Classification of Elements and Periodicity in Properties | NCERT Intext Problems – 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7, 3.8, 3.9, 3.10

Anand Classes brings you detailed solutions for Classification of Elements and Periodicity in Properties | NCERT Intext Problems 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7, 3.8, 3.9, and 3.10. These step-by-step NCERT solutions are explained in simple language with proper reasoning and are highly useful for Class 11 Chemistry, CBSE Exams, JEE, and NEET preparation. Each problem is solved in detail to help students strengthen their understanding of periodic trends, periodic table, and properties of elements. Click the print button to download study material and notes.

NCERT Intext Problems Solutions Chemistry Class 11

Problem 3.1 : What would be the IUPAC name and symbol for the element with atomic number 120?

Solution

From the IUPAC rules, the name of a newly discovered or hypothetical element is derived from its atomic number using the numerical roots.

The atomic number of the element is 120. The digits 1, 2 and 0 correspond to the following roots:

Now, combining these roots in order (1 → 2 → 0) gives the name unbinilium.

The symbol of the element is obtained from the first letter of each root, i.e. Ubn.

Hence, the IUPAC name of the element with atomic number 120 is unbinilium, and its symbol is Ubn.

Problem 3.2 : How would you justify the presence of 18 elements in the 5th period of the Periodic Table?

Solution

For the 5th period, the principal quantum number is $n = 5$.
The possible values of the azimuthal quantum number $l$ are $0, 1, 2, 3$, which correspond to the orbitals $s, p, d,$ and $f$.

However, according to the increasing order of orbital energies, the filling takes place in the following order:

$$5s < 4d < 5p$$

This means that in the 5th period, electrons are filled successively into the $5s$, $4d$, and $5p$ orbitals.

• The $5s$ orbital has 1 orbital and can accommodate $2$ electrons.
• The $4d$ subshell has 5 orbitals and can accommodate $10$ electrons.
• The $5p$ subshell has 3 orbitals and can accommodate $6$ electrons.

Therefore, the total number of orbitals involved is $1 + 5 + 3 = 9$, which together can hold $2 + 10 + 6 = 18$ electrons.

Since each element corresponds to the addition of one electron, there are 18 elements in the 5th period of the Periodic Table.

Problem 3.3 : The elements $Z = 117$ and $Z = 120$ have not yet been discovered. In which family/group would you place these elements and also give the electronic configuration in each case?

Solution

From the position in the extended Periodic Table:

For $Z = 117$:

• It lies just below astatine ($Z = 85$) in Group 17.
• Therefore, it would belong to the halogen family.
• Its electronic configuration will be:

$$[Rn]\, 5f^{14}\, 6d^{10}\, 7s^2\, 7p^5$$

For $Z = 120$:

• It lies just below radium ($Z = 88$) in Group 2.
• Therefore, it would belong to the alkaline earth metals.
• Its electronic configuration will be:

$$[Uuo]\, 8s^2$$

Hence, element 117 is predicted to be a halogen with configuration $[Rn]\, 5f^{14}6d^{10}7s^2 7p^5$, and element 120 is predicted to be an alkaline earth metal with configuration $[Uuo]\, 8s^2$.

Problem 3.4 : Considering the atomic number and position in the Periodic Table, arrange the following elements in the increasing order of metallic character: Si, Be, Mg, Na, P.

Solution

Metallic character increases down a group and decreases along a period from left to right.

• $P$ is a non-metal in Group 15 (Period 3).
• $Si$ is a metalloid in Group 14 (Period 3).
• $Be$ is a metal in Group 2 (Period 2).
• $Mg$ is a metal in Group 2 (Period 3), more metallic than $Be$.
• $Na$ is an alkali metal in Group 1 (Period 3), the most metallic among the given elements.

Therefore, the order of increasing metallic character is:

$$P < Si < Be < Mg < Na$$

Problem 3.5 : Which of the following species will have the largest and the smallest size?
$Mg,\ Mg^{2+},\ Al,\ Al^{3+}$

Solution

Atomic radii decrease across a period in the Periodic Table due to increasing nuclear charge.
Cations are always smaller than their parent atoms because the loss of electrons reduces electron–electron repulsion and increases the effective nuclear charge.
Among isoelectronic species, the one with the higher positive charge will have the smallest radius because the nucleus pulls the same number of electrons more strongly.

• $Mg$ is larger than $Mg^{2+}$.
• $Al$ is larger than $Al^{3+}$.
• Between $Mg^{2+}$ and $Al^{3+}$, both are isoelectronic ($1s^2\,2s^2\,2p^6$), but $Al^{3+}$ has a higher nuclear charge $(Z=13)$ than $Mg^{2+}$ $(Z=12)$, so $Al^{3+}$ is smaller.

Hence, the largest species is $Mg$, and the smallest species is $Al^{3+}$.

Problem 3.6 : The first ionization enthalpy $(\Delta_i H)$ values of the third period elements, $Na$, $Mg$ and $Si$ are respectively $496$, $737$ and $786\ \mathrm{kJ\ mol^{-1}}$. Predict whether the first $(\Delta_i H)$ value for $Al$ will be more close to $575$ or $760\ \mathrm{kJ\ mol^{-1}}$? Justify your answer.

Answer :

1. Given data

• For $Na$: $\Delta_i H = 496\ \mathrm{kJ\ mol^{-1}}$
• For $Mg$: $\Delta_i H = 737\ \mathrm{kJ\ mol^{-1}}$
• For $Si$: $\Delta_i H = 786\ \mathrm{kJ\ mol^{-1}}$

We need to predict $\Delta_i H$ for $Al$.

2. General trend across a period

• As we move across a period (left → right), nuclear charge increases, so electrons are held more tightly, and ionization enthalpy generally increases.
• However, subshell configuration also plays an important role.

3. Electronic configuration

• $Mg$ : $1s^2 2s^2 2p^6 3s^2$
• $Al$ : $1s^2 2s^2 2p^6 3s^2 3p^1$
• The first electron to be removed from $Al$ is a 3p electron, while from $Mg$ it is a 3s electron.

4. Effect of shielding

• The 3p electron in $Al$ is farther from the nucleus and is effectively shielded by the filled $3s^2$ subshell.
• This makes it easier to remove than a $3s$ electron in $Mg$.
• Therefore, $\Delta_i H(Al)$ will be less than that of Mg.

5. Comparison with options

• $\Delta_i H$ of $Mg$ = $737\ \mathrm{kJ\ mol^{-1}}$
• $\Delta_i H$ of $Si$ = $786\ \mathrm{kJ\ mol^{-1}}$
• If $Al$ followed the simple increasing trend, its value should be between $737$ and $786$ (closer to $760$).
• But due to lower stability of 3p electron and greater shielding, the actual value will be significantly lower.

➡️ Thus, the $\Delta_i H$ value of $Al$ will be closer to 575 kJ mol$^{-1}$, not 760.

6. Final Answer

The first ionization enthalpy of $Al$ will be more close to 575 kJ mol$^{-1}$ because the $3p$ electron is more shielded and easier to remove compared to the $3s$ electron of $Mg$.

Problem 3.7 : Which of the following will have the most negative electron gain enthalpy and which the least negative?
$P,\ S,\ Cl,\ F$. Explain your answer.

Answer

1. General trends of electron gain enthalpy

• Across a period (left → right): Electron gain enthalpy becomes more negative due to increasing nuclear charge and decreasing atomic size.
• Down a group (top → bottom): Electron gain enthalpy becomes less negative because atomic size increases and the added electron experiences less attraction from the nucleus.

2. Given elements

• $P$ : Group 15, Period 3 → relatively less negative value.
• $S$ : Group 16, Period 3 → more negative than $P$.
• $Cl$ : Group 17, Period 3 → very high negative value.
• $F$ : Group 17, Period 2 → also high, but slightly less negative than Cl due to smaller size and electron–electron repulsions in the compact $2p$ orbital.

3. Special note about $F$ vs $Cl$

• Although $F$ has higher electronegativity, adding an electron to its small $2p$ orbital causes strong repulsions.
• In $Cl$, the incoming electron enters the larger $3p$ orbital where repulsion is less, making electron gain enthalpy more negative.

4. Order of electron gain enthalpy (according to negative value)

$$Cl > S > P \quad \text{(most negative → less negative)}$$

Numerical trend (approximate, in $\mathrm{kJ\ mol^{-1}}$):

• $Cl \approx -349$
• $F \approx -328$
• $S \approx -200$
• $P \approx -72$

5. Final Answer

• Most negative electron gain enthalpy: $Cl$
• Least negative electron gain enthalpy: $P$

Problem 3.8 : Using the Periodic Table, predict the formulas of compounds which might be formed by the following pairs of elements:
(a) silicon and bromine
(b) aluminium and sulphur

Solution

(a) Silicon is in group 14 and has 4 valence electrons (configuration: $3s^2 3p^2$). Bromine is a halogen in group 17 with 7 valence electrons (configuration: $4s^2 4p^5$). To complete its octet, each bromine atom needs 1 electron. Silicon can share its 4 valence electrons with 4 bromine atoms, forming 4 covalent bonds. Therefore, the compound formed will be $SiBr_4$.

Electron-dot representation:

Silicon ($cdot Si cdot cdot$) forms four covalent bonds with four bromine atoms, each shown as Br with 7 valence electrons.

$$
:Br: – Si – :Br:
\quad | \quad \quad |
:Br: \quad \quad :Br:
$$

This shows that $SiBr_4$ is a covalent compound where Si shares one electron with each Br.

(b) Aluminium belongs to group 13 and has 3 valence electrons (configuration: $3s^2 3p^1$). Sulphur belongs to group 16 and has 6 valence electrons (configuration: $3s^2 3p^4$). Aluminium tends to lose 3 electrons to achieve noble gas configuration, while sulphur needs 2 electrons to complete its octet. Thus, two Al atoms together lose 6 electrons which can be accepted by three S atoms (each gaining 2 electrons). Hence, the compound formed will be $Al_2S_3$.

Electron-dot representation:

Each aluminium atom loses 3 electrons (shown as emptying of valence shell), and each sulphur atom gains 2 electrons, completing its octet.

$$
2Al^{3+} + 3S^{2-} \;\;\rightarrow\;\; Al_2S_3
$$

This shows that $Al_2S_3$ is an ionic compound formed by transfer of electrons from aluminium to sulphur.

Thus, the formulas of the compounds are $SiBr_4$ and $Al_2S_3$.

Problem 3.9 : Are the oxidation state and covalency of Al in $[AlCl(H_2O)_5]^{2+}$ the same?

Solution

No, they are not the same.

To find the oxidation state of Al:
Let the oxidation state of Al be $x$.
Chloride ion contributes $-1$, and each water molecule is neutral. The overall charge on the complex is $+2$.

So,
$$x + (-1) + 0 = +2$$
$$x – 1 = +2$$
$$x = +3$$

Therefore, the oxidation state of Al is $+3$.

To find the covalency of Al:
Covalency is the total number of coordinate or covalent bonds formed by Al in this complex.
Here, Al is bonded to 1 chloride ion and 5 water molecules through coordination bonds.

So, covalency of Al = $1 + 5 = 6$.

Hence, the oxidation state of Al is $+3$ while its covalency is 6.

Problem 3.10 : Show by a chemical reaction with water that $Na_2O$ is a basic oxide and $Cl_2O_7$ is an acidic oxide.

Solution :

1. Reaction of $Na_2O$ with water

• $Na_2O$ is a metal oxide.
• It reacts with water to form sodium hydroxide, a strong base:

$$Na_2O + H_2O \;\rightarrow\; 2NaOH$$

➡️ Since $NaOH$ is a strong base, $Na_2O$ is a basic oxide.

2. Reaction of $Cl_2O_7$ with water

• $Cl_2O_7$ is a non-metal oxide.
• It reacts with water to form perchloric acid, a strong acid:

$$Cl_2O_7 + H_2O \;\rightarrow\; 2HClO_4$$

➡️ Since $HClO_4$ is a strong acid, $Cl_2O_7$ is an acidic oxide.

3. Reason

• Metal oxides (e.g., $Na_2O$) generally form bases with water.
• Non-metal oxides (e.g., $Cl_2O_7$) generally form acids with water.

4. Litmus Test

• $NaOH$ solution turns red litmus → blue $\;\;\Rightarrow\;$ confirms basic nature.
• $HClO_4$ solution turns blue litmus → red $\;\;\Rightarrow\;$ confirms acidic nature.

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