Periodic Table & Periodicity — Page 13 | JEE Chemistry

Q121

The element that does not belong to the same period of the remaining elements (modern periodic table) is:

A Platinum

B Osmium

C Iridium

D Palladium

Correct Answer

Option D

Solution

Let's analyze the periods for each element: Platinum (Pt) has an atomic number of 78, which places it in period 6.

Osmium (Os) has an atomic number of 76, also in period 6.

Iridium (Ir) has an atomic number of 77, which is in period 6 as well.

Palladium (Pd) has an atomic number of 46, placing it in period 5.

Since Palladium is in period 5 while the other elements are in period 6, the element that does not belong to the same period is:

\textbf{Option D: Palladium.}

Q122

A metal M readily forms its sulphate MSO4, which is water-soluble. It forms its oxide MO which becomes inert on heating. It forms an insoluble hydroxide M(OH)2 which is soluble in NaOH solution. The M is :

A Mg

B Ba

C Ca

D Be

Correct Answer

Option D

Solution

Sulphate of alkaline earth metal are sparingly soluble or almost not soluble in water whereas

BeS{O_4}

is soluble in water due to high degree of solvation.

Be{\left( {OH} \right)_2}

is insoluble in water but soluble in

NaOH.

BeO + 2NaOH\,\,

\to \,\,N{a_2}Be{O_2} + {H_2}O

Q123

Which of the following statements are NOT true about the periodic table? A. The properties of elements are function of atomic weights. B. The properties of elements are function of atomic numbers. C. Elements having similar outer electronic configurations are arranged in same period. D. An element's location reflects the quantum numbers of the last filled orbital. E. The number of elements in a period is same as the number of atomic orbitals available in energy level that is being filled. Choose the correct answer from the options given below:

A A, C and E Only

B A and E Only

C B, C and E Only

D D and E Only

Correct Answer

Option A

Solution

Let's analyze each statement one by one:

\textbf{A. The properties of elements are function of atomic weights.}

Originally, elements were arranged by atomic weights, but it was later understood that their properties depend on the atomic number (the number of protons) rather than the atomic weight.

Thus, this statement is NOT true.

\textbf{B. The properties of elements are function of atomic numbers.}

The periodic table is arranged by increasing atomic number, and an element’s properties directly relate to its number of protons and the corresponding electron configuration.

This statement is true.

\textbf{C. Elements having similar outer electronic configurations are arranged in same period.}

Elements that have similar outer electronic configurations are actually grouped in the same column (group) rather than the same row (period).

Therefore, this statement is NOT true.

\textbf{D. An element's location reflects the quantum numbers of the last filled orbital.}

The arrangement of elements corresponds to the electron configurations determined by quantum numbers, especially for the last electron added.

This statement is true.

\textbf{E. The number of elements in a period is same as the number of atomic orbitals available in energy level that is being filled.}

In an energy level, the number of atomic orbitals does not directly equal the number of elements in a period.

For example, period 2 has 1s filled, and then the 2s and 2p orbitals are filled (a total of 4 orbitals), yet there are 8 elements because each orbital can accommodate 2 electrons.

Hence, this statement is NOT true.

So, the statements that are NOT true about the periodic table are A, C, and E.

Thus, the correct answer is: Option A: A, C and E Only.

Q124

Given below are two statements : Statement (I) : The first ionization energy of Pb is greater than that of Sn . Statement (II) : The first ionization energy of Ge is greater than that of Si . In the light of the above statements, choose the correct answer from the options given below :

A Statement I is false but Statement II is true

B Both Statement I and Statement II are false

C Both Statement I and Statement II are true

D Statement I is true but Statement II is false

Correct Answer

Option D

Solution

Order of I.E. (in KJ/mol) :

\begin{array}{lll}C & > & {Si} & > & {Ge} & > & {Sn} & > & {Pb} \\ {1086} & {} & {786} & {} & {761} & {} & {708} & {} & {715} \end{array}

Q125

The atomic number of the element from the following with lowest 1st ionisation enthalpy is :

A 32

B 35

C 19

D 87

Correct Answer

Option D

Solution

To determine which element has the lowest first ionization enthalpy among the given options, we look at the atomic numbers: Atomic number 32 corresponds to Germanium (Ge).

Atomic number 35 corresponds to Bromine (Br).

Atomic number 87 corresponds to Francium (Fr).

Atomic number 19 corresponds to Potassium (K).

The element with the lowest first ionization energy is Francium (Fr) with atomic number 87.

This can be represented by its electron configuration: $\text{Fr} - [\text{Rn}] 7s^1$ .

Francium, being a part of the alkali metal group, has the most loosely bound outer electron compared to the other elements listed, resulting in a lower first ionization enthalpy.

Q126

The group 14 elements

A

and

B

have the first ionisation enthalpy values of 708 and

715 \mathrm{~kJ} \mathrm{~mol}^{-1}

respectively. The above values are lowest among their group members. The nature of their ions

\mathrm{A}^{2+}

and

\mathrm{B}^{4+}

respectively is

A both reducing

B oxidising and reducing

C both oxidising

D reducing and oxidising

Correct Answer

Option D

Solution

Given the ionization energy values for elements $A$ and $B$ in group 14, where $A$ has an ionization energy of 708 kJ/mol and $B$ has 715 kJ/mol, these values are the lowest among their group.

This helps identify the elements as tin (Sn) and lead (Pb), respectively.

In this context: The $\mathrm{A}^{2+}$ ion corresponds to $\mathrm{Sn}^{2+}$ , which acts as a reducing agent.

The $\mathrm{B}^{4+}$ ion corresponds to $\mathrm{Pb}^{4+}$ , which acts as an oxidizing agent.

Therefore, Sn in the +2 oxidation state tends to donate electrons (reduce), while Pb in the +4 oxidation state tends to accept electrons (oxidize), underscoring their respective roles as reducing and oxidizing agents.

Q127

Electronic configuration of four elements A, B, C and D are given below : (A)

1 s^2 2 s^2 2 p^3

(B)

1 s^2 2 s^2 2 p^4

(C)

1 s^2 2 s^2 2 p^5

(D)

1 s^2 2 s^2 2 p^2

Which of the following is the correct order of increasing electronegativity (Pauling's scale)?

A

\mathrm{D}<\mathrm{A}<\mathrm{B}<\mathrm{C}

B

\mathrm{A}<\mathrm{B}<\mathrm{C}<\mathrm{D}

C

\mathrm{A}<\mathrm{C}<\mathrm{B}<\mathrm{D}

D

\mathrm{A}<\mathrm{D}<\mathrm{B}<\mathrm{C}

Correct Answer

Option A

Solution

The electronic configurations of four elements, A, B, C, and D, are presented as follows: (A) $1s^2 2s^2 2p^3$ (B) $1s^2 2s^2 2p^4$ (C) $1s^2 2s^2 2p^5$ (D) $1s^2 2s^2 2p^2$ To determine the correct order of increasing electronegativity according to Pauling's scale, analyze the elements as follows: Nitrogen ( $1s^2 2s^2 2p^3$ ): Electronegativity = 3.0 Oxygen ( $1s^2 2s^2 2p^4$ ): Electronegativity = 3.5 Fluorine ( $1s^2 2s^2 2p^5$ ): Electronegativity = 4.0 Carbon ( $1s^2 2s^2 2p^2$ ): Electronegativity = 2.55 The correct order of increasing electronegativity is: Carbon (D) < Nitrogen (A) < Oxygen (B) < Fluorine (C)

Q128

Choose the incorrect trend in the atomic radii (r) of the elements.

A rRb < rCs

B rAt < rCs

C rBr < rK

D rMg < rAl

Correct Answer

Option D

Solution

In a period (row on the periodic table), the atomic size generally decreases as you move from left to right.

This is because additional electrons are being added to the same outer shell while the number of protons in the nucleus increases, pulling the electron cloud closer to the nucleus and reducing the atomic radii.

Q129

Which of the following atoms has the highest first ionization energy?

A Na

B K

C Sc

D Rb

Correct Answer

Option C

Solution

The first ionization energy is the energy required to remove the most loosely held electron from an atom or a cation.

It is measured in units of electron volts (eV) or kilojoules per mole (kJ/mol).

Among the given options, Scandium (Sc) has the highest first ionization energy because it has the smallest atomic radius and the highest nuclear charge.

As we move from left to right across a period in the periodic table, the atomic radius decreases while the nuclear charge increases.

This results in a stronger attraction between the positively charged nucleus and the negatively charged electrons in the outermost energy level, which requires more energy to remove the most loosely held electron.

The first ionization energy of the given options in increasing order is: Na Therefore, Scandium (Sc) has the highest first ionization energy among the given options.

Q130

The correct order of the solubility of alkaline-earth metal sulphates in water is :

A Mg < Ca < Sr < Ba

B Mg < Sr < Ca < Ba

C Mg > Sr > Ca > Ba

D Mg > Ca > Sr > Ba

Correct Answer

Option D

Solution

The solubility of the sulphates in water decreases down the group.

Mg > Ca > Sr > Ba.

MgSO4 is soluble, but CaSO4 is sparingly soluble, and the sulphates of Sr, Ba and Ra are virtually insoluble.

The significantly high solubility of MgSO4 is due to the high enthalpy of solvation of the smaller Mg2+ ion.