Anand Classes provides comprehensive NCERT Exemplar Solutions for the topic Classification of Elements and Periodicity in Properties for Class 11 Chemistry. Our detailed solutions are designed to help students understand the periodic trends, classifications, and properties of elements clearly and effectively. With step-by-step explanations, solved examples, and practice questions, students can strengthen their concepts and perform better in exams. Click the print button to download study material and notes.
Question 1
Consider the isoelectronic species: $Na^+$, $Mg^{2+}$, $F^-$ and $O^{2-}$. The correct order of increasing length of their radii is
(a) $F^- < O^{2-} < Mg^{2+} < Na^+$
(b) $Mg^{2+} < Na^+ < F^- < O^{2-}$
(c) $O^{2-} < F^- < Na^+ < Mg^{2+}$
(d) $O^{2-} < F^- < Mg^{2+} < Na^+$
Answer: Correct Option (b)
Step 1: Key concept — isoelectronic species
Isoelectronic means all species have the same number of electrons. Here each species has 10 electrons (Ne configuration: $1s^2\,2s^2\,2p^6$). Because electrons are the same, the main factor that controls size is the nuclear charge (number of protons) that pulls those electrons in.
Step 2: Nuclear charge (atomic number, $Z$) for each ion (protons are unchanged when electrons are gained/lost):
- $Mg^{2+}:\ Z=12$
- $Na^{+}:\ Z=11$
- $F^{-}:\ Z=9$
- $O^{2-}:\ Z=8$
Step 3: Why the radii order follows $Z$
- Higher $Z$ → stronger attraction on the same electron cloud → smaller radius.
- Therefore radii increase as $Z$ decreases. So:
$$Mg^{2+} < Na^{+} < F^{-} < O^{2-}$$
Final Answer:
$$\boxed{Mg^{2+} < Na^{+} < F^{-} < O^{2-}}$$
Question 2
Which of the following is not an actinoid?
(a) Curium (Z = 96)
(b) Californium (Z = 98)
(c) Uranium (Z = 92)
(d) Terbium (Z = 65)
Answer: Correct Option (d)
Step 1: Definition
Actinoids = elements with atomic numbers $Z=90$ to $103$.
Step 2: Apply the definition
- Curium (96), Californium (98), Uranium (92) all fall in $90\text{–}103$ → actinoids.
- Terbium (Z = 65) does NOT; Tb belongs to the lanthanoids (4f series).
Final Answer:
$$\boxed{\text{Terbium (Z = 65) — not an actinoid}}$$
Question 3
The order of screening effect of electrons of s, p, d and f orbitals of a given shell of an atom on its outer shell electrons is
(a) $s > p > d > f$
(b) $f > d > p > s$
(c) $p < d < s > f$
(d) $f > p > s > d$
Answer: Correct Option (a)
Step 1: Definition — shielding / screening
Shielding (screening) is how much an electron in an inner orbital reduces the effective nuclear attraction felt by an outer electron.
Step 2: Penetration vs shielding
- Orbitals differ in radial penetration: $s$ electrons penetrate closest to the nucleus (most deeply), then $p$, then $d$, then $f$.
- More penetration = better shielding (because a penetrating electron spends more time between the nucleus and outer electron).
Step 3: Conclusion (for same principal shell)
$$\textbf{Screening (shielding)}: \quad s > p > d > f$$
Important: This ordering is for electrons in the same principal shell; it follows directly from the penetration ability of the orbital types.
Final Answer:
$$\boxed{s > p > d > f}$$
Question 4
The first ionisation enthalpies of Na, Mg, Al and Si are in the order
(a) $Na < Mg > Al < Si$
(b) $Na > Mg > Al > Si$
(c) $Na < Mg < Al < Si$
(d) $Na > Mg > Al < Si$
Answer: Correct Option (a)
Step 1: Electronic configurations
- $Na\ (Z=11):\ [Ne]\,3s^1$
- $Mg\ (Z=12):\ [Ne]\,3s^2$
- $Al\ (Z=13):\ [Ne]\,3s^2\,3p^1$
- $Si\ (Z=14):\ [Ne]\,3s^2\,3p^2$
Step 2: General trend across a period
- Ionisation enthalpy (IE) generally increases left→right because effective nuclear charge increases.
- Exception to watch: removal of a $3p^1$ electron (as in Al) can be easier than removing a $3s$ electron from the preceding Mg atom.
Step 3: Specific explanation for the exception
- Mg has a stable, completely filled $3s^2$ subshell → extra stability, higher IE.
- Al has the electron in $3p$ ($3p^1$) which is slightly higher in energy and less tightly held than the filled $3s$ electrons, so IE(Al) < IE(Mg).
Step 4: Ordering
So the observed order is:
$$Na < Mg > Al < Si$$
Final Answer:
$$\boxed{Na < Mg > Al < Si}$$
Question 5
The electronic configuration of gadolinium (atomic number 64) is
(a) $[Xe]\,4f^5\,5d^3\,6s^2$
(b) $[Xe]\,4f^7\,5d^2\,6s^1$
(c) $[Xe]\,4f^7\,5d^1\,6s^2$
(d) $[Xe]\,4f^8\,5d^0\,6s^2$
Answer: Correct Option (c)
Step 1: Background on lanthanoids
- After La ($Z=57$, $[Xe]\,5d^1 6s^2$) electrons start to fill the $4f$ orbitals.
- Half-filled $4f^7$ configuration has extra stability (exchange/stability effects).
Step 2: What happens at Gd ($Z=64$)
- Eu ($Z=63$) has $4f^7\,6s^2$ (half-filled $4f$). Adding the next electron (to make $Z=64$) is more stable if one electron enters $5d$ rather than pairing in $4f$.
- So Gd is:
$$[Xe]\,4f^7\,5d^1\,6s^2$$
Important: Half-filled $4f^7$ stability and slightly lower energy cost of placing one electron in $5d$ explains this configuration.
Final Answer:
$$\boxed{[Xe]\,4f^7\,5d^1\,6s^2}$$
Question 6
The statement that is not correct for periodic classification of elements is
(a) The properties of elements are periodic function of their atomic numbers
(b) Non-metallic elements are less in number than metallic elements
(c) For transition elements, the 3d-orbitals are filled with electrons after 3p-orbitals and before 4s-orbitals
(d) The first ionisation enthalpies of elements generally increase with increase in atomic number as we go along a period
Answer: Correct Option (c)
Step 1: Filling order reminder
Correct filling order (relevant part) is:
$$3p \;<\; 4s \;<\; 3d$$
which means 4s gets filled before 3d.
Step 2: Why (c) is wrong
- Option (c) implies $3d$ fills before $4s$ — that is incorrect. In neutral atoms 4s is filled prior to 3d (though electrons are often removed from 4s first when ions form).
Final Answer:
$$\boxed{\text{Option (c) is incorrect — 3d fills after 4s}}$$
Question 7
Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy) is
(a) $F > Cl > Br > I$
(b) $F < Cl < Br < I$
(c) $F < Cl > Br > I$
(d) $F < Cl < Br < I$
Answer: Correct Option (c)
Step 1: Trend down the group
- Electron gain enthalpy (E.G.E.) generally becomes less negative down the group (less energy released) because atomic size increases and the incoming electron is farther from the nucleus.
Step 2: The important exception — fluorine vs chlorine
- Fluorine is very small, so when an extra electron is added the added electron experiences strong inter-electronic repulsion in the compact $2p$ orbitals. This reduces the energy released relative to chlorine.
- Chlorine has a more favorable balance of attraction vs repulsion for the added electron and thus more negative E.G.E. than F.
Step 3: Final order (negative values: more negative = more energy released)
$$F < Cl > Br > I$$
Important: F is less negative (releases less energy) than Cl because of high e−–e− repulsion in the small F atom.
Final Answer:
$$\boxed{F < Cl > Br > I}$$
Question 8
The period number in the long form of the periodic table is equal to
(a) magnetic quantum number of any element of the period
(b) atomic number of any element of the period
(c) maximum principal quantum number of any element of the period
(d) maximum azimuthal quantum number of any element of the period
Answer: Correct Option (c)
Step 1: Definition
Each period corresponds to the filling of a new principal quantum number $n$. The highest principal quantum number present for elements in a period equals the period number.
Step 2: Example
- Period 4 includes orbitals with $n=4$ (4s, 4p etc.) though it also contains $3d$ electrons — the maximum $n$ is 4 → period 4.
Final Answer:
$$\boxed{\text{Period number} = \text{maximum principal quantum number } n}$$
Question 9
The elements in which electrons are progressively filled in $4f$-orbital are called
(a) actinoids
(b) transition elements
(c) lanthanoids
(d) halogens
Answer: Correct Option (c)
Step 1: Definition / range
Lanthanoids (lanthanides) are the elements where electrons progressively fill the $4f$ orbitals — typically $Z = 58$ (Ce) to $71$ (Lu).
Final Answer:
$$\boxed{\text{Lanthanoids (elements filling }4f\text{)}}$$
Question 10
Which of the following is the correct order of size of the given species
(a) $I > I^+ > I^-$
(b) $I^+ > I > I^-$
(c) $I > I^- > I^+$
(d) $I^- > I > I^+$
Answer: Correct Option (d)
Step 1: Concept — effect of gain/loss of electron on size
- Anion ($I^-$): more electrons than protons → increased electron–electron repulsion and increased shielding → larger radius than neutral atom.
- Cation ($I^+$): fewer electrons than protons → electrons pulled closer → smaller radius than neutral atom.
Step 2: Order
$$\textbf{I}^- > \textbf{I} > \textbf{I}^+$$
Final Answer:
$$\boxed{I^- > I > I^+}$$
Question 11
The formation of oxide ion $O^{2-}$ (g) from oxygen atom requires first an exothermic and then an endothermic step as shown below
$$O(g) + e^- \rightarrow O^-(g)\quad \Delta H = -141\ \text{kJ mol}^{-1}$$
$$O^-(g) + e^- \rightarrow O^{2-}(g)\quad \Delta H = +780\ \text{kJ mol}^{-1}$$
Thus, the process of formation of $O^{2-}$ in gas phase is unfavourable even though $O^{2-}$ is isoelectronic with neon. It is due to the fact that
(a) oxygen is more electronegative
(b) addition of electron in oxygen results in larger size of the ion
(c) electron repulsion outweighs the stability gained by achieving noble gas configuration
(d) $O^-$ ion has comparatively smaller size than oxygen atom
Answer: Correct Option (c)
Step 1: Energetics explanation
- The first electron addition (to make $O^-$) is exothermic ($\Delta H = -141$ kJ mol$^{-1}$).
- The second electron addition (to make $O^{2-}$) is highly endothermic (+780 kJ mol$^{-1}$) because you must add an electron to an already negatively charged ion — strong electrostatic repulsion must be overcome.
Step 2: Net effect
- The large positive energy for the second step means the overall gas-phase formation of $O^{2-}$ from $O(g)$ is energetically unfavourable, despite the noble-gas electron count.
Important: The increase in electron–electron repulsion on adding the second electron outweighs the stabilization of achieving the noble-gas configuration.
Final Answer:
$$\boxed{\text{Electron repulsion outweighs the stability gained (option (c))}}$$
Question 12 (Comprehension)
In the modern periodic table, elements are arranged in order of
increasing atomic numbers which is related to the electronic
configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic table have been divided into four blocks, viz s, p, d and f.
The modern periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Aufbau principle, the seven periods (1 to 7) have 2, 8, 8, 18, 18, 32 and 32 elements respectively.
The seventh period is still incomplete. To avoid the periodic table
being too long, the two series of f -block elements, called lanthanoids and actinoids are placed at the bottom of the main body of the periodic table
(i) The element with atomic number 57 belongs to
(a) s-block
(b) p-block
(c) d-block
(d) f-block
Answer: Correct Option (c)
Step 1: Experimental configuration and reason
- La (Z = 57) is experimentally observed as $[Xe]\,5d^1\,6s^2$. The 57th electron goes into $5d$ rather than $4f$.
- Reason: placing the electron in $5d$ disturbs the xenon core less than forcing it into an inner $4f$ orbital; the energy balance favors $5d^1 6s^2$.
Important: So La behaves like a d-block element (not f-block) experimentally.
Final Answer:
$$\boxed{\text{La (Z = 57) is a d-block element}}$$
(ii) The last element of the p-block in 6th period is represented by the outermost electronic configuration.
(a) $7s^2 7p^6$
(b) $5d^{10}6s^2 6p^6$
(c) $4f^{14} 5d^{10} 6s^2 6p^6$
(d) $4f^{14} 5d^0 6s^2 6p^6$
Answer: Correct Option (c)
Step 1: Order of filling in 6th period
- 6th period fills $6s$, then $4f$, then $5d$, then $6p$. The last p-block element (the end of the 6th period p-block) thus has full $4f$ (14) and $5d$ (10) plus $6s^2 6p^6$.
Final Answer:
$$\boxed{4f^{14}\,5d^{10}\,6s^2\,6p^6}$$
(iii) Which atomic number cannot be accommodated in the present long form?
(a) 107 (b) 118 (c) 126 (d) 102
Answer: Correct Option (c)
Step 1: Current periodic table extent
- The long form currently includes elements with $Z=1$ to $Z=118$. Anything beyond 118 (e.g., 126) cannot be placed in the present standard long form.
Final Answer:
$$\boxed{126\ \text{cannot be accommodated in current long form}}$$
(iv) The electronic configuration of the element which is just above the element with atomic number 43 in the same group is ………
(a) $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6\,5s^2$
(b) $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6\,5s^2\,5p^6$
(c) $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6\,6s^2$
(d) $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6\,7s^2$
Answer: Correct Option (a)
Step 1: Locate element above Z=43
- Element with $Z=43$ is in the 5th period. The element just above it in the same group is in the 4th period → its atomic number is $43 – 18 = 25$ (Mn).
Step 2: Configuration of Mn ($Z=25$)
$$\text{Mn: } [Ar]\,4s^2\,3d^5 \quad \text{or} \quad 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^5$$
Final Answer:
$$\boxed{\text{Configuration of element above Z=43 is option (a) — Mn}}$$
(v) The elements with atomic numbers 35, 53 and 85 are all ………
(a) noble gases (b) halogens (c) heavy metals (d) light metals
Answer: Correct Option (b)
Step 1: Reason
- Noble gas atomic numbers: $2, 10, 18, 36, 54, 86,ldots$
- The numbers $35=36-1, 53=54-1, 85=86-1$ are one less than noble-gas numbers → therefore they lie in the group just before noble gases → halogens (group 17).
Final Answer:
$$\boxed{\text{Halogens (group 17)}}$$
Question 13
Electronic configuration of four elements A, B, C, and D are given below
A. $1s^2\,2s^2\,2p^6$
B. $1s^2\,2s^2\,2p^4$
C. $1s^2\,2s^2\,2p^6\,3s^1$
D. $1s^2\,2s^2\,2p^5$
Which of the following is the correct order of increasing tendency to gain electron?
(a) $A < C < B < D$
(b) $A < B < C < D$
(c) $D < B < C < A$
(d) $D < A < B < C$
Answer: Correct Option (a)
Step 1: Identify the elements and their chemical behavior
- A: $1s^2,2s^2,2p^6$ → Noble gas (Ne) → virtually zero tendency to gain electrons.
- B: $1s^2\,2s^2\,2p^4$ → Oxygen-like (group 16) → wants 2 electrons to reach noble gas.
- C: $1s^2\,2s^2\,2p^6\,3s^1$ → Sodium-like (Na, group 1) → tends to lose its single 3s electron rather than gain. So low tendency to gain.
- D: $1s^2\,2s^2\,2p^5$ → Fluorine-like (group 17) → one electron short of noble gas → highest tendency to gain.
Step 2: Ranking by tendency to gain (least → greatest)
- Least: A (noble gas, no tendency)
- Next: C (metal, prefers to lose an electron)
- Then: B (needs 2 e− to reach noble gas, moderate tendency)
- Greatest: D (halogen, strongly wants one electron)
Order:
$$A < C < B < D$$
Final Answer:
$$\boxed{A < C < B < D}$$
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