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Chapter 3 Classification Of Elements And Periodicity In Properties
The Periodic Table is considered a cornerstone concept in chemistry. It provides a structured framework for organising the chemical elements, revealing patterns and trends in their properties, which is invaluable for both learning and research. It demonstrates that the elements are not a random collection but exhibit relationships and form families based on their characteristics.
Why Do We Need To Classify Elements ?
Elements are the fundamental building blocks of all matter. In the early 19th century (around 1800), only about 31 elements were known. By the mid-19th century (1865), this number had doubled to 63. Currently, over 118 elements have been identified, with many of the heavier ones being synthesised artificially. Studying the chemistry of each element and the vast number of compounds they form individually is incredibly challenging.
To manage this complexity, scientists sought a systematic method to classify elements based on their properties. Such classification would not only help organise existing knowledge about elements but also potentially predict the properties of new or undiscovered ones, guiding further research.
Genesis Of Periodic Classification
The development of the Periodic Table was a gradual process, built upon the efforts of many scientists who observed and systematised the properties of known elements.
- In the early 1800s, German chemist Johann Dobereiner noticed similarities in the properties of certain groups of three elements, which he called Triads. He observed that the atomic weight of the middle element in a triad was often approximately the average of the atomic weights of the other two, and its properties were intermediate.
Element | Atomic weight | Element | Atomic weight | Element | Atomic weight |
---|---|---|---|---|---|
Li | 7 | Ca | 40 | Cl | 35.5 |
Na | 23 | Sr | 88 | Br | 80 |
K | 39 | Ba | 137 | I | 127 |
However, this "Law of Triads" only worked for a limited number of elements and was not universally accepted.
- In 1862, French geologist A.E.B. de Chancourtois arranged elements by increasing atomic weight on a cylinder (the 'telluric helix'). He noted a periodic recurrence of properties, but his work didn't gain widespread attention.
- In 1865, English chemist John Alexander Newlands proposed the Law of Octaves. He arranged elements in increasing order of atomic weight and observed that the properties of every eighth element were similar to the first, like notes in a musical octave.
Element | Li | Be | B | C | N | O | F |
At. wt. | 7 | 9 | 11 | 12 | 14 | 16 | 19 |
Element | Na | Mg | Al | Si | P | S | Cl |
At. wt. | 23 | 24 | 27 | 29 | 31 | 32 | 35.5 |
Element | K | Ca | |||||
At. wt. | 39 | 40 |
Newlands' law worked reasonably well only for the lighter elements, up to calcium, and wasn't accepted by the scientific community at the time, although he was later recognised for his contribution.
The most significant steps towards the modern Periodic Table were taken by Russian chemist Dmitri Mendeleev and German chemist Lothar Meyer, working independently around 1869. Both found that when elements were arranged by increasing atomic weight, similar physical and chemical properties recurred periodically. Lothar Meyer plotted properties like atomic volume and melting point against atomic weight, showing periodic patterns and noticing variations in the length of these repeating periods. His table from 1868 closely resembled the modern one but was published after Mendeleev's work.
Mendeleev is widely credited with publishing the first clear statement of the Periodic Law:
The properties of the elements are a periodic function of their atomic weights.
Mendeleev arranged elements in rows (periods) and columns (groups) based on increasing atomic weight, placing elements with similar properties in the same vertical column. He used a broader range of properties than Meyer, including the formulas of oxides and hydrides, recognising the fundamental importance of periodicity.
Crucially, Mendeleev did not strictly adhere to the increasing order of atomic weight when it contradicted the placement based on chemical properties. He believed the atomic weight measurements might be inaccurate (which was true in some cases, like for Iodine and Tellurium). He also bravely left gaps in his table, predicting the existence and properties of undiscovered elements that would fit into those positions. For example, he predicted elements he called Eka-Aluminium and Eka-Silicon, later discovered as Gallium and Germanium, whose properties closely matched his predictions.
Property | Eka-aluminium (predicted) | Gallium (found) | Eka-silicon (predicted) | Germanium (found) |
Atomic weight | 68 | 70 | 72 | 72.6 |
Density / (g/cm$^3$) | 5.9 | 5.94 | 5.5 | 5.36 |
Melting point /K | Low | 302.93 | High | 1231 |
Formula of oxide | E$_2$O$_3$ | Ga$_2$O$_3$ | EO$_2$ | GeO$_2$ |
Formula of chloride | ECl$_3$ | GaCl$_3$ | ECl$_4$ | GeCl$_4$ |
Mendeleev's successful predictions solidified the importance of his Periodic Table.
Modern Periodic Law And The Present Form Of The Periodic Table
At the time of Mendeleev, the internal structure of atoms was unknown. Developments in the early 20th century, particularly by English physicist Henry Moseley in 1913, provided a more fundamental basis for the periodic classification. Moseley studied the characteristic X-ray spectra of elements and found that the square root of the frequency of emitted X-rays ($\sqrt{\nu}$) plotted against the atomic number ($Z$) yielded a straight line, whereas a plot against atomic mass did not. This showed that the atomic number is a more fundamental property of an element than its atomic mass.
This led to the refinement of the Periodic Law, now known as the Modern Periodic Law:
The physical and chemical properties of the elements are periodic functions of their atomic numbers.
The atomic number ($Z$) represents the number of protons in the nucleus and also the number of electrons in a neutral atom. Since chemical properties are determined primarily by the electronic configuration, the periodic variation in properties is a direct consequence of the periodic variation in electronic configurations as atomic number increases.
The most widely used form today is the "long form" of the Periodic Table. It organises elements in horizontal rows called periods and vertical columns called groups. Elements in the same group have similar outer electronic configurations and hence similar properties.
According to IUPAC recommendations, groups are numbered 1 through 18, replacing older notations like IA-VIIA, VIII, IB-VIIB, and 0.
There are seven periods. The period number corresponds to the highest principal quantum number ($n$) of the valence shell being filled in that period. The number of elements in each period corresponds to the electron capacity of the orbitals being filled:
- Period 1 ($n=1$): Fills 1s (1 orbital). Maximum 2 electrons. 2 elements (H, He).
- Period 2 ($n=2$): Fills 2s, 2p (1+3=4 orbitals). Maximum 8 electrons. 8 elements (Li to Ne).
- Period 3 ($n=3$): Fills 3s, 3p (1+3=4 orbitals). Maximum 8 electrons. 8 elements (Na to Ar).
- Period 4 ($n=4$): Fills 4s, 3d, 4p (1+5+3=9 orbitals). Maximum 18 electrons. 18 elements (K to Kr). (Note the filling order 4s before 3d).
- Period 5 ($n=5$): Fills 5s, 4d, 5p (1+5+3=9 orbitals). Maximum 18 electrons. 18 elements (Rb to Xe). (Note the filling order 5s before 4d).
- Period 6 ($n=6$): Fills 6s, 4f, 5d, 6p (1+7+5+3=16 orbitals). Maximum 32 electrons. 32 elements (Cs to Rn). The 14 elements filling the 4f orbitals (Lanthanoids) are placed separately below the main table.
- Period 7 ($n=7$): Fills 7s, 5f, 6d, 7p (1+7+5+3=16 orbitals). Maximum 32 electrons. This period is incomplete but contains 32 elements discovered so far, including the Actinoids (filling 5f orbitals), placed separately below the main table.
Lanthanoids and Actinoids are placed in separate panels at the bottom to keep the main body of the Periodic Table manageable and to group elements with very similar properties (within each series) together.
Nomenclature Of Elements With Atomic Numbers > 100
Traditionally, the discoverers of a new element had the privilege of naming it, which was then ratified by IUPAC. However, for elements with atomic numbers greater than 100, synthesised in different labs, competing claims for discovery led to disputes over naming (e.g., element 104 claimed by both American and Soviet scientists). To avoid this, IUPAC established a systematic temporary nomenclature for elements before their discovery is fully verified and a permanent name is agreed upon.
The temporary name is derived directly from the atomic number using numerical roots (Table 3.4) and adding the ending "-ium". The roots are combined in the order of the digits in the atomic number.
Digit | Name | Abbreviation |
0 | nil | n |
1 | un | u |
2 | bi | b |
3 | tri | t |
4 | quad | q |
5 | pent | p |
6 | hex | h |
7 | sept | s |
8 | oct | o |
9 | enn | e |
The temporary symbol is derived from the abbreviations of the roots, using the first letter of each root.
Examples of IUPAC temporary names and symbols for elements with Z > 100 (Table 3.5 shows both temporary and official names):
Atomic Number | Name according to IUPAC nomenclature | Symbol according to IUPAC nomenclature |
101 | Unnilunium | Unu |
104 | Unnilquadium | Unq |
118 | Ununoctium | Uuo |
Once the discovery is verified, a permanent name and one- or two-letter symbol are assigned, often honouring a scientist or place of discovery. Official names have been announced for all elements up to Z=118 (Oganesson).
Problem 3.1. What would be the IUPAC name and symbol for the element with atomic number 120?
Answer:
The atomic number is 120.
Break down the atomic number into its digits: 1, 2, 0.
Find the corresponding numerical roots from Table 3.4:
- Digit 1: root "un", abbreviation "u"
- Digit 2: root "bi", abbreviation "b"
- Digit 0: root "nil", abbreviation "n"
Combine the roots and add "-ium" at the end for the name:
un + bi + nil + ium = Unbinilium
Combine the abbreviations for the symbol:
u + b + n = Ubn
The IUPAC temporary name for the element with atomic number 120 is Unbinilium, and its symbol is Ubn.
Electronic Configurations Of Elements And The Periodic Table
The arrangement of elements in the modern Periodic Table is fundamentally based on their electronic configurations, specifically the quantum numbers of the last electron added to an atom.
(a) Electronic Configurations in Periods
Each period in the Periodic Table corresponds to the filling of electrons into the orbitals of a new principal energy level (shell), indicated by the principal quantum number ($n$) which matches the period number. The number of elements in each period is determined by the total number of electrons that can fill the subshells available in that principal energy level, following the Aufbau principle's energy ordering.
- Period 1 ($n=1$): Fills the 1s orbital. The maximum capacity of the 1s orbital is 2 electrons. Thus, Period 1 has 2 elements (H: $1s^1$, He: $1s^2$).
- Period 2 ($n=2$): Starts filling the 2s orbital, then the 2p orbitals. Orbitals available are 2s (1 orbital) and 2p (3 orbitals). Total orbitals = 4. Maximum capacity = $4 \times 2 = 8$ electrons. Thus, Period 2 has 8 elements (Li: $[He]2s^1$ to Ne: $[He]2s^22p^6$).
- Period 3 ($n=3$): Starts filling the 3s orbital, then the 3p orbitals. Orbitals available are 3s (1 orbital) and 3p (3 orbitals). Total orbitals = 4. Maximum capacity = $4 \times 2 = 8$ electrons. Thus, Period 3 has 8 elements (Na: $[Ne]3s^1$ to Ar: $[Ne]3s^23p^6$).
- Period 4 ($n=4$): Fills 4s, then 3d, then 4p orbitals. Orbitals filled are 4s (1 orbital), 3d (5 orbitals), and 4p (3 orbitals). Total orbitals = $1+5+3=9$. Maximum capacity = $9 \times 2 = 18$ electrons. Thus, Period 4 has 18 elements (K: $[Ar]4s^1$ to Kr: $[Ar]3d^{10}4p^6$). The filling of 3d orbitals starts from Scandium ($Z=21$) and forms the first transition series.
- Period 5 ($n=5$): Fills 5s, then 4d, then 5p orbitals. Orbitals filled are 5s (1 orbital), 4d (5 orbitals), and 5p (3 orbitals). Total orbitals = $1+5+3=9$. Maximum capacity = $9 \times 2 = 18$ electrons. Thus, Period 5 has 18 elements (Rb: $[Kr]5s^1$ to Xe: $[Kr]4d^{10}5p^6$). The filling of 4d orbitals starts from Yttrium ($Z=39$) and forms the second transition series.
- Period 6 ($n=6$): Fills 6s, then 4f, then 5d, then 6p orbitals. Orbitals filled are 6s (1 orbital), 4f (7 orbitals), 5d (5 orbitals), and 6p (3 orbitals). Total orbitals = $1+7+5+3=16$. Maximum capacity = $16 \times 2 = 32$ electrons. Thus, Period 6 has 32 elements (Cs: $[Xe]6s^1$ to Rn: $[Xe]4f^{14}5d^{10}6p^6$). The filling of 4f orbitals starts from Cerium ($Z=58$) and forms the Lanthanoid series, placed separately.
- Period 7 ($n=7$): Fills 7s, then 5f, then 6d, then 7p orbitals. Orbitals available are 7s (1 orbital), 5f (7 orbitals), 6d (5 orbitals), and 7p (3 orbitals). Total orbitals = $1+7+5+3=16$. Maximum capacity = $16 \times 2 = 32$ electrons. This period includes elements up to $Z=118$. The filling of 5f orbitals starts after Actinium ($Z=89$) and forms the Actinoid series, placed separately.
The placement of Lanthanoids and Actinoids separately is a matter of convenience to keep the main table's width manageable and group elements with very similar properties within their respective series.
Problem 3.2. How would you justify the presence of 18 elements in the 5th period of the Periodic Table?
Answer:
The period number corresponds to the principal quantum number ($n$) of the outermost shell being filled. For the 5th period, $n=5$.
According to quantum mechanics, for $n=5$, the possible values of the azimuthal quantum number ($l$) are $0, 1, 2, \dots, n-1$.
So, for $n=5$, the possible $l$ values are 0, 1, 2, and 3.
These correspond to the 5s, 5p, 5d, and 5f subshells.
However, the order of filling of orbitals is determined by the Aufbau principle and the $(n+l)$ rule. The energy ordering for $n=5$ and involved orbitals from lower shells is:
- 4d: $n=4, l=2 \implies n+l=6$
- 5s: $n=5, l=0 \implies n+l=5$
- 5p: $n=5, l=1 \implies n+l=6$
- 5d: $n=5, l=2 \implies n+l=7$
- 5f: $n=5, l=3 \implies n+l=8$
Comparing $(n+l)$ values, the order of filling is 5s (5), then 4d (6), then 5p (6). For 4d and 5p, $(n+l)$ is 6, so we use the $n$ value; 4d has lower $n$ (4) than 5p (5), so 4d is filled before 5p.
Thus, in the 5th period, the orbitals being filled are the 5s, 4d, and 5p orbitals.
- 5s subshell has $2l+1 = 2(0)+1 = 1$ orbital.
- 4d subshell has $2l+1 = 2(2)+1 = 5$ orbitals.
- 5p subshell has $2l+1 = 2(1)+1 = 3$ orbitals.
The total number of orbitals available for filling in the 5th period are $1 + 5 + 3 = 9$ orbitals (5s, five 4d, three 5p). Each orbital can accommodate a maximum of 2 electrons.
Therefore, the maximum number of electrons that can be accommodated in the 5th period is $9 \text{ orbitals} \times 2 \text{ electrons/orbital} = 18 \text{ electrons}$.
Since each element adds one electron, there are 18 elements in the 5th period of the Periodic Table.
(b) Groupwise Electronic Configurations
Elements within the same vertical column (group) in the Periodic Table share similar chemical properties. This similarity is because they have the same number and distribution of electrons in their outermost (valence) shell orbitals. The general valence shell electronic configuration is the same for all elements in a group.
For example, all Group 1 elements (Alkali Metals) have a valence shell electronic configuration of $ns^1$, where $n$ is the period number:
- Li ($Z=3$): $[He]2s^1$ ($n=2$)
- Na ($Z=11$): $[Ne]3s^1$ ($n=3$)
- K ($Z=19$): $[Ar]4s^1$ ($n=4$)
- Rb ($Z=37$): $[Kr]5s^1$ ($n=5$)
- Cs ($Z=55$): $[Xe]6s^1$ ($n=6$)
- Fr ($Z=87$): $[Rn]7s^1$ ($n=7$)
This consistent outer electronic configuration is the fundamental reason behind the similar chemical behavior within a group. While electronic configuration explains periodicity based on atomic number, chemical properties are the observable manifestation of these electron arrangements.
Electronic Configurations And Types Of Elements: S-, P-, D-, F- Blocks
Based on which subshell the last electron enters, elements in the Periodic Table are classified into four blocks: s-block, p-block, d-block, and f-block. This classification is directly linked to the electronic configurations and largely dictates the chemical behavior of the elements.
There are two notable exceptions to this simple block categorisation:
- Helium (He, $Z=2$): Has the configuration $1s^2$. While it technically belongs to the s-block, its properties are very similar to other noble gases (Group 18) due to its completely filled valence shell. Therefore, it's placed in the p-block.
- Hydrogen (H, $Z=1$): Has the configuration $1s^1$. It shares properties with Group 1 (alkali metals) due to having one valence electron, but also with Group 17 (halogens) because it can gain an electron to achieve a stable configuration ($1s^2$). Its chemistry is unique, and it's often placed separately at the top of the Periodic Table.
- General outer electronic configuration: $ns^1$ (Group 1) and $ns^2$ (Group 2).
- They are all reactive metals.
- They have low ionisation enthalpies (energy required to remove an electron).
- They readily lose their valence electron(s) to form cations: $1^+$ ions (Group 1) and $2^+$ ions (Group 2).
- Their metallic character and reactivity generally increase down the group.
- Due to high reactivity, they are never found in elemental form in nature, only in compounds.
- Their compounds are predominantly ionic, except for some compounds of Lithium (Li) and Beryllium (Be) which show more covalent character due to their small size and high charge density.
- General outer electronic configuration: $ns^2np^{1-6}$ in each period.
- The number of electrons in the p subshell increases from 1 in Group 13 ($ns^2np^1$) to 6 in Group 18 ($ns^2np^6$).
- Group 18 elements are the Noble Gases, having a completely filled valence shell ($ns^2np^6$), which is a very stable configuration. This results in very low chemical reactivity.
- Preceding the noble gases are the Halogens (Group 17, $ns^2np^5$) and the Chalcogens (Group 16, $ns^2np^4$). These are important groups of non-metals.
- Halogens and Chalcogens have highly negative electron gain enthalpies (energy released when gaining an electron) and readily gain 1 or 2 electrons, respectively, to achieve the stable noble gas configuration.
- The non-metallic character increases as you move from left to right across a period in the p-block.
- The metallic character increases as you go down a group in the p-block.
- General outer electronic configuration: $(n-1)d^{1-10}ns^{0-2}$. (Some exceptions exist, like Cr and Cu).
- They are all metals.
- They typically form coloured ions in solution.
- They exhibit variable valency or oxidation states.
- Many are paramagnetic (attracted to a magnetic field) due to unpaired electrons in d orbitals.
- Many d-block elements and their compounds are used as catalysts.
- Elements in Group 12 (Zn, Cd, Hg) have $(n-1)d^{10}ns^2$ configuration. Since their d subshells are completely filled in both their elemental state and common oxidation states ($Zn^{2+}, Cd^{2+}, Hg^{2+}$), they do not exhibit most of the characteristic properties of transition elements. They are often considered post-transition metals or sometimes included as transition elements based on the definition of a 'd-block element'.
- Transition metals act as a bridge between the highly reactive s-block metals and the less reactive p-block elements (Groups 13-14), hence their name "Transition Elements".
- The first series is the Lanthanoids (Ce, $Z=58$ to Lu, $Z=71$), filling the 4f orbitals.
- The second series is the Actinoids (Th, $Z=90$ to Lr, $Z=103$), filling the 5f orbitals.
- They are characterised by the filling of the inner f orbitals, specifically $(n-2)f$ orbitals, before the $(n-1)d$ or $ns$ orbitals are completely filled.
- General outer electronic configuration: $(n-2)f^{1-14}(n-1)d^{0-1}ns^2$.
- They are all metals.
- Elements within each series (especially the Lanthanoids) exhibit very similar properties because the f orbitals are deep inside the atom and screen the outer electrons from the increasing nuclear charge effectively, leading to only a small change in size (Lanthanoid contraction and Actinoid contraction).
- The chemistry of Actinoids is more complex than Lanthanoids due to a wider range of accessible oxidation states.
- All Actinoid elements are radioactive.
- Many Actinoids are synthesised artificially in very small quantities. Elements with $Z > 92$ (Uranium) are called Transuranium Elements.
- [He] ($Z=2$): $1s^2$
- [Ne] ($Z=10$): $[He]2s^22p^6$
- [Ar] ($Z=18$): $[Ne]3s^23p^6$
- [Kr] ($Z=36$): $[Ar]4s^23d^{10}4p^6$
- [Xe] ($Z=54$): $[Kr]5s^24d^{10}5p^6$
- [Rn] ($Z=86$): $[Xe]6s^24f^{14}5d^{10}6p^6$
- [Uuo] ($Z=118$): $[Rn]7s^25f^{14}6d^{10}7p^6$
- Metals:
- Comprise over 78% of all known elements.
- Located on the left side of the Periodic Table.
- Usually solid at room temperature (except Mercury, which is liquid; Gallium and Caesium have low melting points).
- Generally have high melting and boiling points.
- Are good conductors of heat and electricity.
- Are malleable (can be hammered into thin sheets).
- Are ductile (can be drawn into wires).
- Metallic character generally increases down a group and decreases across a period from left to right.
- Non-metals:
- Located on the top right-hand side of the Periodic Table.
- Are usually solids or gases at room temperature (Bromine is the only liquid non-metal).
- Generally have low melting and boiling points (exceptions include Boron and Carbon).
- Are poor conductors of heat and electricity (Graphite, an allotrope of carbon, is an exception).
- Solid non-metals are typically brittle.
- Are neither malleable nor ductile.
- Non-metallic character generally increases across a period from left to right and decreases down a group.
- Metalloids (Semi-metals):
- Elements located along the thick zig-zag line separating metals from non-metals in the p-block (e.g., Silicon (Si), Germanium (Ge), Arsenic (As), Antimony (Sb), Tellurium (Te)). Boron (B) and Polonium (Po) are sometimes also considered metalloids.
- They exhibit properties that are intermediate between those of metals and non-metals. For instance, they are often semiconductors.
- Na ($Z=11$): Group 1, Period 3 (Alkali metal, highly metallic)
- Mg ($Z=12$): Group 2, Period 3 (Alkaline earth metal, metallic)
- Al ($Z=13$): Group 13, Period 3 (Metal)
- Si ($Z=14$): Group 14, Period 3 (Metalloid)
- P ($Z=15$): Group 15, Period 3 (Non-metal)
- Be ($Z=4$): Group 2, Period 2 (Alkaline earth metal, metallic)
- Na is most metallic among these.
- Mg is less metallic than Na, but more metallic than Si and P.
- Be is less metallic than Mg, but more metallic than Si and P (as metals are generally more metallic than metalloids/non-metals).
- Si is a metalloid, more metallic than the non-metal P.
- P is the least metallic (most non-metallic) among these.
The S-Block Elements
These are the elements of Group 1 (Alkali Metals) and Group 2 (Alkaline Earth Metals). The last electron enters the outermost s orbital.
The P-Block Elements
These elements are located on the right side of the Periodic Table, from Group 13 to Group 18. The last electron enters the outermost p orbital. The s-block and p-block elements collectively are called Representative Elements or Main Group Elements.
The D-Block Elements (Transition Elements)
These elements are located in the middle of the Periodic Table, from Group 3 to Group 12. They are characterised by the filling of the inner d orbitals, specifically $(n-1)d$ orbitals, before the outermost $ns$ orbital is completely filled.
The F-Block Elements (Inner-Transition Elements)
These elements form two separate rows placed below the main body of the Periodic Table.
Problem 3.3. The elements Z = 117 and 120 have not yet been discovered. In which family / group would you place these elements and also give the electronic configuration in each case.
Answer:
To determine the group and electronic configuration, we need to consider the period and the block where the last electron would enter, based on the Aufbau principle.
We can extend the filling order: $1s \rightarrow 2s \rightarrow 2p \rightarrow 3s \rightarrow 3p \rightarrow 4s \rightarrow 3d \rightarrow 4p \rightarrow 5s \rightarrow 4d \rightarrow 5p \rightarrow 6s \rightarrow 4f \rightarrow 5d \rightarrow 6p \rightarrow 7s \rightarrow 5f \rightarrow 6d \rightarrow 7p \rightarrow 8s \dots$
Let's find the electronic configuration by filling electrons according to this order up to the given atomic numbers.
Noble gas configurations are useful here:
Element with Z = 117:
This element is just before the noble gas Uuo ($Z=118$). The last electron would be the 7th electron added to the 7p subshell (since 7p can hold up to 6 electrons). Elements with the configuration $ns^2np^5$ belong to Group 17 (Halogen family).
The configuration up to $Z=118$ is $[Rn]7s^25f^{14}6d^{10}7p^6$. To get $Z=117$, we remove one electron from the last filled subshell, which is 7p.
Electronic configuration of $Z=117$: $[Rn]7s^25f^{14}6d^{10}7p^5$.
This outer configuration ($7s^27p^5$) is similar to other Group 17 elements ($ns^2np^5$).
Therefore, the element with $Z=117$ would belong to the Halogen family (Group 17).
Element with Z = 120:
This element is two atomic numbers after Uuo ($Z=118$). After the 7p subshell is filled at $Z=118$, the next orbitals to be filled are 8s (according to the Aufbau principle/$(n+l)$ rule, $8+0=8$, while 7d is $7+2=9$). The first two electrons after $Z=118$ would go into the 8s orbital.
Electronic configuration of $Z=118$: $[Rn]7s^25f^{14}6d^{10}7p^6$.
Electronic configuration of $Z=119$: $[Rn]7s^25f^{14}6d^{10}7p^68s^1$ or $[Uuo]8s^1$. This would be in Group 1.
Electronic configuration of $Z=120$: $[Rn]7s^25f^{14}6d^{10}7p^68s^2$ or $[Uuo]8s^2$. This outer configuration ($8s^2$) is similar to other Group 2 elements ($ns^2$).
Therefore, the element with $Z=120$ would be placed in Group 2 (Alkaline Earth Metals).
Metals, Non-Metals And Metalloids
Elements can also be broadly classified based on their physical and chemical properties into Metals, Non-metals, and Metalloids.
Problem 3.4. Considering the atomic number and position in the periodic table, arrange the following elements in the increasing order of metallic character : Si, Be, Mg, Na, P.
Answer:
The metallic character of elements increases down a group and decreases across a period from left to right.
Let's identify the positions of the given elements in the Periodic Table:
Across Period 3 (Na, Mg, Si, P), metallic character decreases from left to right: Na > Mg > Si > P.
Down Group 2 (Be, Mg), metallic character increases: Be < Mg.
Now, let's compare all the given elements (Si, Be, Mg, Na, P).
Na (Group 1) is more metallic than Mg (Group 2) in the same period. Both are more metallic than Si (Group 14) and P (Group 15) in the same period.
Comparing Mg and Be (same group), Mg is below Be, so Mg is more metallic than Be.
Comparing Si and P (same period), Si is to the left of P, so Si is more metallic than P.
Ranking based on position:
Rough order of metallic character:
P (least metallic) < Si < Be < Mg < Na (most metallic).
The increasing order of metallic character is: P < Si < Be < Mg < Na.
Periodic Trends In Properties Of Elements
Many physical and chemical properties of elements show observable patterns or trends as we move across a period or down a group in the Periodic Table. These periodic trends are fundamentally linked to the electronic structure of atoms.
Trends In Physical Properties
We will discuss the periodic trends for several key physical properties:
(a) Atomic Radius
Defining and measuring the size of an atom is challenging because atoms are extremely small ($\approx 10^{-10}$ m) and the electron cloud surrounding the nucleus doesn't have a sharp boundary. Atomic radius is typically estimated based on the distance between the nuclei of bonded atoms:
- Covalent Radius: Half the distance between the nuclei of two identical atoms joined by a single covalent bond in a molecule (used for non-metals). Example: In Cl$_2$, the bond distance is 198 pm, so the covalent radius of Cl is 99 pm.
- Metallic Radius: Half the distance between the nuclei of adjacent atoms in a metallic crystal (used for metals). Example: In solid copper, the distance between adjacent atoms is 256 pm, so the metallic radius of Cu is 128 pm.
- Van der Waals Radius: Half the internuclear distance between two identical non-bonded atoms in adjacent molecules in the solid state (used for noble gases or between non-bonded atoms in molecular solids). These values are generally larger than covalent or metallic radii.
In simplified terms, "Atomic Radius" often refers to either covalent or metallic radius, depending on the element's type. Atomic radii are determined using techniques like X-ray diffraction.
Periodic Trends in Atomic Radius:
- Across a Period (Left to Right): Atomic radius generally decreases.
- Down a Group (Top to Bottom): Atomic radius generally increases.
Atom (Period II) | Li | Be | B | C | N | O | F |
Atomic radius (pm) | 152 | 111 | 88 | 77 | 74 | 66 | 64 |
Atom (Period III) | Na | Mg | Al | Si | P | S | Cl |
Atomic radius (pm) | 186 | 160 | 143 | 117 | 110 | 104 | 99 |
Atom (Group I) | Atomic Radius (pm) | Atom (Group 17) | Atomic Radius (pm) |
---|---|---|---|
Li | 152 | F | 64 |
Na | 186 | Cl | 99 |
K | 231 | Br | 114 |
Rb | 244 | I | 133 |
Cs | 262 | At | 140 |
Noble gases' atomic radii are often listed as Van der Waals radii, which are significantly larger than covalent or metallic radii of other elements in the same period. Comparing them directly requires using consistent types of radii.
(b) Ionic Radius
When an atom loses electrons, it forms a positively charged ion called a cation. When it gains electrons, it forms a negatively charged ion called an anion. Ionic radii are estimated from the distances between ions in ionic crystals.
- Cation vs. Parent Atom: A cation is always smaller than its neutral parent atom. This is because the cation has fewer electrons but the same nuclear charge, leading to a stronger attraction of the remaining electrons to the nucleus and a contraction of the electron cloud. For instance, the radius of $\text{Na}^+$ (95 pm) is much smaller than the atomic radius of Na (186 pm).
- Anion vs. Parent Atom: An anion is always larger than its neutral parent atom. Adding one or more electrons increases electron-electron repulsion within the same valence shell (or adds to a higher shell), effectively increasing the size of the electron cloud and decreasing the effective nuclear charge per electron. For instance, the radius of $\text{F}^-$ (136 pm) is larger than the atomic radius of F (64 pm).
Isoelectronic Species: These are atoms or ions that have the same total number of electrons. Examples include $\text{O}^{2-}$, $\text{F}^-$, Ne, $\text{Na}^+$, $\text{Mg}^{2+}$, and $\text{Al}^{3+}$, all having 10 electrons. In a series of isoelectronic species, the size is determined by the nuclear charge ($Z$).
- Among isoelectronic cations, the one with the higher positive charge (and thus higher nuclear charge) has a smaller radius due to stronger electron-nucleus attraction (e.g., $\text{Mg}^{2+}$ is smaller than $\text{Na}^+$).
- Among isoelectronic anions, the one with the higher negative charge (more electrons, lower $Z_{eff}$) has a larger radius due to increased electron-electron repulsion (e.g., $\text{O}^{2-}$ is larger than $\text{F}^-$).
Thus, for the isoelectronic series $\text{O}^{2-}, \text{F}^-, \text{Ne}, \text{Na}^+, \text{Mg}^{2+}, \text{Al}^{3+}$ (all 10 electrons), the size decreases as the nuclear charge increases: $\text{O}^{2-} > \text{F}^- > \text{Ne} > \text{Na}^+ > \text{Mg}^{2+} > \text{Al}^{3+}$.
Problem 3.5. Which of the following species will have the largest and the smallest size?
Mg, Mg$^{2+}$, Al, Al$^{3+}$.
Answer:
We have four species: neutral atoms Mg and Al, and their cations $\text{Mg}^{2+}$ and $\text{Al}^{3+}$.
First, let's compare the neutral atoms Mg and Al. They are in the same period (Period 3), with Mg ($Z=12$) to the left of Al ($Z=13$). Atomic radii decrease across a period due to increasing effective nuclear charge. So, Mg is larger than Al.
Next, let's compare each atom to its cation. A cation is smaller than its parent atom because electrons are removed, and the remaining electrons are more strongly attracted to the nucleus. So, Mg is larger than $\text{Mg}^{2+}$, and Al is larger than $\text{Al}^{3+}$.
We know Mg > Al and Mg > $\text{Mg}^{2+}$ and Al > $\text{Al}^{3+}$.
Now let's compare the cations $\text{Mg}^{2+}$ and $\text{Al}^{3+}$. Both have 10 electrons (Mg: 12-2=10, Al: 13-3=10). They are isoelectronic species.
Among isoelectronic species, size decreases as nuclear charge ($Z$) increases. Mg has $Z=12$, Al has $Z=13$. So, $\text{Al}^{3+}$ ($Z=13$) has a higher nuclear charge than $\text{Mg}^{2+}$ ($Z=12$). Therefore, $\text{Al}^{3+}$ is smaller than $\text{Mg}^{2+}$.
Putting it all together:
Mg (largest atom) > Al (smaller atom)
Mg > $\text{Mg}^{2+}$
Al > $\text{Al}^{3+}$
$\text{Mg}^{2+}$ > $\text{Al}^{3+}$ (smallest cation)
Considering the species Mg, $\text{Mg}^{2+}$, Al, $\text{Al}^{3+}$, the neutral atoms are larger than their corresponding ions. Mg is larger than Al. $\text{Mg}^{2+}$ is larger than $\text{Al}^{3+}$.
Comparing the four: Mg is larger than Al. Mg is also larger than $\text{Mg}^{2+}$ and $\text{Al}^{3+}$. So Mg is the largest.
Comparing the cations $\text{Mg}^{2+}$ and $\text{Al}^{3+}$, $\text{Al}^{3+}$ is smaller. $\text{Al}^{3+}$ is also smaller than the neutral atoms Al and Mg. So $\text{Al}^{3+}$ is the smallest.
The largest species is Mg, and the smallest species is Al$^{3+}$.
(c) Ionization Enthalpy ($\Delta_i H$)
Ionization enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. It quantifies how easily an atom loses an electron to form a positive ion (cation).
First Ionization Enthalpy ($\Delta_i H_1$): $\text{X(g)} \rightarrow \text{X}^+\text{(g)} + \text{e}^-$
Second Ionization Enthalpy ($\Delta_i H_2$): $\text{X}^+\text{(g)} \rightarrow \text{X}^{2+}\text{(g)} + \text{e}^-$
And so on for successive electrons. Ionization enthalpies are always positive values, indicating energy must be absorbed (endothermic process). Successive ionization enthalpies increase ($\Delta_i H_1 < \Delta_i H_2 < \Delta_i H_3 < \dots$) because it becomes harder to remove an electron from an increasingly positive ion.
Periodic Trends in Ionization Enthalpy:
- Across a Period (Left to Right): Ionization enthalpy generally increases.
- Down a Group (Top to Bottom): Ionization enthalpy generally decreases.
Deviations from the general trend across a period:
- Group 2 vs Group 13: The first ionization enthalpy of Group 13 elements (like B) is slightly lower than that of Group 2 elements (like Be) in the same period. For Be ($1s^22s^2$), the electron is removed from the stable, filled 2s orbital. For B ($1s^22s^22p^1$), the electron is removed from the 2p orbital, which is higher in energy than 2s. The 2p electron is also slightly more shielded from the nucleus by the 2s electrons than the 2s electrons shield each other. This makes it easier to remove the 2p electron from B than a 2s electron from Be.
- Group 15 vs Group 16: The first ionization enthalpy of Group 16 elements (like O) is slightly lower than that of Group 15 elements (like N) in the same period. N ($1s^22s^22p^3$) has a half-filled and stable 2p subshell (according to Hund's rule, each 2p orbital has one electron with parallel spins). O ($1s^22s^22p^4$) has one 2p orbital with two paired electrons, leading to increased electron-electron repulsion compared to the half-filled configuration. This repulsion makes it easier to remove one electron from O than from N.
Problem 3.6. The first ionization enthalpy (D$_i$ H ) values of the third period elements, Na, Mg and Si are respectively 496, 737 and 786 kJ mol–1. Predict whether the first D$_i$ H value for Al will be more close to 575 or 760 kJ mol–1 ? Justify your answer.
Answer:
The given elements are from the third period: Na ($Z=11$), Mg ($Z=12$), Al ($Z=13$), Si ($Z=14$). The values are $\Delta_i H_1$ (Na) = 496 kJ mol$^{-1}$, $\Delta_i H_1$ (Mg) = 737 kJ mol$^{-1}$, $\Delta_i H_1$ (Si) = 786 kJ mol$^{-1}$. We need to predict $\Delta_i H_1$ (Al).
The general trend across a period is increasing ionization enthalpy: Na < Mg < Al < Si < ...
Based on the general trend, the value for Al should be greater than Mg (737 kJ mol$^{-1}$) and less than Si (786 kJ mol$^{-1}$). This would suggest the value should be closer to 760 kJ mol$^{-1}$.
However, there is an exception to this trend when moving from Group 2 to Group 13.
Let's look at the electronic configurations:
- Na: $[Ne]3s^1$
- Mg: $[Ne]3s^2$ (Filled s subshell, stable)
- Al: $[Ne]3s^23p^1$ (Electron removed from p orbital)
- Si: $[Ne]3s^23p^2$
The first electron removed from Mg is from the 3s orbital ($3s^2$). The first electron removed from Al is from the 3p orbital ($3p^1$).
Electrons in s orbitals penetrate closer to the nucleus and are shielded less effectively by inner electrons compared to electrons in p orbitals of the same principal shell. Therefore, the 3p electron in Al is better shielded by the inner core electrons (including the $3s^2$ pair) than the 3s electrons in Mg are shielded from each other. Also, removing an electron from a completely filled subshell (Mg, $3s^2$) is generally more difficult than removing an electron from a p subshell (Al, $3p^1$).
As a result, the first ionization enthalpy of Al is expected to be lower than that of Mg, despite Al having a larger nuclear charge ($Z=13$ vs $Z=12$).
Comparing the given options (575 kJ mol$^{-1}$ and 760 kJ mol$^{-1}$) with the value for Mg (737 kJ mol$^{-1}$), the value 575 kJ mol$^{-1}$ is lower than 737 kJ mol$^{-1}$, while 760 kJ mol$^{-1}$ is higher.
Therefore, the first ionization enthalpy value for Al will be more close to 575 kJ mol$^{-1}$.
Justification: The first electron is removed from the 3p orbital in Al, which is higher in energy and more shielded than the 3s orbital from which the first electron is removed in Mg. Removing the electron from the filled 3s subshell in Mg requires more energy than removing the electron from the $3p^1$ configuration in Al, despite Al having a higher nuclear charge.
(d) Electron Gain Enthalpy ($\Delta_{eg} H$)
Electron gain enthalpy is the enthalpy change that occurs when an electron is added to an isolated gaseous atom in its ground state to form a negative ion (anion).
$\text{X(g)} + \text{e}^- \rightarrow \text{X}^-\text{(g)}$
This process can be exothermic ($\Delta_{eg} H$ is negative, energy is released) or endothermic ($\Delta_{eg} H$ is positive, energy is absorbed).
- If energy is released upon electron gain (exothermic), the process is favourable, and the element has a high electron affinity (tendency to gain electrons). Halogens (Group 17) have very negative $\Delta_{eg} H$ values as gaining an electron completes their valence shell to a stable noble gas configuration.
- If energy is required to force the atom to accept an electron (endothermic), $\Delta_{eg} H$ is positive. Noble gases (Group 18) have positive $\Delta_{eg} H$ values because the added electron must enter a higher, empty energy level, resulting in an unstable configuration and significant electron-electron repulsion.
Periodic Trends in Electron Gain Enthalpy:
- Across a Period (Left to Right): Electron gain enthalpy generally becomes more negative.
Group 1 $\Delta_{eg}H$ Group 16 $\Delta_{eg}H$ Group 17 $\Delta_{eg}H$ Group 0 $\Delta_{eg}H$ H – 73 O – 141 F – 328 He + 48 Li – 60 S – 200 Cl – 349 Ne + 116 Na – 53 Se – 195 Br – 325 Ar + 96 K – 48 Te – 190 I – 295 Kr + 96 - Down a Group (Top to Bottom): Electron gain enthalpy generally becomes less negative. As atomic size increases down a group, the added electron is further from the nucleus and experiences weaker attraction. This results in less energy released or even energy required to add an electron.
Exceptions to the trend down a group (especially in Period 2): The electron gain enthalpies of elements in the second period (like O and F) are often less negative than those of the corresponding elements in the third period (S and Cl). This is because the second-period elements are unusually small. When an electron is added to the relatively small 2p subshell of O or F, it experiences significant repulsion from the electrons already present in the small volume. In contrast, for S or Cl, the added electron enters the larger 3p subshell, where electron-electron repulsion is less significant, leading to a more negative electron gain enthalpy.
Problem 3.7. Which of the following will have the most negative electron gain enthalpy and which the least negative?
P, S, Cl, F.
Explain your answer.
Answer:
The elements are P ($Z=15$), S ($Z=16$), Cl ($Z=17$), and F ($Z=9$).
P, S, and Cl are in the third period (Group 15, 16, and 17 respectively). F is in the second period (Group 17).
General trend across a period (P $\rightarrow$ S $\rightarrow$ Cl): Electron gain enthalpy becomes more negative. So, $\Delta_{eg}H(\text{P}) < \Delta_{eg}H(\text{S}) < \Delta_{eg}H(\text{Cl})$ (values become more negative).
General trend down a group (F $\rightarrow$ Cl): Electron gain enthalpy becomes less negative. So, $\Delta_{eg}H(\text{F}) > \Delta_{eg}H(\text{Cl})$ (values become less negative/more positive). This contradicts the general trend down a group due to the small size of F.
Comparing F and Cl: Cl is in the third period, F in the second period. Due to the small size of F, the incoming electron experiences significant repulsion from the existing electrons in the 2p orbitals. In Cl, the incoming electron enters the larger 3p orbitals, where repulsion is less. Thus, Cl has a more negative electron gain enthalpy than F. (Approximate values: F = -328 kJ/mol, Cl = -349 kJ/mol).
Comparing P, S, Cl (across period 3): $\Delta_{eg}H$ becomes more negative. P (-74 kJ/mol) < S (-200 kJ/mol) < Cl (-349 kJ/mol).
Now comparing all four values: P (-74), S (-200), Cl (-349), F (-328).
The most negative value is -349 kJ/mol (Chlorine).
The least negative value is -74 kJ/mol (Phosphorus).
The element with the most negative electron gain enthalpy is Chlorine (Cl).
The element with the least negative electron gain enthalpy is Phosphorus (P).
Explanation:
- Cl has the most negative value because it is a halogen and is larger than F, reducing electron repulsion for the incoming electron.
- P has the least negative value among these elements because it is on the left side of the period (Group 15), meaning its tendency to accept an electron is lower compared to the more electronegative elements on the right. It also has a half-filled p subshell ($3p^3$), which gives it some stability and reduces the tendency to gain an electron compared to Group 16 and 17 elements.
(e) Electronegativity
Electronegativity is a qualitative measure of the ability of an atom to attract shared electrons towards itself when it is in a chemical compound (i.e., when it is bonded to another atom). Unlike ionization enthalpy or electron gain enthalpy, it is not a property of an isolated atom and cannot be directly measured experimentally. Several scales exist to assign numerical values (e.g., Pauling scale, Mulliken-Jaffe scale, Allred-Rochow scale). The most common is the Pauling scale, where Fluorine is assigned the highest value of 4.0.
Periodic Trends in Electronegativity:
- Across a Period (Left to Right): Electronegativity generally increases.
Atom (Period II) Li Be B C N O F Electronegativity 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Atom (Period III) Na Mg Al Si P S Cl Electronegativity 0.9 1.2 1.5 1.8 2.1 2.5 3.0 - Down a Group (Top to Bottom): Electronegativity generally decreases.
Atom (Group I) Electronegativity Value Atom (Group 17) Electronegativity Value Li 1.0 F 4.0 Na 0.9 Cl 3.0 K 0.8 Br 2.8 Rb 0.8 I 2.5 Cs 0.7 At 2.2
Electronegativity is closely related to non-metallic character (ability to gain electrons) and inversely related to metallic character (tendency to lose electrons). High electronegativity corresponds to high non-metallic character. Low electronegativity corresponds to high metallic character.
Periodic Trends In Chemical Properties
Periodic trends are also evident in the chemical properties of elements, such as valence, oxidation states, and reactivity.
(a) Periodicity of Valence or Oxidation States
The valence of an element, representing its combining capacity, is primarily determined by the number of electrons in its outermost shell. For representative (s-block and p-block) elements, the valence is often equal to the number of valence electrons (Group 1 and 2) or eight minus the number of valence electrons (Group 15-17). Group 18 elements (noble gases) have a valence of 0 (or sometimes 8).
The oxidation state of an element in a compound refers to the apparent charge on an atom based on the assumption that all bonds are ionic, with the more electronegative atom assigned the negative charge. Oxidation states can be positive or negative and often variable, especially for d and f block elements.
Periodic trends in valence (shown by common hydrides and oxides) (Table 3.9):
Group | 1 | 2 | 13 | 14 | 15 | 16 | 17 |
---|---|---|---|---|---|---|---|
Number of valence electrons | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
Valence (common) | 1 | 2 | 3 | 4 | 3, 5 | 2, 6 | 1, 7 |
Formula of hydride (Examples) | LiH, NaH, KH | MgH$_2$, CaH$_2$ | B$_2$H$_6$, AlH$_3$ | CH$_4$, SiH$_4$, GeH$_4$, SnH$_4$ | NH$_3$, PH$_3$, AsH$_3$, SbH$_3$ | H$_2$O, H$_2$S, H$_2$Se, H$_2$Te | HF, HCl, HBr, HI |
Formula of oxide (Examples) | Li$_2$O, Na$_2$O, K$_2$O | MgO, CaO, SrO, BaO | B$_2$O$_3$, Al$_2$O$_3$, Ga$_2$O$_3$, In$_2$O$_3$ | CO$_2$, SiO$_2$, GeO$_2$, SnO$_2$, PbO$_2$ | N$_2$O$_3$, N$_2$O$_5$, P$_4$O$_6$, P$_4$O$_{10}$, As$_2$O$_3$, As$_2$O$_5$, Sb$_2$O$_3$, Sb$_2$O$_5$, Bi$_2$O$_3$ | SO$_3$, SeO$_3$, TeO$_3$ | Cl$_2$O$_7$ |
Many elements, particularly transition and inner-transition metals, exhibit multiple oxidation states due to the involvement of inner d or f electrons in bonding.
Problem 3.8. Using the Periodic Table, predict the formulas of compounds which might be formed by the following pairs of elements;
(a) silicon and bromine (b) aluminium and sulphur.
Answer:
To predict the formula of a binary compound formed between two elements, we can often use their common valencies (or oxidation states).
(a) Silicon (Si) and Bromine (Br):
- Silicon (Si) is in Group 14. Common valence/oxidation state is +4.
- Bromine (Br) is a halogen in Group 17. Common valence/oxidation state is -1.
To form a neutral compound, the total positive charge must balance the total negative charge. If Silicon has a valence of 4 and Bromine has a valence of 1, we need one Si atom and four Br atoms.
Formula: SiBr$_4$.
(b) Aluminium (Al) and Sulphur (S):
- Aluminium (Al) is in Group 13. Common valence/oxidation state is +3.
- Sulphur (S) is in Group 16. Common valence/oxidation state is -2.
To balance the charges, we need to find the least common multiple of the valencies (3 and 2), which is 6. We need a total positive charge of +6 and a total negative charge of -6.
Number of Al atoms = $6 / 3 = 2$. (Total positive charge = $2 \times +3 = +6$).
Number of S atoms = $6 / 2 = 3$. (Total negative charge = $3 \times -2 = -6$).
Formula: Al$_2$S$_3$.
(b) Anomalous Properties of Second Period Elements
The first element in each group of the s-block (Li, Be) and p-block (B, C, N, O, F) exhibits properties that are significantly different from the rest of the elements in their respective groups. This is known as anomalous behaviour.
Reasons for anomalous behaviour of Period 2 elements (Li to F):
- Small Size: They are the smallest elements in their groups.
- High Electronegativity: They have the highest electronegativity values in their groups.
- High Charge/Radius Ratio: For cations, this ratio is very high.
- Absence of d-orbitals in the valence shell: Elements of the second period have only 2s and 2p orbitals (a total of four orbitals) in their valence shell ($n=2$), limiting their maximum covalency to 4. Elements from the third period onwards have empty d orbitals (e.g., 3s, 3p, 3d), allowing them to expand their valence shell and exhibit covalencies greater than 4 (e.g., Al can form $\text{AlF}_6^{3-}$).
- Ability to form $p\pi - p\pi$ multiple bonds: Second-period elements (especially C, N, O, F) can form stable multiple bonds (double or triple bonds) by sideways overlap of their p orbitals with themselves (e.g., C=C, N$\equiv$N) and with other second-period elements (e.g., C=O, C$\equiv$N). Heavier elements in the same group have larger and more diffuse p orbitals, making $p\pi - p\pi$ overlap less effective, thus reducing their tendency to form strong multiple bonds.
Diagonal Relationship: The anomalous behaviour of second-period elements often leads to a similarity in properties between the first element of a group and the second element of the *next* group. This is called the diagonal relationship. Examples: Lithium (Group 1) shows similarities with Magnesium (Group 2), and Beryllium (Group 2) shows similarities with Aluminium (Group 13).
Property | Element | ||
Metallic radius (pm) | Li (152) | Be (111) | B (88) |
Na (186) | Mg (160) | Al (143) | |
Ionic radius M$^+$ / M$^{2+}$ / M$^{3+}$ (pm) | Li$^+$ (76) | Be$^{2+}$ (31) | |
Na$^+$ (102) | Mg$^{2+}$ (72) | Al$^{3+}$ (53.5) |
Li ($Z/r \approx 76/76 = 1.0$) and Mg ($Z/r^2 \approx 12/72^2$? This is charge/radius, not charge/radius squared. Let's use ionic charge and ionic radius $Li^+ (1+, 76pm), Mg^{2+} (2+, 72pm), Al^{3+} (3+, 53.5pm)$ approx charge density $1/76, 2/72, 3/53.5$. The ratio is not simple) exhibit similar charge density ($Z/\text{radius}$ or $Z/\text{radius}^2$ for ions) which leads to comparable polarizing power and similarities in properties, such as forming covalent compounds more readily than other alkali/alkaline earth metals.
Problem 3.9. Are the oxidation state and covalency of Al in [AlCl(H$_2$O)$_5$]$^{2+}$ same ?
Answer:
Let's determine the oxidation state of Al in the complex ion [AlCl(H$_2$O)$_5$]$^{2+}$.
The overall charge of the complex ion is +2.
The ligands are Chloride (Cl) and Water (H$_2$O).
Chloride is a halide and usually has an oxidation state of -1 (in this case, acting as a ligand, its charge is -1).
Water is a neutral molecule, so its charge/oxidation state as a ligand is 0.
Let the oxidation state of Aluminium (Al) be $x$.
Sum of oxidation states = Overall charge of the complex ion
$x$ + (Oxidation state of Cl) + 5 $\times$ (Oxidation state of H$_2$O) = +2
$x$ + (-1) + 5 $\times$ (0) = +2
$x - 1 + 0 = +2$
$x = +2 + 1 = +3$.
So, the oxidation state of Aluminium in [AlCl(H$_2$O)$_5$]$^{2+}$ is +3.
The covalency of the central metal atom in a complex is the number of coordinate bonds formed between the metal ion and the ligands. In [AlCl(H$_2$O)$_5$]$^{2+}$, Al is bonded to one Chloride ligand and five Water ligands.
The number of ligands directly attached to Al is 1 (from Cl) + 5 (from H$_2$O) = 6.
So, the covalency of Aluminium in [AlCl(H$_2$O)$_5$]$^{2+}$ is 6.
The oxidation state of Al is +3, and the covalency is 6. Therefore, they are not the same.
Periodic Trends And Chemical Reactivity
The chemical reactivity of elements is closely related to their tendency to lose or gain electrons, which is reflected in properties like ionization enthalpy and electron gain enthalpy.
- Across a Period (Left to Right): Chemical reactivity is highest at the two extremes (Group 1 and Group 17) and lowest towards the centre (Groups 14-15).
- On the extreme left (Alkali metals, Group 1), elements have low ionization enthalpies and readily lose electrons to form cations. Their reactivity is high due to their strong tendency to lose electrons.
- On the extreme right (Halogens, Group 17), elements have high negative electron gain enthalpies and readily gain electrons to form anions. Their reactivity is high due to their strong tendency to gain electrons. (Noble gases, Group 18, are exceptions as they are generally unreactive due to stable electron configurations).
- Elements in the centre (like Group 14) have intermediate ionization enthalpies and electron gain enthalpies, showing less tendency to lose or gain electrons fully, and often form covalent bonds. They are generally less reactive than elements at the extremes.
- Down a Group (Top to Bottom):
- For metals (like Alkali metals, Group 1), reactivity generally increases down the group. This is because ionization enthalpy decreases as size increases and effective nuclear attraction for valence electrons weakens, making it easier to lose electrons.
- For non-metals (like Halogens, Group 17), reactivity generally decreases down the group. Although electron gain enthalpy becomes less negative, the dominant factor is the decreasing attraction for an incoming electron due to increasing size and shielding. Thus, their ability to gain electrons decreases down the group.
The chemical nature of oxides provides insight into reactivity. Oxides of metals are generally basic, reacting with water to form bases. Oxides of non-metals are generally acidic, reacting with water to form acids. Oxides of elements in the middle of the table can be amphoteric (reacting with both acids and bases) or neutral.
- Oxides of elements on the extreme left of a period (e.g., $\text{Na}_2\text{O}$) are typically most basic.
- Oxides of elements on the extreme right of a period (e.g., $\text{Cl}_2\text{O}_7$) are typically most acidic.
- Oxides in the centre can be amphoteric (e.g., $\text{Al}_2\text{O}_3$) or neutral (e.g., CO, NO).
Problem 3.10. Show by a chemical reaction with water that Na$_2$O is a basic oxide and Cl$_2$O$_7$ is an acidic oxide.
Answer:
Basic oxide: Na$_2$O
Sodium oxide (Na$_2$O) is formed from sodium, an alkali metal on the extreme left of Period 3. Metal oxides are generally basic. When a basic oxide reacts with water, it forms a base (a substance that produces hydroxide ions, OH$^-$).
Reaction of Na$_2$O with water:
$\text{Na}_2\text{O(s)} + \text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)}$
Sodium hydroxide (NaOH) is a strong base. This reaction shows that Na$_2$O is a basic oxide.
Acidic oxide: Cl$_2$O$_7$
Dichlorine heptoxide (Cl$_2$O$_7$) is formed from chlorine, a halogen on the extreme right of Period 3 (before the noble gas). Non-metal oxides are generally acidic. When an acidic oxide reacts with water, it forms an acid (a substance that produces hydronium ions, H$_3$O$^+$, or releases H$^+$).
Reaction of Cl$_2$O$_7$ with water:
$\text{Cl}_2\text{O}_7\text{(l)} + \text{H}_2\text{O(l)} \rightarrow 2\text{HClO}_4\text{(aq)}$
Perchloric acid (HClO$_4$) is a very strong acid. This reaction shows that Cl$_2$O$_7$ is an acidic oxide.
Exercises
Question 3.1 What is the basic theme of organisation in the periodic table?
Answer:
Question 3.2 Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?
Answer:
Question 3.3 What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?
Answer:
Question 3.4 On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Answer:
Question 3.5 In terms of period and group where would you locate the element with Z =114?
Answer:
Question 3.6 Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
Answer:
Question 3.7 Which element do you think would have been named by
(i) Lawrence Berkeley Laboratory
(ii) Seaborg’s group?
Answer:
Question 3.8 Why do elements in the same group have similar physical and chemical properties?
Answer:
Question 3.9 What does atomic radius and ionic radius really mean to you?
Answer:
Question 3.10 How do atomic radius vary in a period and in a group? How do you explain the variation?
Answer:
Question 3.11 What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.
(i) $F^–$
(ii) Ar
(iii) $Mg^{2+}$
(iv) $Rb^+$
Answer:
Question 3.12 Consider the following species :
$N^{3–}, O^{2–}, F^–, Na^+, Mg^{2+}$ and $Al^{3+}$
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.
Answer:
Question 3.13 Explain why cation are smaller and anions larger in radii than their parent atoms?
Answer:
Question 3.14 What is the significance of the terms — ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy?
Hint : Requirements for comparison purposes.
Answer:
Question 3.15 Energy of an electron in the ground state of the hydrogen atom is $–2.18 \times 10^{–18}J$. Calculate the ionization enthalpy of atomic hydrogen in terms of $J \ mol^{–1}$.
Hint: Apply the idea of mole concept to derive the answer.
Answer:
Question 3.16 Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne.
Explain why
(i) Be has higher $\Delta_iH$ than B
(ii) O has lower $\Delta_iH$ than N and F?
Answer:
Question 3.17 How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
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Question 3.18 What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
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Question 3.19 The first ionization enthalpy values (in $kJ \ mol^{–1}$) of group 13 elements are :
B | Al | Ga | In | Tl |
---|---|---|---|---|
801 | 577 | 579 | 558 | 589 |
How would you explain this deviation from the general trend ?
Answer:
Question 3.20 Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) O or F
(ii) F or Cl
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Question 3.21 Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
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Question 3.22 What is the basic difference between the terms electron gain enthalpy and electronegativity?
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Question 3.23 How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
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Question 3.24 Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron
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Question 3.25 Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
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Question 3.26 What are the major differences between metals and non-metals?
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Question 3.27 Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer subshell.
(b) Identify an element that would tend to lose two electrons.
(c) Identify an element that would tend to gain two electrons.
(d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.
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Question 3.28 The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that among group 17 elements is F > CI > Br > I. Explain.
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Question 3.29 Write the general outer electronic configuration of s-, p-, d- and f- block elements.
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Question 3.30 Assign the position of the element having outer electronic configuration (i) $ns^2np^4$ for n=3 (ii) $(n-1)d^2ns^2$ for n=4, and (iii) $(n-2)f^7(n-1)d^1ns^2$ for n=6, in the periodic table.
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Question 3.31 The first ($\Delta_iH_1$) and the second ($\Delta_iH_2$) ionization enthalpies (in $kJ \ mol^{–1}$) and the ($\Delta_{eg}H$) electron gain enthalpy (in $kJ \ mol^{–1}$) of a few elements are given below:
Elements | $\Delta_iH_1$ | $\Delta_iH_2$ | $\Delta_{eg}H$ |
---|---|---|---|
I | 520 | 7300 | –60 |
II | 419 | 3051 | –48 |
III | 1681 | 3374 | –328 |
IV | 1008 | 1846 | –295 |
V | 2372 | 5251 | +48 |
VI | 738 | 1451 | –40 |
Which of the above elements is likely to be :
(a) the least reactive element.
(b) the most reactive metal.
(c) the most reactive non-metal.
(d) the least reactive non-metal.
(e) the metal which can form a stable binary halide of the formula $MX_2$ (X=halogen).
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?
Answer:
Question 3.32 Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.
(a) Lithium and oxygen
(b) Magnesium and nitrogen
(c) Aluminium and iodine
(d) Silicon and oxygen
(e) Phosphorus and fluorine
(f) Element 71 and fluorine
Answer:
Question 3.33 In the modern periodic table, the period indicates the value of :
(a) atomic number
(b) atomic mass
(c) principal quantum number
(d) azimuthal quantum number.
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Question 3.34 Which of the following statements related to the modern periodic table is incorrect?
(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration.
Answer:
Question 3.35 Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons.
Answer:
Question 3.36 The size of isoelectronic species — $F^–$, Ne and $Na^+$ is affected by
(a) nuclear charge (Z)
(b) valence principal quantum number (n)
(c) electron-electron interaction in the outer orbitals
(d) none of the factors because their size is the same.
Answer:
Question 3.37 Which one of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
Answer:
Question 3.38 Considering the elements B, Al, Mg, and K, the correct order of their metallic character is :
(a) B > Al > Mg > K
(b) Al > Mg > B > K
(c) Mg > Al > K > B
(d) K > Mg > Al > B
Answer:
Question 3.39 Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is :
(a) B > C > Si > N > F
(b) Si > C > B > N > F
(c) F > N > C > B > Si
(d) F > N > C > Si > B
Answer:
Question 3.40 Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is :
(a) F > Cl > O > N
(b) F > O > Cl > N
(c) Cl > F > O > N
(d) O > F > N > Cl
Answer: