configuration 4f14 5d10 6s2 6p6.
Teacher's Note:
1. The last element of every period is a noble gas with np6.
2. Options with 7s or 7p belong to the 7th period.
(a) 107
(b) 118
(c) 126
(d) 102
Show Answer & Explanation
Answer: (c) 126
Explanation:
1. The long form of the periodic table contains elements with atomic numbers 1 to 118, ending with the completed seventh period.
2. Element 126 would need a new (eighth) period and a new g-block, so it cannot be accommodated in the present set-up.
Teacher's Note:
1. Element 118 (oganesson) is the last element of period 7.
2. Elements 102 and 107 already have places in the table (No and Bh).
(a) 1s2 2s2 2p6 3s2 3p6 3d5 4s2
(b) 1s2 2s2 2p6 3s2 3p6 3d5 4s3 4p6
(c) 1s2 2s2 2p6 3s2 3p6 3d6 4s2
(d) 1s2 2s2 2p6 3s2 3p6 3d7 4s2
Show Answer & Explanation
Answer: (a) 1s2 2s2 2p6 3s2 3p6 3d5 4s2
Explanation:
1. The fifth period begins with Rb (Z = 37) and ends at Xe (Z = 54), so the element with Z = 43 lies in the 5th period.
2. The 4th period has 18 elements, so the element just above it has atomic number 43 − 18 = 25.
3. The electronic configuration of Z = 25 (Mn) is 1s2 2s2 2p6 3s2 3p6 3d5 4s2.
Teacher's Note:
1. Z = 43 is technetium (Tc), in group 7 below manganese.
2. Option (b) is impossible because an s subshell holds at most 2 electrons.
(a) noble gases
(b) halogens
(c) heavy metals
(d) light metals
Show Answer & Explanation
Answer: (b) halogens
Explanation:
1. Each period ends with a noble gas; the atomic numbers of noble gases are 2, 10, 18, 36, 54 and 86.
2. Elements with atomic numbers 35 (36 − 1), 53 (54 − 1) and 85 (86 − 1) lie just before noble gases, i.e. in group 17.
3. So they are all halogens (Br, I and At).
Teacher's Note:
1. One less than a noble gas atomic number is a halogen; one more is an alkali metal.
2. Learn the noble gas atomic numbers: 2, 10, 18, 36, 54, 86, 118.
Case Study 3: Read the passage given below and answer the questions that follow.
Moseley modified Mendeleev’s periodic law. He stated “physical and chemical properties of elements are the periodic function of their atomic numbers.” It is known as the modern periodic law and is considered the basis of the modern periodic table.
When the elements were arranged in increasing order of atomic numbers, it was observed that the properties of elements were repeated after certain regular intervals of 2, 8, 8, 18, 18 and 32. These numbers are called magic numbers, and periodicity in properties is caused by the repetition of similar electronic configuration. The long form of the periodic table is called Bohr’s periodic table. There are 18 groups and seven periods in this periodic table. The horizontal rows are called periods.
First period (1H – 2He) contains 2 elements. It is the shortest period. Second period (3Li – 10Ne) and third period (11Na – 18Ar) both contain 8 elements each. Fourth period (19K – 36Kr) and fifth period (37Rb – 54Xe) contain 18 elements each. These are long periods.
Sixth period (55Cs – 86Rn) consists of 32 elements and is the longest period, along with the seventh period (87Fr – 118Og), which also has 32 elements.
Elements of group 1 are called alkali metals. Elements of group 2 are called alkaline earth metals. Elements of group 16 are called chalcogens (ore-forming elements). Elements of group 17 are called halogens (sea-salt forming). Elements of group 18 are called noble gases.
Anomalous behaviour of the first element of a group: The first element of a group differs considerably from its congeners (i.e. the rest of the elements of its group). This is due to (i) small size, (ii) high electronegativity and (iii) non-availability of d-orbitals for bonding. Anomalous behaviour is observed among the second row elements (i.e. Li to F).
Assertion (A): Zinc is a d-block element.
Reason (R): Zinc does not form coordination compounds.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. Zinc ([Ar] 3d10 4s2) is a d-block element, so the Assertion is true.
2. Zinc does form coordination compounds, e.g. [Zn(NH3)4]2+, though fewer than other d-block elements, so the Reason is false.
Teacher's Note:
1. Zn is a d-block element but not a typical transition element, as its d subshell is full.
2. Zn2+ readily forms complexes with NH3 and CN−.
Assertion (A): The first ionisation enthalpy of Be is greater than that of B.
Reason (R): 2p-orbital is lower in energy than 2s-orbital.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. Be has the fully filled configuration 1s2 2s2, while B is 1s2 2s2 2p1; the 2p electron of B is easier to remove, so the first IE of Be (899 kJ/mol) is greater than that of B (801 kJ/mol).
2. The Reason is false: the 2p orbital is higher (not lower) in energy than the 2s orbital.
Teacher's Note:
1. Energy order within a shell: ns < np < nd < nf.
2. Be > B and N > O are the two IE exceptions in period 2.
Assertion (A): Outermost electronic configuration of most electronegative element is ns2np7.
Reason (R): Most electronegative elements are halogens.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. A p subshell can hold at most 6 electrons, so ns2np7 is impossible. The outer configuration of the most electronegative elements (halogens) is ns2np5.
2. The Reason is true: halogens, especially fluorine, are the most electronegative elements.
3. So the Assertion is false, but the Reason is true.
Teacher's Note:
1. Halogens are one electron short of a noble gas configuration.
2. Always check that no subshell exceeds its capacity: s 2, p 6, d 10, f 14.
Assertion (A): Mn has less favourable electron affinity than its neighbours on either side.
Reason (R): The magnitude of electron affinity depends on the electronic configuration of the atom.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Mn has the stable configuration [Ar] 3d5 4s2, with a half-filled 3d and a filled 4s subshell, so it does not readily accept an extra electron.
2. So its electron affinity is less favourable than that of Cr and Fe. The magnitude of electron affinity depends on electronic configuration, which explains this.
3. Both A and R are true, and R is the correct explanation of A.
Teacher's Note:
1. Half-filled and fully filled subshells give low electron affinity.
2. N (2p3) shows the same effect in period 2.
Assertion (A): Generally, ionisation enthalpy increases from left to right in a period.
Reason (R): When successive electrons are added to the orbitals in the same principal quantum level, the shielding effect of inner core of electrons does not increase very much to compensate for the increased attraction of electrons to the nucleus.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Across a period, electrons are added to the same shell, so the shielding by inner-core electrons hardly increases while the nuclear charge increases.
2. The effective nuclear charge therefore increases and atomic size decreases, so ionisation enthalpy increases.
3. Both A and R are true, and R is the correct explanation of A.
Teacher's Note:
1. Increasing effective nuclear charge is the key idea for trends across a period.
2. Exceptions (Be > B, N > O) arise from stable configurations.
Case Study 4: Read the passage given below and answer the questions that follow.
In 1913, Henry Moseley, the English physicist, performed an experiment by bombarding 38 different elements, starting from aluminium and ending with gold, with high-speed electrons in vacuum and generated X-rays. He observed that the square root of the frequency (ν) of the X-rays emitted by a metal is proportional to the atomic number and not to the atomic weight of the element, i.e.
√ν = a(Z − b) [where a and b are constants]
Thus, when √ν is plotted against atomic number (Z), a straight line is obtained, but this is not true when √ν is plotted against atomic mass. He postulated that the atomic number is a more fundamental property of an element than its atomic mass.
Thus, he modified Mendeleev’s periodic law and gave the modern periodic law, which states that “The physical and chemical properties of the elements are periodic functions of their atomic numbers,” i.e. when elements are arranged in increasing order of atomic numbers, elements having similar properties are repeated after certain regular intervals.
Assertion (A): Mendeleev arranged elements in horizontal rows and vertical columns.
Reason (R): Mendeleev ignored the order of atomic weight thinking that the atomic measurements might be incorrect.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. Mendeleev arranged elements in horizontal rows and vertical columns in order of increasing atomic weight, so that elements with similar properties occupied the same vertical column (group).
2. He did ignore the order of atomic weights in some places (e.g. Te and I), believing the atomic weight measurements might be incorrect.
3. Both A and R are true, but R is not the correct explanation of A.
Teacher's Note:
1. Mendeleev gave more importance to similarity of properties than to strict atomic weight order.
2. Te (127.6) was placed before I (126.9) for this reason.
Assertion (A): Mendeleev left the gap under aluminium and silicon and called these Eka-aluminium and Eka-silicon, respectively.
Reason (R): Dobereiner arranged elements on the basis of increasing atomic number.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. Gallium and germanium were unknown when Mendeleev published his periodic table. He left gaps under aluminium and silicon and called these elements Eka-aluminium and Eka-silicon.
2. Dobereiner grouped elements into triads on the basis of atomic weights, not atomic numbers, so the Reason is false.
Teacher's Note:
1. Eka-aluminium = gallium; Eka-silicon = germanium.
2. Atomic number was introduced only by Moseley in 1913.
Assertion (A): The horizontal rows in the periodic table are called periods or Mendeleev’s series.
Reason (R): Elements having similar outer electronic configurations in their atoms are arranged in groups/families.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. Horizontal rows in the periodic table are called periods (or series), and elements with similar outer electronic configurations are arranged in vertical columns called groups or families.
2. Both statements are true, but the Reason (about groups) does not explain the Assertion (about periods).
Teacher's Note:
1. Periods are horizontal rows; groups are vertical columns.
2. A true Reason must explain the Assertion for option (a).
Assertion (A): Sixth period is the longest period in the periodic table.
Reason (R): Sixth period involves the filling of all the orbitals of sixth energy level.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. The sixth period has 32 elements (Cs to Rn), so it is one of the longest periods (together with the seventh).
2. The sixth period involves the filling of 6s, 4f, 5d and 6p orbitals, not all the orbitals of the sixth energy level (6d, 6f, … are not filled), so the Reason is false.
Teacher's Note:
1. The orbitals filled in period n are ns, (n − 2)f, (n − 1)d and np.
2. The sixth energy level is never completely filled in any period.
Assertion (A): The elements having 1s2, 2s2, 2p6, 3s2 and 1s2, 2s2 electronic configuration belong to same period.
Reason (R): Both have same outermost electronic configuration.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. 1s2 2s2 2p6 3s2 belongs to the 3rd period, and 1s2 2s2 belongs to the 2nd period.
2. Both have the same outer configuration ns2, so they are in the same group but in different periods.
3. So the Reason is true but the Assertion is false.
Teacher's Note:
1. Same outer configuration means same group.
2. Period = highest principal quantum number.
Case Study 5: Read the passage given below and answer the questions that follow.
Elements in groups 13 (i.e. group IIIA) to 17 (i.e. group VIIA) of the periodic table, together with group 18 (the zero group elements), make up the p-block of the periodic table.
In the elements of the p-block, the last electron enters the outermost p-orbital. They have 3 to 8 electrons in the outermost shell. As there are three p-orbitals, the maximum number of electrons that can be accommodated in a set of p-orbitals is six. Consequently, there are six groups of p-block elements in the periodic table.
In the p-block, all three kinds of elements are found, i.e. metals, non-metals and metalloids. The zig-zag line in the p-block separates the elements that are metals from those that are non-metals. Metals are found on the left of the line and non-metals on the right. Along the line, we find the metalloids. Because of the presence of a wide range of elements, the p-block shows a great deal of variety in properties.
The general valence shell electronic configuration of p-block elements is ns2np1–6 (with the exception of He). The inner core of the electronic configuration may, however, differ.
Assertion (A): The ionisation of s-electrons requires more energy than that for the ionisation of p-electrons of the same shell.
Reason (R): s-electrons are closer to nucleus than p-electrons and hence are more strongly attracted by nucleus.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. s-electrons penetrate closer to the nucleus than p-electrons of the same shell, so they are more strongly attracted by the nucleus.
2. Hence, more energy is needed to remove an s-electron than a p-electron of the same shell.
3. Both A and R are true, and R is the correct explanation of A.
Teacher's Note:
1. Penetration and attraction follow the order s > p > d > f.
2. This explains why IE of B < Be and Al < Mg.
Assertion (A): The first ionisation energy of Al is lower than that of Mg.
Reason (R): Ionic radius of Al is smaller than that of Mg.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. The first IE of Al (577 kJ/mol) is lower than that of Mg (737 kJ/mol) because in Al the electron is removed from 3p1, whereas in Mg it is removed from the fully filled, more penetrating 3s2 subshell.
2. The ionic radius of Al3+ is indeed smaller than that of Mg2+, so the Reason is true, but it does not explain the lower first IE of Al.
3. Both A and R are true, but R is not the correct explanation of A.
Teacher's Note:
1. First IE concerns the neutral atom, so ionic radius is not relevant to it.
2. Removing a 3p electron is easier than removing a 3s electron.
Assertion (A): F is more electronegative than Cl.
Reason (R): F has higher electron affinity than Cl.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. F is the most electronegative element (4.0) and is more electronegative than Cl (3.0), so the Assertion is true.
2. F has a lower (less negative) electron affinity than Cl because of strong electron–electron repulsion in its small 2p subshell, so the Reason is false.
3. The correct reason is that F has a greater tendency than Cl to attract the shared pair of electrons of a covalent bond, owing to its smaller size.
Teacher's Note:
1. Electronegativity and electron affinity do not always follow the same order.
2. Cl has the highest electron affinity; F has the highest electronegativity.
Assertion (A): Boron differs from Al and other members of group 13 in a number of properties.
Reason (R): B does not show anomalous behaviour.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. Boron, the first member of group 13, differs from Al and the other members because of its small size, high electronegativity and absence of d-orbitals, so the Assertion is true.
2. This difference is exactly anomalous behaviour, so the statement that B does not show anomalous behaviour is false.
Teacher's Note:
1. The first element of each group shows anomalous behaviour.
2. B is a non-metal, while the other members of group 13 are metals.
Assertion (A): Carbon forms a large number of stable compounds.
Reason (R): Carbon is less electropositive as compared to other members of group 14.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. Carbon forms a very large number of stable compounds because of its strong tendency for catenation (strong C–C bonds) and its ability to form multiple bonds.
2. Carbon is indeed the least electropositive (most electronegative) member of group 14, so the Reason is also true.
3. But its electronegativity does not explain why carbon forms so many compounds. Both A and R are true, but R is not the correct explanation of A.
Teacher's Note:
1. Catenation and multiple bonding (pπ–pπ) are the reasons for the huge number of carbon compounds.
2. ‘Less electropositive’ means the same as ‘more electronegative’.
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