Periodic Table & Periodicity — Page 10 | JEE Chemistry

Q91

Given below are two statements : Statement I : The decrease in first ionization enthalpy from B to Al is much larger than that from Al to Ga. Statement II : The d orbitals in Ga are completely filled. In the light of the above statements, choose the most appropriate answer from the options given below

A Statement I is correct but statement II is incorrect

B Statement I is incorrect but statement II is correct

C Both the statements I and II are correct

D Both the statements I and II are incorrect

Correct Answer

Option C

Solution

Statement I is correct as the decrease in first ionization enthalpy from B to Al is indeed much larger than from Al to Ga.

This is due to the shielding effect of the electron in the outermost shell and the increased effective nuclear charge as one moves from B to Al and from Al to Ga.

Statement II is also correct, as the d orbitals in Ga are indeed completely filled.

This results in the shielding effect being maximum for Ga and thus, the decrease in first ionization enthalpy from Al to Ga is smaller compared to that from B to Al.

Q92

For elements

\mathrm{B}, \mathrm{C}, \mathrm{N}, \mathrm{Li}, \mathrm{Be}, \mathrm{O}

and

\mathrm{F}

, the correct order of first ionization enthalpy is

A

\mathrm{Li}<\mathrm{Be}<\mathrm{B}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}

B

\mathrm{B}>\mathrm{Li}>\mathrm{Be}>\mathrm{C}>\mathrm{N}>\mathrm{O}>\mathrm{F}

C

\mathrm{Li}<\mathrm{Be}<\mathrm{B}<\mathrm{C}<\mathrm{N}<\mathrm{O}<\mathrm{F}

D

\mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}

Correct Answer

Option D

Solution

The ionization energy of an atom or molecule describes the minimum amount of energy required to remove an electron from the atom or molecule in the gaseous state.

It generally increases across a period (from left to right) on the periodic table because the number of protons is increasing, making the nucleus more positively charged and therefore more strongly attracting the negatively charged electrons.

However, there are exceptions to this trend when half-filled or fully-filled sub-shells are present, which are particularly stable configurations.

The increased stability of these configurations makes it harder to remove an electron, increasing the ionization energy.

The order of first ionization enthalpy for these elements, considering the above factors, should be: Li < B < Be < C < O < N < F This is because: Lithium (Li) has the lowest nuclear charge in this group and thus the lowest ionization energy.

Beryllium (Be) has a fully filled 2s subshell, which increases its ionization energy above that of boron (B).

Carbon (C) has more protons than boron, increasing its ionization energy.

Nitrogen (N) has a half-filled 2p subshell, which increases its ionization energy above that of oxygen (O).

Fluorine (F) has the highest nuclear charge in this group and thus the highest ionization energy.

So the correct answer is:

\mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}

Q93

Given below are two statements : Statement (I) : The metallic radius of Al is less than that of Ga . Statement (II) : The ionic radius of

\mathrm{Al}^{3+}

is less than that of

\mathrm{Ga}^{3+}

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

A Statement I is correct but Statement II is incorrect

B Both Statement I and Statement II are correct

C Statement I is incorrect but Statement II is correct

D Both Statement I and Statement II are incorrect

Correct Answer

Option C

Solution

$\Rightarrow$ The metallic radius order of $\mathrm{Al} \& \mathrm{Ga}$ is (due to poor shielding of d-subshell electrons) $\Rightarrow$ The ionic radius order of $\mathrm{Al}^{+3} \& \mathrm{Ga}^{+3}$ is $\mathrm{B}^{+3}<\mathrm{Al}^{+3}<\mathrm{Ga}^{+3}<\mathrm{In}^{+3}<\mathrm{T} \ell^{+3}$

Q94

The property/properties that show irregularity in first four elements of group-17 is/are (A) Covalent radius (B) Electron affinity (C) Ionic radius (D) First ionization energy Choose the correct answer from the options given below :

A A, B, C and D

B A and C only

C B only

D B and D only

Correct Answer

Option C

Solution

The order of first four elements of group-17 are as follows. $ $\mathrm{F}$ \mathrm{Cl}>\mathrm{F}>\mathrm{Br}>\mathrm{I} $(Electron affinity)$ \mathrm{F} $\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$ ( $\mathrm{I}^{\mathrm{st}}$ ionization energy) Electron affinity order is irregular.

Q95

The solubilities of carbonates decrease down the magnesium group due to a decrease in :

A hydration energy of cations

B inter-ionic attraction

C entropy of solution formation

D lattice energies of solids

Correct Answer

Option A

Solution

As we move down the group, the lattice energies of carbonates remain approximately the same.

However the hydration energies of the metal cation decreases from

\,\,B{e^{ + + }}\,\,

to

\,\,B{a^{ + + }},\,

hence the solubilities of carbonates of the alkaline earth metal decreases down the group mainly due to decreasing hydration energies of the cations from

\,\,B{e^{ + + }}\,\,

to

\,\,B{a^{ + + }}.

Q96

Given below are two statements: one is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

Assertion A : The energy required to form

\mathrm{Mg}^{2+}

from

\mathrm{Mg}

is much higher than that required to produce

\mathrm{Mg}^+

Reason

\mathbf{R}: \mathrm{Mg}^{2+}

is small ion and carry more charge than

\mathrm{Mg}^{+}

In the light of the above statements, choose the correct answer from the options given below.

A A is false but R is true

B Both A and R are correct but R is NOT the correct explanation of A

C Both A and R are correct and R is the correct explanation of A

D A is true but R is false

Correct Answer

Option C

Solution

The correct answer is Option C: Both A and R are correct and R is the correct explanation of A.

Assertion A: The energy required to form Mg2+ from Mg is much higher than that required to produce Mg+.

This is true because it takes more energy to remove two electrons from an atom than it does to remove one electron.

The second electron is more tightly bound to the nucleus than the first electron, so it requires more energy to remove it.

Reason R: Mg2+ is a small ion and carry more charge than Mg+.

This is also true.

Mg2+ has a smaller radius than Mg+ because it has lost two electrons.

The smaller radius means that the protons in the nucleus are more tightly packed together, which makes the charge of the nucleus more effective at attracting the electrons.

This makes it more difficult to remove an electron from Mg2+ than it is to remove an electron from Mg+.

The reason is the correct explanation of the assertion because it explains why it takes more energy to form Mg2+ than it does to form Mg+.

The smaller radius of Mg2+ and the higher charge of the nucleus make it more difficult to remove electrons from Mg2+, which requires more energy.

Q97

The correct order of electronegativity for given elements is:

A

\mathrm{C}>\mathrm{P}>\mathrm{At}>\mathrm{Br}

B

\mathrm{Br}>\mathrm{C}>\mathrm{At}>\mathrm{P}

C

\mathrm{Br}>\mathrm{P}>\mathrm{At}>\mathrm{C}

D

\mathrm{P}>\mathrm{Br}>\mathrm{C}>\mathrm{At}

Correct Answer

Option B

Solution

Electronegativity is a measure of an atom's ability to attract shared electrons to itself.

On the periodic table, electronegativity generally increases as we move from left to right across a period due to increase in effective nuclear charge.

Here are the Pauling electronegativities for the elements: Carbon (C): 2.55 Phosphorus (P): 2.19 Astatine (At): 2.2 Bromine (Br): 2.96 Based on these values, the correct order of electronegativity from highest to lowest is Br > C > At > P.

Therefore, Option B is the correct answer: Br > C > At > P

Q98

Group-13 elements react with

\mathrm{O}_{2}

in amorphous form to form oxides of type

\mathrm{M}_{2} \mathrm{O}_{3}~(\mathrm{M}=

element). Which among the following is the most basic oxide?

A Al

_2

O

_3

B TI

_2

O

_3

C B

_2

O

_3

D Ga

_2

O

_3

Correct Answer

Option B

Solution

As electropositive character increases basic character of oxide increases.

\underbrace{\mathrm{B}_2 \mathrm{O}_3}_{\text{acidic }}<\underbrace{\mathrm{Al}_2 \mathrm{O}_3<\mathrm{Ga}_2 \mathrm{O}_3}_{\text{amphoteric }}<\underbrace{\mathrm{In}_2 \mathrm{O}_3<\mathrm{Tl}_2 \mathrm{O}_3}_{\text{basic }}

Q99

The number of valence electrons present in the metal among

\mathrm{Cr}, \mathrm{Co}, \mathrm{Fe}

and Ni which has the lowest enthalpy of atomisation is :

A 10

B 6

C 9

D 8

Correct Answer

Option B

Solution

To determine the number of valence electrons in the metal with the lowest enthalpy of atomization among Chromium (Cr), Cobalt (Co), Iron (Fe), and Nickel (Ni), we need to examine their respective electron configurations.

Chromium has the electron configuration: $\text{Cr} = [\text{Ar}] 3d^5 4s^1$ Based on this configuration, Chromium has a total of six valence electrons (1 from the 4s orbital and 5 from the 3d orbital).

Chromium is noted for having the lowest enthalpy of atomization among the given metals.

Thus, the number of valence electrons in Chromium, the metal with the lowest enthalpy of atomization, is 6.

Q100

One mole of magnesium nitride on the reaction with an excess of water gives :

A one mole of ammonia

B two moles of nitric acid

C two moles of ammonia

D one mole of nitric acid

Correct Answer

Option C

Solution

M{g_3}{N_2} + 6{H_2}O\,\,\rightleftharpoons\,\,3Mg{\left( {OH} \right)_2} + 2N{H_3}