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Classification of Elements
Periodic Trends & Advanced Practice for JEE/NEET
3.7 Periodic Trends in Physical Properties
Understanding how physical properties change across periods and down groups is crucial for predicting element behavior. This section explores key trends such as atomic/ionic radii, ionization enthalpy, electron gain enthalpy, and electronegativity.
(a) Atomic Radius
Defining the size of an atom is complex due to the diffuse nature of its electron cloud. Atomic radius is generally estimated based on the distance between atoms in a combined state:
- Covalent Radius: Half the distance between two atoms bonded by a single bond in a covalent molecule (e.g., Cl-Cl bond in Cl₂ is 198 pm, so atomic radius of Cl is 99 pm).
- Metallic Radius: Half the internuclear distance between adjacent metal cores in a metallic crystal (e.g., Cu-Cu distance in solid Cu is 256 pm, so metallic radius of Cu is 128 pm).
Trends:
- Across a Period (Left to Right): Decreases. This is due to increasing effective nuclear charge (Z_eff). The outer electrons are in the same valence shell, but the increasing nuclear charge pulls them more tightly towards the nucleus.
- Down a Group (Top to Bottom): Increases. This is due to the increase in principal quantum number (n), meaning valence electrons are in higher energy shells, further from the nucleus. Inner electrons also provide increased shielding, reducing the pull of the nucleus on the outer electrons.
- Noble Gases: Not considered here with covalent radii, as they are monoatomic. Their van der Waals radii are much larger, which is a different measure.
NEET/JEE Practice Question:
Which of the following atoms has the largest atomic radius?
- (a) Li
- (b) Be
- (c) B
- (d) F
Correct Answer: (a) Li
Explanation: Atomic radius decreases across a period from left to right due to an increase in effective nuclear charge. Among elements of the same period (Period 2), Lithium (Li) is the leftmost element, hence it has the largest atomic radius. F is the rightmost element, thus smallest radius.
(b) Ionic Radius
Ionic radii exhibit similar trends to atomic radii, but with key differences depending on whether an ion is a cation or an anion.
- Cation ($X^+$): Always smaller than its parent atom (X). This is because the cation has fewer electrons, but the same nuclear charge, leading to a stronger pull on the remaining electrons. Example: Na (186 pm) vs Na$^+$ (95 pm).
- Anion ($X^-$): Always larger than its parent atom (X). This is due to the addition of one or more electrons, increasing electron-electron repulsion and decreasing the effective nuclear charge felt by the valence electrons, causing the electron cloud to expand. Example: F (64 pm) vs F$^-$ (136 pm).
- Isoelectronic Species: Atoms or ions with the same number of electrons (e.g., O²⁻, F⁻, Na⁺, Mg²⁺ all have 10 electrons). For isoelectronic species, the radius decreases with increasing positive nuclear charge. The greater the nuclear charge, the stronger the attraction for the same number of electrons, leading to a smaller size.
Order of size: O²⁻ > F⁻ > Na⁺ > Mg²⁺ (due to Z: 8, 9, 11, 12 respectively).
NEET/JEE Practice Question:
Which of the following species will have the largest and the smallest size: Mg, Mg²⁺, Al, Al³⁺?
- (a) Largest: Mg, Smallest: Al³⁺
- (b) Largest: Al, Smallest: Mg²⁺
- (c) Largest: Mg²⁺, Smallest: Al
- (d) Largest: Al³⁺, Smallest: Mg
Correct Answer: (a) Largest: Mg, Smallest: Al³⁺
Explanation: Atomic radii decrease across a period (Mg > Al). Cations are smaller than their parent atoms (Mg > Mg²⁺ and Al > Al³⁺). Among isoelectronic species (Mg²⁺ and Al³⁺ both have 10 electrons), the one with the larger positive nuclear charge (Al³⁺, Z=13) will have a smaller radius than Mg²⁺ (Z=12). Therefore, Mg is the largest and Al³⁺ is the smallest.
(c) Ionization Enthalpy ($\Delta_i H$)
Ionization enthalpy is the energy required to remove an electron from an isolated gaseous atom in its ground state. It is always positive (energy is absorbed).
$\text{X(g)} \rightarrow \text{X}^+\text{(g)} + \text{e}^-$ (First Ionization Enthalpy, $\Delta_i H_1$)
$\text{X}^+\text{(g)} \rightarrow \text{X}^{2+}\text{(g)} + \text{e}^-$ (Second Ionization Enthalpy, $\Delta_i H_2$)
Successive ionization enthalpies are always higher (e.g., $\Delta_i H_2 > \Delta_i H_1$) because it’s harder to remove an electron from a positively charged ion.
Factors Affecting Ionization Enthalpy:
- Effective Nuclear Charge (Z_eff): Higher Z_eff means electrons are held more tightly, requiring more energy to remove.
- Atomic Size: Larger atomic size means valence electrons are further from the nucleus, experiencing less attraction, thus easier to remove (lower IE).
- Shielding/Screening Effect: Inner core electrons shield valence electrons from the full nuclear charge. Greater shielding leads to lower IE.
- Penetration Effect of Orbitals: For the same principal quantum number, the order of penetration is s > p > d > f. More penetrating orbitals are closer to the nucleus and held more tightly, so s-electrons are harder to remove than p-electrons from the same shell.
- Stability of Half-filled/Completely Filled Orbitals: Atoms with exactly half-filled or completely filled subshells exhibit extra stability, requiring more energy to remove an electron.
Trends:
- Across a Period (Left to Right): Generally Increases. Increasing Z_eff and decreasing atomic size lead to stronger attraction for valence electrons.
- Exceptions:
- IE of B is less than Be: Be has $1s^22s^2$, electron removed is from stable 2s orbital. B has $1s^22s^22p^1$, electron removed is from 2p orbital. Due to better penetration of 2s over 2p, 2s electron is held more tightly.
- IE of O is less than N: N has $1s^22s^22p^3$ (half-filled p-subshell, stable). O has $1s^22s^22p^4$. Removing an electron from O (one paired p-electron) is easier due to increased electron-electron repulsion, overcoming the effect of increased nuclear charge.
NEET/JEE Practice Question:
The first ionization enthalpy ($\Delta_i H$) values of the third period elements, Na, Mg and Si are respectively 496, 737 and 786 kJ mol⁻¹. Predict whether the first $\Delta_i H$ value for Al will be more close to 575 or 760 kJ mol⁻¹? Justify your answer.
Correct Answer: More close to 575 kJ mol⁻¹.
Explanation: Aluminium (Al) is in Group 13, while Magnesium (Mg) is in Group 2. Normally, ionization enthalpy increases across a period. However, similar to the Be-B anomaly, the electron removed from Al is a 3p electron, while from Mg it is a 3s electron. The 3p electron in Al is more shielded from the nucleus by the inner 3s electrons than the 3s electrons in Mg. This makes it easier to remove the 3p electron from Al compared to the 3s electron from Mg, leading to a lower ionization enthalpy for Al than Mg (575 kJ mol⁻¹ is lower than Mg’s 737 kJ mol⁻¹).
(d) Electron Gain Enthalpy ($\Delta_{eg} H$)
Electron gain enthalpy is the enthalpy change when an electron is added to a neutral gaseous atom to form a negative ion.
$\text{X(g)} + \text{e}^- \rightarrow \text{X}^-\text{(g)}$
This process can be exothermic ($\Delta_{eg}H$ is negative, energy released) or endothermic ($\Delta_{eg}H$ is positive, energy absorbed).
- Exothermic: Group 17 elements (halogens) have highly negative $\Delta_{eg}H$ values because they achieve a stable noble gas configuration by gaining an electron.
- Endothermic: Noble gases have large positive $\Delta_{eg}H$ values because adding an electron requires it to enter a higher, unstable energy level.
Trends:
- Across a Period (Left to Right): Generally becomes more negative. Effective nuclear charge increases, making it easier for an atom to attract an additional electron.
- Down a Group (Top to Bottom): Generally becomes less negative. Atomic size increases, meaning the added electron is further from the nucleus and experiences less attraction.
- Anomalies:
- Electron gain enthalpy of O or F is less negative than that of S or Cl, respectively. This is because O (2p) and F (2p) are very small, and the incoming electron experiences significant repulsion from existing electrons in the small 2p orbitals. For S (3p) and Cl (3p), the larger atomic size means less electron-electron repulsion for the incoming electron. Thus, S and Cl have more negative electron gain enthalpies than O and F.
NEET/JEE Practice Question:
Which of the following will have the most negative electron gain enthalpy and which the least negative? P, S, Cl, F. Explain your answer.
- (a) Most negative: Cl; Least negative: P
- (b) Most negative: F; Least negative: P
- (c) Most negative: Cl; Least negative: F
- (d) Most negative: S; Least negative: P
Correct Answer: (a) Most negative: Cl; Least negative: P
Explanation: Electron gain enthalpy generally becomes more negative across a period. So, for P, S, Cl (Period 3), it increases from P to Cl. For F (Period 2), though it is a halogen, its smaller size leads to higher electron-electron repulsion, making its electron gain enthalpy less negative than that of Cl. P is on the left, making it the least negative. Thus, the order is P (least negative) < S < F < Cl (most negative).
(e) Electronegativity
Electronegativity is a qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself. Unlike IE and EGE, it is not a directly measurable quantity, but scales (like the Pauling scale, where Fluorine is assigned 4.0) have been developed.
Trends:
- Across a Period (Left to Right): Increases. As atomic radius decreases and effective nuclear charge increases, the attraction for shared electrons becomes stronger.
- Down a Group (Top to Bottom): Decreases. As atomic radius increases, the valence electrons are further from the nucleus, and the attraction for shared electrons weakens.
Relationship with Metallic/Non-metallic Character:
- Electronegativity is directly related to non-metallic properties (strong tendency to gain electrons).
- Electronegativity is inversely related to metallic properties.
- Thus, increasing electronegativity across a period is accompanied by an increase in non-metallic properties (and decrease in metallic properties).
- Decreasing electronegativity down a group is accompanied by a decrease in non-metallic properties (and increase in metallic properties).
3.7.2 Periodic Trends in Chemical Properties
The periodicity in electronic configuration manifests in distinct chemical properties, including valence, unique behavior of second-period elements, and reactivity patterns.
(a) Periodicity of Valence or Oxidation States
The valence of representative elements is typically equal to the number of electrons in their outermost orbitals or eight minus that number. Oxidation state is a more frequently used term, representing the charge an atom acquires based on electronegativity considerations in a compound.
Examples:
- In OF₂: Fluorine (most electronegative) is -1. Oxygen (2s²2p⁴) shares two electrons, exhibiting +2 oxidation state.
- In Na₂O: Oxygen (more electronegative) accepts two electrons (one from each Na), showing -2. Sodium (3s¹) loses one electron, showing +1.
NEET/JEE Practice Question:
Using the Periodic Table, predict the formulas of compounds which might be formed by the following pairs of elements: (i) silicon and bromine (ii) aluminium and sulphur.
Correct Formulas:
(i) Silicon and Bromine: Silicon is Group 14, valence 4. Bromine is Group 17, valence 1. Formula: SiBr₄.
(ii) Aluminium and Sulphur: Aluminium is Group 13, valence 3. Sulphur is Group 16, valence 2. Formula: Al₂S₃.
(b) Anomalous Properties of Second Period Elements
The first element of each group in s- and p-blocks (Li, Be, B, C, N, O, F) often shows properties different from other members of their respective groups. This is termed anomalous behavior.
Reasons for Anomalous Behavior:
- Small Size: Extremely small atomic and ionic radii.
- Large Charge/Radius Ratio: Leads to high charge density.
- High Electronegativity: Strong tendency to attract electrons.
- Absence of d-orbitals in Valence Shell: The second period elements only have 2s and 2p orbitals (max 4 valence orbitals) available for bonding. This limits their maximum covalency to 4 (e.g., Boron can form [BF₄]⁻, but not [BF₆]³⁻). Subsequent members (e.g., Al) can expand their valence shell using d-orbitals (e.g., [AlF₆]³⁻).
- Ability to Form Multiple Bonds (p-block): First members of p-block (C, N, O, F) show greater ability to form $p\pi-p\pi$ multiple bonds (C=C, C≡C, N=N, N≡N, C=O, C≡N, N=O). Heavier elements form weaker $p\pi-p\pi$ bonds.
Diagonal Relationship:
The first element of a group often resembles the second element of the *next* group diagonally. Examples: Li resembles Mg, and Be resembles Al, due to similar charge/radius ratios and electronegativities.
NEET/JEE Practice Question:
Are the oxidation state and covalency of Al in [AlCl(H₂O)₅]²⁺ the same?
Answer: No, they are not the same.
Explanation:
- Oxidation state of Al: In [AlCl(H₂O)₅]²⁺, H₂O is neutral (0), Cl is -1. Let oxidation state of Al be x. So, x + (-1) + 5(0) = +2. Thus, x = +3. The oxidation state of Al is +3.
- Covalency of Al: Covalency refers to the number of bonds an atom forms. In this complex, Al is bonded to one Cl atom and five H₂O molecules, forming a total of 1 + 5 = 6 bonds. Therefore, the covalency of Al is 6.
3.7.3 Periodic Trends and Chemical Reactivity
Chemical reactivity is a manifestation of electronic configuration, and it shows distinct periodic trends.
Trends Across a Period:
- Reactivity: Highest at the two extremes (Group 1 metals, Group 17 non-metals) and lowest in the center.
- Extreme Left (Alkali Metals): High reactivity due to ease of electron loss (low ionization enthalpy) to form cations. They are strong reducing agents.
- Extreme Right (Halogens): High reactivity due to high electron gain enthalpy (strong tendency to gain electrons) to form anions. They are strong oxidizing agents.
- Metallic Character: Decreases from left to right.
- Non-metallic Character: Increases from left to right.
- Nature of Oxides:
- Left (e.g., Na₂O): Most basic oxides (react with water to form strong bases: Na₂O + H₂O → 2NaOH).
- Right (e.g., Cl₂O₇): Most acidic oxides (react with water to form strong acids: Cl₂O₇ + H₂O → 2HClO₄).
- Center (e.g., Al₂O₃, As₂O₃): Amphoteric oxides (behave as acidic with bases and basic with acids).
- Some (e.g., CO, NO, N₂O): Neutral oxides (no acidic or basic properties).
Trends Down a Group:
- Metallic Character: Increases down the group (due to increasing atomic size and decreasing ionization enthalpy).
- Non-metallic Character: Decreases down the group.
- Reactivity of Metals (e.g., Alkali Metals): Increases down the group.
- Reactivity of Non-metals (e.g., Halogens): Decreases down the group.
NEET/JEE Practice Question:
Show by a chemical reaction with water that Na₂O is a basic oxide and Cl₂O₇ is an acidic oxide.
Reactions:
Basic Oxide (Na₂O): Na₂O + H₂O → 2NaOH (Sodium Hydroxide, a strong base)
Acidic Oxide (Cl₂O₇): Cl₂O₇ + H₂O → 2HClO₄ (Perchloric acid, a strong acid)
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