The radii of ${\rm{F}},{{\rm{F}}^ - },{\rm{O}}$ and ${{\rm{O}}^{2 - }}$are in the order of:

${{\rm{O}}^{2 - }} > {{\rm{F}}^ - } > O > F$

${{\rm{O}}^{2 - }} > {{\rm{F}}^ - } > F > O$

${{\rm{F}}^ - } > {{\rm{O}}^{2 - }} > F > O$

${{\rm{O}}^{2 - }} > O > {{\rm{F}}^ - } > F$

${O^{2 - }}$ and $S{i^{4 + }}$ are iso electronic ions. If the ionic radius of ${O^{2 - }}$ is $1\,\,\mathop {\rm{A}}\limits^{\rm{o}} $, the ionic radius of $S{i^{4 + }}$ is

$0.41\,\mathop {\rm{A}}\limits^{\rm{o}} $

$1\,.4\,\mathop {\rm{A}}\limits^{\rm{o}} $

$2.8\mathop {\rm{A}}\limits^{\rm{o}} $

$1.5\mathop {\rm{A}}\limits^{\rm{o}} $

Oxygen is divalent, but sulphur exhibits variable valency of $2,{\rm{ }}4{\rm{ }} and {\rm{ }}6$, because:

Sulphur is less electronegative than oxygen

Sulphur is bigger atom than oxygen

Ionisation potential of sulphur is more than oxygen

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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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151 Hard

151 of 407

Analyzing +4 | -1

The decreasing order of the second ionization energy of

K

\mathrm{Ca}

and

\mathrm{Ba}

is:

View answer and explanation

Correct option: A

\mathrm{K} > Ca > Ba

Option A (

\mathrm{K} > Ca > Ba

) is correct because after the first ionization, K attains a noble gas configuration, requiring significantly more energy for the second ionization. Ca and Ba lose electrons from the s-orbital.

152 Easy

152 of 407

Remembering +4 | -1

An element with atomic number 21 is a

View answer and explanation

Correct option: C

Transition element

Option C (Transition element) is correct because elements with atomic number 21 (Scandium) belong to the transition metal series.

153 Easy

153 of 407

Application +4 | -1

Which of the following combinations contain only iso electronic ions

View answer and explanation

Correct option: C

{P^{3 - }},{S^{2 - }},C{l^ - },Ar

Option C (

{P^{3 - }},{S^{2 - }},C{l^ - },Ar

) is correct because all the ions and the atom have 18 electrons each, making them isoelectronic.

154 Medium

154 of 407

Remembering +4 | -1

The common oxidation state exhibited by inner transition elements usually in their compound is

View answer and explanation

Correct option: B

+ 3

Option B is correct because the most common oxidation state exhibited by inner transition elements in their compounds is +3, due to the ease of losing the two s-electrons and one f-electron.

155 Hard

155 of 407

Understanding +4 | -1

The radii of

\mathrm{F},{\mathrm{F}^ - },\mathrm{O}

and

{\mathrm{O}^{2 - }}

are in the order of:

View answer and explanation

Correct option: D

{\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > O > F

Option D (

{\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > O > F

) is correct because anions are larger than their neutral atoms, and for isoelectronic species, the species with lower nuclear charge has a larger radius.

156 Medium

156 of 407

Remembering +4 | -1

The set that represents the pair of neutral oxides nitrogen is

View answer and explanation

Correct option: A

NO

and

{N_2}O

Option A (

NO

and

{N_2}O

) is correct because both nitric oxide (

NO

) and nitrous oxide (

{N_2}O

) are neutral oxides.

157 Hard

157 of 407

Understanding +4 | -1

Which of the following statements are correct regarding electronegativity?

View answer and explanation

Correct option: C

Decrease in electronegativity down a group in the periodic table confirms decreases in non-metallic properties

Option C (Decrease in electronegativity down a group in the periodic table confirms decreases in non-metallic properties) is correct because electronegativity is directly related to non-metallic character; as electronegativity decreases, non-metallic properties decrease.

158 Easy

158 of 407

Remembering +4 | -1

Which of the following element is metalloid?

View answer and explanation

Correct option: C

Option C is correct. Germanium (Ge) is a metalloid, possessing properties of both metals and nonmetals.

159 Easy

159 of 407

Understanding +4 | -1

An element with atomic number of 50 belongs to

View answer and explanation

Correct option: B

p – block

Option B (p – block) is correct because an element with atomic number 50 is Tin, which is a p-block element.

160 Medium

160 of 407

Analyzing +4 | -1

Which of the following is different from other three oxides

View answer and explanation

Correct option: A

MgO

Option A (MgO) is correct because it is a basic oxide, while SnO, ZnO, and PbO are amphoteric oxides.

161 Easy

161 of 407

Remembering +4 | -1

Ionization potential is lowest for:

View answer and explanation

Correct option: D

Alkali metals

Option D (Alkali metals) is correct because alkali metals have the lowest ionization potential due to their electronic configuration.

162 Medium

162 of 407

Analyzing +4 | -1

The atomic number of elements A, B, C and D are Z-1, Z+1, and Z+2, respectively. If ‘B’ is a noble gas, choose the correct answer from the following statements.
V) ‘A’ has higher electron affinity
VI) ‘C’ exists in +2 oxidation state
VII) ‘D’ is an alkaline earth metal

View answer and explanation

Correct option: C

I and III

Option C (I and III) is correct because if B is a noble gas (Z+1), then A (Z-1) has a high electron affinity and D (Z+2) is an alkaline earth metal.

163 Easy

163 of 407

Understanding +4 | -1

The element with atomic number 36 belongs to …block in the Periodic Table.

View answer and explanation

Correct option: A

Option A (p) is correct because elements with atomic number 36 (Krypton) belongs to the p-block of the periodic table.

164 Medium

164 of 407

Application +4 | -1

The atomic masses of

Li

and

K

are 7 and 39, respectively. According to law of triads the atomic mass of

Na

will be

View answer and explanation

Correct option: A

Option A (23) is correct because, according to the Law of Triads, the atomic mass of Na is approximately the average of the atomic masses of Li and K: (7+39)/2 = 23.

165 Medium

165 of 407

Analyzing +4 | -1

Which of the following has the largest size

View answer and explanation

Correct option: A

{N^{ - 3}}

Option A (

{N^{ - 3}})

is correct because among isoelectronic species, the size increases with increasing negative charge.

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166 Hard

166 of 407

Understanding +4 | -1

The second electron gain enthalpies (in kJ/mol) of oxygen and Sulphur respectively are

View answer and explanation

Correct option: D

+780, +590

Option D (+780, +590) is correct because the second electron gain enthalpy is always positive due to electron-electron repulsion, and oxygen is smaller than sulfur, leading to greater repulsion.

167 Medium

167 of 407

Remembering +4 | -1

Metal used as catalyst in the hydrogenation of vegetable oils

View answer and explanation

Correct option: C

Nickel

Option C (Nickel) is correct because Nickel is commonly used as a catalyst in the hydrogenation of vegetable oils.

168 Medium

168 of 407

Application +4 | -1

Among the following, pair of elements belonging to same
period is

View answer and explanation

Correct option: C

Z = 55 and Z = 77

Option C (Z = 55 and Z = 77) is correct because both elements with atomic numbers 55 (Cesium) and 77 (Iridium) belong to the 6th period.

169 Easy

169 of 407

Analyzing +4 | -1

Among the following elements, the one having the highest ionisation energy is

View answer and explanation

Correct option: C

\left[ {Ne} \right]\,\,3{s^2}\,\,3{p^3}

Option C is correct because half-filled p orbitals (

p^3

) have extra stability, leading to a higher ionization energy.

170 Medium

170 of 407

Understanding +4 | -1

Which of the following sets represents the collection of isoelectronic species?

View answer and explanation

Correct option: D

{\mathrm{K}^ + },\mathrm{C}{\mathrm{a}^{2 + }},\mathrm{S}{\mathrm{c}^{3 + }},\mathrm{C}{\mathrm{l}^ - }

Option D (

{\mathrm{K}^ + },\mathrm{C}{\mathrm{a}^{2 + }},\mathrm{S}{\mathrm{c}^{3 + }},\mathrm{C}{\mathrm{l}^ - }

) is correct because all the species have the same number of electrons (18), making them isoelectronic.

171 Medium

171 of 407

Understanding +4 | -1

The ionization potential order for which set is correct?

View answer and explanation

Correct option: B

B>Li>K

Option B (B>Li>K) is correct because ionization potential generally increases across a period and decreases down a group.

172 Medium

172 of 407

Application +4 | -1

{O^{2 - }}

and

S{i^{4 + }}

are iso electronic ions. If the ionic radius of

{O^{2 - }}

1\,\,\mathop \mathrm{A}\limits^\mathrm{o}

, the ionic radius of

S{i^{4 + }}

View answer and explanation

Correct option: B

0.41\,\mathop \mathrm{A}\limits^\mathrm{o}

Option A (

M{g^{2 + }}

) is correct because

M{g^{2 + }}

has a higher nuclear charge and fewer electrons than Ne, resulting in a smaller ionic radius.

173 Hard

173 of 407

Understanding +4 | -1

Which of the following isoelectronic ions has lowest ionisation energy?

View answer and explanation

Correct option: D

{S^{2 - }}

Option D (

{S^{2 - }}

) is correct because for isoelectronic ions, the ion with the least nuclear charge has the lowest ionization energy.

174 Medium

174 of 407

Understanding +4 | -1

The set representing the correct order of first ionisation potential is:

View answer and explanation

Correct option: B

\mathrm{Be} > Mg > Ca

Option B (

\mathrm{Be} > Mg > Ca

) is correct because ionization energy decreases down a group, and Be, Mg, and Ca are in Group 2.

175 Easy

175 of 407

Remembering +4 | -1

Which group of periodic table contains no metal?

View answer and explanation

Correct option: C

VIIA

Option C (VIIA) is correct because Group VIIA (Halogens) consists entirely of non-metals.

176 Medium

176 of 407

Understanding +4 | -1

The element with the electronic configuration as

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

represents a

View answer and explanation

Correct option: C

Metalloid

Option C (Metalloid) is correct because the electronic configuration

[Ar]3d^{10}4s^24p^3

corresponds to Arsenic (As), which is a metalloid.

177 Medium

177 of 407

Understanding +4 | -1

A sudden large jump between the value of first and second ionisation energies of elements would be associated with which of the following electronic configurations?

View answer and explanation

Correct option: A

1{s^2},2{s^2}2{p^6},3{s^1}

Option A (

1{s^2},2{s^2}2{p^6},3{s^1}

) is correct because removing the single electron from the 3s orbital results in a stable noble gas configuration, making the second ionization energy significantly higher.

178 Medium

178 of 407

Application +4 | -1

The pair, in which ions are isoelectronic with

A{l^{3 + }}

is:

View answer and explanation

Correct option: D

{O^{2 - }}\,\,{\text{and}}\,\,M{g^{2 + }}

Option D (

{O^{2 - }}\,{\text{and}}\,M{g^{2 + }})

is correct because both ions have the same number of electrons as

A{l^{3 + }}

179 Medium

179 of 407

Understanding +4 | -1

Oxygen is divalent, but sulphur exhibits variable valency of

2,\mathrm{ }4\mathrm{ } and \mathrm{ }6

, because:

View answer and explanation

Correct option: D

Of the presence of

d

-orbitals in sulphur

Option D (Of the presence of

d

-orbitals in sulphur) is correct because sulphur has available d-orbitals in its valence shell, allowing it to expand its octet and exhibit variable valency.

180 Easy

180 of 407

Remembering +4 | -1

What would be the IUPAC name for the element with atomic number 115

View answer and explanation

Correct option: C

UuP

Option C (UuP) is correct because the IUPAC symbol for element 115 (Ununpentium) is Uup, which is often shortened to UuP.

Revision notes and key takeaways

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