The metallic radius of sodium is $1.86\,\,\mathop {\rm{A}}\limits^{\rm{o}} ,$ where as its cation radius from its vapour state is

$1.54\,\,\mathop {\rm{A}}\limits^{\rm{o}} $

$1.86\,\,\mathop {\rm{A}}\limits^{\rm{o}} $

$1.94\,\,\mathop {\rm{A}}\limits^{\rm{o}} $

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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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271 Medium

271 of 407

Understanding +4 | -1

Which of the following statements is wrong?

View answer and explanation

Correct option: D

Non-metals are more than metals.

Option D (Non-metals are more than metals) is the wrong statement because metals constitute a larger portion of the periodic table than non-metals.

272 Easy

272 of 407

Application +4 | -1

Which pair of the atomic numbers represents s-block elements?

View answer and explanation

Correct option: A

3, 12

Option A is correct because elements with atomic numbers 3 (Lithium) and 12 (Magnesium) have their valence electrons in s-orbitals, making them s-block elements.

273 Medium

273 of 407

Understanding +4 | -1

Which of the following statements about modern periodic table is incorrect?

View answer and explanation

Correct option: A

The

p

-block has 6 column because a maximum of 6 electrons can occupy all the orbitals in the

p

-subshell

The correct answer is B (The

d

-block has 8 columns because a maximum of 8 electrons can occupy all the orbitals in the

d

-subshell) because the d-block has 10 columns, not 8, as it can accommodate 10 electrons.

274 Medium

274 of 407

Analyzing +4 | -1

The electronic configuration,

1{s^2},\;2{s^2}2{p^6},\;3{s^2}\;3{p^6}3{d^9}

represents a

View answer and explanation

Correct option: D

Metallic cation

Option D is correct because the electronic configuration indicates a loss of electrons, resulting in a positive charge, thus a metallic cation.

275 Easy

275 of 407

Application +4 | -1

The approximate size of the nucleus of

\mathop {\mathop {Ni}\limits_{28} }\limits^{64}

View answer and explanation

Correct option: C

5fm

Option C (5fm) is correct because the nuclear radius is proportional to

A^{1/3}

, where A is the mass number. Using the formula

R = R_0 A^{1/3}

with

R_0 \approx 1.2 fm

, we get approximately 5 fm.

276 Easy

276 of 407

Remembering +4 | -1

The lanthanide contraction refers to

View answer and explanation

Correct option: A

radius of the series

Option A is correct because lanthanide contraction refers to the gradual decrease in the atomic and ionic radii of the lanthanide elements as the atomic number increases.

277 Easy

277 of 407

Remembering +4 | -1

Which one of the elements has the maximum electron affinity?

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Correct option: B

Option B (Cl) is correct because chlorine has the highest electron affinity due to its size and effective nuclear charge, which allows it to accept an electron more readily than fluorine, bromine, or iodine.

278 Medium

278 of 407

Remembering +4 | -1

Which of the following groups in modern periodic table have largest number of elements

View answer and explanation

Correct option: C

III B

Option C (III B) is correct because Group IIIB (also known as Group 3) has the largest number of elements due to the inclusion of Lanthanides and Actinides.

279 Hard

279 of 407

Remembering +4 | -1

Which of the following is an amphoteric oxide?

View answer and explanation

Correct option: C

\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}

Option C (

\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}

) is correct because aluminum oxide is a well-known amphoteric oxide, reacting with both acids and bases.

280 Easy

280 of 407

Remembering +4 | -1

The atomic weight of ‘Be’ and ‘In’ were corrected by Mendeleev using the formula

View answer and explanation

Correct option: C

Atomic weight = Equivalent weight

\times

valency

Option C (Atomic weight = Equivalent weight × valency) is correct because Mendeleev used this formula to correct the atomic weights of elements like Be and In.

281 Medium

281 of 407

Understanding +4 | -1

The correct ionic radii order is:

View answer and explanation

Correct option: A

{\mathrm{N}^{3 - }} > {\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > N{\mathrm{a}^ + } > M{\mathrm{g}^{2 + }} > A{\mathrm{l}^{3 + }}

Option A (

{\mathrm{N}^{3 - }} > {\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > N{\mathrm{a}^ + } > M{\mathrm{g}^{2 + }} > A{\mathrm{l}^{3 + }}

) is correct because it correctly orders the isoelectronic species based on their nuclear charge; higher the nuclear charge, smaller the ionic radii.

282 Easy

282 of 407

Remembering +4 | -1

The elements which occupy peaks in the atomic volume curves are

View answer and explanation

Correct option: B

Na, K, Rb

Option B (Na, K, Rb) is correct because alkali metals (Na, K, Rb) have the largest atomic volumes in their respective periods, thus occupying the peaks in the atomic volume curves.

283 Hard

283 of 407

Understanding +4 | -1

The correct order of ionic radii for the ions,

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

View answer and explanation

Correct option: A

{P^{3 - }} > {S^{2 - }} > C{l^ - } > {K^ + } > C{a^{2 + }}

Option A (

{P^{3 - }} > {S^{2 - }} > C{l^ - } > {K^ + } > C{a^{2 + }}

) is correct because for isoelectronic species, ionic radii decrease with increasing nuclear charge.

284 Medium

284 of 407

Remembering +4 | -1

Which of the following set of oxides are amphoteric

View answer and explanation

Correct option: D

A{l_2}{O_3},ZnO,Sn{O_2},S{b_4}{O_6}

Option D (

A{l_2}{O_3},ZnO,Sn{O_2},S{b_4}{O_6}

) is correct because all the oxides in this set can react with both acids and bases.

285 Hard

285 of 407

Application +4 | -1

The 71st electron of an element

X

with an atomic number of 71 enters into the orbital

View answer and explanation

Correct option: C

5d

The correct answer is C (

5d

) because after filling the

4f

orbitals, the 71st electron of element X (Lutetium) enters the

5d

orbital.

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286 Medium

286 of 407

Understanding +4 | -1

The metallic radius of sodium is

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

where as its cation radius from its vapour state is

View answer and explanation

Correct option: C

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

Option C is correct because the cation radius is smaller than the metallic radius due to increased effective nuclear charge and removal of an electron.

287 Medium

287 of 407

Understanding +4 | -1

The correct order of ionic radii is:

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because in each case, the neutral atom is larger than its positive ions, and the negative ion is larger than the neutral atom.

288 Medium

288 of 407

Remembering +4 | -1

The IUPAC name of an element with atomic number 119 is

View answer and explanation

Correct option: B

ununennium

Option B (ununennium) is correct because it follows the IUPAC nomenclature rules for naming elements with atomic number 119.

289 Hard

289 of 407

Application +4 | -1

General electronic configuration of actionoids is

(n-2) f^{1-14}(n-1) d^{0-2} n s^2

. Which of the following actinoids have one electron in

6d

orbital?

View answer and explanation

Correct option: C

{Cm}

(Atomic no. 96)

Option C is correct because Curium (Cm) with atomic number 96 has an electronic configuration that includes one electron in the 6d orbital.

290 Hard

290 of 407

Understanding +4 | -1

The set representing the correct order of ionic radius is

View answer and explanation

Correct option: B

N{a^ + } > L{i^ + } > M{g^{2 + }} > B{e^{2 + }}

Option B (

N{a^ + } > L{i^ + } > M{g^{2 + }} > B{e^{2 + }}

) is correct because ionic radius decreases with increasing nuclear charge for isoelectronic species and increases down a group.

291 Hard

291 of 407

Analyzing +4 | -1

The successive ionisation energy values for an element ‘X’ are given below XV. Ist ionisation energy =410 kJ

\mathrm{mo}{\mathrm{l}^{ - 1}}

XVI. 2nd ionisation energy =820 kJ

\mathrm{mo}{\mathrm{l}^{ - 1}}

XVII. 3rd ionisation energy =1100 kJ

\mathrm{mo}{\mathrm{l}^{ - 1}}

XVIII. 4th ionisation energy =1500 kJ

\mathrm{mo}{\mathrm{l}^{ - 1}}

XIX. 5th ionisation energy =3200 kJ

\mathrm{mo}{\mathrm{l}^{ - 1}}

Find out the number of valence electron for the atom ‘X ’

View answer and explanation

Correct option: A

Option A (4) is correct because the large jump in ionization energy between the 4th and 5th ionization energies indicates that the element has 4 valence electrons.

292 Easy

292 of 407

Remembering +4 | -1

Which of the following shows diagonal relationship

View answer and explanation

Correct option: A

boron and silicon

Option A is correct because boron and silicon exhibit a diagonal relationship due to their similar electronegativities and polarizing power.

293 Easy

293 of 407

Understanding +4 | -1

Among

O,\mathrm{ }C,\mathrm{ }F,\mathrm{ }Cl,\mathrm{ }Br

the correct order of increasing atomic radii is:

View answer and explanation

Correct option: A

\mathrm{F} < O < C < Cl < Br

Option A (

\mathrm{F} < O < C < Cl < Br

) is correct because atomic radii increase down a group and generally decrease across a period.

294 Medium

294 of 407

Analyzing +4 | -1

Maximum I.E is for

View answer and explanation

Correct option: B

L{i^ + }

Option B (

Li^+

) is correct because removing an electron from

Li^+

(electronic configuration

1s^2

) requires breaking a stable noble gas configuration, which needs maximum energy.

295 Medium

295 of 407

Remembering +4 | -1

Which of the following metals exhibits more than one oxidation state?

View answer and explanation

Correct option: D

Option D (Fe) is correct because Iron can exhibit multiple oxidation states such as +2 and +3.

296 Easy

296 of 407

Understanding +4 | -1

The electron with z = 117 and z = 120 belong to ______ ______ family respectively.

View answer and explanation

Correct option: C

Halogen family and alkali earth metals

Option C (Halogen family and alkali earth metals) is correct because element with Z=117 belongs to the halogen family and Z=120 belongs to alkaline earth metals.

297 Hard

297 of 407

Application +4 | -1

The electronic configuration of an element is

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

. The atomic number and the group number of the element '

\mathrm{X}

' which is just below the above element in the periodic table are respectively

View answer and explanation

Correct option: C

33 and 15

Option C (33 and 15) is correct. Adding one more shell to the given configuration results in an element with atomic number 33, which is in group 15.

298 Medium

298 of 407

Understanding +4 | -1

The periodic table of elements does not

View answer and explanation

Correct option: B

Tell us about the arrangement of atoms in a molecule

Option B (Tell us about the arrangement of atoms in a molecule) is correct because the periodic table primarily deals with elements and their properties, not molecular structure.

299 Hard

299 of 407

Analyzing +4 | -1

The third ionisation enthalpy is minimum for:

View answer and explanation

Correct option: D

Fe

Option D (Fe) has the minimum third ionization enthalpy because after losing two electrons, it attains a relatively stable d6 configuration, making it easier to remove the third electron compared to the other options.

300 Easy

300 of 407

Understanding +4 | -1

An element whose IUPAC name is ununtrium (Uut), belongs to

View answer and explanation

Correct option: B

P – block

Option B (P – block) is correct because Ununtrium (Uut), now known as Moscovium (Mc), belongs to group 15, which is in the p-block.

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