Anand Classes presents NCERT Solutions for Chemistry Class 11 Chapter 3 – Classification of Elements and Periodicity in Properties (Questions 3.6, 3.7, 3.8, 3.9, 3.10). These solutions provide detailed explanations about the periodic table, atomic number determination, element naming, similarities in group properties, atomic radius, ionic radius, and periodic trends in atomic size. The step-by-step answers make it easier for students to understand and prepare effectively for exams. Click the print button to download study material and notes.
NCERT 3.6: Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
Answer:
The atomic number is 17 (the element is chlorine, Cl).
Detailed explanation:
The third period corresponds to principal quantum number $n=3$, so only the $3s$ and $3p$ subshells are filled in this period. That means the third period contains 2 s-block elements and 6 p-block elements, a total of 8 elements. The third period therefore starts at $Z = 11$ (sodium, Na) and ends at $Z = 18$ (argon, Ar).
Listing them with their $Z$ values:
- $Z = 11$: Na (3s¹) — group 1
- $Z = 12$: Mg (3s²) — group 2
- $Z = 13$: Al (3p¹) — group 13
- $Z = 14$: Si (3p²) — group 14
- $Z = 15$: P (3p³) — group 15
- $Z = 16$: S (3p⁴) — group 16
- $Z = 17$: Cl (3p⁵) — group 17
- $Z = 18$: Ar (3p⁶) — group 18
Explanation by counting:
After the two s-block elements ($Z=11,12$), the p-block elements occupy $Z=13$ to $Z=18$, which correspond respectively to groups 13–18. Therefore group 17 in the third period corresponds to $Z = 17$.
Simple arithmetic check (step-by-step):
The last s-block atomic number is $12$. Group 13 starts at $Z=13$ (first p-element), so group 17 is the 5th p-element:
$$Z = 12 + 5 = 17.$$
Electron configuration (shorthand) of the element with $Z=17$:
$$[Ne]\;3s^2\,3p^5$$
Final note:
The element with $Z=17$ is chlorine (Cl), a halogen (group 17) — a reactive non-metal with high electronegativity and typical halogen chemistry.
Question 3.7: Which element do you think would have been named by
(i) Lawrence Berkeley Laboratory (ii) Seaborg’s group?
Answer:
(i) The Lawrence Berkeley Laboratory in California has contributed to the discovery of many transuranium elements. Two elements named in its honor are:
- Lawrencium (Lr, Z = 103), named after Ernest O. Lawrence, the inventor of the cyclotron and founder of the Lawrence Berkeley Laboratory.
- Berkelium (Bk, Z = 97), named after the city of Berkeley, where the laboratory is located.
(ii) Seaborg’s group at the University of California was instrumental in the discovery of several heavy elements. One element was directly named to honor Glenn T. Seaborg, the famous American chemist who contributed greatly to actinide chemistry. That element is:
- Seaborgium (Sg, Z = 106).
Thus, the names are:
- Lawrence Berkeley Laboratory → Lawrencium (Z = 103) and Berkelium (Z = 97)
- Seaborg’s group → Seaborgium (Z = 106)
NCERT 3.8: Why do elements in the same group have similar physical and chemical properties?
Answer:
Reason for Similar Properties :
Elements in the same group of the periodic table have similar physical and chemical properties because they have the same number of electrons in their outermost shell (valence shell). Since chemical behavior mainly depends on the valence electrons, similar configurations result in similar properties.
Example: Group 1 Elements (Alkali Metals)
All alkali metals have the general configuration ns¹. Each of them has one electron in the outermost shell, which they readily lose to form M⁺ ions. This explains why all alkali metals are soft, highly reactive, good reducing agents, and form similar compounds like oxides, hydroxides, and halides.
Example: Group 17 Elements (Halogens)
All halogens have the outer electronic configuration ns²np⁵. With seven valence electrons, they need only one more electron to complete their octet. This explains why halogens are highly electronegative, reactive, and form salts such as NaCl, KBr, etc. All halogens also exist naturally as diatomic molecules (Cl₂, Br₂, I₂).
Physical Properties Similarity
Because of similar electronic configurations, elements in the same group also exhibit similarities in physical properties such as atomic radius trends, ionization enthalpy patterns, and electronegativity.
Conclusion
Thus, the repetition of the same type of valence shell electronic configuration down a group is the fundamental reason why elements in the same group have similar physical and chemical properties.
NCERT 3.9 : What does atomic radius and ionic radius really mean to you?
Answer.
Atomic radius literally means the size of the atom. It can be measured either by X–ray diffraction or by spectroscopic methods.
In case of non–metals, atomic radius is called covalent radius. It is defined as one–half the distance between the nuclei of two covalently bonded atoms of the same element in a molecule.
For example, the internuclear distance between two chlorine atoms in a chlorine molecule is $198 \ \text{pm}$. Therefore, the covalent radius of a chlorine atom is:
$$
r_{\text{cov}} = \frac{198}{2} = 99 \ \text{pm} \ (0.99 \ \text{Å})
$$
In case of metals, atomic radius is called metallic radius. It is defined as one–half the distance between two adjacent atoms in the crystal lattice.
For example, the distance between two adjacent copper atoms in solid copper is $256 \ \text{pm}$. Therefore, the metallic radius of copper is:
$$
r_{\text{met}} = \frac{256}{2} = 128 \ \text{pm} \ (1.28 \ \text{Å})
$$
Ionic Radius
Ionic radius means the size of the ion. An ion can be either a cation or an anion.
- The size of a cation is always smaller than that of the parent atom because loss of one or more electrons increases the effective nuclear charge ($Z_{\text{eff}}$). As a result, the force of attraction increases and the ionic size decreases.
- The size of an anion is always larger than that of the parent atom because the addition of one or more electrons decreases the effective nuclear charge ($Z_{\text{eff}}$). As a result, the force of attraction decreases and the ionic size increases.
Examples
- Sodium:
- Atomic radius of Na = $186 \ \text{pm}$
- Ionic radius of Na⁺ = $95 \ \text{pm}$
$$
r_{\text{Na}^+} < r_{\text{Na}}
$$
- Fluorine:
- Atomic radius of F = $72 \ \text{pm}$
- Ionic radius of F⁻ = $136 \ \text{pm}$
$$
r_{\text{F}^-} > r_{\text{F}}
$$
NCERT 3.10 : How do atomic radius vary in a period and in a group? How do you explain the variation?
Answer
The atomic radius increases down the group. This is because a new energy shell (i.e. principal quantum number increases by unity) is added at each succeeding element while the number of electrons in the valence shell remains the same. In other words, the electrons in the valence shell of each succeeding element lie farther and farther away from the nucleus. As a result, the force of attraction of the nucleus for the valence electrons decreases and hence the atomic size increases.
In contrast, the atomic size decreases as we move from left to right in a period. This is because within a period the outer electrons remain in the same shell but the nuclear charge increases by one unit at each succeeding element. Due to this increased nuclear charge, the attraction of the nucleus for the outer electrons increases and hence the atomic size decreases.
Variation in a Group (Top to Bottom)
- As we move down a group in the periodic table:
- A new shell (principal quantum number $n$) is added for each element.
- Electrons of the valence shell lie farther from the nucleus.
- Although nuclear charge increases, the shielding effect (or screening effect) due to inner electrons reduces the effective nuclear attraction.
- Therefore, atomic radius increases.
Example (Group 1 elements):
$$
r_{\text{Li}} < r_{\text{Na}} < r_{\text{K}} < r_{\text{Rb}} < r_{\text{Cs}}
$$
Variation in a Period (Left to Right)
- As we move across a period from left to right:
- Electrons are added to the same shell (no increase in principal quantum number).
- Nuclear charge increases by +1 for each successive element.
- This increased nuclear attraction pulls the outer electrons closer.
- Hence, atomic radius decreases across a period.
Example (Period 2 elements):
$$
r_{\text{Li}} > r_{\text{Be}} > r_{\text{B}} > r_{\text{C}} > r_{\text{N}} > r_{\text{O}} > r_{\text{F}} > r_{\text{Ne}}
$$
Key Points
- Down a group: Atomic radius increases due to addition of shells.
- Across a period: Atomic radius decreases due to increased nuclear charge.
- Exceptions: Noble gases have exceptionally small radii (measured as van der Waals radius), while transition elements show smaller variations because of poor shielding by $d$-electrons.
Importance of This Trend
- Explains why alkali metals (Group 1) are the most reactive metals (large size, low ionization energy).
- Explains why halogens (Group 17) are highly reactive non-metals (small size, high electron affinity).
- Plays a key role in determining bond lengths, electronegativity, and chemical reactivity of elements.
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