Periodic Table & Periodicity — Page 22 | JEE Chemistry

Q211

The correct order of the ionic radii of O2–, N3–, F– , Mg2+, Na+ and Al3+ is :

A N3– < O2– < F– < Na+ < Mg2+ < Al3+

B N3– < F– < O2– < Mg2+ < Na+ < Al3+

C Al3+ < Na+ < Mg2+ < O2– < F– < N3–

D Al3+ < Mg2+ < Na+ < F– < O2– < N3–

Correct Answer

Option D

Solution

For isoelectronic species, as the no. of protons increases, size of ions decreases.

$\therefore$ Correct order of size for isoelectronic species Al3+ < Mg2+ < Na+ < F– < O2– < N3–

Q212

The five successive ionization enthalpies of an element are 800, 2427, 3658, 25024 and 32824 kJ mol–1. The number of valence electrons in the element is :

A 2

B 3

C 4

D 5

Correct Answer

Option B

Solution

There is a sudden jump after 3rd I.E. due to attainment of noble gas configuration.

So, the number of valence electrons in this element are 3.

Q213

Outer electronic configuration of Gd (Atomic no : 64) is -

A 4f85d06s2

B 4f45d46s2

C 4f75d16s2

D 4f35d56s2

Correct Answer

Option C

Solution

Electronic configuration of Gd(64) : [Xe]4f75s25p65d16s2 So the outer electronic configuration will be : 4f75d16s2 Shortcut Method : Gd belongs to 4f series which is also called lanthanide series.

In this lanthanide series for Cerium (Ce), Gadolinium (Gd) and Lutetium (Lu) outer d orbital has 1 electron and for all other lanthanide series elements outer d orbital has 0 electron.

From this you can easily say option (C) is correct.

Q214

Aluminium is usually found in +3 oxidation state. In contrast, thallium exists in + 1 and + 3 oxidation states. This is due to :

A inert pair effect

B diagonal relationship

C lattice structure

D lanthanoid contraction

Correct Answer

Option A

Solution

Electronic configuration of Thallium (Tl) is = [Xe]4f14 5d10 6s2 6p1 Here because of poor sheilding of 4f and 5d orbital on 6s orbital, electrons of 6s orbital is more tightly held by the nucleus and perticipate less in bond formation.

This is called as innert pair effect.

And that is why Tl become stable in +1 oxidation state.

Q215

Match List I with List II .tg .tg LIST I (Element) LIST II (Electronic Configuration) A.

\mathrm{N}

I.

[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^5

B.

\mathrm{S}

II.

[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^4

C.

\mathrm{Br}

III.

[\mathrm{He}] 2 \mathrm{~s}^2 2 \mathrm{p}^3

D.

\mathrm{Kr}

IV.

[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^6

Choose the correct answer from the options given below:

A A-IV, B-III, C-II, D-I

B A-I, B-IV, C-III, D-II

C A-III, B-II, C-I, D-IV

D A-II, B-I, C-IV, D-III

Correct Answer

Option C

Solution

To match each element with its correct electronic configuration, let's consider the electronic configurations of each element based on their positions in the periodic table: Nitrogen (N): It is in the 2nd period and belongs to the p-block, specifically group 15.

The electronic configuration of nitrogen is

1\text{s}^2 2\text{s}^2 2\text{p}^3

. Looking at the options in List II, this matches with the electronic configuration labeled 'III' (

[\mathrm{He}] 2 \mathrm{s}^2 2 \mathrm{p}^3

), which reflects the electronic configuration of nitrogen starting after the helium core.

Sulfur (S): Sulfur is in the 3rd period, also in the p-block, specifically group 16.

Its electronic configuration is

1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^4

. In the given list, this corresponds to 'II' (

[\mathrm{Ne}] 3 \mathrm{s}^2 3 \mathrm{p}^4

), which captures the configuration of sulfur starting after the neon core.

Bromine (Br): Bromine is located in the 4th period and in the p-block, belonging to group 17.

Its electronic configuration is

1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^{10} 4\text{s}^2 4\text{p}^5

, which is correctly represented by 'I' (

[\mathrm{Ar}] 3 \mathrm{d}^{10} 4 \mathrm{s}^2 4 \mathrm{p}^5

), showing the configuration after the argon core.

Krypton (Kr): Krypton falls in the 4th period and is a noble gas situated in the p-block at group 18.

Its electronic configuration is

1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^{10} 4\text{s}^2 4\text{p}^6

, which aligns with 'IV' (

[\mathrm{Ar}] 3 \mathrm{d}^{10} 4 \mathrm{s}^2 4 \mathrm{p}^6

), clearly depicting krypton's electronic configuration post-argon.

Therefore, the correct match is: A-III (Nitrogen matches with

[\mathrm{He}] 2 \mathrm{s}^2 2 \mathrm{p}^3

) B-II (Sulfur matches with

[\mathrm{Ne}] 3 \mathrm{s}^2 3 \mathrm{p}^4

) C-I (Bromine matches with

[\mathrm{Ar}] 3 \mathrm{d}^{10} 4 \mathrm{s}^2 4 \mathrm{p}^5

) D-IV (Krypton matches with

[\mathrm{Ar}] 3 \mathrm{d}^{10} 4 \mathrm{s}^2 4 \mathrm{p}^6

) Thus, the correct answer is Option C: A-III, B-II, C-I, D-IV.

Q216

Match List I with List II .tg .tg LIST I (Elements) LIST II (Properties in their respective groups) A.

\mathrm{Cl,S}

I. Elements with highest electronegativity B.

\mathrm{Ge,As}

II. Elements with largest atomic size C.

\mathrm{Fr,Ra}

III. Elements which show properties of both metals and non-metal D.

\mathrm{F,O}

IV. Elements with highest negative electron gain enthalpy Choose the correct answer from the options given below:

A A-III, B-II, C-I, D-IV

B A-II, B-I, C-IV, D-III

C A-II, B-III, C-IV, D-I

D A-IV, B-III, C-II, D-I

Correct Answer

Option D

Solution

To match the elements from List I with their corresponding properties in List II, let's analyze each pair: A.

\mathrm{Cl,S}

These elements are known for having high negative electron gain enthalpy (energy released when an electron is added to an atom).

Chlorine (

\mathrm{Cl}

) and sulfur (

\mathrm{S}

) are both elements that release a significant amount of energy when they gain an electron.

Hence, they match with IV: Elements with highest negative electron gain enthalpy.

B.

\mathrm{Ge,As}

Germanium (

\mathrm{Ge}

) and arsenic (

\mathrm{As}

) are metalloids, meaning they show properties of both metals and non-metals.

Hence, they match with III: Elements which show properties of both metals and non-metal.

C.

\mathrm{Fr,Ra}

Francium (

\mathrm{Fr}

) and radium (

\mathrm{Ra}

) are elements that have very large atomic sizes, being at the bottom of their respective groups on the periodic table.

Hence, they match with II: Elements with largest atomic size.

D.

\mathrm{F,O}

Fluorine (

\mathrm{F}

) and oxygen (

\mathrm{O}

) are elements with the highest electronegativity, meaning they strongly attract electrons toward themselves.

Hence, they match with I: Elements with highest electronegativity.

Summarizing, we have: A-IV, B-III, C-II, D-I The correct option is: Option D A-IV, B-III, C-II, D-I

Q217

Match the following items in column I with the corresponding items in column II. .tg .tg Item I Item II

List - I		List - II
(A)	Portland cement ingredient	(i)	Na2CO3.10H2O
(B)	Castner-Kellner process	(ii)	Mg(HCO3)2
(C)	Solvay process	(iii)	NaOH
(D)	Temporary hardness	(iv)	Ca3Al2O6

A (i)

\to

(D); (ii)

\to

(A); (iii)

\to

(B); (iv)

\to

(C)

B (i)

\to

(B); (ii)

\to

(C); (iii)

\to

(A); (iv)

\to

(D)

C (i)

\to

(C); (ii)

\to

(B); (iii)

\to

(D); (iv)

\to

(A)

D (i)

\to

(C); (ii)

\to

(D); (iii)

\to

(B); (iv)

\to

(A)

Correct Answer

Option D

Solution

Na2CO3.10H2O $\to$ Solvay process Mg(HCO3)2 $\to$ Temporary hardness NaOH $\to$ Castner-kellner cell Ca3Al2O6 $\to$ Portland cement

Q218

Match with : $$

List - I		List - II
(c)	K	(iii)	coolant in fast breeder nuclear reactor
(d)	Cs	(iv)	treatment of cancer
()		(v)	bearings for motor engines

A (a) - (v), (b) - (i), (c) - (ii), (d) - (iv)

B (a) - (v), (b) - (ii), (c) - (iv), (d) - (i)

C (a) - (iv), (b) - (iii), (c) - (i), (d) - (ii)

D (a) - (v), (b) - (iii), (c) - (ii), (d) - (i)

Correct Answer

Option D

Solution

Li makes alloy with Lead to make white metal bearings for motor engines Liquid Na metal is used s coolant in fast breeder nuclear reactor K is a very absorbent of CO2 Cs is used in making photoelectric cell

Q219

Match List I with List II .tg .tg List-I List-II A. Chlorophyll I.

\mathrm{Na_2CO_3}

B. Soda ash II.

\mathrm{CaSO_4}

C. Dentistry, Ornamental work III.

\mathrm{Mg^{2+}}

D. Used in white washing IV.

\mathrm{Ca(OH)_2}

Choose the correct answer from the options given below :

A A-III, B-I, C-II, D-IV

B A-II, B-III, C-IV, D-I

C A-II, B-I, C-III, D-IV

D A-III, B-IV, C-I, D-II

Correct Answer

Option A

Solution

Chlorophyll contains $\mathrm{Mg}^{2+}$ Ions (A - III) Soda ash is $\mathrm{Na}_{2} \mathrm{Co}_{3}$ (B - I) Dentistry; ornamental work $-\mathrm{CaSO}_{4}\quad(\mathrm{C}-\mathrm{II})$ Used in white washing $-\mathrm{Ca}(\mathrm{OH})_{2} \quad(\mathrm{D}-\mathrm{IV})$

Q220

Match List-I with List-II: .tg .tg List - I List - II A. K I. Thermonuclear reactions B. KCl II. Fertilizer C. KOH III. Sodium potassium pump D. Li IV. Absorbent of CO

_2

Choose the correct answer from the options given below :

A A-IV, B-III, C-I, D-II

B A-III, B-II, C-IV, D-I

C A-III, B-IV, C-II, D-I

D A-IV, B-I, C-III, D-II

Correct Answer

Option B

Solution

Let's consider each item in List-I and match it with an appropriate item from List-II: A.

K (Potassium): Potassium ions are crucial for the operation of the sodium-potassium pump in nerve cells.

So, A matches with III.

B.

KCl (Potassium Chloride): Potassium chloride is widely used in agriculture as a fertilizer.

So, B matches with II.

C.

KOH (Potassium Hydroxide): Potassium hydroxide is a strong base, but it is not typically used in a sodium-potassium pump, as a fertilizer, for thermonuclear reactions, or as a CO₂ absorbent.

However, as per the given options and its closest possible use, KOH is often used to absorb CO₂ in certain applications like rebreathing gas masks.

So, C matches with IV.

D.

Li (Lithium): Lithium is used in thermonuclear reactions, specifically in fusion reactions.

So, D matches with I.

Therefore, the correct option is Option B: A-III, B-II, C-IV, D-I.