Periodic Table & Periodicity — Page 9 | JEE Chemistry

Q81

Given below are two statements : Statement I : The correct order of first ionization enthalpy values of

\mathrm{Li}, \mathrm{Na}, \mathrm{F}

and

\mathrm{Cl}

is

\mathrm{Na} Statement II : The correct order of negative electron gain enthalpy values of

\mathrm{Li}, \mathrm{Na}, \mathrm{F}

and

\mathrm{Cl}

is

\mathrm{Na} In the light of the above statements, choose the correct answer from the options given below :

A Statement I is true but Statement II is false

B Both Statement I and Statement II are false

C Statement I is false but Statement II is true

D Both Statement I and Statement II are true

Correct Answer

Option D

Solution

First ionization enthalpy of

\mathrm{F}>\mathrm{Cl}

as first IE decreases down the group. Similarly, first IE of

\mathrm{Li}>\mathrm{Na}

.

So, statement-I is correct.

Electron gain enthalpy of given elements are negative.

Considering the magnitude the given order is correct.

Thus, statement-II is correct

Q82

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R: Assertion A:

\mathrm{H}_2 \mathrm{Te}

is more acidic than

\mathrm{H}_2 \mathrm{~S}

. Reason R: Bond dissociation enthalpy of

\mathrm{H}_2 \mathrm{Te}

is lower than

\mathrm{H}_2 \mathrm{~S}

. In the light of the above statements, choose the most appropriate from the options given below:

A Both

A

and

R

are true and

R

is the correct explanation of

A

.

B Both

A

and

R

are true but

R

is NOT the correct explanation of

A

.

C

A

is false but

R

is true.

D

A

is true but

R

is false.

Correct Answer

Option A

Solution

Due to lower Bond dissociation enthalpy of

\mathrm{H}_2 \mathrm{Te}

it ionizes to give

\mathrm{H}^{+}

more easily than

\mathrm{H}_2 \mathrm{S}

.

Q83

Given two statements below : Statement I : In

\mathrm{Cl}_{2}

molecule the covalent radius is double of the atomic radius of chlorine. Statement II : Radius of anionic species is always greater than their parent atomic radius. Choose the most appropriate answer from options given below :

A Both Statement I and Statement II are correct.

B Both Statement I and Statement II are incorrect.

C Statement I is correct but Statement II is incorrect.

D Statement I is incorrect but Statement II is correct.

Correct Answer

Option D

Solution

The given statements pertain to the concepts of atomic and ionic radii in chemical bonding and periodic properties.

Let's analyze each statement individually for clarity.

Statement I: In

\mathrm{Cl}_{2}

molecule the covalent radius is double of the atomic radius of chlorine.

The atomic radius of an element like chlorine refers to the size of its atoms, typically measured when the element is in its gas phase and not bonded to anything else.

The covalent radius, on the other hand, is a measure of the size of an atom that forms part of a single covalent bond - it's essentially half the distance between two atoms bonded together.

In the case of a

\mathrm{Cl}_{2}

molecule, the covalent bond is formed between two chlorine atoms.

Therefore, the distance from the nucleus of one chlorine atom to the nucleus of the other (the bond length) is essentially twice the covalent radius of chlorine.

This makes the statement true only if correctly interpreted: the covalent bond length is double the covalent radius, but it tends to be conflated with the concept of the atomic radius.

However, it is important to clarify that the atomic radius and covalent radius are different measures.

In a diatomic molecule like

\mathrm{Cl}_2

, saying the covalent radius is "double of the atomic radius" is inaccurate.

The precise statement should be that the bond length (twice the covalent radius) is not directly twice the atomic radius but rather the distance between the nuclei of the bonding atoms.

Therefore, this statement as phrased is misleading or incorrect.

Statement II: Radius of anionic species is always greater than their parent atomic radius.

This statement is based on the idea that when an atom gains electrons and becomes an anion, the increased electron-electron repulsions in the electron cloud cause it to expand.

This is generally true across the periodic table.

For example, when chlorine gains an electron to become

\mathrm{Cl}^{-}

, its radius increases due to the addition of an extra electron which increases repulsion among electrons and expands the electron cloud.

This principle holds for virtually all anions compared to their parent atoms, so Statement II is correct.

Given the analysis, the correct option is: Option D: Statement I is incorrect but Statement II is correct.

Q84

In which of the following pairs, electron gain enthalpies of constituent elements are nearly the same or identical? (A) Rb and Cs (B) Na and K (C) Ar and Kr (D) I and At Choose the correct answer from the options given below :

A (A) and (B) only

B (B) and (C) only

C (A) and (C) only

D (C) and (D) only

Correct Answer

Option C

Solution

$\mathrm{Rb} \,\& \,\mathrm{Cs}$ have nearly same electron gain enthalpy electron gain enthalpy $=-46\, \mathrm{kj} / \mathrm{ml}$ Ar & Kr have same $\Delta \mathrm{H}_{\mathrm{eq}} .$ Value is $+96 \,\mathrm{kj} / \mathrm{ml}$

Q85

The first ionization enthalpy of Na, Mg and Si, respectively, are : 496, 737 and

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

. The first ionization enthalpy (

\mathrm{kJ} \,\mathrm{mol}^{-1}

) of

\mathrm{Al}

is :

A 487

B 768

C 577

D 856

Correct Answer

Option C

Solution

I.E.

$\mathrm{Na}<\mathrm{Al}<\mathrm{Mg}<\mathrm{Si}$ $\because 496<$ I.E.

(Al) $<737$ Option (C), matches the condition i.e.

I.E.

$(\mathrm{Al})=577 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Q86

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R) Assertion (A): The first ionization enthalpy of

3 \mathrm{~d}

series elements is more than that of group 2 metals Reason (R): In 3d series of elements successive filling of d-orbitals takes place. In the light of the above statements, choose the correct answer from the options given below :

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B (A) is true but (R) is false

C Both (A) and (R) are true but

(\mathbf{R})

is not the correct explanation of (A)

D (A) is false but (R) is true

Correct Answer

Option D

Solution

From Sc to Mn ionization energy is less than that of Mg.

Reason (R) : The statement "In $3d$ series of elements successive filling of $d$ -orbitals takes place" is indeed true.

It describes the characteristic of $3d$ series elements where there is successive filling of $d$ orbitals as we move across the period.

Q87

The correct increasing order of the ionic radii is

A

\mathrm{Cl^- < Ca^{2+} < K^+ < S^{2-}}

B

\mathrm{K^+ < S^{2-} < Ca^{2+} < Cl^-}

C

\mathrm{Ca^{2+} < K^+ < Cl^- < S^{2-}}

D

\mathrm{S^{2-} < Cl^- < Ca^{2+} < K^+}

Correct Answer

Option C

Solution

In isoelectronic species size $\propto \dfrac{1}{Z}$

\mathrm{Ca}^{2+}<\mathrm{K}^{+}<\mathrm{Cl}^{-}<\mathrm{S}^{2-} \text{ : Size }

$\begin{array}{lllll}Z: & 20 & 19 & 17 & 18\end{array}$

Q88

Bond dissociation energy of "E-H" bond of the "

\mathrm{H}_{2} \mathrm{E}

" hydrides of group 16 elements (given below), follows order. A.

\mathrm{O}

B.

\mathrm{S}

C. Se D.

\mathrm{Te}

Choose the correct from the options given below:

A

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

B

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

C

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

D

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

Correct Answer

Option A

Solution

The correct order of bond strength is

\mathrm{H_2O > H_2S > H_2Se > H_2Te}

Q89

Given below are two statements: Statement - I: Along the period, the chemical reactivity of the elements gradually increases from group 1 to group 18 . Statement - II: The nature of oxides formed by group 1 elements is basic while that of group 17 elements is acidic. In the light of the above statements, choose the most appropriate 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 Statement I is True But Statement II is False

D Both Statement I and Statement II are True

Correct Answer

Option A

Solution

Chemical reactivity of elements decreases along the period therefore statement - I is false.

Group - 1 elements from basic nature oxides while group - 17 elements form acidic oxides therefore statement - II is true.

Q90

Given below are two statements:Statement I: H2Se is more acidic than H2TeStatement II: H2Se has higher bond enthalpy for dissociation than H2TeIn the light of the above statements, choose the correct answer from the options given below:

A Both Statement I and Statement II are true

B Statement I is true but Statement II is false

C Both Statement I and Statement II are false

D Statement I is false but Statement II is true

Correct Answer

Option D

Solution

To determine the relative acidity and bond enthalpy of H₂Se and H₂Te, let's examine both: Acidic Strength: The acidity of hydrides increases as we move down the group in the periodic table.

This is because the bond strength between hydrogen and the central atom decreases, making it easier for the hydrogen ions to dissociate.

Therefore, H₂Te, being further down the group than H₂Se, is more acidic: $\mathrm{H}_2\mathrm{Se} Bond Enthalpy: Bond enthalpy refers to the energy required to dissociate a bond. The bond enthalpy decreases down the group due to weaker bonds forming as the atomic size increases. Hence, H₂Se has a higher bond enthalpy compared to H₂Te:$ \Delta_{\text{dis}} \mathrm{H}: \mathrm{H}_2\mathrm{Se} > \mathrm{H}_2\mathrm{Te} \text{H}_2\mathrm{Se} \, [276 \, \text{KJ/mol}] > \text{H}_2\mathrm{Te} \, [238 \, \text{KJ/mol}] $ Thus, H₂Se is indeed less acidic than H₂Te, but it has a higher bond enthalpy for dissociation.