Identify the correct order of solubility of ${\rm{N}}{{\rm{a}}_2}{\rm{S}},{\rm{CuS}}$ and ${\rm{ZnS}}$ in aqueous medium:

${\rm{N}}{{\rm{a}}_2}{\rm{S}} > ZnS > CuS$

${\rm{CuS}} > ZnS > N{{\rm{a}}_2}{\rm{S}}$

${\rm{ZnS}} > N{{\rm{a}}_2}{\rm{S}} > CuS$

${\rm{N}}{{\rm{a}}_2}{\rm{S}} > CuS > ZnS$

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

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

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

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

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

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JEE • CHEMISTRY • CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

Classification of elements and periodicity in properties practice · Hard

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1 Hard

1 of 15

Remembering +4 | -1

The pair of amphoteric hydroxides is

View answer and explanation

Correct option: D

\mathrm{Be}{\left( {\mathrm{OH}} \right)_2},\mathrm{Zn}{\left( {\mathrm{OH}} \right)_2}

Option D (

\mathrm{Be}{\left( {\mathrm{OH}} \right)_2},\mathrm{\;Zn}{\left( {\mathrm{OH}} \right)_2}

) is correct because both beryllium hydroxide and zinc hydroxide exhibit amphoteric behavior.

2 Hard

2 of 15

Understanding +4 | -1

The radii of

F,{F^ - },O

and

{O^{ - 2}}

are in the order of

View answer and explanation

Correct option: D

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

Option D is correct because anions are larger than their corresponding neutral atoms, and size decreases across a period, so O2- > F- > O > F.

3 Hard

3 of 15

Understanding +4 | -1

The charge/size ratio of a cation determines its polarising power. Which one of the following sequences represents the increasing order of the polarising power of the cationic species,

{\mathrm{K}^ + },\mathrm{C}{\mathrm{a}^{2 + }},\mathrm{M}{\mathrm{g}^{2 + }},\mathrm{B}{\mathrm{e}^{2 + }}

View answer and explanation

Correct option: C

{\mathrm{K}^ + } < C{\mathrm{a}^{2 + }} < M{\mathrm{g}^{2 + }} < B{\mathrm{e}^{2 + }}

Option C (

{\mathrm{K}^ + } < C{\mathrm{a}^{2 + }} < M{\mathrm{g}^{2 + }} < B{\mathrm{e}^{2 + }}

) is correct because the polarizing power increases with increasing charge and decreasing size.

4 Hard

4 of 15

Analyzing +4 | -1

The

{\mathrm{I}^{\mathrm{st}}}

IEs of four consecutive elements present in the second period of Periodic Table are 8.3, 11.3, 14.5 and 13.6 eV respectively. Which of these is the IE of nitrogen?

View answer and explanation

Correct option: C

14.5

Option C (14.5) is correct because nitrogen has a higher ionization energy than its neighbors due to its stable half-filled p-subshell.

5 Hard

5 of 15

Application +4 | -1

The energy required to convert all atoms present in

1.2 \mathrm{~g}

magnesium to

{Mg}^{2+}

ions if

\mathrm{IE}

, and

\mathrm{IE}_2

of magnesium are

120 \mathrm{~kJ} \mathrm{~mol}^{-1}

and

240 \mathrm{~kJ} \mathrm{~mol}^{-1}

respectively

View answer and explanation

Correct option: A

18 \mathrm{~kJ}

Option A (

18 \mathrm{~kJ}

) is correct because the total ionization energy for converting 0.05 moles of Mg to Mg²⁺ is calculated by summing IE₁ and IE₂ and then multiplying by the number of moles.

6 Hard

6 of 15

Understanding +4 | -1

Considering the elements

\mathrm{B}, \mathrm{C}, \mathrm{N}, \mathrm{F}

and

\mathrm{Si}

, the correct order of their non-metallic character is

View answer and explanation

Correct option: C

>

>

>

>

Option C (F

>

>

>

>

Si) is correct because non-metallic character increases across a period and decreases down a group.

7 Hard

7 of 15

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 is correct because there is a significant jump in ionization energy between the 4th and 5th ionization energies, indicating 4 valence electrons.

8 Hard

8 of 15

Analyzing +4 | -1

The third ionisation enthalpy is minimum for:

View answer and explanation

Correct option: D

Fe

Option D (

Fe

) is correct because

Fe^{2+}

has a

3d^6

configuration, and removing the third electron results in the stable

3d^5

configuration, requiring less energy.

9 Hard

9 of 15

Analyzing +4 | -1

The correct order of ionic radius is

View answer and explanation

Correct option: D

{\mathrm{P}^{3 + }} > {\mathrm{P}^{5 + }}

The correct answer is D (

{\mathrm{P}^{3 + }} > {\mathrm{P}^{5 + }})

because ionic radius is inversely proportional to effective nuclear charge, and P³⁺ has a lower effective nuclear charge than P⁵⁺.

10 Hard

10 of 15

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 the ionization potential of Ga is similar to Al due to poor shielding by the d-electrons in Ga.

11 Hard

11 of 15

Analyzing +4 | -1

Out of

Ne,\,\,N{a^ + },\,\,O,\,\,Na

and

N,

the species having maximum and minimum ionization energies respectively are

View answer and explanation

Correct option: C

N{a^ + },\,\,Na

The correct answer is C (

N{a^ + },\,Na

) because

Na^+

has a higher nuclear charge and smaller size, leading to a higher ionization energy, while Na has the lowest.

12 Hard

12 of 15

Understanding +4 | -1

Identify the correct order of solubility of

\mathrm{N}{\mathrm{a}_2}\mathrm{S},\mathrm{CuS}

and

\mathrm{ZnS}

in aqueous medium:

View answer and explanation

Correct option: D

\mathrm{N}{\mathrm{a}_2}\mathrm{S} > ZnS > CuS

Option D (

\mathrm{N}{\mathrm{a}_2}\mathrm{S} > ZnS > CuS

) is correct because sodium salts are highly soluble, and the

K_{sp}

value of

\mathrm{CuS}

is less than

\mathrm{ZnS}

, making

\mathrm{ZnS}

more soluble than

\mathrm{CuS}

13 Hard

13 of 15

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 is correct because anions are larger than their neutral atoms, and among isoelectronic species, the size decreases with increasing nuclear charge.

14 Hard

14 of 15

Application +4 | -1

{\mathrm{I}^{\mathrm{st}}}\mathrm{\;and\;I}{\mathrm{I}^{\mathrm{nd}}}\mathrm{\;IE}

of Mg are 7.646 and 15.035 eV respectively. The amount of energy needed to convert all the atoms of magnesium into  

\mathrm{M}{\mathrm{g}^{2 + }}

ions present in 12 mg of magnesium vapours is [Given, 1eV = 96.5 kJ

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

]

View answer and explanation

Correct option: C

1.1

Option C (1.1) is correct because it is the result of calculating the total ionization energy for all magnesium atoms.

15 Hard

15 of 15

Understanding +4 | -1

The basic character of

\mathrm{MgO}, \mathrm{BaO}, \mathrm{Na}_2 \mathrm{O}

and

\mathrm{FeO}

follow the order

View answer and explanation

Correct option: C

\mathrm{FeO} < \mathrm{MgO} < \mathrm{BaO} < \mathrm{Na}_2 \mathrm{O}

Option C is correct because basic character increases down a group and decreases across a period, with Na2O being the most basic and FeO the least.

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