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Chapter 3 Classification Of Elements And Periodicity In Properties
Why Do We Need To Classify Elements ?
By the early 19th century, the number of known elements was increasing rapidly (from 31 in 1800 to 63 by 1865, and currently over 114). With such a large number, studying the individual properties of each element and their countless compounds became extremely challenging.
Scientists recognized the need for a systematic way to organize this vast amount of information. Classifying elements based on similarities in their properties would not only simplify their study but also help in understanding the relationships between different elements and potentially predict the properties of new or undiscovered ones.
The goal was to find a way to organize the elements that would rationalize the known chemical facts and provide a framework for further research and discovery.
Genesis Of Periodic Classification
The idea of classifying elements and observing trends in their properties developed gradually through the efforts of several scientists.
Early Attempts at Classification
Johann Dobereiner (early 1800s):
- Observed similarities within groups of three elements, which he called Triads.
- Noted that in a Triad, the atomic weight of the middle element was approximately the average of the atomic weights of the other two elements.
- Also found that the properties of the middle element were intermediate between those of the other two.
| 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 |
($\text{Average wt of Li and K} = (7+39)/2 = 23$, which is the wt of Na. Similarly for other triads). Dobereiner's Law of Triads was limited, as it only applied to a few elements and was initially seen as a coincidence.
A.E.B. de Chancourtois (1862):
- Arranged elements by increasing atomic weight in a cylindrical fashion (telluric helix).
- Observed a periodic repetition of properties along the cylinder's vertical lines.
- His work did not gain widespread recognition at the time.
John Alexander Newlands (1865):
- Proposed the Law of Octaves.
- Arranged elements in increasing order of atomic weights and noticed that properties of every eighth element were similar to those of the first element, analogous to musical octaves.
| 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 well only for lighter elements up to Calcium and was not initially accepted by the scientific community. However, he was later recognized for his contribution.
Contribution Of Mendeleev And Lothar Meyer
The foundation of the modern Periodic Table is largely credited to Dmitri Mendeleev (Russian chemist) and Lothar Meyer (German chemist), working independently around 1869.
- Both observed that when elements are arranged in increasing order of their atomic weights, similarities in physical and chemical properties recur at regular intervals (periodicity).
- Lothar Meyer plotted physical properties (like atomic volume, melting point, boiling point) against atomic weight and found repeating patterns. He also noticed that the length of the repeating pattern changed for different properties. His table, developed by 1868, closely resembled the modern one but was published later than Mendeleev's.
Dmitri Mendeleev is generally considered the father of the Periodic Table because he published his findings first and made bold predictions based on his classification.
- Mendeleev stated the first Periodic Law: "The properties of the elements are a periodic function of their atomic weights."
- He arranged elements in a table with horizontal rows (called series) and vertical columns (called groups) based on increasing atomic weights, placing elements with similar properties in the same group.
- Mendeleev considered a wider range of physical and chemical properties for classification, particularly focusing on the empirical formulas of compounds formed with hydrogen and oxygen (hydrides and oxides).
- Recognizing the importance of grouping elements with similar properties, he sometimes ignored the strict order of atomic weights (e.g., placing Iodine after Tellurium despite its lower atomic weight) assuming the measured atomic weights might be incorrect.
- Crucially, he left gaps in his table for elements he believed were yet to be discovered. He even predicted the properties of these unknown elements (e.g., Eka-Aluminium and Eka-Silicon, later discovered as Gallium and Germanium, respectively).
| Property | Eka-aluminium (predicted) | Gallium (found) | Eka-silicon (predicted) | Germanium (found) |
|---|---|---|---|---|
| Atomic weight | 68 | 70 | 72 | 72.6 |
| Density/(g/cm³) | 5.9 | 5.94 | 5.5 | 5.36 |
| Melting point/K | Low | 302.93 | High | 1231 |
| Formula of oxide | E₂O₃ | Ga₂O₃ | EO₂ | GeO₂ |
| Formula of chloride | ECl₃ | GaCl₃ | ECl₄ | GeCl₄ |
The subsequent discovery of these elements with properties remarkably close to his predictions solidified the acceptance of Mendeleev's Periodic Table.
Modern Periodic Law And The Present Form Of The Periodic Table
While Mendeleev's table was based on atomic weight, the understanding of atomic structure at the beginning of the 20th century led to a crucial refinement.
Modern Periodic Law
In 1913, Henry Moseley studied the characteristic X-ray spectra of elements. He found that the frequency of emitted X-rays was related to the element's atomic number (Z) rather than its atomic mass. Specifically, a plot of $\sqrt{\nu}$ vs Z yielded a straight line.
This experimental evidence showed that the atomic number (Z) is a more fundamental property of an element than its atomic mass.
Based on Moseley's work, the Periodic Law was modified, leading to the Modern Periodic Law:
"The physical and chemical properties of the elements are periodic functions of their atomic numbers."
Since the atomic number equals the number of protons in the nucleus (and the number of electrons in a neutral atom), the Periodic Law is essentially a consequence of the periodic recurrence of similar electronic configurations in the outermost shell of atoms, which are responsible for their chemical and physical properties.
The Long Form Of The Periodic Table
The most widely used form of the Periodic Table today is the "long form". In this table:
- Elements are arranged in increasing order of their atomic numbers.
- Horizontal rows are called periods. There are seven periods. The period number corresponds to the highest principal quantum number ($n$) of the valence shell electrons in the elements of that period.
- Vertical columns are called groups or families. There are eighteen groups, numbered 1 to 18 according to IUPAC recommendations (replacing the older IA-VIIA, VIII, IB-VIIB, 0 notation). Elements in the same group have similar outermost (valence shell) electronic configurations and thus exhibit similar chemical properties.
The number of elements in each period corresponds to the number of atomic orbitals available in the shell being filled:
- Period 1 ($n=1$): 1s orbital. Holds 2 electrons. 2 elements (H, He).
- Period 2 ($n=2$): 2s, 2p orbitals. Holds 2+6 = 8 electrons. 8 elements.
- Period 3 ($n=3$): 3s, 3p orbitals. Holds 2+6 = 8 electrons. 8 elements.
- Period 4 ($n=4$): 4s, 3d, 4p orbitals. Holds 2+10+6 = 18 electrons. 18 elements.
- Period 5 ($n=5$): 5s, 4d, 5p orbitals. Holds 2+10+6 = 18 electrons. 18 elements.
- Period 6 ($n=6$): 6s, 4f, 5d, 6p orbitals. Holds 2+14+10+6 = 32 electrons. 32 elements.
- Period 7 ($n=7$): 7s, 5f, 6d, 7p orbitals. Can hold up to 32 elements (incomplete). Includes synthetic elements.
The 14 elements filling the 4f orbitals (Lanthanoids) and the 14 elements filling the 5f orbitals (Actinoids) in periods 6 and 7, respectively, are placed separately at the bottom of the table to maintain its structure and highlight the similar chemistry within these series (called inner-transition elements).
Nomenclature Of Elements With Atomic Numbers > 100
Traditionally, the discoverer(s) of a new element had the privilege of naming it. However, for very heavy, unstable elements synthesized in different labs, this led to disputes over discovery claims and naming rights (e.g., element 104, claimed by both American and Soviet scientists).
To avoid controversy, IUPAC developed a systematic nomenclature for newly discovered elements with atomic numbers above 100. This temporary name is derived directly from the atomic number using numerical roots.
IUPAC Numerical Roots:
| 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 |
Naming Rule: Combine the roots for the digits of the atomic number in order, and add the suffix "-ium". The symbol is the first letter of each root.
Example: Element with Z = 101
- Digits are 1, 0, 1.
- Roots are un, nil, un.
- Name: unnilunium
- Symbol: Unu
Once a discovery is verified and officially recognized by IUPAC, a permanent name and symbol are assigned, often honoring a scientist or place of discovery.
| Atomic Number | IUPAC Temporary Name | Symbol | IUPAC Official Name | Official Symbol |
|---|---|---|---|---|
| 101 | Unnilunium | Unu | Mendelevium | Md |
| 102 | Unnilbium | Unb | Nobelium | No |
| 103 | Unniltrium | Unt | Lawrencium | Lr |
| 104 | Unnilquadium | Unq | Rutherfordium | Rf |
| 105 | Unnilpentium | Unp | Dubnium | Db |
| 106 | Unnilhexium | Unh | Seaborgium | Sg |
| 107 | Unnilseptium | Uns | Bohrium | Bh |
| 108 | Unniloctium | Uno | Hassium | Hs |
| 109 | Unnilennium | Une | Meitnerium | Mt |
| 110 | Ununnillium | Uun | Darmstadtium | Ds |
| 111 | Unununium | Uuu | Rontgenium | Rg |
| 112 | Ununbium | Uub | Copernicium | Cn |
| 113 | Ununtrium | Uut | Nihonium | Nh |
| 114 | Ununquadium | Uuq | Flerovium | Fl |
| 115 | Ununpentium | Uup | Moscovium | Mc |
| 116 | Ununhexium | Uuh | Livermorium | Lv |
| 117 | Ununseptium | Uus | Tennessine | Ts |
| 118 | Ununoctium | Uuo | Oganesson | Og |
Example 1. What would be the IUPAC name and symbol for the element with atomic number 120?
Answer:
Atomic number = 120.
Digits are 1, 2, 0.
Roots are un, bi, nil.
Combine the roots and add "-ium": un + bi + nil + ium = unbinilium.
The symbol is the first letter of each root: Ubn.
So, the IUPAC temporary name is unbinilium and the symbol is Ubn.
Electronic Configurations Of Elements And The Periodic Table
The electronic configuration of an atom, describing the arrangement of electrons in its orbitals, is the fundamental basis for its position and properties in the modern Periodic Table. The quantum numbers ($n, l, m_l, m_s$) characterize the electrons and their orbitals, and the Periodic Table visually organizes elements based on these characteristics.
Electronic Configurations In Periods
The period number ($n$) in the Periodic Table corresponds to the principal quantum number of the valence shell (outermost energy level) being filled. As you move across a period, electrons are added to orbitals within this valence shell or sometimes to inner shells that become energetically accessible (like 3d filling after 4s).
The number of elements in a period is determined by the total number of electrons that can be accommodated in all the orbitals of the energy levels being filled in that period.
- Period 1 (n=1): Starts with filling the 1s orbital. Maximum 2 electrons in 1s ($1s^1$ to $1s^2$). 2 elements (H, He).
- Period 2 (n=2): Starts with 2s ($2s^1, 2s^2$), followed by 2p ($2p^1$ to $2p^6$). Maximum 2 in 2s + 6 in 2p = 8 electrons. 8 elements.
- Period 3 (n=3): Starts with 3s ($3s^1, 3s^2$), followed by 3p ($3p^1$ to $3p^6$). Maximum 2 in 3s + 6 in 3p = 8 electrons. 8 elements.
- Period 4 (n=4): Starts with 4s ($4s^1, 4s^2$). Then the 3d orbitals become energetically favorable and are filled (3d¹ to 3d¹⁰ - transition series). Finally, 4p orbitals are filled (4p¹ to 4p⁶). Maximum 2 in 4s + 10 in 3d + 6 in 4p = 18 electrons. 18 elements.
- Period 5 (n=5): Similar to Period 4. Starts with 5s ($5s^1, 5s^2$), then 4d (4d¹ to 4d¹⁰ - transition series), and finally 5p (5p¹ to 5p⁶). Maximum 2 in 5s + 10 in 4d + 6 in 5p = 18 electrons. 18 elements.
- Period 6 (n=6): Starts with 6s ($6s^1, 6s^2$). Then 4f orbitals are filled (4f¹ to 4f¹⁴ - Lanthanoids). Then 5d (5d¹ to 5d¹⁰ - remaining transition elements). Finally, 6p (6p¹ to 6p⁶). Maximum 2 in 6s + 14 in 4f + 10 in 5d + 6 in 6p = 32 electrons. 32 elements.
- Period 7 (n=7): Similar to Period 6. Starts with 7s ($7s^1, 7s^2$), then 5f (5f¹ to 5f¹⁴ - Actinoids), then 6d (6d¹ to 6d¹⁰), and finally 7p (7p¹ to 7p⁶). Incomplete, but can theoretically accommodate 32 elements.
The Lanthanoids (4f series) and Actinoids (5f series) are placed separately at the bottom because their filling involves inner f-orbitals, leading to similar properties within each series and keeping the main body of the table structured according to s, p, and d block filling.
Example 2. How would you justify the presence of 18 elements in the 5th period of the Periodic Table?
Answer:
The 5th period corresponds to the principal quantum number $n=5$.
According to the Aufbau principle and $(n+l)$ rule, the orbitals filled in the 5th period are:
- 5s ($n=5, l=0$): Holds up to 2 electrons.
- 4d ($n=4, l=2$): Holds up to 10 electrons. ($n+l = 4+2 = 6$, lower than $5+1$ for 5p, but 5s ($5+0=5$) is lower energy).
- 5p ($n=5, l=1$): Holds up to 6 electrons. ($n+l = 5+1 = 6$, higher $n$ than 4d, so 4d fills first).
The filling order is 5s, then 4d, then 5p.
The total number of orbitals available for filling in the 5th period is 1 (in 5s) + 5 (in 4d) + 3 (in 5p) = 9 orbitals.
Since each orbital can hold a maximum of 2 electrons (Pauli exclusion principle), the maximum number of electrons that can be accommodated in these orbitals is $9 \times 2 = 18$ electrons.
Therefore, the 5th period contains 18 elements, corresponding to the filling of the 5s, 4d, and 5p subshells.
Groupwise Electronic Configurations
Elements in the same vertical column (group) of the Periodic Table have similar chemical properties because they have the same number of valence electrons and the same general outer electronic configuration.
For example, Group 1 elements (Alkali Metals) all have one electron in their outermost s subshell ($ns^1$).
| Atomic number | Symbol | Electronic configuration | Outer electronic configuration |
|---|---|---|---|
| 3 | Li | 1s² 2s¹ | 2s¹ |
| 11 | Na | 1s² 2s² 2p⁶ 3s¹ | 3s¹ |
| 19 | K | 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ | 4s¹ |
| 37 | Rb | [Kr] 5s¹ | 5s¹ |
| 55 | Cs | [Xe] 6s¹ | 6s¹ |
| 87 | Fr | [Rn] 7s¹ | 7s¹ |
This identical outermost electronic configuration leads to similar chemical behavior throughout the group.
Electronic Configurations And Types Of Elements : S-, P-, D-, F- Blocks
Based on which atomic orbital receives the last electron during the building up of electronic configuration, elements are broadly classified into four blocks: s-block, p-block, d-block, and f-block.
Exceptions:
- Hydrogen (H): Has only one electron ($1s^1$). Can be placed in Group 1 due to $s^1$ configuration or with Group 17 (halogens) as it can gain an electron to achieve a noble gas configuration. Due to its unique properties, it is often placed separately.
- Helium (He): Has configuration $1s^2$. It is in the s-block by electronic configuration, but its properties are very similar to other noble gases (Group 18) because of its completely filled valence shell. Hence, it is placed in the p-block with Group 18 elements.
The S-Block Elements
- Located on the left side of the Periodic Table.
- Comprise Group 1 (Alkali Metals) and Group 2 (Alkaline Earth Metals).
- General outer electronic configuration is $ns^1$ (Group 1) or $ns^2$ (Group 2), where $n$ is the period number.
- All are reactive metals.
- Have low ionization enthalpies and readily lose their valence electron(s) to form positive ions ($+1$ for Group 1, $+2$ for Group 2).
- Metallic character and reactivity increase down a group.
- Not found pure in nature due to high reactivity.
- Form predominantly ionic compounds (except for some compounds of Li and Be, which show covalent character).
The P-Block Elements
- Located on the right side of the Periodic Table.
- Comprise Group 13 to Group 18.
- Together with s-block elements, they are called Representative Elements or Main Group Elements.
- General outer electronic configuration varies from $ns^2np^1$ (Group 13) to $ns^2np^6$ (Group 18) in each period.
- Includes metals, non-metals, and metalloids.
- Non-metallic character increases from left to right across a period.
- Metallic character increases down a group.
- Group 18 elements are Noble Gases with a stable $ns^2np^6$ configuration (or $1s^2$ for He). They have high ionization enthalpies and positive electron gain enthalpies, making them very unreactive.
- Group 17 (Halogens, $ns^2np^5$) and Group 16 (Chalcogens, $ns^2np^4$) are important non-metal groups that readily gain one or two electrons, respectively, to achieve stable noble gas configuration, having highly negative electron gain enthalpies.
The D-Block Elements (Transition Elements)
- Located in the middle of the Periodic Table.
- Comprise Group 3 to Group 12.
- Characterized by the filling of the $(n-1)d$ orbitals (the d subshell of the inner shell).
- General outer electronic configuration is $(n-1)d^{1-10}ns^{0-2}$ (with some exceptions like Pd ($4d^{10}5s^0$) and Cr ($3d^54s^1$), Cu ($3d^{10}4s^1$)).
- All are metals.
- Typically form colored ions, exhibit variable valency (oxidation states), are often paramagnetic, and are used as catalysts.
- Elements in Group 12 (Zn, Cd, Hg) with $(n-1)d^{10}ns^2$ configuration in their common oxidation states do not show typical transition metal properties as their d orbitals are completely filled. They are sometimes considered transition metals, but often grouped separately.
- Transition metals act as a bridge between the highly reactive s-block metals and the less reactive p-block elements.
The F-Block Elements (Inner-Transition Elements)
- Located in two separate rows at the bottom of the Periodic Table.
- Comprise the Lanthanoids (Ce to Lu, filling 4f orbitals) and Actinoids (Th to Lr, filling 5f orbitals).
- Characterized by the filling of the $(n-2)f$ orbitals (the f subshell two shells inside the outermost shell).
- General outer electronic configuration is $(n-2)f^{1-14}(n-1)d^{0-1}ns^2$.
- All are metals.
- Properties within each series are quite similar due to the filling of inner f-orbitals (which shield the outer electrons effectively).
- Actinoids are radioactive, and many are synthetic. Their chemistry is more complex than lanthanoids due to the larger number of oxidation states.
- Elements after Uranium (Z=92) are called Transuranium elements.
Example 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 figure out which orbitals are being filled.
For Z = 117:
- The period number is determined by the highest principal quantum number ($n$) of the valence shell. Element 118 (Oganesson) is the last element of Period 7, completing the 7p subshell. So, Z=117 is in Period 7.
- The electronic configuration follows the Aufbau order: [Rn] 7s² 5f¹⁴ 6d¹⁰ 7p⁵. (Rn is Z=86, then 32 more electrons fill 7s, 5f, 6d, 7p).
- The last electron enters the 7p subshell. Elements ending in $np^5$ belong to Group 17 (Halogens).
Element Z=117 would be in Group 17 (Halogen Family). Its electronic configuration would be [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁵.
For Z = 120:
- Element 118 completes Period 7. So, Z=119 starts Period 8, and Z=120 is also in Period 8 ($n=8$).
- The Aufbau filling order after 7p⁶ (element 118) goes to the 8s orbital.
- The electronic configuration starts with the previous noble gas (element 118, Ununoctium, Uuo). So, Z=120 will have configuration [Uuo] 8s².
- Elements ending in $ns^2$ belong to Group 2 (Alkaline Earth Metals).
Element Z=120 would be in Group 2 (Alkaline Earth Metals). Its electronic configuration would be [Uuo] 8s².
Metals, Non-Metals And Metalloids
Elements can also be broadly classified based on their general physical and chemical properties.
- Metals:
- Comprise over 78% of elements, located on the left side of the Periodic Table.
- Usually solid at room temperature (except Mercury). Ga and Cs have low melting points.
- Generally have high melting and boiling points.
- Good conductors of heat and electricity.
- Are malleable (can be hammered into sheets) and ductile (can be drawn into wires).
- Tend to lose electrons easily to form positive ions.
- Non-metals:
- Located on the top right side of the Periodic Table.
- Usually solids or gases at room temperature (except Bromine, which is liquid).
- Generally have low melting and boiling points (except Boron and Carbon).
- Poor conductors of heat and electricity.
- Solid non-metals are often brittle.
- Tend to gain electrons easily to form negative ions.
- Metalloids (Semi-metals):
- Elements located along the zigzag line separating metals and non-metals (e.g., Si, Ge, As, Sb, Te, Po, At).
- Exhibit properties intermediate between those of metals and non-metals.
- Often act as semiconductors.
The metallic character of elements increases as you move down a group and decreases as you move from left to right across a period. Conversely, non-metallic character increases across a period and decreases down a group.
Example 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:
Locate the elements in the periodic table:
- Na (Z=11): Period 3, Group 1
- Mg (Z=12): Period 3, Group 2
- Al (Z=13): Period 3, Group 13 (not in the list, but helpful for context)
- Si (Z=14): Period 3, Group 14
- P (Z=15): Period 3, Group 15
- Be (Z=4): Period 2, Group 2
Metallic character increases down a group and decreases across a period (left to right).
Comparing elements in Period 3: Na (Group 1), Mg (Group 2), Si (Group 14), P (Group 15).
Order of decreasing metallic character in Period 3 is Na > Mg > Si > P.
Comparing Be (Period 2, Group 2) and Mg (Period 3, Group 2): Metallic character increases down Group 2, so Mg > Be.
Combining these observations:
P is the most non-metallic among the given elements in Period 3. Si is next.
In Group 2, Mg is more metallic than Be.
Comparing across periods and groups: Na is the most metallic (Group 1). P is the least metallic (Group 15). Mg and Be are in Group 2, with Mg below Be, making Mg more metallic than Be. Si is in between metals and non-metals.
Relative order: Na (most metallic) > Mg > Be > Si > P (least metallic).
Therefore, 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 gradual changes and predictable patterns as you move across periods or down groups in the Periodic Table. These systematic variations are called periodic trends and are directly linked to the electronic configuration of the atoms.
Trends In Physical Properties
We will discuss periodic trends in atomic and ionic radii, ionization enthalpy, electron gain enthalpy, and electronegativity.
Atomic Radius
Defining the exact size of an atom is difficult because the electron cloud doesn't have a sharp boundary. Atomic size is often estimated from the distance between the centers of bonded atoms.
- Covalent Radius: Half the distance between the nuclei of two identical atoms joined by a single covalent bond (e.g., in a Cl₂ molecule, Cl-Cl distance is 198 pm, so covalent radius of Cl is 99 pm).
- Metallic Radius: Half the internuclear distance between two adjacent metal atoms in a metallic crystal lattice (e.g., Cu-Cu distance in solid copper is 256 pm, so 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. This is typically larger than covalent or metallic radii.
Generally, the term "Atomic Radius" is used to refer to covalent or metallic radius.
Periodic Trends in Atomic Radius:
- Across a Period (Left to Right): Atomic radius generally decreases.
Explanation: As you move across a period, electrons are added to the same valence shell. The nuclear charge (number of protons) increases, leading to a stronger attraction between the nucleus and the valence electrons. While inner electrons do provide some shielding, the effect of increased nuclear charge is more significant, pulling the electron cloud closer and reducing the atomic size.
- Down a Group (Top to Bottom): Atomic radius generally increases.
Explanation: As you move down a group, the principal quantum number ($n$) of the valence shell increases, meaning the valence electrons are in shells further away from the nucleus. Additionally, the number of inner core electrons increases, providing increased shielding of the valence electrons from the nuclear charge. This shielding effect outweighs the increase in nuclear charge, reducing the effective nuclear attraction on the valence electrons and causing the atomic size to increase.
Noble gases' radii are typically given as van der Waals radii, which are larger than covalent radii, making direct comparison within a period tricky.
Ionic Radius
The ionic radius is the effective distance from the nucleus of an ion to its outermost electrons. It is determined from distances between ions in ionic crystals.
- Cations: Positively charged ions formed by losing electrons. A cation is always smaller than its parent neutral atom.
Explanation: The nucleus has the same positive charge, but there are fewer electrons. The effective nuclear charge per electron increases, pulling the remaining electrons closer to the nucleus.
- Anions: Negatively charged ions formed by gaining electrons. An anion is always larger than its parent neutral atom.
Explanation: The nucleus has the same positive charge, but there are more electrons. The increased electron-electron repulsion within the same shell (or addition to an outer shell) outweighs the nuclear attraction, causing the electron cloud to expand.
Isoelectronic Species: These are atoms or ions that have the same number of electrons. For isoelectronic species, ionic size is determined by the nuclear charge (number of protons).
- Among isoelectronic species, the one with the highest positive charge (largest Z) will have the smallest size (electrons are pulled more strongly by the nucleus).
- Among isoelectronic species, the one with the highest negative charge (lowest Z) will have the largest size (greatest electron-electron repulsion).
Example of isoelectronic species with 10 electrons: O²⁻, F⁻, Ne, Na⁺, Mg²⁺, Al³⁺.
Their increasing order of nuclear charge (Z): O (8) < F (9) < Ne (10) < Na (11) < Mg (12) < Al (13).
Their decreasing order of size: O²⁻ > F⁻ > Ne > Na⁺ > Mg²⁺ > Al³⁺.
Example 5. Which of the following species will have the largest and the smallest size? Mg, Mg²⁺, Al, Al³⁺.
Answer:
We have neutral atoms Mg and Al, and their cations Mg²⁺ and Al³⁺.
Atomic radii decrease across a period (Mg is to the left of Al in Period 3). So, Atomic radius of Mg > Atomic radius of Al.
Cations are smaller than their parent atoms. So, Mg²⁺ is smaller than Mg, and Al³⁺ is smaller than Al.
Mg (Z=12) has 12 electrons. Mg²⁺ has 10 electrons.
Al (Z=13) has 13 electrons. Al³⁺ has 10 electrons.
Mg²⁺ and Al³⁺ are isoelectronic species (both have 10 electrons).
Compare Mg²⁺ and Al³⁺: Al³⁺ has a higher nuclear charge (+13) than Mg²⁺ (+12) for the same number of electrons. Therefore, Al³⁺ is smaller than Mg²⁺.
Combining these:
- Mg is larger than Al (across period).
- Mg is larger than Mg²⁺ (cation smaller than atom).
- Al is larger than Al³⁺ (cation smaller than atom).
- Mg²⁺ is larger than Al³⁺ (isoelectronic ions, higher charge = smaller size).
The largest species will be one of the neutral atoms. Since Mg is to the left of Al, Mg is larger than Al. So, Mg is the largest.
The smallest species will be one of the cations. Mg²⁺ and Al³⁺ are isoelectronic. Al³⁺ has a higher nuclear charge (+13) compared to Mg²⁺ (+12), pulling the 10 electrons more strongly. Therefore, Al³⁺ is the smallest.
Order of increasing size: Al³⁺ < Mg²⁺ < Al < Mg.
Ionization Enthalpy
Ionization Enthalpy ($\Delta_i H$) is the minimum energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. It is a measure of how difficult it is to remove an electron.
$$ \text{X(g)} \rightarrow \text{X⁺(g)} + \text{e⁻} \quad (\text{First Ionization Enthalpy, } \Delta_i H_1) $$Subsequent ionization enthalpies ($\Delta_i H_2$, $\Delta_i H_3$, etc.) are the energies required to remove the second, third, etc., electrons. $\Delta_i H_2 > \Delta_i H_1$ because removing an electron from a positively charged ion is harder.
Ionization enthalpies are always positive (endothermic process).
Periodic Trends in Ionization Enthalpy:
- Across a Period (Left to Right): Ionization enthalpy generally increases.
Explanation: As atomic number increases across a period, the effective nuclear charge increases while the valence electrons are in the same shell. This stronger attraction to the nucleus makes it harder to remove an electron, requiring more energy.
Exceptions exist:
- Group 13 elements (e.g., B) have lower $\Delta_i H_1$ than Group 2 (e.g., Be). This is because Group 2 has a filled $ns^2$ configuration, making it stable. Group 13 has an electron in the $np^1$ orbital, which is slightly higher in energy and less penetrating than the $ns^2$ electrons, making it easier to remove.
- Group 16 elements (e.g., O) have lower $\Delta_i H_1$ than Group 15 (e.g., N). Group 15 has a stable half-filled $np^3$ configuration (according to Hund's rule, each p orbital has one electron). Group 16 has $np^4$, meaning one p orbital has paired electrons, creating electron-electron repulsion that makes it slightly easier to remove one of the paired electrons.
- Down a Group (Top to Bottom): Ionization enthalpy generally decreases.
Explanation: As you move down a group, the valence electrons are in higher energy shells, further from the nucleus. The increased number of inner core electrons provides greater shielding of the valence electrons from the nuclear charge. This increased distance and shielding reduce the effective nuclear attraction, making it easier to remove the outermost electron, requiring less energy.
Alkali metals (Group 1) have the lowest ionization enthalpies in their periods, correlating with their high reactivity as they easily lose an electron. Noble gases (Group 18) have the highest ionization enthalpies due to their very stable closed electron shell configurations.
Example 6. The first ionization enthalpy ($\Delta_i H_1$) values of the third period elements, Na, Mg and Si are respectively 496, 737 and 786 kJ mol⁻¹. Predict whether the first $\Delta_i H_1$ value for Al will be more close to 575 or 760 kJ mol⁻¹ ? Justify your answer.
Answer:
The elements are Na (Z=11), Mg (Z=12), Al (Z=13), Si (Z=14), all in the third period.
The trend across a period is generally increasing ionization enthalpy: Na < Mg < Al < Si (with exceptions).
The given values are: Na (496), Mg (737), Si (786).
The value for Al should be between Mg and Si, but we must consider the exception for Group 13.
The electronic configurations are:
- Na: [Ne] 3s¹
- Mg: [Ne] 3s² (full 3s subshell)
- Al: [Ne] 3s² 3p¹
- Si: [Ne] 3s² 3p²
Removing the first electron from Mg involves breaking the stable, filled 3s² subshell. Removing the first electron from Al involves removing the single electron from the 3p¹ orbital.
The 3p electron in Al is slightly higher in energy and is shielded by the inner [Ne] core and the filled 3s² subshell. It is less strongly held by the nucleus compared to a 3s electron in the same shell.
Therefore, it is easier to remove the 3p electron from Al than a 3s electron from Mg, even though Al has a higher nuclear charge.
This suggests that the ionization enthalpy of Al should be lower than that of Mg.
Given options are 575 kJ mol⁻¹ and 760 kJ mol⁻¹.
Since $\Delta_i H_1$ for Mg is 737 kJ mol⁻¹, the value for Al should be less than 737 kJ mol⁻¹.
Thus, the first ionization enthalpy for Al will be more close to 575 kJ mol⁻¹.
Electron Gain Enthalpy
Electron Gain Enthalpy ($\Delta_{eg} H$) is the enthalpy change that occurs when an electron is added to an isolated gaseous atom in its ground state to form a gaseous anion.
$$ \text{X(g)} + \text{e⁻} \rightarrow \text{X⁻(g)} \quad (\Delta_{eg} H) $$Electron gain enthalpy can be exothermic (energy released, $\Delta_{eg} H$ is negative) or endothermic (energy absorbed, $\Delta_{eg} H$ is positive). A more negative $\Delta_{eg} H$ indicates that the atom has a greater tendency to accept an electron.
Examples:
- Halogens (Group 17) have very negative $\Delta_{eg} H$ because gaining one electron gives them a stable noble gas configuration.
- Noble gases (Group 18) have large positive $\Delta_{eg} H$ because adding an electron requires it to enter a new, higher energy shell, 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.
Explanation: As effective nuclear charge increases across a period, the nucleus has a stronger attraction for an incoming electron. This makes it easier for the atom to gain an electron, and more energy is released (more negative $\Delta_{eg} H$).
- Down a Group (Top to Bottom): Electron gain enthalpy generally becomes less negative.
Explanation: As atomic size increases down a group, the incoming electron is further from the nucleus and experiences weaker attraction. The increased shielding also reduces the effective nuclear charge felt by the incoming electron. This makes it harder to add an electron, and less energy is released (less negative $\Delta_{eg} H$).
Exceptions exist, particularly for the second period elements (O, F). Their $\Delta_{eg} H$ values are less negative than those of the corresponding third period elements (S, Cl). This is because the second period elements are smaller. When an electron is added to a small 2p orbital, it experiences significant repulsion from the existing electrons in that relatively compact shell. For the larger 3p orbitals in the third period, the added electron occupies a larger space, and electron-electron repulsion is reduced, making electron addition more favorable (more negative $\Delta_{eg} H$). Thus, Cl has a more negative $\Delta_{eg} H$ than F, and S has a more negative $\Delta_{eg} H$ than O.
Example 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:
Locate the elements in the periodic table:
- F (Z=9): Period 2, Group 17 (Halogen)
- Cl (Z=17): Period 3, Group 17 (Halogen)
- P (Z=15): Period 3, Group 15
- S (Z=16): Period 3, Group 16
Electron gain enthalpy generally becomes more negative across a period (P $\rightarrow$ S $\rightarrow$ Cl) and less negative down a group (F $\rightarrow$ Cl, S $\rightarrow$ Se, etc.). Noble gases (like Ne) have positive $\Delta_{eg} H$, making them the least likely to gain electrons.
Comparing P, S, Cl in Period 3: Moving from left to right, $\Delta_{eg} H$ becomes more negative. So, $\Delta_{eg} H$ of P < $\Delta_{eg} H$ of S < $\Delta_{eg} H$ of Cl (increasingly negative).
Comparing F and Cl in Group 17: Cl is below F. While $\Delta_{eg} H$ generally becomes less negative down a group, there's an exception for the second period elements. F (Period 2) is smaller than Cl (Period 3). The incoming electron in F experiences more electron-electron repulsion in the compact 2p subshell than in the larger 3p subshell of Cl. Thus, Cl has a more negative electron gain enthalpy than F.
Comparing with P (Group 15): Halogens (Group 17) readily accept an electron to achieve noble gas configuration, so their $\Delta_{eg} H$ values are generally much more negative than elements in Group 15.
Therefore, among P, S, Cl, F:
- Cl has the most negative electron gain enthalpy (highest tendency to accept an electron).
- P, being to the left and not needing just one electron for stability, will have the least negative (or possibly even positive, though typically negative) electron gain enthalpy among the non-metals listed from Period 3. P needs 3 electrons to achieve noble gas configuration, whereas S needs 2 and Cl needs 1. Accepting the first electron is generally less favorable further from Group 17 or 16. Group 15 elements (like P) tend to have less negative $\Delta_{eg}H$ than Group 16 and 17 elements in the same period.
Thus, the element with the most negative electron gain enthalpy is Chlorine (Cl).
The element with the least negative electron gain enthalpy is Phosphorus (P).
Order of increasing (less negative to more negative) electron gain enthalpy: P < S < F < Cl.
Electronegativity
Electronegativity is a qualitative measure of the ability of an atom in a chemical compound to attract a shared pair of electrons towards itself. It is not a directly measurable quantity but is calculated based on various scales (e.g., Pauling scale, Mulliken scale).
On the widely used Pauling scale, Fluorine (F) is assigned the highest electronegativity value (4.0).
Electronegativity values are not constant for an element but vary depending on the atoms it is bonded to and its oxidation state.
Periodic Trends in Electronegativity:
- Across a Period (Left to Right): Electronegativity generally increases.
Explanation: As atomic size decreases and effective nuclear charge increases across a period, the nucleus has a stronger pull on the shared electrons in a bond.
- Down a Group (Top to Bottom): Electronegativity generally decreases.
Explanation: As atomic size increases and the distance between the nucleus and valence shell increases down a group, the nucleus's pull on the shared electrons weakens due to increased distance and shielding.
Relationship with Metallic and Non-metallic Character:
- Non-metals tend to gain electrons and are generally highly electronegative. Electronegativity is directly related to non-metallic character.
- Metals tend to lose electrons and have low electronegativity. Electronegativity is inversely related to metallic character.
As electronegativity increases across a period, non-metallic character increases, and metallic character decreases. As electronegativity decreases down a group, metallic character increases, and non-metallic character decreases.
Periodic Trends In Chemical Properties
The chemical properties of elements are determined by their electronic configurations, particularly the valence electrons. Periodic trends in fundamental properties like ionization enthalpy and electron gain enthalpy directly influence chemical reactivity and the nature of compounds formed.
Periodicity Of Valence Or Oxidation States
The valence of a representative element is often related to the number of electrons in its outermost shell or eight minus that number. The term oxidation state is frequently used and refers to the charge an atom appears to have in a compound based on electronegativity differences.
General Trend in Valence (for Representative Elements):
| Group | 1 | 2 | 13 | 14 | 15 | 16 | 17 | 18 |
|---|---|---|---|---|---|---|---|---|
| Number of valence electrons | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Valence (Common) | 1 | 2 | 3 | 4 | 3,5 | 2,6 | 1,7 | 0,8 |
Elements in Group 1 typically show a valence of 1 (losing one $ns^1$ electron). Group 2 shows 2 (losing two $ns^2$ electrons). Group 13 shows 3 (losing $ns^2np^1$). Group 14 shows 4 (losing $ns^2np^2$). Groups 15, 16, and 17 often show valence equal to 8 minus the number of valence electrons (gaining electrons to achieve octet), or variable valency by involving d orbitals.
Oxidation state considers the charge based on electronegativity differences. In OF₂, Oxygen is less electronegative than Fluorine, so Oxygen has a positive oxidation state (+2), while F is -1. In Na₂O, Oxygen is more electronegative than Sodium, so Oxygen is -2, and Na is +1.
Transition elements (d-block) and inner-transition elements (f-block) often exhibit variable valency/oxidation states due to the involvement of inner d or f electrons in bonding.
Example 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 stable binary compound, we can use the typical valencies (combining capacity) or oxidation states of the elements.
(a) Silicon (Si) and Bromine (Br):
- Silicon (Si) is in Group 14, Period 3. It typically has a valence of 4.
- Bromine (Br) is in Group 17 (Halogens), Period 4. It typically has a valence of 1 (gain 1 electron) when bonding with less electronegative elements like Si.
To balance the valencies (4 for Si, 1 for Br), we need 4 Br atoms for every 1 Si atom.
Formula = Si₁Br₄ = SiBr₄.
(b) Aluminium (Al) and Sulphur (S):
- Aluminium (Al) is in Group 13, Period 3. It typically has a valence of 3 (lose 3 electrons).
- Sulphur (S) is in Group 16, Period 3. It typically has a valence of 2 (gain 2 electrons) when bonding with more electropositive elements like Al.
To balance the valencies (3 for Al, 2 for S), we need 2 Al atoms ($2 \times 3 = 6$) and 3 S atoms ($3 \times 2 = 6$).
Formula = Al₂S₃ = Al₂S₃.
Anomalous Properties Of Second Period Elements
The first element of each group in the s-block and p-block (Li, Be, B, C, N, O, F) exhibits some properties that are different from the other members of their respective groups. This is known as the anomalous behavior of second period elements.
Examples:
- Lithium (Li) and Beryllium (Be) form compounds with more covalent character compared to other alkali and alkaline earth metals, which form predominantly ionic compounds.
- The properties of the first element of a group are often more similar to the second element of the next group. This is called the diagonal relationship (e.g., Li resembles Mg, Be resembles Al).
Reasons for anomalous behavior:
- Small Size: Second period elements are significantly smaller than other elements in their group.
- Large Charge/Radius Ratio: Due to small size, they have a high concentration of charge relative to their size, leading to greater polarizing power.
- High Electronegativity: They are generally more electronegative than heavier elements in their group.
- Absence of d-orbitals: The valence shell of second period elements (n=2) has only s and p orbitals (2s, 2p), totaling 4 valence orbitals. This limits their maximum covalency to 4. Heavier elements in the same groups have access to vacant d-orbitals in their valence shell (e.g., 3s, 3p, 3d for Period 3), allowing them to expand their valence shell and show covalencies greater than 4 (e.g., P can form PCl₅).
- Ability to form $p\pi - p\pi$ multiple bonds: Second period elements (especially C, N, O) can form stable double and triple bonds with themselves (e.g., C=C, N≡N) and with other second period elements (e.g., C=O, C≡N). Heavier elements form weaker $p\pi - p\pi$ bonds and prefer single bonds or $d\pi - p\pi$ or $d\pi - d\pi$ bonding.
Example 9. Are the oxidation state and covalency of Al in [AlCl(H₂O)₅]²⁺ same ?
Answer:
In the complex ion [AlCl(H₂O)₅]²⁺:
- The overall charge on the complex is +2.
- Water (H₂O) is a neutral ligand (oxidation state = 0).
- Chloride (Cl) is a ligand with a charge of -1 (oxidation state = -1).
- Let the oxidation state of Aluminium (Al) be $x$.
The sum of oxidation states must equal the charge on the complex:
$x + (\text{oxidation state of Cl}) + 5 \times (\text{oxidation state of H₂O}) = +2$
$x + (-1) + 5 \times (0) = +2$
$x - 1 = +2$
$x = +3$
So, the oxidation state of Al is +3.
The covalency of Al in this complex is the number of bonds Al forms with the ligands. Al is bonded to one Cl atom and five H₂O molecules, totaling $1 + 5 = 6$ bonds.
So, the covalency of Al is 6.
Therefore, the oxidation state (+3) and the covalency (6) of Al in [AlCl(H₂O)₅]²⁺ are not the same.
Periodic Trends And Chemical Reactivity
The chemical reactivity of an element is closely related to its tendency to lose or gain electrons, which in turn depends on properties like ionization enthalpy and electron gain enthalpy.
Trends in Reactivity:
- Across a Period: Chemical reactivity is generally high at the extremes (left and right) and lowest in the center.
- Left side (Alkali Metals): Highly reactive, easily lose electrons (low $\Delta_i H$). Reactivity increases down the group.
- Right side (Halogens): Highly reactive (except noble gases), easily gain electrons (highly negative $\Delta_{eg} H$). Reactivity decreases down the group.
- Center (e.g., Group 14): Lower reactivity compared to extremes.
- Down a Group:
- For Metals (e.g., Alkali Metals, Alkaline Earth Metals): Reactivity increases down the group due to decreasing ionization enthalpy (easier to lose electrons).
- For Non-metals (e.g., Halogens): Reactivity decreases down the group due to less negative electron gain enthalpy (harder to gain electrons).
Metals on the left side tend to form basic oxides (e.g., Na₂O + H₂O $\rightarrow$ NaOH). Non-metals on the right side tend to form acidic oxides (e.g., Cl₂O₇ + H₂O $\rightarrow$ HClO₄). Elements in the center can form amphoteric oxides (react with both acids and bases, e.g., Al₂O₃) or neutral oxides (e.g., CO, NO).
Summary of Major Periodic Trends:
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?
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 $\Delta_i H$ than B
(ii) O has lower $\Delta_i H$ 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?
Answer:
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?
Answer:
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
Answer:
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.
Answer:
Question 3.22. What is the basic difference between the terms electron gain enthalpy and electronegativity?
Answer:
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?
Answer:
Question 3.24. Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron
Answer:
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.
Answer:
Question 3.26. What are the major differences between metals and non-metals?
Answer:
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.
Answer:
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 > Cl > Br > I. Explain.
Answer:
Question 3.29. Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Answer:
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.
Answer:
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 | $\Delta H_1$ | $\Delta H_2$ | $\Delta_{eg}H$ |
|---|---|---|---|
| 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.
Answer:
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: