1. Which statement correctly expresses the Modern Periodic Law?
ⓐ. The physical and chemical properties of elements are periodic functions of their atomic numbers.
ⓑ. The physical and chemical properties of elements are periodic functions of their atomic masses.
ⓒ. The physical and chemical properties of elements are periodic functions of their neutron numbers.
ⓓ. The physical and chemical properties of elements are periodic functions of their mass numbers.
Correct Answer: The physical and chemical properties of elements are periodic functions of their atomic numbers.
Explanation: The Modern Periodic Law establishes atomic number as the fundamental basis of periodic classification.
Atomic number represents the number of protons present in the nucleus of an atom.
For a neutral atom, it also determines the total number of electrons.
As atomic number increases, electrons occupy available orbitals according to the principles governing electronic configuration.
The resulting arrangements of outer electrons show recurring patterns.
These recurring valence-shell configurations produce similarities in the chemical behaviour of elements at different positions in the periodic table.
Consequently, many physical and chemical properties exhibit periodic variations rather than changing continuously in one direction throughout the table.
Atomic mass was the principal basis of Mendeleev's earlier periodic law, but it is not the governing variable in the modern law.
The essential distinction is that atomic number determines elemental identity and electronic structure, whereas atomic mass does not independently determine the periodic arrangement.
2. An ion \(X^-\) contains 17 protons, 20 neutrons and 18 electrons. Which statement about the corresponding neutral atom \(X\) is correct?
ⓐ. Its atomic number is 18 and it contains 18 electrons.
ⓑ. Its atomic number is 17 and it contains 17 electrons.
ⓒ. Its atomic number is 17 and it contains 18 electrons.
ⓓ. Its atomic number is 18 and it contains 17 electrons.
Correct Answer: Its atomic number is 17 and it contains 17 electrons.
Explanation: \(\textbf{Nuclear identity:}\)
Atomic number is determined exclusively by the number of protons in the nucleus.
The given ion contains 17 protons, so its atomic number is 17.
The presence of 20 neutrons gives the mass number \(17+20=37\), but does not alter the atomic number.
\(\textbf{Effect of electron gain:}\)
The ion contains 18 electrons because the neutral atom has gained one electron.
For a neutral atom, the electron count equals the proton count.
Therefore, the corresponding neutral atom contains 17 electrons.
Its electronic configuration is \(1s^2\,2s^2\,2p^6\,3s^2\,3p^5\).
The extra electron in \(X^-\) does not convert the element into the element having atomic number 18.
Formation of an ordinary ion changes electronic population, not nuclear charge or elemental identity.
Thus, periodic classification is based on the proton number of the element rather than the temporary electron count of its ion.
3. Argon and potassium have the following approximate atomic masses and atomic numbers:
| Element | Atomic number | Atomic mass |
|---|
| Argon | 18 | 39.95 |
| Potassium | 19 | 39.10 |
Which arrangement and justification are consistent with the Modern Periodic Law?
ⓐ. Potassium precedes argon because potassium has the smaller atomic mass.
ⓑ. Argon precedes potassium because argon has the smaller atomic number.
ⓒ. Potassium precedes argon because potassium has the larger atomic number.
ⓓ. Argon precedes potassium because argon has the larger atomic mass.
Correct Answer: Argon precedes potassium because argon has the smaller atomic number.
Explanation: \(\textbf{Selecting the governing property:}\)
The Modern Periodic Law requires elements to be arranged in increasing order of atomic number.
Atomic mass is not the deciding criterion when atomic-mass order and atomic-number order disagree.
Argon has atomic number 18, whereas potassium has atomic number 19.
Therefore, argon must precede potassium in the modern arrangement.
\(\textbf{Electronic interpretation:}\)
Argon has the electronic configuration \(1s^2\,2s^2\,2p^6\,3s^2\,3p^6\).
Its third shell is complete, giving it a stable noble-gas electronic configuration.
Potassium contains one additional proton and, in its neutral atom, one additional electron.
That electron enters the \(4s\) orbital, producing the configuration \([\mathrm{Ar}]\,4s^1\).
Thus, potassium begins a new period despite having a slightly smaller atomic mass than argon.
\(\textbf{Final inference:}\)
The change from a completed \(n=3\) shell to occupation of the \(n=4\) shell is consistent with the atomic-number sequence.
An atomic-mass-based ordering would reverse these two elements and fail to represent their correct modern periodic positions.
4. Two neutral elements \(P\) and \(Q\) have atomic numbers 4 and 12, respectively. Which conclusion follows from their electronic configurations?
ⓐ. They belong to the same period but different groups.
ⓑ. They belong to different groups and different periods.
ⓒ. They belong to the same group but different periods.
ⓓ. They belong to the same group and the same period.
Correct Answer: They belong to the same group but different periods.
Explanation: The atomic number of \(P\) is 4, so its electronic configuration is \(1s^2\,2s^2\).
Its outermost occupied shell has principal quantum number \(n=2\).
Therefore, \(P\) belongs to the second period.
The atomic number of \(Q\) is 12, giving the configuration \(1s^2\,2s^2\,2p^6\,3s^2\).
Its outermost occupied shell has principal quantum number \(n=3\), placing it in the third period.
Both configurations terminate in \(ns^2\), where \(n\) represents the outermost shell.
This common valence-shell pattern is characteristic of group 2 elements.
Therefore, the two elements occupy the same group but different periods.
Their similar outer electronic configurations explain their broad similarity in chemical behaviour.
However, the presence of different numbers of occupied shells means that their atomic sizes and other properties need not have identical values.
The question illustrates how electronic configuration simultaneously explains group similarity and period difference.
5. A species \(P^+\) contains 10 electrons, while a neutral atom \(Q\) also contains 10 electrons. The nucleus of \(P^+\) contains 11 protons, whereas that of \(Q\) contains 10 protons. Which conclusion is correct?
ⓐ. \(P\) and \(Q\) have identical atomic numbers because their electron counts are equal.
ⓑ. \(P\) and \(Q\) occupy the same periodic position because they are isoelectronic.
ⓒ. \(P\) and \(Q\) belong to the same element because their outer shells are complete.
ⓓ. \(P\) and \(Q\) occupy different periodic positions despite the equal electron counts of the given species.
Correct Answer: \(P\) and \(Q\) occupy different periodic positions despite the equal electron counts of the given species.
Explanation: \(\textbf{Identifying the species:}\)
An ion containing 11 protons has atomic number 11, irrespective of its electron count.
Its neutral atom contains 11 electrons, whereas \(P^+\) contains only 10 because one electron has been removed.
The neutral atom \(Q\), having 10 protons, has atomic number 10 and contains 10 electrons.
Consequently, \(P^+\) and \(Q\) are isoelectronic species.
Both possess the electronic configuration \(1s^2\,2s^2\,2p^6\).
\(\textbf{Periodic-position distinction:}\)
The modern periodic table assigns positions using atomic numbers, not the number of electrons remaining after ion formation.
Neutral \(P\) has configuration \([\mathrm{Ne}]\,3s^1\), whereas neutral \(Q\) has configuration \(1s^2\,2s^2\,2p^6\).
Their neutral atoms therefore have different valence-shell configurations.
Neutral \(P\) belongs to group 1 and period 3, while \(Q\) belongs to group 18 and period 2.
Electron removal can make two species isoelectronic without making their parent elements identical.
Thus, equality of electronic configurations for an ion and an atom does not establish equality of periodic positions.
6. Three nuclides are represented as follows:
\[
P={}^{40}_{18}X,\qquad Q={}^{40}_{19}Y,\qquad R={}^{41}_{19}Y
\]
Which statement correctly identifies their relationships and periodic positions?
ⓐ. \(P\) and \(Q\) are isotopes, while \(Q\) and \(R\) occupy different periodic positions.
ⓑ. \(P\) and \(Q\) are isobars, while \(Q\) and \(R\) occupy the same periodic position.
ⓒ. \(P\) and \(R\) are isotopes, while \(P\) and \(Q\) occupy the same periodic position.
ⓓ. \(P\) and \(Q\) occupy the same periodic position, while \(Q\) and \(R\) are isobars.
Correct Answer: \(P\) and \(Q\) are isobars, while \(Q\) and \(R\) occupy the same periodic position.
Explanation: \(\textbf{Comparing nuclear compositions:}\)
The notation \({}^{A}_{Z}X\) specifies mass number \(A\) and atomic number \(Z\).
Nuclides \(P\) and \(Q\) both have mass number 40 but possess different atomic numbers, 18 and 19.
They are therefore isobars rather than isotopes.
Nuclides \(Q\) and \(R\) both have atomic number 19 but possess mass numbers 40 and 41.
They are isotopes of the same element.
\(\textbf{Applying periodic classification:}\)
Since atomic number determines the identity and periodic position of an element, \(Q\) and \(R\) must occupy the same position.
Their neutral atoms contain the same number of electrons and have the same electronic configuration.
Their different neutron numbers do not change their group or period.
In contrast, \(P\) and \(Q\) are distinct elements with different electronic configurations.
The equal mass numbers of \(P\) and \(Q\) cannot make them occupy the same periodic position.
This establishes why classification based on nuclear charge is more fundamental than classification based on mass number.
The correct relationship is therefore the isobaric pairing \(P,Q\) and the common periodic position of \(Q,R\).
7. Four elements have atomic numbers 3, 11, 19 and 37. They all belong to the same group of the modern periodic table. What is the most appropriate inference from their atomic-number sequence?
ⓐ. Their atomic-number differences are constant because their valence configurations are similar.
ⓑ. Their atomic-number differences are unequal because their valence configurations are different.
ⓒ. Their atomic-number differences are constant because each successive element begins a new period.
ⓓ. Their atomic-number differences are unequal although their valence configurations follow the same pattern.
Correct Answer: Their atomic-number differences are unequal although their valence configurations follow the same pattern.
Explanation: \(\textbf{Recognising the recurring configuration:}\)
The elements with atomic numbers 3, 11, 19 and 37 are lithium, sodium, potassium and rubidium.
Their respective outer configurations are \(2s^1\), \(3s^1\), \(4s^1\) and \(5s^1\).
Thus, each possesses the characteristic \(ns^1\) valence-shell configuration.
This recurring outer electronic pattern accounts for their placement in group 1.
\(\textbf{Examining atomic-number intervals:}\)
The successive atomic-number differences are:
\[
11-3=8,\qquad 19-11=8,\qquad 37-19=18
\]
The final difference is larger because the fourth period includes filling of the \(3d\) subshell before the next group 1 element appears.
Consequently, successive members of one family need not be separated by identical atomic-number intervals.
Their similar chemical behaviour arises from recurring valence configurations, not from a fixed numerical separation between atomic numbers.
\(\textbf{Final inference:}\)
Periodicity means recurrence of characteristic electronic arrangements and corresponding properties.
It does not mean that the same pattern must reappear after an unchanging number of elements throughout the table.
The given sequence demonstrates how unequal period lengths coexist with recurring group behaviour.
8. The electronic arrangements of four neutral main-group elements are given below:
| Element | Shell-wise electronic arrangement |
|---|
| \(P\) | 2, 8, 1 |
| \(Q\) | 2, 8, 7 |
| \(R\) | 2, 8, 8, 1 |
| \(S\) | 2, 8, 18, 7 |
Which pairing correctly identifies two different groups, each represented by elements in successive periods?
ⓐ. \(P,Q\) and \(R,S\)
ⓑ. \(P,S\) and \(Q,R\)
ⓒ. \(P,R\) and \(Q,S\)
ⓓ. \(P,R\) and \(Q,R\)
Correct Answer: \(P,R\) and \(Q,S\)
Explanation: \(\textbf{Identifying the valence patterns:}\)
Element \(P\) has one electron in its outermost shell.
Element \(R\) also has one outermost electron, although its electron occupies the fourth shell rather than the third.
Thus, \(P\) and \(R\) possess related outer \(ns^1\) configurations and belong to group 1.
Element \(Q\) has seven electrons in its outermost shell.
Element \(S\) also has seven outermost electrons, giving both elements the characteristic \(ns^2np^5\) configuration.
Therefore, \(Q\) and \(S\) belong to group 17.
\(\textbf{Checking the periods:}\)
The highest occupied principal quantum number is 3 for both \(P\) and \(Q\).
The highest occupied principal quantum number is 4 for both \(R\) and \(S\).
Hence, \(P,R\) occupy successive periods within group 1, while \(Q,S\) occupy successive periods within group 17.
\(\textbf{Why the distinction matters:}\)
Elements in the same period can have very different valence configurations and chemical behaviour.
Conversely, elements in different periods may exhibit related chemical behaviour when their outer electronic configurations recur.
The correct pairing therefore requires simultaneous consideration of valence-electron pattern and occupied-shell number.
9. Consider the following Assertion and Reason.
Assertion (A): Two elements belonging to the same group may exhibit similar chemical behaviour without having identical numerical values of all their physical properties.
Reason (R): Elements of the same group have related valence-shell electronic configurations, but their numbers of occupied shells and inner-electron arrangements may differ.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains Assertion.
ⓑ. Both Assertion and Reason are true, but Reason does not correctly explain Assertion.
ⓒ. Assertion is true, but Reason is false.
ⓓ. Assertion is false, but Reason is true.
Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Explanation: The Assertion is true because periodic similarity does not imply complete equality of properties.
Elements belonging to the same group generally possess similar outer electronic configurations.
This similarity contributes to their characteristic chemical behaviour, including their broad valency patterns and types of compounds formed.
However, their atoms can contain different numbers of occupied electron shells.
Their inner-electron configurations and the shielding of outer electrons can also differ.
These differences influence properties such as atomic size and the attraction experienced by valence electrons.
Consequently, the numerical magnitudes of many physical properties change on descending a group.
The Reason correctly identifies both the origin of chemical similarity and an important source of differences among group members.
It therefore explains why the Assertion is valid.
A common misconception is that periodic recurrence means exact repetition of every physical and chemical property.
In reality, periodic classification groups elements according to recurring patterns of properties, not perfect identity.
Hence, both statements are true, and the Reason provides the correct explanation of the Assertion.
10. Tellurium has a slightly greater atomic mass than iodine, although tellurium has atomic number 52 and iodine has atomic number 53. Mendeleev placed iodine with the halogens rather than allowing strict atomic-mass order to determine every position. Which explanation correctly connects this decision with the modern periodic classification?
ⓐ. Mendeleev followed atomic mass without exception, while modern classification follows neutron number.
ⓑ. Mendeleev used proton number for iodine, while modern classification uses atomic mass for tellurium.
ⓒ. Mendeleev recognised chemical similarity, while modern classification establishes the order through atomic number.
ⓓ. Mendeleev assigned positions through ionic charge, while modern classification uses naturally occurring isotopes.
Correct Answer: Mendeleev recognised chemical similarity, while modern classification establishes the order through atomic number.
Explanation: \(\textbf{Historical classification:}\)
Mendeleev's periodic arrangement was principally based on increasing atomic masses.
However, he recognised that strictly following atomic-mass order did not always preserve the grouping of elements with similar chemical properties.
Iodine resembles the halogen family in its characteristic chemical behaviour.
Placing iodine according to its slightly smaller atomic mass alone would have disturbed its appropriate chemical grouping relative to tellurium.
Mendeleev therefore gave priority to chemical similarities in such exceptional placements.
\(\textbf{Modern explanation:}\)
The modern arrangement uses atomic number as its fundamental ordering variable.
Tellurium has atomic number 52, whereas iodine has atomic number 53.
Accordingly, tellurium naturally precedes iodine in increasing atomic-number order.
Their ground-state outer configurations also explain their different group memberships.
Tellurium has the characteristic group 16 outer configuration \(5s^2\,5p^4\), while iodine has the group 17 configuration \(5s^2\,5p^5\).
\(\textbf{Consequence:}\)
The modern law removes the need to treat the tellurium–iodine order as an exception to its governing numerical sequence.
It simultaneously preserves the chemically meaningful grouping that Mendeleev recognised.
The essential improvement is the replacement of atomic mass by atomic number as the fundamental basis of periodic classification.
11. Two neutral elements have the following ground-state electronic configurations:
\[
P=[\mathrm{Ne}]\,3s^2\,3p^4
\]
\[
Q=[\mathrm{Ar}]\,3d^{10}\,4s^2\,4p^4
\]
Which assignment of their positions in the modern periodic table is correct?
ⓐ. \(P\): Period 3, Group 14; \(Q\): Period 4, Group 14
ⓑ. \(P\): Period 3, Group 16; \(Q\): Period 3, Group 16
ⓒ. \(P\): Period 4, Group 16; \(Q\): Period 4, Group 16
ⓓ. \(P\): Period 3, Group 16; \(Q\): Period 4, Group 16
Correct Answer: \(P\): Period 3, Group 16; \(Q\): Period 4, Group 16
Explanation: \(\textbf{Position of the first element:}\)
For \(P\), the highest occupied principal quantum number is \(n=3\).
Therefore, \(P\) belongs to the third period.
Its valence-shell configuration is \(3s^2\,3p^4\), giving six electrons in the outermost shell.
For an ordinary p-block element, the configuration \(ns^2np^4\) corresponds to group 16.
Thus, \(P\) belongs to group 16.
\(\textbf{Position of the second element:}\)
For \(Q\), the highest occupied principal quantum number is \(n=4\), so it belongs to the fourth period.
The filled \(3d^{10}\) subshell is part of its inner electronic configuration and does not make the element a third-period element.
Its outermost configuration is \(4s^2\,4p^4\).
This again corresponds to six valence electrons and group 16.
\(\textbf{Combined conclusion:}\)
The two elements therefore share the same group but occupy different periods.
Their related outer configurations explain their broad similarity in chemical behaviour.
The additional filled \(3d\) subshell in \(Q\) affects its detailed electronic structure without changing its group-16 classification.
The decisive procedure is to identify the highest occupied shell for the period and the appropriate valence-shell pattern for the group.
12. Consider two neutral elements \(X\) and \(Y\):
\[
X=[\mathrm{Ne}]\,3s^2\,3p^1
\]
\[
Y=[\mathrm{Ar}]\,3d^{10}\,4s^2\,4p^1
\]
Which statement correctly describes their classification?
ⓐ. Both belong to group 13 and the p-block, but occupy different periods.
ⓑ. Both belong to group 3 and the d-block, but occupy different periods.
ⓒ. \(X\) belongs to the p-block, whereas \(Y\) belongs to the d-block.
ⓓ. \(X\) belongs to group 13, whereas \(Y\) belongs to group 11.
Correct Answer: Both belong to group 13 and the p-block, but occupy different periods.
Explanation: \(\textbf{Determining the group:}\)
The outer electronic configuration of \(X\) is \(3s^2\,3p^1\).
Its outermost shell therefore contains three electrons, giving the group-13 pattern \(ns^2np^1\).
Element \(Y\) has outer electronic configuration \(4s^2\,4p^1\).
This is also the characteristic valence configuration of group 13.
Therefore, both elements belong to the same group.
\(\textbf{Determining the block:}\)
Block classification depends on the subshell associated with the differentiating electron in the ground-state electronic configuration.
Both elements have a partially occupied outer p-subshell and are classified as p-block elements.
The filled \(3d^{10}\) subshell in \(Y\) does not make it a d-block element.
\(\textbf{Comparing periodic positions:}\)
The highest occupied principal quantum number is 3 for \(X\) and 4 for \(Y\).
Thus, \(X\) belongs to period 3 and \(Y\) belongs to period 4.
This example demonstrates why the mere presence of d-electrons is insufficient for identifying an element as a d-block element.
Both the outer configuration and the subshell being filled must be interpreted correctly.
Their related valence-shell configurations explain their common group placement despite different inner electronic structures.
13. The fourth period of the modern periodic table begins with filling of the \(4s\) subshell and ends after filling of the \(4p\) subshell. The \(3d\) subshell is filled between them. How many elements are present in this period?
ⓐ. 8
ⓑ. 18
ⓒ. 32
ⓓ. 16
Correct Answer: 18
Explanation: \(\textbf{Identifying the filling sequence:}\)
The fourth period begins when the \(4s\) subshell starts receiving electrons.
After the \(4s\) subshell, the \(3d\) subshell is filled before the \(4p\) subshell.
The relevant filling sequence is:
\[
4s\ \longrightarrow\ 3d\ \longrightarrow\ 4p
\]
\(\textbf{Subshell capacities:}\)
An s-subshell contains one orbital and can accommodate two electrons.
A d-subshell contains five orbitals and can accommodate ten electrons.
A p-subshell contains three orbitals and can accommodate six electrons.
Therefore, the numbers of electrons accommodated are:
\[
4s:2,\qquad 3d:10,\qquad 4p:6
\]
\(\textbf{Calculating the period length:}\)
Each successive element contributes one additional electron to the ground-state atomic electronic configuration.
Hence, the number of elements in the period equals the total number of electrons accommodated across these subshells.
\[
N=2+10+6
\]
\[
N=18
\]
\(\textbf{Consistency check:}\)
The expression \(2n^2\) gives the maximum electron capacity of a complete shell with principal quantum number \(n\).
Using \(n=4\) to obtain 32 would not give the actual length of the fourth period.
The fourth period involves filling \(4s\), \(3d\) and \(4p\), not all possible subshells of the fourth shell.
Its length is therefore 18 elements.
\(\textbf{Final answer:}\)
\[
\boxed{18\ \text{elements}}
\]
14. An element has the ground-state electronic configuration \([\mathrm{Ar}]\,3d^6\,4s^2\). Which assignment correctly identifies its period and group in the modern periodic table?
ⓐ. Period 3, Group 8
ⓑ. Period 4, Group 6
ⓒ. Period 4, Group 8
ⓓ. Period 3, Group 6
Correct Answer: Period 4, Group 8
Explanation: \(\textbf{Determining the period:}\)
The configuration contains electrons in both the \(3d\) and \(4s\) subshells.
The highest occupied principal quantum number is 4 because the \(4s\) subshell is occupied.
Therefore, the element belongs to the fourth period.
The fact that its d-electrons occupy a subshell with \(n=3\) does not make it a third-period element.
\(\textbf{Determining the group:}\)
The element has an incomplete \(3d\) subshell and belongs to the d-block.
For the usual d-block group assignment, electrons in the relevant \((n-1)d\) and \(ns\) subshells are counted together.
Here, the subshell populations are \(3d^6\) and \(4s^2\).
The relevant total is:
\[
6+2=8
\]
Thus, the element belongs to group 8.
\(\textbf{Checking the identification:}\)
The configuration corresponds to iron, whose atomic number is 26.
Its periodic position is fourth period, eighth group.
A common error is to use the principal quantum number of the d-subshell to assign the period.
Another error is to count only the two \(4s\) electrons when assigning a d-block group.
Both errors are avoided by treating period identification and group identification as distinct operations.
The correct assignment is therefore period 4 and group 8.
15. Hydrogen has electronic configuration \(1s^1\), while helium has electronic configuration \(1s^2\). Which conclusion best explains their exceptional positions in periodic classification?
ⓐ. Both belong to group 1 because their electrons occupy the \(1s\) subshell.
ⓑ. Helium belongs to group 2, while hydrogen belongs exclusively to group 17.
ⓒ. Both belong to group 18 because neither atom contains p-electrons.
ⓓ. Helium exhibits noble-gas behaviour, while hydrogen cannot be treated as an ordinary alkali metal.
Correct Answer: Helium exhibits noble-gas behaviour, while hydrogen cannot be treated as an ordinary alkali metal.
Explanation: Hydrogen has a single electron in the \(1s\) subshell and therefore resembles group 1 elements in having an \(ns^1\) configuration.
However, hydrogen is a non-metal and displays chemical behaviour that differs significantly from that of typical alkali metals.
It can lose an electron in suitable chemical situations, showing one aspect of group 1 similarity.
It can also gain an electron to achieve the helium electronic configuration, showing another similarity with group 17 elements.
Consequently, its periodic position requires special consideration rather than classification by a single superficial resemblance.
Helium contains two electrons in the \(1s\) subshell, completely filling the first principal shell.
Although its configuration is of the \(1s^2\) type, helium is placed in group 18.
Its closed electronic shell and characteristic noble-gas behaviour justify this placement.
The remaining noble gases generally possess outer configurations \(ns^2np^6\), but helium does not require p-electrons to complete its first shell.
Thus, identical subshell letters in configurations do not always imply identical group placement.
Hydrogen and helium illustrate the importance of interpreting electronic configuration together with characteristic chemical behaviour.
Their exceptional positions do not contradict the Modern Periodic Law; they reflect the special electronic structure of the first period.
16. Three neutral elements have the following ground-state configurations:
\[
P=[\mathrm{Ne}]\,3s^2
\]
\[
Q=[\mathrm{Ar}]\,4s^2
\]
\[
R=[\mathrm{Ar}]\,3d^{10}\,4s^2
\]
A student assigns all three elements to group 2 because each configuration terminates in \(ns^2\). Which corrected group assignment is appropriate?
ⓐ. \(P:2,\ Q:12,\ R:12\)
ⓑ. \(P:2,\ Q:2,\ R:12\)
ⓒ. \(P:2,\ Q:2,\ R:2\)
ⓓ. \(P:12,\ Q:2,\ R:12\)
Correct Answer: \(P:2,\ Q:2,\ R:12\)
Explanation: \(\textbf{Classifying the first two elements:}\)
Element \(P\) has the configuration \([\mathrm{Ne}]\,3s^2\), which corresponds to magnesium.
Its electrons beyond the noble-gas core occupy the \(3s\) subshell.
Magnesium is an s-block element belonging to group 2.
Element \(Q\) has the configuration \([\mathrm{Ar}]\,4s^2\), corresponding to calcium.
It is also an s-block element with the characteristic group-2 configuration \(ns^2\).
\(\textbf{Examining the third element:}\)
Element \(R\) has the configuration \([\mathrm{Ar}]\,3d^{10}\,4s^2\), corresponding to zinc.
Its configuration contains a completely filled \(3d\) subshell in addition to the \(4s^2\) electrons.
Zinc occupies group 12 in the d-block region of the periodic table.
Therefore, its group cannot be assigned simply by observing the final \(ns^2\) component.
\(\textbf{Why the shortcut fails:}\)
For ordinary s-block elements, an outer \(ns^2\) configuration is characteristic of group 2.
For d-block elements, group identification generally requires consideration of the relevant \((n-1)d\) and \(ns\) electron populations.
For zinc, these populations give:
\[
3d^{10}+4s^2\ \Rightarrow\ 10+2=12
\]
The group assignment is therefore 12, not 2.
\(\textbf{Final inference:}\)
The configurations of magnesium, calcium and zinc demonstrate that the same \(ns^2\) component does not guarantee identical group placement across different blocks.
The complete electronic configuration must be interpreted in its appropriate block context.
Hence, \(P\) and \(Q\) belong to group 2, whereas \(R\) belongs to group 12.
17. Consider the following Assertion and Reason.
Assertion (A): The third period contains only eight elements, although the maximum electron capacity of the third principal shell is eighteen.
Reason (R): During the filling of ground-state electronic configurations, the \(4s\) subshell begins to be occupied before the \(3d\) subshell.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains Assertion.
ⓑ. Both Assertion and Reason are true, but Reason does not correctly explain Assertion.
ⓒ. Assertion is true, but Reason is false.
ⓓ. Assertion is false, but Reason is true.
Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
Explanation: The Assertion is true because the maximum electron capacity of a principal shell is not necessarily equal to the number of elements in the corresponding period.
The third principal shell contains the \(3s\), \(3p\) and \(3d\) subshells.
Their maximum electron capacities are 2, 6 and 10, respectively.
Consequently, the third shell can accommodate a maximum of eighteen electrons.
However, the third period begins with sodium and ends with argon.
Across these elements, the \(3s\) and \(3p\) subshells become occupied, accounting for eight elements.
The \(3d\) subshell remains unoccupied throughout the third period.
The Reason is also true because the \(4s\) subshell is occupied before the \(3d\) subshell in the relevant ground-state filling sequence.
After argon, the next electron enters \(4s\), beginning the fourth period with potassium.
Filling of the \(3d\) subshell occurs subsequently within the fourth period.
Thus, the actual length of a period depends on the sequence of subshell occupation rather than the complete theoretical capacity of the shell having the same principal quantum number.
The Reason directly explains why the third period ends after eight elements despite the availability of additional orbitals within the third shell.
Therefore, both statements are true, and the Reason correctly explains the Assertion.
18. Calcium has atomic number 20, while gallium has atomic number 31. Both belong to the fourth period of the modern periodic table. Which option correctly identifies the number of elements lying strictly between them, the subshell predominantly involved in filling across these intervening elements, and their period?
ⓐ. Ten elements; \(3d\) subshell; fourth period
ⓑ. Ten elements; \(4d\) subshell; fourth period
ⓒ. Nine elements; \(3d\) subshell; fourth period
ⓓ. Ten elements; \(3d\) subshell; third period
Correct Answer: Ten elements; \(3d\) subshell; fourth period
Explanation: \(\textbf{Locating the boundary elements:}\)
Calcium has the ground-state electronic configuration:
\[
\mathrm{Ca}:[\mathrm{Ar}]\,4s^2
\]
It occupies group 2 and completes the initial s-block portion of the fourth period.
Gallium has the configuration:
\[
\mathrm{Ga}:[\mathrm{Ar}]\,3d^{10}\,4s^2\,4p^1
\]
Gallium begins the p-block portion of the fourth period.
\(\textbf{Counting the intervening elements:}\)
The atomic numbers lying strictly between 20 and 31 are 21 through 30.
Therefore:
\[
N=31-20-1
\]
\[
N=10
\]
\(\textbf{Identifying the subshell:}\)
These ten positions are occupied by scandium through zinc.
They constitute the first d-block series, associated with the progressive occupation of the \(3d\) subshell.
Chromium and copper have recognised ground-state configuration exceptions, but these do not change the number of elements or their d-block positions.
The \(3d\) subshell contains five orbitals and has a maximum capacity of ten electrons.
Although the relevant d-subshell has principal quantum number 3, these elements belong to period 4 because their neutral ground-state configurations occupy the fourth principal shell.
Thus, the intervening region consists of ten fourth-period d-block elements associated with \(3d\) filling.
19. An element \(X\) belongs to group 15 and period 5 of the modern periodic table. Which combination correctly gives its atomic number and ground-state electronic configuration?
ⓐ. \(Z=33;\ [\mathrm{Ar}]\,3d^{10}\,4s^2\,4p^3\)
ⓑ. \(Z=53;\ [\mathrm{Kr}]\,4d^{10}\,5s^2\,5p^5\)
ⓒ. \(Z=51;\ [\mathrm{Kr}]\,4d^{10}\,5s^2\,5p^5\)
ⓓ. \(Z=51;\ [\mathrm{Kr}]\,4d^{10}\,5s^2\,5p^3\)
Correct Answer: \(Z=51;\ [\mathrm{Kr}]\,4d^{10}\,5s^2\,5p^3\)
Explanation: \(\textbf{Reconstructing the outer configuration:}\)
An ordinary group-15 element belongs to the p-block and has the characteristic valence-shell configuration \(ns^2np^3\).
Since the given element belongs to period 5, its outermost principal quantum number is \(n=5\).
Therefore, its outer configuration must be:
\[
5s^2\,5p^3
\]
\(\textbf{Accounting for the intervening d-subshell:}\)
The preceding noble gas is krypton, which has atomic number 36.
Between the beginning of period 5 and the group-15 position, the \(4d\) subshell becomes completely occupied.
Thus, the relevant complete configuration is:
\[
[\mathrm{Kr}]\,4d^{10}\,5s^2\,5p^3
\]
\(\textbf{Determining the atomic number:}\)
For a neutral atom, the total number of electrons equals its atomic number.
The krypton core contributes 36 electrons.
The additional contributions are 10 electrons from \(4d\), 2 from \(5s\) and 3 from \(5p\).
Hence:
\[
Z=36+10+2+3
\]
\[
Z=51
\]
\(\textbf{Checking the periodic position:}\)
The highest occupied principal quantum number is 5, confirming period 5.
The outer \(5s^2\,5p^3\) arrangement confirms group 15.
The differentiating electron is associated with the p-subshell, confirming p-block membership.
The element is antimony, and the group, period, electronic configuration and atomic number are mutually consistent.
20. Four elements are assigned the periodic positions shown below.
| Element | Atomic number | Period | Group | Block |
|---|
| \(P\) | 12 | 3 | 2 | s |
| \(Q\) | 17 | 3 | 17 | p |
| \(R\) | 30 | 4 | 12 | d |
| \(S\) | 36 | 5 | 18 | p |
Which element has an incorrectly assigned periodic position, even though its stated group and block are correct?
ⓐ. Element \(P\)
ⓑ. Element \(Q\)
ⓒ. Element \(R\)
ⓓ. Element \(S\)
Correct Answer: Element \(S\)
Explanation: \(\textbf{Checking the first two assignments:}\)
Element \(P\) has atomic number 12 and the configuration \([\mathrm{Ne}]\,3s^2\).
It belongs to the third period, group 2 and the s-block, so its complete assignment is correct.
Element \(Q\) has atomic number 17 and the configuration \([\mathrm{Ne}]\,3s^2\,3p^5\).
It belongs to the third period, group 17 and the p-block, so its assignment is also correct.
\(\textbf{Examining the d-block entry:}\)
Element \(R\) has atomic number 30 and the configuration \([\mathrm{Ar}]\,3d^{10}\,4s^2\).
It is zinc, occupying group 12 in the fourth-period d-block region.
A completely filled \(3d\) subshell does not remove zinc from its established d-block position.
Therefore, the assignment of \(R\) is correct.
\(\textbf{Testing the remaining position:}\)
Element \(S\) has atomic number 36 and the configuration:
\[
[\mathrm{Ar}]\,3d^{10}\,4s^2\,4p^6
\]
This is krypton, whose highest occupied principal quantum number is 4.
Krypton belongs to group 18 because it has a completed outer \(4s^2\,4p^6\) configuration.
It belongs to the p-block and completes the fourth period.
Therefore, its group and block are correctly identified, but its stated period number is wrong.
This example shows why a periodic-position assignment must simultaneously satisfy atomic number, highest occupied shell and block-specific electronic configuration.
The incorrect assignment is that of element \(S\).
