Dobereiner's Triads: Classification of Elements into Groups of Three with Similar Properties
Introduction to Early Classification Attempts
Before the periodic table was fully developed, several scientists attempted to classify elements based on their physical and chemical properties. One of the earliest attempts was made by Johann Wolfgang Döbereiner, a German chemist, who observed that certain elements could be grouped into sets of three based on their similarities in properties. These sets became known as Döbereiner's Triads.
Döbereiner’s triads were an early stepping stone in the development of the periodic law and the eventual arrangement of elements into the periodic table. While Döbereiner's attempt was limited in scope, it laid the groundwork for later scientists, such as Mendeleev, who eventually organized all known elements in a comprehensive and logical system.
What are Döbereiner's Triads?
Döbereiner’s triads refer to a classification system in which elements with similar chemical properties were grouped into sets of three. These triads followed a distinct mathematical relationship: the atomic mass of the middle element in the triad was approximately the average of the atomic masses of the other two elements.
Characteristics of Döbereiner's Triads:
1. Grouping Based on Similar Properties: Elements in each triad exhibited similar chemical and physical properties.
2. Mathematical Relationship: The atomic mass of the middle element was roughly the arithmetic mean of the atomic masses of the other two elements.
3. Chemical Reactivity: The elements in a triad showed gradual changes in reactivity, with the middle element’s reactivity often being intermediate between the other two.
Examples of Döbereiner's Triads
1. Alkali Metal Triad: Lithium (Li), Sodium (Na), Potassium (K)
- Atomic Masses:
- Lithium (Li): 6.9
- Sodium (Na): 23
- Potassium (K): 39.1
- Mathematical Relationship:
The average atomic mass of lithium and potassium is approximately 23, which is close to the atomic mass of sodium.
\[
\text{Average} = \frac{6.9 + 39.1}{2} = 23
\]
- Chemical Properties: All three elements are highly reactive alkali metals that form strong bases when combined with water (alkalies). Their reactivity increases down the group (Li < Na < K).
2. Halogen Triad: Chlorine (Cl), Bromine (Br), Iodine (I)
- Atomic Masses:
- Chlorine (Cl): 35.5
- Bromine (Br): 80
- Iodine (I): 127
- Mathematical Relationship:
The average atomic mass of chlorine and iodine is approximately 80, which is close to the atomic mass of bromine.
\[
\text{Average} = \frac{35.5 + 127}{2} = 81.25
\]
- Chemical Properties: All three elements are halogens, which means they react with metals to form salts. Their reactivity decreases down the group (Cl > Br > I).
3. Alkaline Earth Metal Triad: Calcium (Ca), Strontium (Sr), Barium (Ba)
- Atomic Masses:
- Calcium (Ca): 40
- Strontium (Sr): 87.6
- Barium (Ba): 137
- Mathematical Relationship:
The average atomic mass of calcium and barium is approximately 87.6, which is close to the atomic mass of strontium.
\[
\text{Average} = \frac{40 + 137}{2} = 88.5
\]
- Chemical Properties: All three elements are reactive alkaline earth metals, forming strong bases with water, and their reactivity increases down the group.
Explanation of the Triads' Significance
Although Döbereiner's system could only classify a few elements into triads, it was significant because it hinted at a relationship between atomic masses and chemical properties. This relationship was an early precursor to the idea of periodicity, which was later developed in full by Dmitri Mendeleev.
Döbereiner’s triads demonstrated that properties of elements change in a systematic way with increasing atomic mass. This idea is the foundation of the modern periodic table, where elements are arranged by increasing atomic number and show repeating trends in properties.
Limitations of Döbereiner’s Triads
1. Limited Number of Elements:
Döbereiner could only classify a few elements into triads. For example, elements such as iron (Fe), gold (Au), and platinum (Pt) did not fit into any triad, limiting the scope of the system.
2. No Explanation for Atomic Mass Variations:
Döbereiner’s model did not provide an explanation for why the atomic mass of the middle element was the average of the other two, nor did it predict how properties change within a triad.
3. Inability to Classify All Known Elements:
As more elements were discovered, it became clear that not all elements could be grouped into triads, highlighting the need for a more comprehensive system of classification.
Relation to Modern Periodic Table
In modern chemistry, elements are no longer classified into triads. Instead, they are organized in the periodic table according to increasing atomic number. However, Döbereiner’s triads provided early evidence that chemical properties are related to atomic mass (and later atomic number), helping to pave the way for the development of the periodic law and the periodic table.
For example, the alkali metals (Group 1 elements) today are known to have similar properties due to having one electron in their outermost shell. The triad of lithium, sodium, and potassium fits well into this group, demonstrating that Döbereiner’s observations were valid and insightful even with the limited knowledge of his time.
Real-Life Applications and Examples
1. Chemical Reactivity:
The alkali metals (Li, Na, K) show an increase in reactivity down the group. This pattern is used in chemical industries to predict reactions. For instance, potassium reacts more vigorously with water than sodium, a fact that can be predicted by their position in the triad.
2. Halogens in Pharmaceuticals:
Halogens like chlorine, bromine, and iodine are frequently used in the pharmaceutical industry. Understanding their reactivity (which decreases down the group) is essential for their safe and effective use in medical treatments.
Conclusion
Döbereiner’s triads were a significant early attempt to classify elements based on their atomic masses and properties. Although the system was limited, it laid the groundwork for the periodic classification of elements. The systematic increase in atomic mass and the gradual changes in chemical properties within the triads foreshadowed the periodic trends that would later be formalized in the periodic table.
IUPAC Nomenclature System
The International Union of Pure and Applied Chemistry (IUPAC) is responsible for creating standardized rules for naming chemical elements and compounds. The IUPAC naming system is used globally to ensure that elements and compounds are named systematically, enabling scientists from different parts of the world to communicate effectively. The IUPAC system applies to both newly discovered elements and compounds, ensuring consistency.
Systematic Naming of Elements with Atomic Numbers ≥ 100
For elements with atomic numbers 100 and above, IUPAC has introduced a systematic naming convention using Latin roots for numbers. This systematic nomenclature is used as a temporary placeholder until a formal name is decided, based on the element’s discovery or honorific name (usually a scientist).
The system uses the following Latin and Greek numeral roots for naming:
| Number | Root |
|--------|--------|
| 0 | nil |
| 1 | un |
| 2 | bi |
| 3 | tri |
| 4 | quad |
| 5 | pent |
| 6 | hex |
| 7 | sept |
| 8 | oct |
| 9 | enn |
General Formula for Naming:
\[
\text{Name of element} = \text{Root of the digits of atomic number} + \text{“ium”}
\]
Examples of IUPAC Naming for Higher Atomic Numbers
1. Element 104:
- IUPAC Name: Unnilquadium (from Un=1, Nil=0, Quad=4)
- Symbol: Unq
- Official Name: Rutherfordium (Rf)
2. Element 105:
- IUPAC Name: Unnilpentium (from Un=1, Nil=0, Pent=5)
- Symbol: Unp
- Official Name: Dubnium (Db)
3. Element 106:
- IUPAC Name: Unnilhexium (from Un=1, Nil=0, Hex=6)
- Symbol: Unh
- Official Name: Seaborgium (Sg)
4. Element 107:
- IUPAC Name: Unnilseptium (from Un=1, Nil=0, Sept=7)
- Symbol: Uns
- Official Name: Bohrium (Bh)
5. Element 108:
- IUPAC Name: Unniloctium (from Un=1, Nil=0, Oct=8)
- Symbol: Uno
- Official Name: Hassium (Hs)
6. Element 109:
- IUPAC Name: Unnilennium (from Un=1, Nil=0, Enn=9)
- Symbol: Une
- Official Name: Meitnerium (Mt)
These names are used as placeholders until an element’s discovery is confirmed, and it is given a permanent name, typically in honor of a scientist, place, or characteristic of the element.
---
Special Case: Unnilunium (Element 101, Now Known as Mendelevium)
Element 101 was originally named Unnilunium following the IUPAC systematic nomenclature rules. The name comes from the Latin roots for its atomic number:
- Un = 1
- Nil = 0
- Un = 1
Thus, Unnilunium (UnUn) was the temporary IUPAC name for this element. However, once its discovery was confirmed, the element was renamed Mendelevium (Md) in honor of Dmitri Mendeleev, the creator of the periodic table.
Properties of Mendelevium:
- Symbol: Md
- Atomic Number: 101
- Group: Actinides
- Electron Configuration: \([Rn] 5f^{13} 7s^2\)
- First Ionization Energy: 6.58 eV
- Oxidation States: +2, +3
- Discovery: Synthesized by bombarding einsteinium \((_Z^{253}Es\)) with alpha particles in 1955 by a team led by Albert Ghiorso at the University of California, Berkeley.
Synthesis Reaction:
\[
\ _{99}^{253}Es + _{2}^{4}He \rightarrow _{101}^{256}Md + 2n
\]
Applications:
Mendelevium is a synthetic element that does not occur naturally. Due to its radioactivity, it has no large-scale industrial applications and is used primarily for scientific research, especially in the study of heavy elements and nuclear properties.
Other IUPAC Naming Examples Relevant to JEE Advanced
1. Element 110:
- IUPAC Name: Ununnilium (Un=1, Nil=0)
- Symbol: Uun
- Official Name: Darmstadtium (Ds)
2. Element 111:
- IUPAC Name: Unununium (Un=1, Un=1, Un=1)
- Symbol: Uuu
- Official Name: Roentgenium (Rg)
3. Element 112:
- IUPAC Name: Ununbium (Un=1, Un=1, Bi=2)
- Symbol: Uub
- Official Name: Copernicium (Cn)
4. Element 113:
- IUPAC Name: Ununtrium (Un=1, Un=1, Tri=3)
- Symbol: Uut
- Official Name: Nihonium (Nh)
5. Element 114:
- IUPAC Name: Ununquadium (Un=1, Un=1, Quad=4)
- Symbol: Uuq
- Official Name: Flerovium (Fl)
The IUPAC nomenclature provides a systematic way to name newly discovered elements based on their atomic number. For JEE Advanced, it is essential to be familiar with the IUPAC rules and understand how the temporary naming system works, especially for elements with atomic numbers greater than 100. You should also be familiar with the permanent names of these elements and their properties, particularly the ones that are part of the actinide and transactinide series.
Atomic Radii Trend in Alkali Metals
What is Atomic Radius?
Atomic radius is defined as the distance from the nucleus of an atom to the outermost shell of electrons. This is a crucial property that helps to understand the size of an atom and its chemical behavior, particularly when it comes to bonding and reactivity. However, since atoms do not have sharply defined boundaries, the atomic radius is often considered the average distance between the nuclei of two atoms in a homonuclear diatomic molecule.
There are different types of atomic radii:
1. Covalent Radius: Half of the distance between the nuclei of two identical atoms bonded by a covalent bond.
2. Van der Waals Radius: Half the distance between two non-bonded atoms when they are closest to each other.
3. Metallic Radius: Half the distance between the nuclei of two adjacent atoms in a metallic lattice.
General Trends in Atomic Radius
1. Across a Period:
- Atomic radius decreases across a period from left to right in the periodic table. This is due to the increasing nuclear charge, which pulls the electron cloud closer to the nucleus without a corresponding increase in shielding.
2. Down a Group:
- Atomic radius increases as you move down a group due to the addition of new electron shells (energy levels). As new shells are added, the distance between the nucleus and the outermost electron increases, and electron shielding reduces the effect of the nucleus’s pull on the outer electrons.
Atomic Radii Trend in Alkali Metals (Group 1 Elements)
Alkali metals, which include lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr), exhibit a significant increase in atomic radii as you move down the group. These elements have one electron in their outermost shell, and as more electron shells are added, the outermost electron becomes farther away from the nucleus, causing the atomic size to increase.
| Element | Atomic Radius (pm) |
|-------------|-------------------------|
| Lithium (Li) | 152 pm |
| Sodium (Na) | 186 pm |
| Potassium (K)| 231 pm |
| Rubidium (Rb)| 244 pm |
| Cesium (Cs) | 262 pm |
| Francium (Fr)| ~270 pm |
Reasons for the Increase in Atomic Radii down the Group:
1. Addition of Energy Levels:
- Each successive element in Group 1 has an additional electron shell compared to the element above it. For example, sodium has three shells (\(1s^2 2s^2 2p^6 3s^1\)) compared to lithium's two shells (\(1s^2 2s^1\)), which increases the atomic radius significantly.
2. Increased Electron Shielding:
- As more electron shells are added, the inner shells "shield" the outermost electron from the full attractive force of the nucleus. This causes the outer electron to be less tightly bound, further increasing the atomic radius.
3. Decreased Effective Nuclear Charge:
- While the number of protons in the nucleus increases down the group, the shielding effect offsets the increase in nuclear charge. The effective nuclear charge felt by the outermost electron remains relatively low, allowing the atom to expand.
Comparison of Alkali Metals’ Atomic Radii with Other Groups
Compared to other groups, alkali metals have the largest atomic radii in their respective periods. This is because they have fewer electrons in the outermost shell and experience less effective nuclear charge than elements with more valence electrons. The increased distance between the nucleus and the outermost electron also makes alkali metals highly reactive, as they can easily lose their outermost electron.
Example Comparison:
- In Period 2, lithium (Li) has an atomic radius of 152 pm, while fluorine (F), at the end of the period, has a much smaller atomic radius of 64 pm. This stark difference illustrates the trend of decreasing atomic radius across a period.
Effects of Large Atomic Radius in Alkali Metals
1. Lower Ionization Energy:
As atomic radii increase down the group, the outermost electron is farther from the nucleus and more easily removed. This explains why alkali metals have low ionization energies and why ionization energy decreases down the group.
- Example:
The ionization energy of lithium is 520 kJ/mol, while for cesium, it drops to 375 kJ/mol.
2. Increasing Reactivity:
The large atomic radii of alkali metals, combined with their low ionization energy, make them highly reactive. Reactivity increases down the group because the outermost electron is more easily lost, facilitating the formation of positive ions (cations).
- Example:
Potassium reacts more vigorously with water compared to sodium, which in turn reacts more vigorously than lithium.
3. Metallic Character:
As the atomic radius increases, the metallic character of alkali metals also increases. Larger atoms are more likely to lose their outer electrons and form metallic bonds, making them better conductors of heat and electricity.
Related Periodic Trends
- Electronegativity: Decreases down the group as atomic size increases.
- Electron Affinity: Becomes less negative down the group due to the larger atomic size.
- Ionic Radius: Increases down the group, and the cations formed by alkali metals (e.g., \(Li^+\), \(Na^+\)) are smaller than their parent atoms because they lose their outermost shell during ionization.
Real-Life Application of Atomic Radii Trends
1. Reactivity with Water:
- Alkali metals react with water to produce hydroxides and hydrogen gas. The reactivity of these metals increases as their atomic radii increase. For instance, sodium reacts more vigorously with water than lithium, while potassium reacts even more explosively.
- Reaction:
\[
2Na + 2H_2O \rightarrow 2NaOH + H_2 \uparrow
\]
2. Usage in Batteries:
- Lithium is widely used in batteries because of its small atomic size, which allows for high charge density. Larger alkali metals like potassium and sodium are less suitable for such applications due to their larger atomic radii and reactivity.
Heavy Water is Deuterium Oxide (D\(_2\)O): Occurs in Compounds Where Hydrogen is Replaced by Deuterium
Introduction to Deuterium
Deuterium, symbolized as \(D\), is an isotope of hydrogen. Unlike ordinary hydrogen, which consists of one proton and one electron, deuterium has one proton, one neutron, and one electron. This extra neutron makes deuterium approximately twice as heavy as hydrogen, hence the term “heavy hydrogen.”
Deuterium occurs naturally in small quantities in water. For every 6,400 ordinary hydrogen atoms, there is approximately 1 deuterium atom. When deuterium atoms replace the hydrogen atoms in water, the result is heavy water, chemically known as deuterium oxide (D\(_2\)O).
What is Heavy Water (D\(_2\)O)?
Heavy water is a form of water in which the two hydrogen atoms in the molecule are replaced with deuterium atoms. Since deuterium is heavier than hydrogen, heavy water has distinct physical and chemical properties compared to regular water.
- Chemical Formula: D\(_2\)O (Deuterium Oxide)
- Molecular Structure: Similar to ordinary water (H\(_2\)O), but with deuterium atoms instead of hydrogen.
| Property | H\(_2\)O (Regular Water) | D\(_2\)O (Heavy Water) |
|--||-|
| Boiling Point | 100°C | 101.4°C |
| Freezing Point | 0°C | 3.8°C |
| Density | 1.00 g/cm³ | 1.11 g/cm³ |
Properties of Heavy Water
1. Physical Properties:
- Higher Boiling and Freezing Points: Due to the heavier deuterium atoms, heavy water has slightly higher boiling and freezing points compared to ordinary water.
- Greater Density: D\(_2\)O is denser than regular water. This makes it heavier and is the reason for its name “heavy water.”
- Increased Bond Strength: The O-D bond in heavy water is stronger than the O-H bond in regular water due to the mass difference between deuterium and hydrogen.
2. Chemical Properties:
- Isotopic Substitution: In chemical reactions, deuterium behaves similarly to hydrogen, but the rate of reactions involving D\(_2\)O can be slower than reactions with H\(_2\)O because of the greater mass of deuterium. This phenomenon is known as the kinetic isotope effect.
- Neutron Moderation: Heavy water is not radioactive itself, but it has unique properties in nuclear reactors due to deuterium’s ability to moderate neutrons without absorbing them.
Formation and Occurrence of Heavy Water
Heavy water occurs naturally, though in very small amounts, as a part of natural water (H\(_2\)O) due to the presence of deuterium in the hydrogen content of water. For every 6,400 molecules of H\(_2\)O, one molecule of D\(_2\)O can be found. Although the concentration of heavy water in natural sources is low, it can be concentrated and extracted through electrolysis or distillation processes.
Uses of Heavy Water (D\(_2\)O)
1. Nuclear Reactors:
- Heavy water is primarily used as a neutron moderator in certain types of nuclear reactors, such as the CANDU (Canada Deuterium Uranium) reactors. In these reactors, heavy water slows down fast-moving neutrons, making it easier for nuclear fission to occur without capturing the neutrons.
- Heavy water is preferred over regular water in some reactors because deuterium does not absorb neutrons as readily as ordinary hydrogen does.
2. Biological and Chemical Research:
- Due to the mass difference between hydrogen and deuterium, heavy water is used in isotope tracing studies. Researchers can use D\(_2\)O to study the pathways of water in biological systems because the body processes D\(_2\)O similarly to H\(_2\)O.
- Heavy water is used in experiments to study reaction rates and the kinetic isotope effect, where chemical reactions involving deuterium proceed at different rates compared to those involving hydrogen.
3. Hydrogen Bonding Studies:
- Heavy water is valuable for studying hydrogen bonding and its effects on molecular structures. The stronger O-D bond in heavy water is used to model and study hydrogen bonds in other chemical systems.
Differences Between H\(_2\)O and D\(_2\)O
While D\(_2\)O and H\(_2\)O are chemically very similar, the substitution of deuterium for hydrogen in water results in subtle but significant differences in physical and chemical properties.
1. Boiling and Freezing Points:
- D\(_2\)O has a higher boiling point (101.4°C) and freezing point (3.8°C) compared to H\(_2\)O. This makes heavy water behave differently in temperature-sensitive environments.
2. Density:
- Heavy water is denser than regular water, making it about 11% heavier. This property is used to distinguish between the two in separation processes.
3. Kinetic Isotope Effect:
- The presence of deuterium affects the rate of chemical reactions. Reactions involving D\(_2\)O often proceed more slowly than those involving H\(_2\)O due to the greater mass of deuterium, which leads to stronger bonds and a higher activation energy for reactions.
Biological Effects of Heavy Water
While D\(_2\)O is not radioactive or acutely toxic, consuming large quantities of heavy water (i.e., replacing more than 20% of the body's water with D\(_2\)O) can interfere with normal cellular processes and inhibit cell division, leading to biological effects. This is because deuterium forms stronger bonds than hydrogen, and biological processes evolved to work with the lighter hydrogen isotope.
- Short-Term Use: Small quantities of heavy water are generally harmless and are used in scientific research. However, prolonged exposure or ingestion in large amounts can cause disturbances in cellular metabolism.
Example: Heavy Water in CANDU Reactors
In the Canadian CANDU reactors, heavy water plays a critical role as a moderator. The reaction chain for fission in such reactors can be summarized as:
\[
^{235}U + n \rightarrow ^{92}Kr + ^{141}Ba + 3n + \text{energy}
\]
In this reaction, the heavy water moderates the speed of the emitted neutrons, making them more effective in sustaining the chain reaction. Without heavy water, the reactor would not be able to maintain a steady rate of nuclear fission.
Production of Heavy Water
1. Electrolysis:
- The electrolysis of water involves passing an electric current through water to separate oxygen and hydrogen. Since deuterium is heavier than hydrogen, it tends to accumulate in the remaining water, allowing for the extraction of heavy water over time.
2. Distillation:
- Heavy water can also be separated from regular water by fractional distillation, where the higher boiling point of D\(_2\)O allows for its collection at a different stage than H\(_2\)O.
Related Concept: Deuterium Exchange in Organic Compounds
In organic chemistry, deuterium can replace hydrogen in organic molecules, leading to deuterated compounds. These compounds are often used in NMR (Nuclear Magnetic Resonance) spectroscopy and other analytical techniques. Deuterium-labeled compounds help scientists trace reaction pathways and study mechanisms in greater detail.
Conclusion
Heavy water (D\(_2\)O) is a crucial substance in both nuclear and chemical research. Its unique properties, resulting from the substitution of deuterium for hydrogen, make it valuable in a wide range of applications, from moderating nuclear reactions to studying biological processes. Understanding the differences between regular water and heavy water, and how isotopic substitution affects physical and chemical properties, is essential for mastering the concepts required for JEE Advanced.
Ionic Compounds Causing Hardness in Water: Presence of Ions Like Ca\(^{2+}\) and Mg\(^{2+}\)
What is Water Hardness?
Water hardness refers to the concentration of certain dissolved minerals, primarily calcium (\(Ca^{2+}\)) and magnesium (\(Mg^{2+}\)) ions, in water. These ions, when present in significant quantities, react with soap to form an insoluble substance called "soap scum" and can lead to the buildup of scale in pipes and boilers. Hard water does not lather well with soap, making it less effective for cleaning purposes.
Water hardness is generally classified into two types:
1. Temporary Hardness:
Caused by the presence of bicarbonate ions \((HCO_3^-)\) of calcium and magnesium, which can be removed by boiling the water. Boiling causes the bicarbonates to decompose into insoluble carbonates, which can be removed through filtration.
2. Permanent Hardness:
Caused by the presence of sulfate \((SO_4^{2-})\), chloride \((Cl^-)\), and nitrate \((NO_3^-)\) salts of calcium and magnesium. Permanent hardness cannot be removed by boiling and requires chemical treatment for softening.
Ions Responsible for Hardness
1. Calcium Ions (Ca\(^{2+}\)):
- Calcium ions are a primary cause of water hardness. These ions come from the dissolution of calcium-containing minerals like limestone \((CaCO_3)\), gypsum \((CaSO_4 \cdot 2H_2O)\), and dolomite \((CaMg(CO_3)_2)\) into water.
- When calcium ions are present in water, they react with soap (fatty acid salts) to form insoluble calcium salts, which precipitate as soap scum.
Chemical Equation:
\[
2C_{17}H_{35}COO^-Na^+ + Ca^{2+} \rightarrow (C_{17}H_{35}COO)_2Ca \, (\text{insoluble soap scum}) + 2Na^+
\]
2. Magnesium Ions (Mg\(^{2+}\)):
- Like calcium, magnesium ions also contribute to water hardness. Magnesium ions are found in minerals such as dolomite \((CaMg(CO_3)_2)\) and magnesite \((MgCO_3)\).
- Magnesium ions similarly react with soap to form insoluble compounds, reducing soap’s effectiveness in hard water.
Chemical Equation:
\[
2C_{17}H_{35}COO^-Na^+ + Mg^{2+} \rightarrow (C_{17}H_{35}COO)_2Mg \, (\text{insoluble soap scum}) + 2Na^+
\]
Effects of Water Hardness
1. Formation of Scale:
- The most notable effect of water hardness is the formation of scale in pipes, boilers, and other heating systems. Scale is primarily composed of insoluble salts, such as calcium carbonate (\(CaCO_3\)) and magnesium carbonate (\(MgCO_3\)), which precipitate when hard water is heated or left to stand.
- Example of Scale Formation:
When hard water containing calcium bicarbonate \((Ca(HCO_3)_2)\) is heated, the bicarbonate decomposes into calcium carbonate, carbon dioxide, and water:
\[
Ca(HCO_3)_2 \xrightarrow{\Delta} CaCO_3 \, (\text{insoluble scale}) + CO_2 + H_2O
\]
- This scaling reduces the efficiency of heating systems, increasing energy consumption, and can cause blockages in pipes, leading to costly repairs in industrial and domestic settings.
2. Reduced Efficiency of Soap:
- Hard water reduces the cleaning power of soap by forming insoluble compounds with the soap molecules. This results in the formation of soap scum, which does not dissolve in water, leading to inefficient cleaning. More soap is required to achieve the same cleaning effect in hard water compared to soft water.
Temporary and Permanent Hardness
1. Temporary Hardness:
- Temporary hardness is mainly caused by the presence of bicarbonates of calcium and magnesium. It can be removed by boiling the water, which causes the bicarbonates to decompose into insoluble carbonates that can be filtered out.
- Example Reaction:
\[
Ca(HCO_3)_2 \xrightarrow{\Delta} CaCO_3 \, (\text{precipitate}) + CO_2 + H_2O
\]
- Key Process: Boiling drives off carbon dioxide and converts soluble bicarbonates into insoluble carbonates, which can then be removed by filtration.
2. Permanent Hardness:
- Permanent hardness is caused by the presence of sulfates, chlorides, and nitrates of calcium and magnesium. Unlike temporary hardness, permanent hardness cannot be removed by boiling and requires chemical treatment for softening.
- Treatment Methods:
- Ion Exchange: Involves replacing \(Ca^{2+}\) and \(Mg^{2+}\) ions with sodium \((Na^+)\) or potassium \((K^+)\) ions through ion exchange resins.
- Addition of Chelating Agents: Compounds like EDTA (Ethylenediaminetetraacetic acid) can be used to bind calcium and magnesium ions, preventing them from precipitating as scale.
Softening Hard Water
To make hard water more suitable for domestic or industrial use, several methods are employed to remove calcium and magnesium ions:
1. Boiling:
- Effective for removing temporary hardness by precipitating bicarbonates of calcium and magnesium as carbonates.
2. Ion Exchange Method:
- One of the most common methods for softening both temporary and permanent hard water. It involves exchanging \(Ca^{2+}\) and \(Mg^{2+}\) ions in hard water with \(Na^+\) or \(K^+\) ions. This is done using an ion-exchange resin.
Chemical Reaction:
\[
Ca^{2+} + 2Na^+ \, (\text{resin}) \rightarrow Na_2 + Ca^{2+} \, (\text{on resin})
\]
3. Lime-Soda Process:
- Calcium hydroxide (lime) and sodium carbonate (soda) are added to hard water to precipitate calcium as calcium carbonate and magnesium as magnesium hydroxide.
Reactions:
\[
Ca^{2+} + CO_3^{2-} \rightarrow CaCO_3 \, (\text{precipitate})
\]
\[
Mg^{2+} + 2OH^- \rightarrow Mg(OH)_2 \, (\text{precipitate})
\]
Health Impacts of Hard Water
1. Mineral Intake:
- Although hard water can be inconvenient in terms of cleaning and industrial processes, it does provide essential minerals like calcium and magnesium, which are beneficial for health, particularly for bones and teeth.
2. Health Risks:
- Very high concentrations of calcium and magnesium, however, can contribute to kidney stone formation in some individuals, although this is generally rare from drinking water alone.
Industrial and Domestic Problems Associated with Hard Water
1. Industrial Issues:
- Hard water leads to scale buildup in boilers and cooling towers, reducing their efficiency and increasing maintenance costs. The scale formed on the interior surfaces of pipes reduces the flow of water and decreases the efficiency of heat transfer in industrial equipment, leading to higher energy consumption and frequent system failures.
2. Domestic Problems:
- In homes, hard water leads to soap scum buildup in sinks, bathtubs, and washing machines. It can also leave spots on dishes, glassware, and fixtures, requiring the use of water softeners or more cleaning agents to counteract the effects of hardness.
Conclusion
Water hardness, caused by the presence of calcium and magnesium ions, has important implications for domestic, industrial, and biological processes. Understanding the chemistry behind hardness and how it can be removed or mitigated is essential for both practical applications and theoretical knowledge.
Periodic Table
Introduction to the Periodic Table
The periodic table is a tabular arrangement of chemical elements, organized by their atomic number, electron configuration, and recurring chemical properties. It is one of the most important tools in chemistry, providing a framework to predict the behavior of elements and their compounds. The table is arranged in rows called periods and columns called groups or families.
The modern periodic table is based on the periodic law, which states that "the properties of elements are a periodic function of their atomic numbers." This means that elements with similar chemical properties appear at regular intervals when arranged by increasing atomic number.
Development of the Periodic Table
1. Early Classifications:
- Döbereiner’s Triads (1817): Johann Döbereiner grouped elements into triads based on similarities in properties. The atomic mass of the middle element in a triad was roughly the average of the other two.
- Newlands’ Law of Octaves (1865): John Newlands noticed that every eighth element had properties similar to the first when arranged by increasing atomic mass. However, this law was only applicable to lighter elements.
2. Mendeleev’s Periodic Table (1869):
- Dmitri Mendeleev organized elements into a table based on increasing atomic mass. Mendeleev’s periodic law stated that "the properties of elements are a periodic function of their atomic masses." One of his significant contributions was leaving gaps for undiscovered elements and accurately predicting their properties.
3. Modern Periodic Table:
- The modern periodic table, developed after the discovery of atomic numbers, is arranged according to increasing atomic number (rather than atomic mass). This resolved many inconsistencies in Mendeleev’s table.
Structure of the Periodic Table
The periodic table is divided into periods (horizontal rows) and groups (vertical columns). Elements in the same group exhibit similar chemical properties because they have the same number of valence electrons.
1. Periods:
- The horizontal rows in the periodic table are called periods. There are 7 periods in the periodic table, and each period corresponds to the filling of a particular electron shell.
- First Period: Contains 2 elements (Hydrogen and Helium).
- Second and Third Periods: Each contains 8 elements.
- Fourth and Fifth Periods: Each contains 18 elements.
- Sixth Period: Contains 32 elements, including the lanthanides.
- Seventh Period: Contains 32 elements, including the actinides.
2. Groups:
- The vertical columns are called groups, and the periodic table has 18 groups. Elements in the same group have similar chemical properties due to having the same number of valence electrons.
- Group 1: Alkali metals (Li, Na, K, etc.), highly reactive metals.
- Group 2: Alkaline earth metals (Be, Mg, Ca, etc.), slightly less reactive than Group 1.
- Group 17: Halogens (F, Cl, Br, etc.), highly reactive non-metals.
- Group 18: Noble gases (He, Ne, Ar, etc.), inert gases with complete valence electron shells.
Classification of Elements in the Periodic Table
1. s-Block Elements:
- Groups 1 and 2 (Alkali metals and Alkaline earth metals).
- These elements have their valence electrons in the s-orbital.
- Example: Sodium (\(Na\)) has an electron configuration of \(1s^2 2s^2 2p^6 3s^1\).
2. p-Block Elements:
- Groups 13 to 18.
- These elements have their valence electrons in the p-orbital.
- Example: Chlorine (\(Cl\)) has an electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^5\).
- Includes non-metals, metalloids, and some metals. Group 18 contains noble gases, which are chemically inert.
3. d-Block Elements:
- Groups 3 to 12 (Transition metals).
- These elements have their valence electrons in the d-orbital.
- Transition metals are known for their ability to form complex ions, variable oxidation states, and colored compounds.
- Example: Iron (\(Fe\)) has an electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2\).
4. f-Block Elements:
- Consists of the Lanthanides and Actinides, which are placed separately at the bottom of the periodic table.
- These elements have their valence electrons in the f-orbital.
- Example: Uranium (\(U\)) has an electron configuration of \( [Rn]5f^3 6d^1 7s^2\).
Periodic Trends in the Periodic Table
1. Atomic Radius:
- Across a Period: Atomic radius decreases from left to right due to increasing nuclear charge, which pulls the electron cloud closer to the nucleus.
- Down a Group: Atomic radius increases due to the addition of new electron shells.
2. Ionization Energy:
- Across a Period: Ionization energy increases as the atomic radius decreases, making it more difficult to remove an electron.
- Down a Group: Ionization energy decreases as the atomic radius increases, making it easier to remove an electron.
3. Electronegativity:
- Across a Period: Electronegativity increases as atomic size decreases, leading to a stronger attraction for electrons.
- Down a Group: Electronegativity decreases as atomic size increases.
4. Electron Affinity:
- Across a Period: Electron affinity becomes more negative as atoms are more eager to accept electrons.
- Down a Group: Electron affinity becomes less negative as the added electron is farther from the nucleus.
5. Metallic and Non-Metallic Character:
- Across a Period: Metallic character decreases, while non-metallic character increases.
- Down a Group: Metallic character increases, while non-metallic character decreases.
Groups of Elements in the Periodic Table
1. Alkali Metals (Group 1):
- These elements have one valence electron, making them highly reactive.
- React vigorously with water to produce hydrogen gas and form strong bases.
- Example: Sodium (\(Na\)).
2. Alkaline Earth Metals (Group 2):
- These elements have two valence electrons and are less reactive than alkali metals.
- They form oxides and hydroxides that are basic in nature.
- Example: Magnesium (\(Mg\)).
3. Halogens (Group 17):
- Non-metals with seven valence electrons, making them highly reactive.
- Form salts when reacting with metals.
- Example: Chlorine (\(Cl\)).
4. Noble Gases (Group 18):
- Chemically inert elements with a complete octet of electrons, making them non-reactive under normal conditions.
- Example: Argon (\(Ar\)).
Modern Applications of the Periodic Table
1. Predicting Chemical Reactions:
- By using the periodic table, scientists can predict the reactivity of elements, their compounds, and their interactions with other elements. For example, Group 1 elements (alkali metals) are expected to form ionic bonds with halogens (Group 17).
2. Industrial Use of Transition Metals:
- Transition metals like iron, copper, and platinum are widely used in catalysis, construction, and electrical industries. Their variable oxidation states and ability to form colored compounds make them crucial in multiple industries.
3. Lanthanides and Actinides:
- Lanthanides are used in high-tech industries for making strong magnets and phosphors for lighting.
- Actinides, such as uranium and plutonium, are used in nuclear reactors and nuclear weapons.
Conclusion
The periodic table is a powerful tool for understanding the properties and behaviors of elements. The periodic trends in atomic size, ionization energy, electronegativity, and reactivity provide a systematic way to predict the chemical behavior of elements, making the periodic table essential for solving complex chemical problems in JEE Advanced.
Q 1. The law of triad is applicable to a group of; |
(D) \(\mathrm{H}, \mathrm{O}, \mathrm{N}\); |
(C) \(\mathrm{Na}, \mathrm{K}, \mathrm{Rb}\); |
(B) \(\mathrm{C}, \mathrm{N}, \mathrm{O}\); |
(A) \(\mathrm{Cl}, \mathrm{Br}, \mathrm{I}\); |
Q 2. \(2 \mathrm{Fe}^{2+}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{xA}+\mathrm{yB}\)\n(in basic medium)\n\(2 \mathrm{MnO}_{4}^{-}+6 \mathrm{H}^{+}+5 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{x}^{\prime} \mathrm{C}+\mathrm{y}^{\prime} \mathrm{D}+\mathrm{z}^{\prime} \mathrm{E}\)\n(in acidic medium)\nThe sum of the stoichiometric coefficients \(\mathrm{x}, \mathrm{y}, \mathrm{x}^{\prime}, \mathrm{y}^{\prime}\) and \(\mathrm{z}^{\prime}\) for products \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}\) and \(\mathrm{E}\) respectively, is; |
(D)34; |
(C)18; |
(B)14; |
(A)19; |
Q 3. Very pure hydrogen gas is prepared by the action of pure dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) on; |
(D) aluminium; |
(C) magnesium ribbon; |
(B) sodium hydride; |
(A) water; |
Q 4. Assertion\n\(\mathrm{PbCl}_{2}\) is more stable than \(\mathrm{PbCl}_{4}\)\nReason\n\(\mathrm{PbCl}_{4}\) is a powerful oxidising agent; |
(D) Assertion is incorrect but Reason is correct; |
(C) Assertion is correct but Reason is incorrect; |
(B) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion; |
(A) Both Assertion and Reason are correct and Reason is the correct explanation for Assertion; |
Q 5. In the increasing order of their sizes, arrange the following ions \(\mathrm{N}^{3-}, \mathrm{Na}^{+}, \mathrm{F}^{-}, \mathrm{Mg}^{2+}\) and \(\mathrm{O}^{2-}\); |
(D) \(\mathrm{N}^{3}>\mathrm{O}^{2}>\mathrm{F}>\mathrm{Na}^{+}<\mathrm{Mg}^{2+}\); |
(C) \(\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}<\mathrm{O}^{2-}<\mathrm{N}^{3-}\); |
(B) \(\mathrm{N}^{3-}<\mathrm{F}>\mathrm{O}^{2}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}\); |
(A) \(\mathrm{Mg}^{2+}>\mathrm{Na}^{+}>\mathrm{F}>\mathrm{O}^{2-}<\mathrm{N}^{3-}\); |
Q 6. Which one of the following relations is correct with respect to first (I) and second (II) ionization potentials of sodium and Magnesium?; |
(D) \(\mathrm{II}_{\mathrm{Na}}>\mathrm{II}_{\mathrm{Mg}}\); |
(C) \(\mathrm{II}_{\mathrm{Mg}}>\mathrm{II}_{\mathrm{Na}}\); |
(B) \(\mathrm{I}_{\mathrm{Mg}}>\mathrm{II}_{\mathrm{Na}}\); |
(A) \(I_{\mathrm{Na}}>I_{\mathrm{Mg}}\); |
Q 7. Number of electron-rich hydrides among the following are: \(\mathrm{CH}_{4}, \mathrm{NH}_{3}, \mathrm{PH}_{3}, \mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{BH}_{3}, \mathrm{HF}, \mathrm{AIH}_{3}, \mathrm{AsH}_{3}\). Q9. When electric current is passed through an ionic hydride in the molten state,\n(1) Hydrogen is liberated at the cathode\n(2) Hydrogen is liberated at the anode\n(3) Hydride ion migrates towards cathode\n(4) No reaction takes place\nA commercial sample of hydrogen peroxide marked as 100 volume \(\mathrm{H}_{2} \mathrm{O}_{2}\), it means that-; |
(A) \(1 \mathrm{ml}\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) will give \(100 \mathrm{mlO}_{2}\) at STP; |
(B) \(1 \mathrm{~L}\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) will give \(100 \mathrm{mlO}_{2}\) at STP; |
(C) \(1 \mathrm{~L}\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) will give \(22.4 \mathrm{~L}\) at STP; |
(D) \(1 \mathrm{ml}\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) will give 1 mole of \(\mathrm{O}_{2}\) at STP; |
Q 8. Element with electronic configuration \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}\) \(3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2} 5 p^{3}\) belongs to the following group of the periodic table; |
(a) 2nd; |
(b) 5th; |
(c) 3rd; |
(d) 7th; |
Q 9. According to Newlands theory, when elements are arranged in the order of increasing atomic weight which number element had similar properties to the first element; |
(a) third; |
(b) seventh; |
(c) eighth; |
(d) sixth; |
Q 10. Who developed long form of the periodic table?; |
(b) Neils Bohr; |
(c) Mendeleev; |
(d) Moseley; |
(a) Lothar Meyer; |
Ionization Enthalpy
Definition of Ionization Enthalpy
Ionization enthalpy (also known as ionization energy) is the amount of energy required to remove the outermost (or valence) electron from an isolated gaseous atom to form a cation. It is a measure of the tendency of an atom to resist losing electrons. Ionization enthalpy is expressed in kilojoules per mole (kJ/mol).
- General Formula:
\[
M(g) \rightarrow M^+(g) + e^-
\]
In this process, energy must be supplied to overcome the attraction between the negatively charged electron and the positively charged nucleus.
First, Second, and Third Ionization Enthalpies
1. First Ionization Enthalpy:
The energy required to remove the first electron from a neutral atom in the gaseous state.
\[
M(g) \rightarrow M^+(g) + e^-
\]
Example:
\[
Na(g) \rightarrow Na^+(g) + e^-
\quad \Delta H = 496 \, \text{kJ/mol}
\]
2. Second Ionization Enthalpy:
The energy required to remove an electron from a singly charged cation.
\[
M^+(g) \rightarrow M^{2+}(g) + e^-
\]
The second ionization enthalpy is always higher than the first because the electron is being removed from a positively charged ion, where the remaining electrons are held more tightly by the nucleus.
3. Third Ionization Enthalpy:
The energy required to remove an electron from a doubly charged cation.
\[
M^{2+}(g) \rightarrow M^{3+}(g) + e^-
\]
Similarly, the third ionization enthalpy is even higher than the second because the attraction between the remaining electrons and the nucleus becomes stronger as more electrons are removed.
Factors Affecting Ionization Enthalpy
1. Atomic Size (Atomic Radius):
- As the size of an atom increases, the outermost electron is farther from the nucleus and experiences less electrostatic attraction. Therefore, less energy is required to remove it.
- Example: The ionization energy of cesium (\(Cs\)) is lower than that of lithium (\(Li\)) because cesium has a much larger atomic radius.
2. Nuclear Charge:
- The greater the number of protons in the nucleus, the stronger the attraction between the nucleus and the electrons. A higher nuclear charge means a higher ionization enthalpy.
- Example: Helium has a higher ionization energy than hydrogen due to its greater nuclear charge.
3. Shielding Effect (Electron Shielding):
- Inner electrons shield the outermost electron from the full effect of the nucleus's positive charge. The more inner electron shells there are, the weaker the attraction between the nucleus and the outermost electron, resulting in lower ionization enthalpy.
- Example: In sodium (\(Na\)) and potassium (\(K\)), although potassium has a higher nuclear charge, its ionization energy is lower due to the shielding effect of the inner electron shells.
4. Penetration of Electrons:
- Electrons in s-orbitals are closer to the nucleus and more tightly bound than electrons in p, d, or f orbitals. Therefore, s-electrons have higher ionization enthalpy compared to electrons in other orbitals.
- Example: The 2s electrons in oxygen are more tightly bound than the 2p electrons due to greater penetration, resulting in higher ionization energy for the 2s electrons.
5. Stable Electron Configurations:
- Atoms with half-filled or fully filled orbitals (e.g., \(p^3\) or \(p^6\)) have higher ionization enthalpy because they are more stable.
- Example: Nitrogen (\(1s^2 2s^2 2p^3\)) has a higher ionization energy than oxygen (\(1s^2 2s^2 2p^4\)) because nitrogen has a half-filled, more stable 2p sublevel.
Trends in Ionization Enthalpy
1. Across a Period:
- Ionization enthalpy generally increases across a period (from left to right) in the periodic table. This is due to the increasing nuclear charge as more protons are added to the nucleus, while the electron shielding effect remains relatively constant. The electrons are drawn closer to the nucleus, making them harder to remove.
Example:
- Ionization energy increases from sodium (\(Na\)) to chlorine (\(Cl\)) in Period 3:
- \(Na: 496 \, \text{kJ/mol}\)
- \(Cl: 1251 \, \text{kJ/mol}\)
2. Down a Group:
- Ionization enthalpy decreases down a group as the atomic size increases. The outermost electron is farther from the nucleus, and the shielding effect of inner electrons increases, reducing the attraction between the nucleus and the valence electron.
Example:
- Ionization energy decreases from lithium (\(Li\)) to cesium (\(Cs\)) in Group 1:
- \(Li: 520 \, \text{kJ/mol}\)
- \(Cs: 375 \, \text{kJ/mol}\)
Anomalies in Ionization Enthalpy Trends
1. Group 13 and Group 14 Elements:
- There is a slight decrease in ionization enthalpy from Boron (B) to Aluminum (Al) and from Gallium (Ga) to Indium (In). This is due to the poor shielding effect of the d- and f-electrons, which increases the effective nuclear charge and pulls electrons closer to the nucleus, making it harder to remove them.
2. Anomalous Behavior in Group 2 and Group 13:
- The ionization energy of Beryllium (Be) is higher than that of Boron (B), despite Be coming before B in the periodic table. This anomaly arises because beryllium has a filled 2s sublevel, which is more stable and requires more energy to remove an electron compared to the partially filled 2p sublevel in boron.
Successive Ionization Enthalpies
- Successive Ionization Energies: After the first electron is removed, removing additional electrons requires progressively more energy. This is because the atom becomes positively charged after the first ionization, making the remaining electrons more strongly attracted to the nucleus.
- First Ionization Energy: Energy to remove the first electron.
- Second Ionization Energy: Energy to remove the second electron, which is always higher than the first.
- Third Ionization Energy: Even higher energy required to remove the third electron.
Example: Ionization energies for sodium (\(Na\)):
\[
\text{First}: 496 \, \text{kJ/mol}, \quad \text{Second}: 4562 \, \text{kJ/mol}, \quad \text{Third}: 6912 \, \text{kJ/mol}
\]
The second ionization energy is much higher because after the first electron is removed, sodium forms a stable \(Na^+\) ion, and removing another electron from the stable \(2p^6\) configuration requires significantly more energy.
Applications of Ionization Enthalpy
1. Predicting Chemical Behavior:
- Elements with low ionization energies tend to form cations easily and are more reactive. For example, alkali metals (Group 1) have low ionization energies, making them highly reactive.
- Noble gases (Group 18) have very high ionization energies, which explains their inertness.
2. Formation of Ionic Compounds:
- Ionization energy helps explain why certain elements form ionic bonds. Metals with low ionization energies, like sodium (\(Na\)) and magnesium (\(Mg\)), easily lose electrons and form cations, while non-metals with high electron affinity accept these electrons, forming anions.
3. Trends in Bond Formation:
- Elements with low ionization energy (such as metals) form ionic bonds by losing electrons. Elements with high ionization energy (such as non-metals) tend to form covalent bonds, where electrons are shared rather than transferred.
Conclusion
Ionization enthalpy is a fundamental concept in chemistry, influencing an element's ability to form bonds and undergo chemical reactions. Understanding trends in ionization enthalpy helps predict the reactivity of elements, their tendency to form ions, and their behavior in both ionic and covalent bonding.
Q 1. The atomic number of Unnilunium is; |
(A)105; |
(B)104; |
(C)101; |
(D)110; |
Q 2. Number of electron-rich hydrides among the following are: \(\mathrm{CH}_{4}, \mathrm{NH}_{3}, \mathrm{PH}_{3}, \mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{BH}_{3}, \mathrm{HF}, \mathrm{AIH}_{3}, \mathrm{AsH}_{3}\). Q9. When electric current is passed through an ionic hydride in the molten state,\n(1) Hydrogen is liberated at the cathode\n(2) Hydrogen is liberated at the anode\n(3) Hydride ion migrates towards cathode\n(4) No reaction takes place\nA commercial sample of hydrogen peroxide marked as 100 volume \(\mathrm{H}_{2} \mathrm{O}_{2}\), it means that-; |
(A) \(1 \mathrm{ml}\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) will give \(100 \mathrm{mlO}_{2}\) at STP; |
(B) \(1 \mathrm{~L}\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) will give \(100 \mathrm{mlO}_{2}\) at STP; |
(C) \(1 \mathrm{~L}\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) will give \(22.4 \mathrm{~L}\) at STP; |
(D) \(1 \mathrm{ml}\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) will give 1 mole of \(\mathrm{O}_{2}\) at STP; |
Q 3. Which one of the following relations is correct with respect to first (I) and second (II) ionization potentials of sodium and Magnesium?; |
(A) \(I_{\mathrm{Na}}>I_{\mathrm{Mg}}\); |
(B) \(\mathrm{I}_{\mathrm{Mg}}>\mathrm{II}_{\mathrm{Na}}\); |
(C) \(\mathrm{II}_{\mathrm{Mg}}>\mathrm{II}_{\mathrm{Na}}\); |
(D) \(\mathrm{II}_{\mathrm{Na}}>\mathrm{II}_{\mathrm{Mg}}\); |
Q 4. In the increasing order of their sizes, arrange the following ions \(\mathrm{N}^{3-}, \mathrm{Na}^{+}, \mathrm{F}^{-}, \mathrm{Mg}^{2+}\) and \(\mathrm{O}^{2-}\); |
(A) \(\mathrm{Mg}^{2+}>\mathrm{Na}^{+}>\mathrm{F}>\mathrm{O}^{2-}<\mathrm{N}^{3-}\); |
(B) \(\mathrm{N}^{3-}<\mathrm{F}>\mathrm{O}^{2}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}\); |
(C) \(\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}<\mathrm{O}^{2-}<\mathrm{N}^{3-}\); |
(D) \(\mathrm{N}^{3}>\mathrm{O}^{2}>\mathrm{F}>\mathrm{Na}^{+}<\mathrm{Mg}^{2+}\); |
Q 5. Assertion\n\(\mathrm{PbCl}_{2}\) is more stable than \(\mathrm{PbCl}_{4}\)\nReason\n\(\mathrm{PbCl}_{4}\) is a powerful oxidising agent; |
(A) Both Assertion and Reason are correct and Reason is the correct explanation for Assertion; |
(B) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion; |
(C) Assertion is correct but Reason is incorrect; |
(D) Assertion is incorrect but Reason is correct; |
Q 6. Very pure hydrogen gas is prepared by the action of pure dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) on; |
(A) water; |
(B) sodium hydride; |
(C) magnesium ribbon; |
(D) aluminium; |
Q 7. \(2 \mathrm{Fe}^{2+}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{xA}+\mathrm{yB}\)\n(in basic medium)\n\(2 \mathrm{MnO}_{4}^{-}+6 \mathrm{H}^{+}+5 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{x}^{\prime} \mathrm{C}+\mathrm{y}^{\prime} \mathrm{D}+\mathrm{z}^{\prime} \mathrm{E}\)\n(in acidic medium)\nThe sum of the stoichiometric coefficients \(\mathrm{x}, \mathrm{y}, \mathrm{x}^{\prime}, \mathrm{y}^{\prime}\) and \(\mathrm{z}^{\prime}\) for products \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}\) and \(\mathrm{E}\) respectively, is; |
(A)19; |
(B)14; |
(C)18; |
(D)34; |
Q 8. The long form of periodic table consists of; |
(a) seven periods and eight groups; |
(b) seven periods and eighteen groups; |
(c) eight periods and eighteen groups; |
(d) eighteen periods and eight groups; |
Q 9. Lothar Meyer plotted the physical properties such as atomic volume, melting point and X against atomic weight Here, \(\mathrm{X}\) refers to; |
(a) mass; |
(b) boiling point; |
(c) surface tension; |
(d) None of these; |
Q 10. At present, how many elements are known; |
(a) 110; |
(b) 112; |
(c) 113; |
(d) 118; |
Atomic Radius
Definition of Atomic Radius
Atomic radius refers to the distance between the nucleus of an atom and its outermost electron shell. It is a key concept in understanding the size of an atom and its influence on various chemical and physical properties, such as reactivity, ionization energy, and bond formation. Since atoms do not have well-defined boundaries, the atomic radius is often considered as the effective distance from the nucleus to the boundary within which the electrons can be found.
The atomic radius is typically measured in picometers (pm), where \(1 \, \text{pm} = 10^{-12} \, \text{m}\).
Types of Atomic Radii
1. Covalent Radius:
- Defined as half the distance between the nuclei of two identical atoms bonded by a single covalent bond. This radius is commonly used for non-metals, where atoms share electrons to form covalent bonds.
- Example: In a chlorine molecule (\(Cl_2\)), the covalent bond length is 198 pm, so the covalent radius of chlorine is 99 pm.
2. Metallic Radius:
- Defined as half the distance between the nuclei of two adjacent atoms in a metallic lattice. This is typically used for metals, where atoms are arranged in a crystal lattice and electrons are delocalized.
- Example: The metallic radius of sodium (\(Na\)) is 186 pm.
3. Van der Waals Radius:
- Refers to half the distance between the nuclei of two non-bonded atoms in a molecule or crystal. This radius applies to noble gases and molecules where atoms are only weakly interacting.
- Example: The Van der Waals radius of neon (\(Ne\)) is 154 pm.
Trends in Atomic Radius
The atomic radius shows distinct periodic trends as we move across periods and down groups in the periodic table. These trends are influenced by the number of electron shells, the nuclear charge, and the electron shielding effect.
1. Across a Period (Left to Right):
- Atomic Radius Decreases: As we move across a period from left to right, the atomic radius decreases. This is because, within the same period, electrons are added to the same shell, but the number of protons in the nucleus increases. The increased nuclear charge pulls the electron cloud closer to the nucleus, reducing the size of the atom.
Example:
- Lithium (Li): 152 pm
- Fluorine (F): 64 pm
- Reason: Increased nuclear charge pulls electrons closer to the nucleus.
2. Down a Group (Top to Bottom):
- Atomic Radius Increases: As we move down a group, the atomic radius increases. This is because each successive element has an additional electron shell, increasing the distance between the nucleus and the outermost electrons. The shielding effect from inner shells reduces the effective nuclear charge felt by the outermost electron, allowing the atom to expand.
Example:
- Fluorine (F): 64 pm
- Chlorine (Cl): 99 pm
- Bromine (Br): 114 pm
- Iodine (I): 133 pm
- Reason: Each element down the group has an additional electron shell, increasing atomic size.
Factors Affecting Atomic Radius
1. Nuclear Charge:
- As the number of protons in the nucleus increases, the positive charge on the nucleus increases, pulling electrons closer to the nucleus and reducing the atomic radius.
- Example: Across Period 2, atomic radius decreases from lithium (3 protons) to fluorine (9 protons) because of the increased nuclear charge.
2. Number of Electron Shells:
- The more electron shells an atom has, the larger the atomic radius. As more shells are added, the outermost electrons are farther from the nucleus.
- Example: Potassium (\(K\)) has a larger atomic radius than sodium (\(Na\)) because potassium has more electron shells.
3. Electron Shielding:
- Inner electron shells shield the outermost electrons from the full attractive force of the nucleus. The more electron shells present, the greater the shielding effect, which allows the outer electrons to be less tightly held by the nucleus.
- Example: In cesium (\(Cs\)), the outermost electron is well shielded by the inner electron shells, resulting in a large atomic radius.
4. Electron-Electron Repulsion:
- Electrons in the same shell repel each other, causing the electron cloud to expand slightly. This is a minor effect compared to the attraction between the nucleus and electrons.
Anomalies in Atomic Radius Trends
1. Transition Metals (d-Block Elements):
- The atomic radius of transition metals does not change significantly across a period. This is because the addition of electrons occurs in the inner d-orbital, which only marginally affects the overall size of the atom. The increased nuclear charge is balanced by the electron shielding effect of the d-electrons.
Example:
- The atomic radii of iron (\(Fe\)), cobalt (\(Co\)), and nickel (\(Ni\)) are nearly the same despite moving across the period.
2. Lanthanide Contraction:
- In the lanthanide series, there is a gradual decrease in atomic radius across the series due to the poor shielding effect of the f-electrons. This contraction leads to smaller than expected atomic radii for elements following the lanthanides (the post-lanthanide elements).
Example:
- The atomic radius of hafnium (\(Hf\)) is smaller than that of zirconium (\(Zr\)) despite being in the same group.
Comparing Ionic Radii and Atomic Radii
1. Cations (Positively Charged Ions):
- Cations are smaller than their parent atoms because they lose one or more electrons, leading to a decrease in electron-electron repulsion and allowing the remaining electrons to be pulled closer to the nucleus.
- Example: The radius of sodium ion (\(Na^+\)) is smaller than that of neutral sodium atom (\(Na\)).
2. Anions (Negatively Charged Ions):
- Anions are larger than their parent atoms because they gain one or more electrons, increasing electron-electron repulsion and causing the electron cloud to expand.
- Example: The radius of chloride ion (\(Cl^-\)) is larger than that of neutral chlorine atom (\(Cl\)).
Applications of Atomic Radius
1. Reactivity of Metals:
- Metals with larger atomic radii tend to lose electrons more easily and are thus more reactive. For example, alkali metals like potassium (\(K\)) and cesium (\(Cs\)) have large atomic radii and readily lose their outermost electron, making them highly reactive.
2. Reactivity of Non-metals:
- Non-metals with smaller atomic radii tend to gain electrons more easily due to the stronger pull of the nucleus on the incoming electron. For example, fluorine (\(F\)) has a small atomic radius and a high electronegativity, making it highly reactive.
3. Bond Length:
- The atomic radius plays a role in determining bond length in molecules. Larger atoms form longer bonds, while smaller atoms form shorter bonds.
- Example: The bond length in \(HCl\) is longer than in \(HF\) due to the larger atomic radius of chlorine compared to fluorine.
Conclusion
The concept of atomic radius is fundamental to understanding periodic trends and the behavior of elements. By studying the trends and factors that influence atomic size, students can predict the reactivity of elements, their bonding tendencies, and their behavior in chemical reactions.
Q 1. Atomic radii of alkali metals follow the order:; |
(A) \(\mathrm{Li}>\mathrm{Na}>\mathrm{K}>\mathrm{Cs}\); |
(B) \(\mathrm{K}>\mathrm{Cs}>\mathrm{Li}>\mathrm{Na}\); |
(C) \(\mathrm{Na}>\mathrm{K}>\mathrm{Cs}>\mathrm{Li}\); |
(D) \(\mathrm{Cs}>\mathrm{K}>\mathrm{Na}>\mathrm{Li}\); |
Q 2. Dalda is prepared from oils by; |
(A) Oxidation; |
(B) Reduction; |
(C) Hydrolysis; |
(D) Distillation; |
Q 3. Which of the following element shows maximum valency?; |
(A) Carbon; |
(B) Barium; |
(C) Nitrogen; |
(D) Sulphur; |
Q 4. The first \(\left(\Delta_{\mathrm{i}} \mathrm{H}_{1}\right)\) and the second \(\left(\Delta_{\mathrm{i}} \mathrm{H}_{2}\right)\) ionization enthalpies ( in \(\mathrm{kJmol}^{-1}\) ) and the \(\Delta \mathrm{e}_{\mathrm{g}} \mathrm{H}\) electron gain enthalpy (in \(\mathrm{kJ}\) mol -1 ) of a few elements are given below :\n\begin{array}{|c|c|c|c|}\n\hline\nElements & \(\Delta \mathrm{H}_{1}\) & \(\Delta \mathrm{H}_{2}\) & \(\Delta_{\mathrm{eg}} \mathrm{H}\) \\\n\hline\n\(\mathrm{I}\) & \(52 \mathrm{O}\) & 7300 & -60 \\\n\hline\n\(\mathrm{II}\) & 419 & 3051 & -48 \\\n\hline\n\(\mathrm{III}\) & 1681 & 3374 & -328 \\\n\hline\n\(\mathrm{IV}\) & 1008 & 1846 & -295 \\\n\hline\n\(\mathrm{V}\) & 2372 & 5251 & +48 \\\n\hline\n\(\mathrm{VI}\) & 738 & 1451 & -40 \\\n\hline\n\end{array}\nWhich of the above likely to be the least reactive non-metal?; |
(A)I; |
(B)II; |
(C)III; |
(D)IV; |
Q 5. The correct increasing order of electronegativity for \(\mathrm{Sn}, \mathrm{Sn}^{2+}\) and \(\mathrm{Sn}^{4+}\) species is:; |
(A) \(\mathrm{Sn}<\mathrm{Sn}^{2+}<\mathrm{Sn}^{4+}\); |
(B) \(\mathrm{Sn}^{4+}<\mathrm{Sn}^{2+}<\mathrm{Sn}\); |
(C) \(\mathrm{Sn}=\mathrm{Sn}^{2+}=\mathrm{Sn}^{4}\); |
(D) \(\mathrm{Sn}<\mathrm{Sn}^{2+}=\mathrm{Sn}^{4+}\); |
Q 6. In context with the industrial preparation of hydrogen from water gas \(\left(\mathrm{CO}+\mathrm{H}_{2}\right)\), which of the following is the correct statement?; |
(A) \(\mathrm{CO}\) is oxidised to \(\mathrm{CO}_{2}\) with steam, in the presence of a catalyst followed by the dissolution of \(\mathrm{CO}_{2}\) in water under ssure.; |
(B) \(\mathrm{CO}\) and \(\mathrm{H}_{2}\) are fractionally separated using the differences in their densities.; |
(C) \(\mathrm{CO}\) is removed by absorption in aqueous \(\mathrm{Cu}_{2} \mathrm{Cl}_{2}\) solution.; |
(D) \(\mathrm{H}_{2}\) is removed through occlusion with \(\mathrm{Pd}\).; |
Q 7. The reason of bleaching properties of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is; |
(A) Unstable nature; |
(B) Acidic nature; |
(C) Reducing nature; |
(D) Oxidising nature; |
Q 8. All the members in a group in long form of periodic table have the same; |
(a) valence; |
(b) number of valence electrons; |
(c) chemical properties; |
(d) All of the above; |
Q 9. The most significant contribution towards the development of periodic table was made by; |
(a) Mendeleev; |
(b) Avogadro; |
(c) Dalton; |
(d) Cavendish; |
Q 10. Which of the scientists given below discovered that periodic table should be based on the atomic number?; |
(a) Mendeleev; |
(b) Newlands; |
(c) Moseley; |
(d) Lothar Meyer; |
Periodic Trends
Introduction to Periodic Trends
Periodic trends refer to the predictable patterns observed in the properties of elements when arranged in the periodic table. These trends arise due to the regular variation in atomic structure, specifically the number of protons (atomic number) and the arrangement of electrons in atoms. Understanding periodic trends is fundamental to predicting the behavior of elements in chemical reactions, their bonding characteristics, and their reactivity.
These trends include:
- Atomic Radius
- Ionization Enthalpy
- Electronegativity
- Electron Affinity
- Metallic and Non-Metallic Character
1. Atomic Radius
Trend Across a Period:
- Atomic radius decreases from left to right across a period. As the atomic number increases, the number of protons in the nucleus increases, which increases the nuclear charge. The increased nuclear charge pulls the electron cloud closer to the nucleus, reducing the size of the atom.
Example:
- In Period 2, the atomic radius decreases from lithium (152 pm) to fluorine (64 pm).
Trend Down a Group:
- Atomic radius increases down a group. This is due to the addition of new electron shells as we move down a group, which increases the distance between the nucleus and the outermost electrons. Although the nuclear charge increases, the effect is offset by increased electron shielding, causing the outer electrons to experience less pull from the nucleus.
Example:
- In Group 17, the atomic radius increases from fluorine (64 pm) to iodine (133 pm).
2. Ionization Enthalpy
Trend Across a Period:
- Ionization enthalpy increases from left to right across a period. As the atomic radius decreases and the nuclear charge increases, it becomes more difficult to remove an electron from the atom. Elements on the right side of the periodic table, such as fluorine and oxygen, have high ionization energies because their valence electrons are tightly held by the nucleus.
Example:
- The first ionization energy of sodium (\(Na\)) is 496 kJ/mol, while that of chlorine (\(Cl\)) is 1251 kJ/mol.
Trend Down a Group:
- Ionization enthalpy decreases down a group. As the atomic radius increases and the outermost electron is farther from the nucleus, it becomes easier to remove the electron. Additionally, the shielding effect from inner electrons reduces the effective nuclear charge experienced by the valence electrons.
Example:
- The first ionization energy of lithium (\(Li\)) is 520 kJ/mol, while that of cesium (\(Cs\)) is only 375 kJ/mol.
3. Electronegativity
Trend Across a Period:
- Electronegativity increases from left to right across a period. As the atomic radius decreases and the nuclear charge increases, the ability of an atom to attract electrons in a chemical bond also increases. Elements on the right side of the periodic table, such as fluorine, oxygen, and nitrogen, are highly electronegative and tend to attract electrons strongly.
Example:
- Fluorine (\(F\)) is the most electronegative element with an electronegativity value of 3.98, while lithium (\(Li\)) has a much lower electronegativity of 0.98.
Trend Down a Group:
- Electronegativity decreases down a group. As the atomic size increases and the electron cloud extends further from the nucleus, the ability of the atom to attract electrons in a bond decreases.
Example:
- In Group 17, electronegativity decreases from fluorine (\(F\), 3.98) to iodine (\(I\), 2.66).
4. Electron Affinity
Trend Across a Period:
- Electron affinity becomes more negative (increases in magnitude) from left to right across a period. As the atomic radius decreases and the nuclear charge increases, atoms on the right side of the periodic table (especially halogens) are more eager to gain electrons to complete their octet. A more negative electron affinity value indicates that the atom releases more energy when gaining an electron.
Example:
- Chlorine (\(Cl\)) has a highly negative electron affinity of -349 kJ/mol, while sodium (\(Na\)) has a much less negative electron affinity of -53 kJ/mol.
Trend Down a Group:
- Electron affinity becomes less negative (decreases in magnitude) down a group. As the atomic radius increases, the added electron is farther from the nucleus and experiences less nuclear attraction. As a result, atoms are less likely to gain an electron, and the energy released is smaller.
Example:
- In Group 17, the electron affinity becomes less negative from chlorine (\(Cl\), -349 kJ/mol) to iodine (\(I\), -295 kJ/mol).
5. Metallic and Non-Metallic Character
Metallic Character:
- Trend Across a Period: Metallic character decreases from left to right across a period. As we move from metals on the left side (such as sodium and magnesium) to non-metals on the right side (such as sulfur and chlorine), the tendency to lose electrons and form positive ions (a characteristic of metals) decreases.
- Example: Sodium (\(Na\)) is a highly metallic element, while sulfur (\(S\)) and chlorine (\(Cl\)) are non-metals.
- Trend Down a Group: Metallic character increases down a group. As the atomic radius increases, it becomes easier for atoms to lose electrons and form positive ions. This is why alkali metals such as potassium and cesium are more metallic than elements higher up in the group like lithium.
- Example: Cesium (\(Cs\)) is more metallic than lithium (\(Li\)).
Non-Metallic Character:
- Trend Across a Period: Non-metallic character increases from left to right across a period. Non-metals are elements that tend to gain electrons and form negative ions. As we move across a period, elements become more non-metallic as their tendency to gain electrons increases.
- Example: Oxygen (\(O\)) and fluorine (\(F\)) are highly non-metallic elements.
- Trend Down a Group: Non-metallic character decreases down a group. As the atomic radius increases, the ability of atoms to attract and gain electrons decreases, making elements less non-metallic.
- Example: Fluorine (\(F\)) is more non-metallic than iodine (\(I\)).
6. Reactivity Trends
1. Reactivity of Metals:
- Trend Across a Period: The reactivity of metals decreases from left to right across a period. This is because metallic reactivity is associated with the ability to lose electrons, and ionization energy increases across a period, making it more difficult for metals to lose electrons.
Example: Sodium (\(Na\)) is more reactive than magnesium (\(Mg\)).
- Trend Down a Group: The reactivity of metals increases down a group. This is because ionization energy decreases, making it easier for metals to lose electrons and participate in chemical reactions.
Example: Potassium (\(K\)) is more reactive than lithium (\(Li\)).
2. Reactivity of Non-Metals:
- Trend Across a Period: The reactivity of non-metals increases from left to right across a period. Non-metals tend to gain electrons, and as electronegativity increases across a period, so does the reactivity of non-metals.
Example: Fluorine (\(F\)) is more reactive than oxygen (\(O\)).
- Trend Down a Group: The reactivity of non-metals decreases down a group. As atomic size increases and electronegativity decreases, non-metals become less reactive.
Example: Fluorine (\(F\)) is more reactive than iodine (\(I\)).
Applications of Periodic Trends
1. Predicting Chemical Reactivity:
- Periodic trends allow us to predict which elements will react and how they will react with one another. For example, alkali metals are highly reactive with water, while noble gases are inert.
2. Bond Formation:
- Elements with low ionization energies and large atomic radii tend to form ionic bonds, while elements with high electronegativity and small atomic radii tend to form covalent bonds.
3. Industrial Applications:
- Understanding trends in electronegativity and reactivity helps industries choose the right materials for chemical reactions, catalysis, and the production of compounds like fertilizers and pharmaceuticals.
Q 1. Which of the following statements about heavy water are correct?\n(1) Heavy water is used as a moderator in nuclear reactors.\n(2) Heavy water is more associated than ordinary water.\n(3) Heavy water is a more effective solvent than ordinary water.; |
(A) (b) and (c); |
(B) (a) and (b); |
(C) (a) and (c); |
(D) (a), (b) and (c); |
Q 2. Which of the following atomic and physical properties of hydrogen is false?; |
(A) Hydrogen \(>\) Deuterium \(>\) Tritium; (melting point \(/ K\) ); |
(B) Hydrogen \(<\) Deuterium \(<\) Tritium; (boiling point \(/ K\) ); |
(C) Hydrogen \(<\) Deuterium \(<\) Tritium; (density \(/ g L^{-1}\) ); |
(D) Hydrogen > Deuterium > Tritium; ( \() \%\) relative abundance); |
Q 3. The correct decreasing order of electropositive character among the following elements is; |
(A) \(\mathrm{Fe}>\mathrm{Sc}>\mathrm{Rb}>\mathrm{Br}>\mathrm{Te}>\mathrm{F}>\mathrm{Ca}\); |
(B) \(\mathrm{Ca}>\mathrm{Rb}>\mathrm{Sc}>\mathrm{Fe}>\mathrm{Te}>\mathrm{F}>\mathrm{Br}\); |
(C) \(\mathrm{Rb}>\mathrm{Ca}>\mathrm{Sc}>\mathrm{Fe}>\mathrm{Br}>\mathrm{Te}>\mathrm{F}\); |
(D) \(\mathrm{Rb}>\mathrm{Ca}>\mathrm{Sc}>\mathrm{Fe}>\mathrm{Te}>\mathrm{Br}>\mathrm{F}\); |
Q 4. Atomic radii of alkali metals follow the order:; |
(D) \(\mathrm{Cs}>\mathrm{K}>\mathrm{Na}>\mathrm{Li}\); |
(C) \(\mathrm{Na}>\mathrm{K}>\mathrm{Cs}>\mathrm{Li}\); |
(B) \(\mathrm{K}>\mathrm{Cs}>\mathrm{Li}>\mathrm{Na}\); |
(A) \(\mathrm{Li}>\mathrm{Na}>\mathrm{K}>\mathrm{Cs}\); |
Q 5. Find the correct order of electron gain enthalpy \(\left(\Delta_{\mathrm{eg}} \mathrm{H}\right)\) of the given elements.; |
(A) \(\mathrm{O}>\mathrm{B}>\mathrm{C}>\mathrm{N}\); |
(B) \(\mathrm{O}>\mathrm{C}>\mathrm{N}>\mathrm{B}\); |
(C) \(\mathrm{O}>\mathrm{C}>\mathrm{B}>\mathrm{N}\); |
(D) \(\mathrm{O}>\mathrm{N}>\mathrm{C}>\mathrm{B}\); |
Q 6. Hydrogen can be prepared by the action of dil. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) on; |
(A) \(\mathrm{Cu}\); |
(B) \(\mathrm{Fe}\); |
(C) \(\mathrm{Pb}\); |
(D) \(\mathrm{Hg}\); |
Q 7. Permanent hardness can be removed by adding; |
(A) \(\mathrm{Cl}_{2}\); |
(B) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\); |
(C) \(\mathrm{Ca}(\mathrm{OCl}) \mathrm{Cl}\); |
(D) \(\mathrm{K}_{2} \mathrm{CO}_{3}\); |
Q 8. Johann Doberiner gave the idea of trends among physical and X of several groups of three elements Here, X refers to; |
(a) atomic number; |
(b) atomic mass; |
(c) chemical properties; |
(d) None of these; |
Q 9. Noble gases were included in Mendeleev's periodic table in the; |
(a) 1st group; |
(b) 7th group; |
(c) 8th group; |
(d) None of these; |
Q 10. How many elements are there in \(6^{\text {th }}\) period of periodic table?; |
(a) 18; |
(b) 8; |
(c) 30; |
(d) 32; |
Electron Configuration
Definition of Electron Configuration
Electron configuration refers to the distribution of electrons in an atom’s orbitals. The arrangement of electrons follows specific rules based on quantum mechanics, which governs how electrons occupy energy levels (shells) and sublevels (orbitals). The electron configuration of an atom determines its chemical properties, reactivity, and place in the periodic table.
The electron configuration provides a systematic way of understanding the arrangement of electrons in terms of principal energy levels (n), sublevels (s, p, d, f), and the number of electrons in each sublevel.
Principles Governing Electron Configuration
1. Aufbau Principle:
- Electrons are added to orbitals in order of increasing energy levels. This principle states that lower-energy orbitals are filled first before moving to higher-energy orbitals.
- Example: In the electron configuration of carbon (\(1s^2 2s^2 2p^2\)), the 1s orbital is filled before the 2s orbital, and the 2s orbital is filled before the 2p orbital.
2. Pauli Exclusion Principle:
- No two electrons in an atom can have the same set of quantum numbers. This means each orbital can hold a maximum of two electrons, and these electrons must have opposite spins.
- Example: In the 1s orbital of hydrogen, there can only be two electrons, and they must have opposite spins (\(m_s = +1/2\) and \(m_s = -1/2\)).
3. Hund’s Rule:
- When electrons occupy orbitals of equal energy (degenerate orbitals), they fill them singly first, with parallel spins, before pairing up. This minimizes electron repulsion and makes the atom more stable.
- Example: In the configuration of oxygen (\(1s^2 2s^2 2p^4\)), the 2p sublevel has three degenerate orbitals. The first three electrons will occupy each orbital singly with parallel spins, and the fourth electron will pair up in one of the orbitals.
Notation for Electron Configuration
Electron configuration is written in a shorthand notation where:
- Numbers represent the energy level (\(n\)).
- Letters represent the type of sublevel (s, p, d, f).
- Superscripts represent the number of electrons in that sublevel.
Example:
- Carbon (C): \(1s^2 2s^2 2p^2\)
This means there are two electrons in the 1s orbital, two electrons in the 2s orbital, and two electrons in the 2p sublevel.
- Chlorine (Cl): \(1s^2 2s^2 2p^6 3s^2 3p^5\)
Chlorine has a total of 17 electrons distributed across the orbitals in increasing energy.
Electron Configuration of Elements in the Periodic Table
The periodic table is arranged in such a way that electron configurations follow a predictable pattern based on the element's position:
1. s-Block Elements (Groups 1 and 2):
- Elements in the s-block have their valence electrons in the s-orbital.
- Example: Sodium (\(Na\)) has an electron configuration of \(1s^2 2s^2 2p^6 3s^1\). Its single valence electron is in the 3s orbital.
2. p-Block Elements (Groups 13 to 18):
- Elements in the p-block have their valence electrons in the p-orbital.
- Example: Chlorine (\(Cl\)) has an electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^5\). It has five valence electrons in the 3p sublevel.
3. d-Block Elements (Transition Metals, Groups 3 to 12):
- Transition metals have their valence electrons in the d-orbital.
- Example: Iron (\(Fe\)) has an electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2\). The 3d sublevel is partially filled, which gives transition metals their characteristic properties like variable oxidation states and the ability to form colored compounds.
4. f-Block Elements (Lanthanides and Actinides):
- Elements in the f-block have their valence electrons in the f-orbital.
- Example: Uranium (\(U\)) has an electron configuration of \( [Rn] 5f^3 6d^1 7s^2\). These elements are characterized by their complex electron configurations and their use in nuclear chemistry.
Exceptions to the Electron Configuration Rules
Some elements exhibit exceptions to the general electron configuration rules, especially in the d-block and f-block. These exceptions occur because of the relative energies of the orbitals and the additional stability provided by half-filled or fully filled sublevels.
1. Chromium (Cr):
- Expected Configuration: \( [Ar] 3d^4 4s^2 \)
- Actual Configuration: \( [Ar] 3d^5 4s^1 \)
- Reason: A half-filled 3d sublevel provides extra stability, so one electron from the 4s orbital moves to the 3d orbital.
2. Copper (Cu):
- Expected Configuration: \( [Ar] 3d^9 4s^2 \)
- Actual Configuration: \( [Ar] 3d^{10} 4s^1 \)
- Reason: A fully filled 3d sublevel is more stable, so one electron from the 4s orbital moves to the 3d orbital.
These exceptions are crucial for understanding the chemical behavior of transition metals, particularly their variable oxidation states and coordination chemistry.
Electron Configuration and Periodic Properties
1. Atomic Size:
As we move across a period, electrons are added to the same energy level, but the increasing nuclear charge pulls the electron cloud closer, decreasing the atomic radius. The electron configuration helps explain why atomic size decreases across a period.
2. Ionization Energy:
The electron configuration shows how tightly electrons are held by the nucleus. Elements with filled or half-filled sublevels (e.g., noble gases, nitrogen) have higher ionization energies due to the stability of their configurations.
3. Reactivity:
The reactivity of an element is closely related to its electron configuration. Metals with only a few valence electrons (e.g., alkali metals) are highly reactive because they tend to lose electrons easily. Non-metals with nearly complete valence shells (e.g., halogens) are highly reactive because they tend to gain electrons.
4. Formation of Ions:
The electron configuration also explains how elements form ions. Metals lose electrons to achieve a noble gas configuration (e.g., \(Na^+ \) with \(1s^2 2s^2 2p^6\)), while non-metals gain electrons to complete their valence shells (e.g., \(Cl^-\) with \(1s^2 2s^2 2p^6 3s^2 3p^6\)).
Shorthand Notation for Electron Configuration
To simplify electron configurations, especially for elements with many electrons, we can use noble gas shorthand notation. In this notation, the electron configuration of the nearest noble gas with a lower atomic number is written in square brackets, followed by the configuration of the remaining electrons.
Examples:
- Sodium (Na): \( [Ne] 3s^1 \)
- Chlorine (Cl): \( [Ne] 3s^2 3p^5 \)
- Calcium (Ca): \( [Ar] 4s^2 \)
- Iron (Fe): \( [Ar] 3d^6 4s^2 \)
This shorthand notation simplifies the electron configurations of elements, particularly for elements with large numbers of electrons.
Applications of Electron Configuration
1. Chemical Bonding:
- Electron configuration helps explain why atoms form certain types of bonds. For example, elements with nearly full or nearly empty valence shells will form ionic bonds by gaining or losing electrons, while elements with half-filled or similar electron configurations often form covalent bonds.
2. Transition Metals and Catalysis:
- The electron configuration of transition metals (d-block elements) is crucial for understanding their catalytic activity, as their ability to adopt multiple oxidation states makes them effective catalysts.
3. Magnetic Properties:
- Electron configurations are also used to explain the magnetic properties of materials. Elements or ions with unpaired electrons (e.g., transition metals like iron) exhibit paramagnetism, while those with all paired electrons are diamagnetic.
Q 1. Which of the following ions will cause hardness of sample water?; |
(A) \(\mathrm{Na}^{+}\); |
(B) \(\mathrm{K}^{+}\); |
(C) \(\mathrm{Ca}^{2+}\); |
(D) \(\mathrm{Li}^{+}\); |
Q 2. Which of the following statements is not true for hydrogen:\n It exists as a diatomic molecule.\n It has one electron in the outermost shell.\n It can lose an electron to form a cation which can freely exist\n It cannot form ionic compounds.; |
(A) 1; |
(B) 2; |
(C) 3; |
(D) 4; |
Q 3. \(\mathrm{A} \rightarrow \mathrm{A}^{-}+\mathrm{e}, \mathrm{E}_{1}\) and \(\mathrm{A}^{+} \rightarrow \mathrm{A}^{2+}+\mathrm{e}, \mathrm{E}_{2}\). The energy required to pull out the two electrons are \(\mathrm{E}_{1}\) and \(\mathrm{E}_{2}\) respectively. The correct relationship between two energy would be :; |
(A) \(\mathrm{E}_{1}<\mathrm{E}_{2}\); |
(B) \(\mathrm{E}_{1}>\mathrm{E}_{2}\); |
(C) \(\mathrm{E}_{1}=\mathrm{E}_{2}\); |
(D) \(\mathrm{E}_{1} \neq \mathrm{E}_{2}\); |
Q 4. The atomic number of Unnilunium is; |
(A)105; |
(B)104; |
(C)101; |
(D)110; |
Q 5. \(\mathrm{A} \rightarrow \mathrm{A}^{-}+\mathrm{e}, \mathrm{E}_{1}\) and \(\mathrm{A}^{+} \rightarrow \mathrm{A}^{2+}+\mathrm{e}, \mathrm{E}_{2}\). The energy required to pull out the two electrons are \(\mathrm{E}_{1}\) and \(\mathrm{E}_{2}\) respectively. The correct relationship between two energy would be :; |
(A) \(\mathrm{E}_{1}<\mathrm{E}_{2}\); |
(B) \(\mathrm{E}_{1}>\mathrm{E}_{2}\); |
(C) \(\mathrm{E}_{1}=\mathrm{E}_{2}\); |
(D) \(\mathrm{E}_{1} \neq \mathrm{E}_{2}\); |
Q 6. Which of the following statements is not true for hydrogen:\n It exists as a diatomic molecule.\n It has one electron in the outermost shell.\n It can lose an electron to form a cation which can freely exist\n It cannot form ionic compounds.; |
(A) 1; |
(B) 2; |
(C) 3; |
(D) 4; |
Q 7. Which of the following ions will cause hardness of sample water?; |
(A) \(\mathrm{Na}^{+}\); |
(B) \(\mathrm{K}^{+}\); |
(C) \(\mathrm{Ca}^{2+}\); |
(D) \(\mathrm{Li}^{+}\); |
Q 8. Which of the following is the correct set of elements to Dobereiner's triads?; |
(a) \(\underset{7}{\mathrm{Li}} \underset{23}{\mathrm{Na}} \underset{39}{\mathrm{~K}}\); |
(b) \(\begin{array}{ccc}\mathrm{Br} & \mathrm{Cl} & \mathrm{I} \\ 80 & 35.5 & 127\end{array}\); |
(c) \(\begin{array}{ccc}\mathrm{Fe} & \mathrm{Ni} & \mathrm{Co} \\ 55.85 & 58.71 & 58.93\end{array}\); |
(d) Data insufficient; |
Q 9. Mendeleev classified elements in; |
(a) increasing order of atomic groups; |
(b) eight periods and eight groups; |
(c) seven periods and nine groups; |
(d) eight periods and seven groups; |
Q 10. Modern periodic table is based on the atomic number of the elements The experiment which proved the significance of the atomic number was; |
(a) Mulliken's oil drop experiment; |
(b) Mosley's work on X-ray spectra; |
(c) Bragg's work on X-ray diffraction; |
(d) Discovery of X-rays by Rontgen; |
Electron Gain Enthalpy
Definition of Electron Gain Enthalpy
Electron gain enthalpy (also known as electron affinity) refers to the amount of energy released (or absorbed) when an electron is added to a neutral atom in its gaseous state to form a negative ion. It reflects the atom’s tendency to accept an electron and become an anion. The more negative the electron gain enthalpy, the greater the tendency of the atom to gain an electron.
- Process:
\[
X(g) + e^- \rightarrow X^-(g)
\]
Here, \(X(g)\) is the neutral atom, \(e^-\) is the added electron, and \(X^-(g)\) is the resulting anion.
- Units: Electron gain enthalpy is typically expressed in kilojoules per mole (kJ/mol).
Significance of Electron Gain Enthalpy
- A negative electron gain enthalpy means that the atom releases energy when it gains an electron, indicating that the process is energetically favorable.
- A positive electron gain enthalpy means that energy must be supplied for the atom to gain an electron, indicating that the process is not energetically favorable.
For most non-metals, electron gain enthalpy is negative, as these elements readily accept electrons to achieve a stable configuration. However, for noble gases and certain metals, the electron gain enthalpy can be positive.
First and Second Electron Gain Enthalpy
1. First Electron Gain Enthalpy:
The energy change when one electron is added to a neutral gaseous atom to form a uninegative ion.
\[
X(g) + e^- \rightarrow X^-(g)
\]
2. Second Electron Gain Enthalpy:
The energy change when an additional electron is added to an already negatively charged ion. Since the added electron experiences repulsion from the negatively charged ion, the second electron gain enthalpy is usually positive (endothermic).
\[
X^-(g) + e^- \rightarrow X^{2-}(g)
\]
Example:
- The first electron gain enthalpy of oxygen (\(O\)) is negative because oxygen releases energy when gaining an electron:
\[
O(g) + e^- \rightarrow O^-(g) \quad \Delta H = -141 \, \text{kJ/mol}
\]
- The second electron gain enthalpy of oxygen is positive because energy is required to add an electron to an already negatively charged \(O^-\) ion:
\[
O^-(g) + e^- \rightarrow O^{2-}(g) \quad \Delta H = +744 \, \text{kJ/mol}
\]
Factors Affecting Electron Gain Enthalpy
1. Atomic Size:
As the atomic radius increases, the distance between the nucleus and the added electron increases, resulting in a weaker attraction. Therefore, larger atoms tend to have less negative electron gain enthalpy.
- Example: Chlorine (\(Cl\)) has a more negative electron gain enthalpy than iodine (\(I\)) because chlorine is smaller and its nucleus can attract the added electron more strongly.
2. Nuclear Charge:
A higher nuclear charge increases the attraction between the nucleus and the added electron, resulting in more negative electron gain enthalpy.
- Example: Fluorine (\(F\)) has a highly negative electron gain enthalpy due to its high nuclear charge, which pulls the added electron closer.
3. Electron Shielding:
Inner electrons shield the outermost electrons from the full effect of the nuclear charge. In elements with more electron shells, the outer electrons are less tightly bound due to the shielding effect, leading to less negative electron gain enthalpy.
- Example: Electron gain enthalpy becomes less negative as we move down Group 17 because electron shielding increases.
4. Electron Configuration:
Elements with stable electron configurations (e.g., noble gases, filled subshells, or half-filled subshells) resist gaining electrons, resulting in positive or less negative electron gain enthalpy.
- Example: Neon (\(Ne\)) and argon (\(Ar\)) have positive electron gain enthalpies because they already have stable configurations and do not readily accept additional electrons.
Trends in Electron Gain Enthalpy
1. Across a Period (Left to Right):
- Electron gain enthalpy becomes more negative across a period. As we move from left to right, the atomic size decreases, and the nuclear charge increases, making it easier for the atom to attract an electron. Non-metals, especially halogens, have highly negative electron gain enthalpies.
- Example:
- The electron gain enthalpy of chlorine (\(Cl\)) is -349 kJ/mol, while that of sodium (\(Na\)) is -53 kJ/mol.
2. Down a Group (Top to Bottom):
- Electron gain enthalpy becomes less negative down a group. As the atomic radius increases, the added electron is farther from the nucleus, and the attraction between the nucleus and the electron decreases. Consequently, elements lower in the group are less likely to accept electrons.
- Example:
- The electron gain enthalpy of chlorine (\(Cl\)) is more negative than that of iodine (\(I\)).
Anomalies in Electron Gain Enthalpy
1. Fluorine vs. Chlorine:
- Although fluorine is smaller than chlorine, its electron gain enthalpy is less negative than chlorine's. This is because the small size of fluorine leads to significant electron-electron repulsion in its compact electron cloud, which reduces the energy released when fluorine gains an electron.
- Example:
- Fluorine: \(-328 \, \text{kJ/mol}\)
- Chlorine: \(-349 \, \text{kJ/mol}\)
2. Noble Gases:
- Noble gases have highly positive electron gain enthalpies. Since their outermost shells are completely filled, they do not tend to gain electrons, and energy must be supplied to force an additional electron into their electron cloud.
- Example:
- Neon (\(Ne\)) and argon (\(Ar\)) have positive electron gain enthalpies, indicating that they resist electron addition.
Comparison of Electron Gain Enthalpy and Ionization Energy
- Electron Gain Enthalpy is the energy change when an electron is added to an atom, while Ionization Energy is the energy required to remove an electron from an atom.
- Across a Period: Both ionization energy and electron gain enthalpy increase across a period as atomic size decreases and nuclear charge increases.
- Down a Group: Both ionization energy and electron gain enthalpy decrease down a group as atomic size increases and electron shielding reduces nuclear attraction.
Applications of Electron Gain Enthalpy
1. Predicting Reactivity of Non-Metals:
Non-metals, particularly halogens, have highly negative electron gain enthalpies, indicating their strong tendency to gain electrons and form anions. This makes halogens highly reactive.
- Example: Chlorine (\(Cl\)) and fluorine (\(F\)) are highly reactive non-metals due to their negative electron gain enthalpies.
2. Formation of Anions:
Elements with highly negative electron gain enthalpies are likely to form anions by gaining electrons. For example, oxygen forms \(O^{2-}\) and chlorine forms \(Cl^-\) in ionic compounds.
- Example: In sodium chloride (\(NaCl\)), chlorine gains an electron to form \(Cl^-\), and the electron gain enthalpy plays a role in stabilizing the ionic bond.
3. Industrial Use of Halogens:
The strong electron affinity of halogens makes them useful in industries for disinfection (e.g., chlorine in water treatment) and as oxidizing agents in chemical processes.
Q 1. Permanent hardness can be removed by adding; |
(A) \(\mathrm{Cl}_{2}\); |
(B) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\); |
(C) \(\mathrm{Ca}(\mathrm{OCl}) \mathrm{Cl}\); |
(D) \(\mathrm{K}_{2} \mathrm{CO}_{3}\); |
Q 2. Hydrogen can be prepared by the action of dil. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) on; |
(A) \(\mathrm{Cu}\); |
(B) \(\mathrm{Fe}\); |
(C) \(\mathrm{Pb}\); |
(D) \(\mathrm{Hg}\); |
Q 3. Find the correct order of electron gain enthalpy \(\left(\Delta_{\mathrm{eg}} \mathrm{H}\right)\) of the given elements.; |
(A) \(\mathrm{O}>\mathrm{B}>\mathrm{C}>\mathrm{N}\); |
(B) \(\mathrm{O}>\mathrm{C}>\mathrm{N}>\mathrm{B}\); |
(C) \(\mathrm{O}>\mathrm{C}>\mathrm{B}>\mathrm{N}\); |
(D) \(\mathrm{O}>\mathrm{N}>\mathrm{C}>\mathrm{B}\); |
Q 4. The law of triad is applicable to a group of; |
(A) \(\mathrm{Cl}, \mathrm{Br}, \mathrm{I}\); |
(B) \(\mathrm{C}, \mathrm{N}, \mathrm{O}\); |
(C) \(\mathrm{Na}, \mathrm{K}, \mathrm{Rb}\); |
(D) \(\mathrm{H}, \mathrm{O}, \mathrm{N}\); |
Q 5. The correct decreasing order of electropositive character among the following elements is; |
(A) \(\mathrm{Fe}>\mathrm{Sc}>\mathrm{Rb}>\mathrm{Br}>\mathrm{Te}>\mathrm{F}>\mathrm{Ca}\); |
(B) \(\mathrm{Ca}>\mathrm{Rb}>\mathrm{Sc}>\mathrm{Fe}>\mathrm{Te}>\mathrm{F}>\mathrm{Br}\); |
(C) \(\mathrm{Rb}>\mathrm{Ca}>\mathrm{Sc}>\mathrm{Fe}>\mathrm{Br}>\mathrm{Te}>\mathrm{F}\); |
(D) \(\mathrm{Rb}>\mathrm{Ca}>\mathrm{Sc}>\mathrm{Fe}>\mathrm{Te}>\mathrm{Br}>\mathrm{F}\); |
Q 6. Which of the following atomic and physical properties of hydrogen is false?; |
(A) Hydrogen \(>\) Deuterium \(>\) Tritium; (melting point \(/ K\) ); |
(B) Hydrogen \(<\) Deuterium \(<\) Tritium; (boiling point \(/ K\) ); |
(C) Hydrogen \(<\) Deuterium \(<\) Tritium; (density \(/ g L^{-1}\) ); |
(D) Hydrogen > Deuterium > Tritium; ( \() \%\) relative abundance); |
Q 7. Which of the following statements about heavy water are correct?\n(1) Heavy water is used as a moderator in nuclear reactors.\n(2) Heavy water is more associated than ordinary water.\n(3) Heavy water is a more effective solvent than ordinary water.; |
(A) (b) and (c); |
(B) (a) and (b); |
(C) (a) and (c); |
(D) (a), (b) and (c); |
Q 8. On which of the following Dobereiner's Triad law is not applicable?; |
(a) \(\mathrm{Cl}, \mathrm{Br}, \mathrm{I}\); |
(b) \(\mathrm{Ca}, \mathrm{Sr}, \mathrm{B}\); |
(c) \(\mathrm{F}, \mathrm{Cl}, \mathrm{Br}\); |
(d) \(\mathrm{Li}, \mathrm{Na}, \mathrm{K}\); |
Q 9. Select the correct chronological order for the discoveries of the following scientists Dobereiner, Newlands, Chancourtois, Mendeleev; |
(a) Chancourtois, Dobereiner, Newlands, Mendeleev; |
(b) Dobereiner, Chancourtois, Newlands, Mendeleev; |
(c) Dobereiner, Newlands, Chancourtois, Mendeleev; |
(d) Chancourtois, Newlands, Dobereiner, Mendeleev; |
Q 10. The period number in the periodic table corresponds to the A principal quantum number (n) of the elements Here, A refers to; |
(a) lowest; |
(b) highest; |
(c) middle; |
(d) None of these; |
Difference Between Electronegativity and Electron Gain Enthalpy
Although electronegativity and electron gain enthalpy both describe an atom’s tendency to attract electrons, they refer to different concepts and are measured in distinct ways. Here’s a detailed explanation of how they differ:
1. Definition
- Electronegativity:
Electronegativity is the tendency of an atom to attract a shared pair of electrons toward itself in a chemical bond. It is a dimensionless property and is measured on a relative scale, such as the Pauling scale. Electronegativity applies to atoms in covalent or ionic bonds.
Example:
- Fluorine (\(F\)) has the highest electronegativity (3.98 on the Pauling scale), meaning it strongly attracts shared electrons in a bond.
- Electron Gain Enthalpy:
Electron gain enthalpy refers to the amount of energy released or absorbed when an isolated gaseous atom gains an electron to form a negative ion. It is an absolute value measured in kilojoules per mole (kJ/mol). It applies to atoms in the gaseous state, not in a chemical bond.
Example:
- Chlorine (\(Cl\)) has a highly negative electron gain enthalpy (\(-349 \, \text{kJ/mol}\)), meaning it releases a significant amount of energy when it gains an electron.
2. Application
- Electronegativity:
- Describes an atom’s ability to attract electrons in a bond.
- Applies when atoms are part of a molecule, influencing bond polarity and molecule reactivity.
Example:
- In a hydrogen chloride molecule (HCl), chlorine is more electronegative than hydrogen, so the electron pair in the H-Cl bond is pulled closer to chlorine, resulting in a polar covalent bond.
- Electron Gain Enthalpy:
- Describes the energy change when an electron is added to a neutral gaseous atom.
- Applies to isolated atoms, not to atoms in a chemical bond.
Example:
- The electron gain enthalpy of oxygen (\(-141 \, \text{kJ/mol}\)) describes the energy released when an oxygen atom gains an electron to form an \(O^-\) ion.
3. Measurement and Units
- Electronegativity:
- Relative property with no units, typically measured on scales like the Pauling scale, where fluorine has the highest value (3.98).
- Electron Gain Enthalpy:
- Absolute value measured in kilojoules per mole (kJ/mol). It can be positive (when energy is required to add an electron) or negative (when energy is released upon gaining an electron).
4. Trend in the Periodic Table
- Electronegativity:
- Increases across a period from left to right, as atomic size decreases and nuclear charge increases, enhancing an atom's ability to attract electrons in a bond.
- Decreases down a group, as atomic size increases and the outer electrons are farther from the nucleus, reducing the atom's ability to attract electrons in a bond.
Example:
- Electronegativity increases from lithium (\(Li\)) to fluorine (\(F\)) across Period 2.
- Electron Gain Enthalpy:
- Becomes more negative across a period, as atoms more readily accept electrons due to higher nuclear charge and smaller atomic size.
- Becomes less negative down a group, as the added electron is farther from the nucleus and experiences less attraction, making it harder for the atom to gain an electron.
Example:
- Chlorine (\(Cl\)) has a more negative electron gain enthalpy than iodine (\(I\)) because chlorine’s smaller size allows it to attract electrons more strongly.
5. Key Differences
| Feature | Electronegativity | Electron Gain Enthalpy |
|||--|
| Definition | Tendency to attract shared electrons in a bond | Energy change when an electron is added to a gaseous atom |
| Measurement | Dimensionless, relative scale (Pauling scale) | Absolute value, measured in kJ/mol |
| Context | Applies to atoms in a bond | Applies to isolated gaseous atoms |
| Trend Across a Period | Increases from left to right | Becomes more negative from left to right |
| Trend Down a Group | Decreases from top to bottom | Becomes less negative from top to bottom |
| Examples | Fluorine has the highest electronegativity | Chlorine has the most negative electron gain enthalpy |
Examples for Clarification
1. Fluorine (F):
- Electronegativity: Fluorine has the highest electronegativity (\(3.98\)), meaning it strongly attracts electrons in bonds.
- Electron Gain Enthalpy: Fluorine’s electron gain enthalpy is \(-328 \, \text{kJ/mol}\), meaning it releases energy when it gains an electron. However, due to its small size, electron-electron repulsion makes its electron gain enthalpy less negative than chlorine.
2. Chlorine (Cl):
- Electronegativity: Chlorine’s electronegativity is \(3.16\), meaning it strongly attracts electrons in bonds, though not as strongly as fluorine.
- Electron Gain Enthalpy: Chlorine has a very negative electron gain enthalpy (\(-349 \, \text{kJ/mol}\)), making it more likely to gain an electron than fluorine, as it experiences less electron-electron repulsion.
Conclusion
Although both electronegativity and electron gain enthalpy relate to an atom's ability to attract electrons, they describe different phenomena. Electronegativity applies to atoms in bonds and measures their pull on shared electrons, while electron gain enthalpy describes the energy change when an isolated atom gains an electron. Understanding both concepts is essential for predicting chemical behavior, bonding patterns, and reactivity in elements.
Let me know if you need further clarification or additional examples!
Electronegativity
Definition of Electronegativity
Electronegativity is a measure of the ability of an atom in a molecule to attract shared electrons toward itself in a chemical bond. It is a dimensionless quantity and is commonly measured on scales like the Pauling scale, which assigns fluorine the highest value of 3.98 because it attracts electrons most strongly.
Electronegativity plays a crucial role in determining the polarity of bonds and the reactivity of elements in compounds. It explains why some elements tend to form ionic bonds by transferring electrons, while others form covalent bonds by sharing electrons unequally.
Key Concepts Related to Electronegativity
1. Bond Polarity:
- Electronegativity differences between two atoms in a bond determine whether the bond is non-polar covalent, polar covalent, or ionic.
- Non-polar Covalent Bond: When two atoms have identical or nearly identical electronegativities, they share electrons equally.
Example: The bond between two hydrogen atoms (\(H-H\)) is non-polar because both atoms have the same electronegativity.
- Polar Covalent Bond: When two atoms have different electronegativities, the electron pair is shared unequally, and the bond is polar.
Example: The bond between hydrogen and chlorine in HCl is polar because chlorine is more electronegative, pulling the electrons closer to itself.
- Ionic Bond: When the electronegativity difference is very large (usually greater than 1.7), one atom completely transfers its electron to the other, resulting in the formation of ions and an ionic bond.
Example: The bond between sodium (\(Na\)) and chlorine (\(Cl\)) in sodium chloride (\(NaCl\)) is ionic.
2. Electronegativity Scales:
- The most widely used scale for electronegativity is the Pauling scale, developed by Linus Pauling. Other scales include the Mulliken scale and the Allred-Rochow scale, but the Pauling scale remains the standard.
- On the Pauling scale, fluorine is the most electronegative element (3.98), and cesium and francium are among the least electronegative elements (0.7).
Factors Affecting Electronegativity
1. Atomic Size:
- As the size of an atom decreases, the distance between the nucleus and the bonding electrons decreases. A smaller atom has a stronger pull on shared electrons, resulting in a higher electronegativity.
- Example: Fluorine is the smallest halogen, and thus it has the highest electronegativity (3.98), while iodine, a larger halogen, has a lower electronegativity (2.66).
2. Nuclear Charge:
- A higher nuclear charge means that the nucleus exerts a stronger attractive force on the bonding electrons, increasing the atom's electronegativity. As the number of protons in the nucleus increases across a period, the electronegativity increases.
- Example: As we move across Period 2 from lithium to fluorine, the number of protons increases, and so does electronegativity.
3. Electron Shielding:
- Inner electrons shield the outer electrons from the full effect of the nuclear charge. More electron shielding (more inner electron shells) results in lower electronegativity because the nucleus’s pull on bonding electrons is reduced.
- Example: Electronegativity decreases down a group due to increased electron shielding, which reduces the nucleus’s ability to attract bonding electrons.
Trends in Electronegativity
1. Across a Period (Left to Right):
- Electronegativity increases from left to right across a period in the periodic table. This is because atomic size decreases and nuclear charge increases, leading to a stronger attraction for bonding electrons.
- Example: In Period 2, the electronegativity increases from lithium (\(Li\), 0.98) to fluorine (\(F\), 3.98), with fluorine being the most electronegative element in the entire periodic table.
2. Down a Group (Top to Bottom):
- Electronegativity decreases down a group. As the atomic size increases and the number of electron shells increases, the added electrons experience more shielding. This reduces the effective nuclear charge felt by the outermost electrons, making it harder for the atom to attract shared electrons.
- Example: In Group 17, the electronegativity decreases from fluorine (\(F\), 3.98) to iodine (\(I\), 2.66).
Anomalies in Electronegativity Trends
1. Noble Gases:
- Noble gases like helium, neon, and argon are generally not assigned electronegativity values on the Pauling scale because they do not usually form bonds under normal conditions. This is because their electron configurations are stable, and they rarely attract additional electrons to form bonds.
2. Fluorine vs. Chlorine:
- Fluorine is more electronegative than chlorine despite both being in the same group. This anomaly is due to fluorine’s small size and high nuclear charge, which allows it to strongly attract bonding electrons despite electron-electron repulsion within its small atomic radius.
Comparison of Electronegativity with Other Properties
1. Electronegativity vs. Electron Gain Enthalpy:
- Electronegativity refers to the tendency of an atom to attract electrons in a bond, whereas electron gain enthalpy refers to the energy change when an isolated atom gains an electron.
- Both properties increase across a period and decrease down a group, but electronegativity applies to atoms in a bonding context, while electron gain enthalpy applies to isolated atoms.
2. Electronegativity vs. Ionization Energy:
- Ionization energy is the energy required to remove an electron from an atom, while electronegativity is the tendency to attract shared electrons in a bond.
- Both increase across a period and decrease down a group. High electronegativity often correlates with high ionization energy, as both indicate a strong attraction between the nucleus and electrons.
Applications of Electronegativity
1. Predicting Bond Type:
- Electronegativity is key in predicting the type of bond formed between two atoms. A large difference in electronegativity (>1.7) results in an ionic bond, while a small difference results in a covalent bond (polar or non-polar).
- Example: The bond in sodium chloride (\(NaCl\)) is ionic due to the large electronegativity difference between sodium (\(0.93\)) and chlorine (\(3.16\)), whereas the bond in oxygen (\(O_2\)) is non-polar covalent because both oxygen atoms have the same electronegativity.
2. Determining Molecular Polarity:
- Electronegativity helps determine whether a molecule is polar or non-polar. A molecule with atoms of different electronegativities is likely to be polar, meaning it has a partial positive and partial negative charge.
- Example: In water (\(H_2O\)), oxygen is more electronegative than hydrogen, creating a polar molecule with a partial negative charge on the oxygen atom and partial positive charges on the hydrogen atoms.
3. Reactivity of Elements:
- Highly electronegative elements, such as fluorine, oxygen, and nitrogen, are highly reactive because they have a strong tendency to attract electrons.
- Example: Fluorine is highly reactive due to its high electronegativity, reacting vigorously with almost all elements, including metals and non-metals.
Electronegativity and Molecular Shape
Electronegativity not only influences bond polarity but also plays a significant role in determining molecular geometry and shape. In a molecule, atoms with different electronegativities can create a dipole, affecting the spatial arrangement of atoms in the molecule. The unequal sharing of electrons in polar covalent bonds leads to the creation of polar molecules, influencing properties like boiling point, solubility, and reactivity.
Example:
- In water (\(H_2O\)), oxygen's higher electronegativity pulls electrons closer, resulting in a bent molecular shape with a polar nature, contributing to water's high boiling point and solvent properties.
Q 1. The reason of bleaching properties of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is; |
(A) Unstable nature; |
(B) Acidic nature; |
(C) Reducing nature; |
(D) Oxidising nature; |
Q 2. In context with the industrial preparation of hydrogen from water gas \(\left(\mathrm{CO}+\mathrm{H}_{2}\right)\), which of the following is the correct statement?; |
(A) \(\mathrm{CO}\) is oxidised to \(\mathrm{CO}_{2}\) with steam, in the presence of a catalyst followed by the dissolution of \(\mathrm{CO}_{2}\) in water under ssure.; |
(B) \(\mathrm{CO}\) and \(\mathrm{H}_{2}\) are fractionally separated using the differences in their densities.; |
(C) \(\mathrm{CO}\) is removed by absorption in aqueous \(\mathrm{Cu}_{2} \mathrm{Cl}_{2}\) solution.; |
(D) \(\mathrm{H}_{2}\) is removed through occlusion with \(\mathrm{Pd}\).; |
Q 3. The correct increasing order of electronegativity for \(\mathrm{Sn}, \mathrm{Sn}^{2+}\) and \(\mathrm{Sn}^{4+}\) species is:; |
(A) \(\mathrm{Sn}<\mathrm{Sn}^{2+}<\mathrm{Sn}^{4+}\); |
(B) \(\mathrm{Sn}^{4+}<\mathrm{Sn}^{2+}<\mathrm{Sn}\); |
(C) \(\mathrm{Sn}=\mathrm{Sn}^{2+}=\mathrm{Sn}^{4}\); |
(D) \(\mathrm{Sn}<\mathrm{Sn}^{2+}=\mathrm{Sn}^{4+}\); |
Q 4. The first \(\left(\Delta_{\mathrm{i}} \mathrm{H}_{1}\right)\) and the second \(\left(\Delta_{\mathrm{i}} \mathrm{H}_{2}\right)\) ionization enthalpies ( in \(\mathrm{kJmol}^{-1}\) ) and the \(\Delta \mathrm{e}_{\mathrm{g}} \mathrm{H}\) electron gain enthalpy (in \(\mathrm{kJ}\) mol -1 ) of a few elements are given below :\n\begin{array}{|c|c|c|c|}\n\hline\nElements & \(\Delta \mathrm{H}_{1}\) & \(\Delta \mathrm{H}_{2}\) & \(\Delta_{\mathrm{eg}} \mathrm{H}\) \\\n\hline\n\(\mathrm{I}\) & \(52 \mathrm{O}\) & 7300 & -60 \\\n\hline\n\(\mathrm{II}\) & 419 & 3051 & -48 \\\n\hline\n\(\mathrm{III}\) & 1681 & 3374 & -328 \\\n\hline\n\(\mathrm{IV}\) & 1008 & 1846 & -295 \\\n\hline\n\(\mathrm{V}\) & 2372 & 5251 & +48 \\\n\hline\n\(\mathrm{VI}\) & 738 & 1451 & -40 \\\n\hline\n\end{array}\nWhich of the above likely to be the least reactive non-metal?; |
(A)I; |
(B)II; |
(C)III; |
(D)IV; |
Q 5. Which of the following element shows maximum valency?; |
(A) Carbon; |
(B) Barium; |
(C) Nitrogen; |
(D) Sulphur; |
Q 6. Dalda is prepared from oils by; |
(A) Oxidation; |
(B) Reduction; |
(C) Hydrolysis; |
(D) Distillation; |
Q 7. In 1800, only X elements were known Here, X refers to; |
(a) 63; |
(b) 31; |
(c) 114; |
(d) 92; |
Q 8. Newlands could classify elements only upto -; |
(a) copper; |
(b) chlorine; |
(c) calcium; |
(d) chromium; |
Q 9. The molecular formula of chloride of Eka-Aluminium and Eka-Silicon respectively are; |
(a) \(\mathrm{GaCl}_{3}\) and \(\mathrm{SiO}_{4}\); |
(b) \(\mathrm{GaCl}_{3}\) and \(\mathrm{AlCl}_{3}\); |
(c) \(\mathrm{AlCl}_{3}\) and \(\mathrm{SiCl}_{4}\); |
(d) \(\mathrm{GaCl}_{3}\) and \(\mathrm{GeCl}_{4}\); |
Q 10. The symbol and IUPAC name for the element with atomic number 120 , respectively are; |
(a) Ubn and unbinilium; |
(b) Ubn and unbiunium; |
(c) Ubn and unnibium; |
(d) Ubn and unnilium; |
Dobereiner's triads: Classification of elements into groups of three with similar properties. |
IUPAC Nomenclature: Unnilunium (Element 101, now known as Mendelevium). |
Atomic radii trend in alkali metals: Increase down the group due to increasing number of energy levels. |
Heavy water is deuterium oxide (D2O): Occurs in compounds where hydrogen is replaced by deuterium. |
Ionic compounds causing hardness in water: Presence of ions like Ca2+ and Mg2+. |
Periodic Table |
Ionization Enthalpy |
Atomic Radius |
Periodic Trends |
Electron Configuration |
Electron Gain Enthalpy |
Electronegativity |