Consider the ions ${{\rm{K}}^ + },{\rm{\;}}{{\rm{S}}^{2 - }},{\rm{C}}{{\rm{l}}^ - }{\rm{\;and\;C}}{{\rm{a}}^{2 + }}$. The radii of these ionic species follow the order

${{\rm{S}}^{2 - }} > C{{\rm{l}}^ - } > {{\rm{K}}^ + } > {\rm{C}}{{\rm{a}}^{2 + }}{\rm{\;}}$

${\rm{C}}{{\rm{a}}^{2 + }} > {{\rm{K}}^ + } > C{{\rm{l}}^ - } > {{\rm{S}}^{2 - }}$

${\rm{C}}{{\rm{l}}^ - } > {{\rm{S}}^{2 - }} > {{\rm{K}}^ + } > {\rm{C}}{{\rm{a}}^{2 + }}$

${\rm{C}}{{\rm{a}}^{2 + }} > C{{\rm{l}}^ - } > K > {{\rm{S}}^{2 - }}$

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NEET • CHEMISTRY • CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

Classification of elements and periodicity in properties practice · Mixed difficulty

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1 Medium

1 of 407

Remembering +4 | -1

The elements with zero electron affinity are

View answer and explanation

Correct option: B

Beryllium and Helium

Option B (Beryllium and Helium) is correct because both beryllium and helium have stable, filled electron configurations, making it energetically unfavorable to add an electron.

2 Medium

2 of 407

Analyzing +4 | -1

S-I:

\mathrm{Cu}^{+2}

paramagnetic.

S-II: All the orbitals present in

\mathrm{Cu}^{+2}

ion are doubly occupied.

View answer and explanation

Correct option: C

Only I is true

Option C is correct because S-I is true (Cu+2 is paramagnetic due to the presence of unpaired electrons), but S-II is false (not all orbitals in Cu+2 are doubly occupied).

3 Easy

3 of 407

Understanding +4 | -1

Which of the following is the most electropositive element?

View answer and explanation

Correct option: C

Option C (Mg) is correct because electropositivity increases down and to the left of the periodic table, and Mg is more to the left and down compared to P, S, and Al.

4 Medium

4 of 407

Remembering +4 | -1

The law of Octaves is not applicable to:

View answer and explanation

Correct option: D

Se,Te,As

Option D (Se,Te,As) is correct because the Law of Octaves, proposed by Newlands, states that every eighth element has similar properties. This law was applicable up to Calcium (Ca). Se, Te, and As are beyond Calcium, making the law inapplicable.

5 Medium

5 of 407

Understanding +4 | -1

Assertion (A): Mendeleev's arranged elements in horizontal rows and vertical columns.
Reason (R): Mendeleev's ignored the order of atomic weight thinking that the atomic measurements might be incorrect.

View answer and explanation

Correct option: B

Both Assertion and Reason are correct; and Reason is not the correct explanation of Assertion

Option B (Both Assertion and Reason are correct; and Reason is not the correct explanation of Assertion) is correct because Mendeleev did arrange elements in rows and columns and sometimes ignored atomic weight order, but the latter doesn't explain the former.

6 Medium

6 of 407

Analyzing +4 | -1

Which of the following electronic configuration does not belong to same block as others?

View answer and explanation

Correct option: C

[{Kr}] 5 \mathrm{s}^2

Option C ([Kr] 5s²) is correct because it represents an s-block element, while the others represent d-block elements.

7 Easy

7 of 407

Analyzing +4 | -1

Identify the least stable ion among the following

View answer and explanation

Correct option: B

B{e^ - }

The correct answer is B (

B{e^ - }

) because Beryllium (

Be

) has a stable electronic configuration (2s²) and adding an electron to form

Be^-

is energetically unfavorable.

8 Easy

8 of 407

Application +4 | -1

An element has E.C.

1{S^2}\,\,2{S^2}\,\,2{P^6}\,\,3{S^2}\,\,3{P^3}

predict their period, group and block

View answer and explanation

Correct option: A

period = 3, block = p, group = 15

Option A (period = 3, block = p, group = 15) is correct because the element's electronic configuration ends in 3p3, indicating the 3rd period, p-block, and group 15.

9 Hard

9 of 407

Understanding +4 | -1

{K^ + }\,\,\,{F^ - }

ionic radius of

{F^ - }

is more while atomic radius of

{K^ + }

View answer and explanation

Correct option: B

More than

{F^ - }

Option B (More than

{F^ - }

) is correct because the potassium ion has fewer electrons than the fluoride ion, but more protons, resulting in a smaller ionic radius, but the atomic radius of K is larger than F-

10 Hard

10 of 407

Understanding +4 | -1

Consider the following statements:
I. Metals comprise more than

78 \%

of all known elements and appear on the left side of the periodic table.
II. Most non-metallic solids are brittle and are neither malleable nor ductile.
III. The elements become more metallic as we go down a group.
IV. The non-metallic character decreases as one goes from left to right across the periodic table.
Choose the option with all correct statements:

View answer and explanation

Correct option: A

I, II and III

Option A (I, II and III) is correct because metals are abundant on the left side of the periodic table, non-metallic solids are brittle, and metallic character increases down a group.

11 Medium

11 of 407

Understanding +4 | -1

Identify the correct order in which the covalent radius of the following elements increases (I)

Ti

(II)

Ca

(III)

Sc

View answer and explanation

Correct option: D

(I), (III), (II)

Option D (I), (III), (II) is correct because covalent radius generally increases down a group and decreases across a period, with exceptions due to electronic configurations and effective nuclear charge.

12 Easy

12 of 407

Understanding +4 | -1

The correct order according to size is

View answer and explanation

Correct option: C

{O^{2 - }} > {O^ - } > O

Option C (

{O^{2 - }} > {O^ - } > O

) is correct because as the negative charge increases, the electron-electron repulsion also increases, leading to an increase in size.

13 Easy

13 of 407

Application +4 | -1

Element with electronic configuration

1 s^2 2 s^2 2 p^6 3 s^2 3 p^6

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

belongs to the following group of the periodic table

View answer and explanation

Correct option: B

$5 th$

Option B ($5 th$) is correct because the electronic configuration ends with

5p^3

, which indicates that the element belongs to the 5th group (Group 15) of the p-block elements.

14 Easy

14 of 407

Analyzing +4 | -1

On which of the following Dobereiner's Triad law is not applicable?

View answer and explanation

Correct option: C

{F}, {Cl}, {Br}

The correct answer is C (

{F}, {Cl}, {Br}

) because the atomic mass of Chlorine is not approximately the average of Fluorine and Bromine.

15 Easy

15 of 407

Application +4 | -1

The electronegativity of A and B are 0.7 and 4.0 respectively. The bond formed in AB and

{B_2}

molecule will be

View answer and explanation

Correct option: A

Ionic and covalent resp

Option A (Ionic and covalent resp) is correct because the large electronegativity difference between A and B leads to ionic bonding in AB, while identical atoms in B2 form a covalent bond.

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16 Easy

16 of 407

Understanding +4 | -1

Which one of the following combinations represents a metallic element?

View answer and explanation

Correct option: A

2, 8, 2

Option A is correct because the electronic configuration 2, 8, 2 represents an element with 2 valence electrons, which is characteristic of metallic elements in Group 2 (alkaline earth metals).

17 Medium

17 of 407

Remembering +4 | -1

Which among the following is/are amphoteric oxides

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because

BeO

SnO

, and

ZnO

are all known to exhibit amphoteric behavior, reacting with both acids and bases.

18 Medium

18 of 407

Analyzing +4 | -1

Which one of the following is smallest in size

View answer and explanation

Correct option: C

Option C (

\mathrm{N}{\mathrm{a}^ + }

) is correct because it has the highest nuclear charge among the isoelectronic species, leading to the smallest ionic radius.

19 Medium

19 of 407

Understanding +4 | -1

Chemical properties of

Li

and

Mg

are similar because:

View answer and explanation

Correct option: B

Shows diagonal relationship

The correct answer is B (Shows diagonal relationship) because Lithium (Li) and Magnesium (Mg) exhibit similar chemical properties due to their diagonal relationship in the periodic table.

20 Easy

20 of 407

Remembering +4 | -1

IUPAC name of an element with atomic number 110 is

View answer and explanation

Correct option: A

Ununnillium

Option A (Ununnillium) is correct because it follows the IUPAC nomenclature rules for naming elements with atomic number 110.

21 Medium

21 of 407

Analyzing +4 | -1

The ionic radius of ‘Cr’ is minimum in which of the following compounds?

View answer and explanation

Correct option: B

{\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4}

Option B (

{\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4}

) is correct because Cr has the highest oxidation state (+6) in this compound, leading to the smallest ionic radius.

22 Medium

22 of 407

Application +4 | -1

If you are given Avogadro's number of atoms of a gas '

X

'. If half of the atoms are converted into

X_{(g)}^{+}

by energy

\Delta {H}

. The IE of

{X}

View answer and explanation

Correct option: A

\frac{2 \Delta {H}}{\mathrm{N}_{\mathrm{A}}}

Option A (

\frac{2 \Delta {H}}{\mathrm{N}_{\mathrm{A}}}

) is correct because ionization energy is the energy required to remove an electron from one atom, and the formula correctly scales the total energy by the number of ionized atoms.

23 Medium

23 of 407

Application +4 | -1

Total no. of elements in 6th period and 3rd group of modern periodic table.

View answer and explanation

Correct option: D

Option D (15) is correct. The 6th period and 3rd group contains Lanthanum (La) and the 14 Lanthanides, totaling 15 elements.

24 Hard

24 of 407

Understanding +4 | -1

The pair of elements having approximately equal ionisation potential is

View answer and explanation

Correct option: A

Al, Ga

Option A (Al, Ga) is correct because of the lanthanide contraction and shielding effects, which cause Ga to have a slightly higher effective nuclear charge, resulting in a similar ionization potential to Al.

25 Medium

25 of 407

Application +4 | -1

An element having electronic configuration

1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^1

forms

View answer and explanation

Correct option: B

Basic oxide

Option B (Basic oxide) is correct because the electronic configuration indicates the element is an alkali metal, which forms basic oxides.

26 Easy

26 of 407

Understanding +4 | -1

Which of the following element can exhibit more than one oxidation state

View answer and explanation

Correct option: A

The correct answer is A (Cr) because Chromium (Cr) can exhibit multiple oxidation states due to its electronic configuration.

27 Medium

27 of 407

Understanding +4 | -1

The ionic radii of

L{i^ + },\,B{e^{ + 2}}

and

B{e^{ + 3}}

follow the order

View answer and explanation

Correct option: D

L{i^ + } > B{e^{ + 2}} > {B^{3 + }}

Option D (

L{i^ + } > B{e^{ + 2}} > {B^{3 + }}

) is correct because for isoelectronic ions, the ionic radius decreases with increasing nuclear charge.

28 Easy

28 of 407

Understanding +4 | -1

Consider the ions

{\mathrm{K}^ + },\mathrm{\;}{\mathrm{S}^{2 - }},\mathrm{C}{\mathrm{l}^ - }\mathrm{\;and\;C}{\mathrm{a}^{2 + }}

. The radii of these ionic species follow the order

View answer and explanation

Correct option: D

{\mathrm{S}^{2 - }} > C{\mathrm{l}^ - } > {\mathrm{K}^ + } > \mathrm{C}{\mathrm{a}^{2 + }}\mathrm{\;}

Option D (

{\mathrm{S}^{2 - }} > C{\mathrm{l}^ - } > {\mathrm{K}^ + } > {\mathrm{Ca}^{2 + }}\mathrm{\;}

) is correct because ionic radii decrease with increasing nuclear charge and increase with increasing negative charge.

29 Medium

29 of 407

Understanding +4 | -1

The correct order of ionic radii is represented in

View answer and explanation

Correct option: B

A{l^ + } > A{l^{2 + }} > A{l^{3 + }}

Option B (

A{l^ + } > A{l^{2 + }} > A{l^{3 + }}

) is correct because as the positive charge increases, the ionic radii decreases due to increased effective nuclear charge.

30 Easy

30 of 407

Remembering +4 | -1

The outer electronic configuration of transition element is

View answer and explanation

Correct option: A

\left( {n - 1} \right){d^{1 - 10}}\,\,n{s^{0 - 2}}

Option A (

\left( {n - 1} \right){d^{1 - 10}}\,\,n{s^{0 - 2}}

) is correct because it accurately represents the general outer electronic configuration of transition elements, allowing for variations in the d and s orbital filling.

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