${\rm{BC}}{{\rm{l}}_3} + {{\rm{H}}_2}{\rm{O}} \to {\rm{X}},$ the products formed in the reaction are

${{\rm{H}}_3}{\rm{B}}{{\rm{O}}_3} + {\rm{HCl}}$

${{\rm{B}}_2}{{\rm{O}}_3} + {\rm{HOCl}}$

${{\rm{B}}_2}{{\rm{H}}_6} + {\rm{HCl}}$

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JEE • CHEMISTRY • 13 TH GROUP ELEMENTS (P-BLOCK)

13 th group elements (p-block) practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-60 of 163 questions

31 Easy

31 of 163

Remembering +4 | -1

Magnalium contains

View answer and explanation

Correct option: A

\mathrm{Al} + \mathrm{Mg}

Option A (

\mathrm{Al} + \mathrm{Mg}

) is correct because Magnalium is an alloy primarily composed of aluminum and magnesium.

32 Easy

32 of 163

Understanding +4 | -1

\mathrm{BC}{\mathrm{l}_3} + {\mathrm{H}_2}\mathrm{O} \to \mathrm{X},

the products formed in the reaction are

View answer and explanation

Correct option: B

{\mathrm{H}_3}\mathrm{B}{\mathrm{O}_3} + \mathrm{HCl}

Option B (

{\mathrm{H}_3}\mathrm{B}{\mathrm{O}_3} + \mathrm{HCl}

) is correct because BCl3 reacts with water to form boric acid and hydrochloric acid.

33 Easy

33 of 163

Application +4 | -1

An element

'X'

which occurs in the first short period has an outer electronic structure

{s^2}{p^1}

. What is the formula and acid-base character of its oxides?

View answer and explanation

Correct option: C

{X_2}{\mathrm{O}_3}

, acidic

Option C (

{X_2}{\mathrm{O}_3}

, acidic) is correct because the element X belongs to group 13 and forms an acidic oxide since it is a non-metal.

34 Easy

34 of 163

Remembering +4 | -1

B-10 isotopes

View answer and explanation

Correct option: A

Absorbs neutron

Option A (Absorbs neutron) is correct because Boron-10 has a high neutron absorption cross-section and is used in nuclear reactors for neutron capture.

35 Easy

35 of 163

Understanding +4 | -1

Which one of the following elements of unable to form

\mathrm{MF}_\mathrm{6}^{\mathrm{3 - }}

ion ?

View answer and explanation

Correct option: C

\mathrm{B}

Option C (B) is correct because boron has a small size and only four valence orbitals, limiting its coordination number to a maximum of 4. Therefore, it cannot form

MF_6^{3-}

ions.

36 Easy

36 of 163

Remembering +4 | -1

Molecular weight of anhydrous aluminium chloride is:

View answer and explanation

Correct option: B

267

Option B (267) is correct because the molecular weight of anhydrous aluminium chloride (

\mathrm{A}{\mathrm{l}_2}\mathrm{C}{\mathrm{l}_6}

) is 267.

37 Medium

37 of 163

Understanding +4 | -1

Which of the following oxide is of acidic nature?

View answer and explanation

Correct option: A

{B}_2 {O}_3

Option A (

{B}_2 {O}_3

) is correct because boron oxide is acidic due to the small size of boron, as explained by Fajan's rule.

38 Medium

38 of 163

Remembering +4 | -1

Which of the following is pseudoalum?

View answer and explanation

Correct option: C

\mathrm{MnS}{\mathrm{O}_4} \cdot \mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3} \cdot 24{\mathrm{H}_2}\mathrm{O}

Option C (

\mathrm{MnS}{\mathrm{O}_4} \cdot \mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3} \cdot 24{\mathrm{H}_2}\mathrm{O}

) is correct because a pseudoalum has a divalent cation (like Mn²⁺) in place of the monovalent cation.

39 Easy

39 of 163

Analyzing +4 | -1

Assertion: Boron does not form a stable compound with oxygen.

Reason: Boron has a smaller size and higher electronegativity than oxygen.

View answer and explanation

Correct option: A

Both Assertion and Reason are true and Reason is the correct explanation for Assertion

Option A (Both Assertion and Reason are true and Reason is the correct explanation for Assertion) is correct because boron's small size and high electronegativity prevent it from forming stable compounds with oxygen.

40 Medium

40 of 163

Remembering +4 | -1

Amphoteric hydroxide from the following is

View answer and explanation

Correct option: A

{Al}({OH})_3

Option A (

{Al}({OH})_3

) is correct because aluminum hydroxide is amphoteric, meaning it can react with both acids and bases.

41 Easy

41 of 163

Remembering +4 | -1

The liquefied metal which expands on solidification is:

View answer and explanation

Correct option: A

{Ga}

Option A (

{Ga}

) is correct because Gallium is a unique metal that expands upon solidification, unlike most metals which contract.

42 Medium

42 of 163

Understanding +4 | -1

Taking stability as the factor, which one of the following represents correct relationship?

View answer and explanation

Correct option: B

\text{TlI} > \text{T}\ell\text{I}_3

Option B (

\text{TlI} > \text{T}\ell\text{I}_3

) is correct because thallium(I) iodide is more stable than thallium(III) iodide due to the inert pair effect.

43 Medium

43 of 163

Understanding +4 | -1

Aluminium chloride exists as dimer,

\mathrm{A}{\mathrm{l}_2}\mathrm{C}{\mathrm{l}_6}

, in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives

View answer and explanation

Correct option: B

{\left[ {\mathrm{Al}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{3 + }} + 3\mathrm{C}{\mathrm{l}^ - }

Option B is correct because in water, aluminum chloride undergoes hydrolysis to form the hexaaquaaluminum(III) ion,

{\left[ {\mathrm{Al}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{3 + }}

, and chloride ions.

44 Medium

44 of 163

Understanding +4 | -1

The correct order of the first ionization enthalpy is

View answer and explanation

Correct option: D

\mathrm{Tl} > \mathrm{Ga} > \mathrm{Al}

Option C (

\mathrm{B} > \mathrm{Al} > \mathrm{Ga}

) is correct because ionization enthalpy generally decreases down a group, but Aluminum has a lower ionization enthalpy than Boron and Gallium due to electronic configuration and shielding effects.

45 Medium

45 of 163

Remembering +4 | -1

In group-13, thallium shows +1 oxidation state, while other members show +3 oxidation state, this is due to

View answer and explanation

Correct option: C

Inert pair effect

Option C (Inert pair effect) is correct because the inert pair effect causes the +1 oxidation state to be more stable than the +3 oxidation state for heavier elements like thallium in group 13.

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46 Medium

46 of 163

Understanding +4 | -1

Potash alum is water soluble and ionises in aqueous solution to give:

View answer and explanation

Correct option: C

Three types of ions

Option D (Four types of ions) is correct because Potash alum,

KAl(SO_4)_2 \cdot 12H_2O

, dissociates into

K^+

Al^{3+}

, and

SO_4^{2-}

ions in water.

47 Easy

47 of 163

Remembering +4 | -1

The catalyst used in Friedel-Craft’s reaction is:

View answer and explanation

Correct option: C

Anhydrous aluminium chloride

Option C (Anhydrous aluminium chloride) is correct because anhydrous aluminium chloride (

AlCl_3

) is the catalyst used in Friedel-Crafts reactions.

48 Easy

48 of 163

Remembering +4 | -1

Eka aluminium is:

View answer and explanation

Correct option: A

Gallium

Option A (Gallium) is correct because Mendeleev predicted the existence of gallium and referred to it as eka-aluminum, based on its position in the periodic table.

49 Medium

49 of 163

Remembering +4 | -1

Given below are two statements:

Statement I: The boron-10

\left({ }^{10} \mathrm{~B}\right)

isotope has high ability to absorb neutrons.

Statement II: On a weight-to-weight basis, the electrical conductivity of 'Al' is twice that of 'Cu'.

In the light of the above statements, choose the most appropriate answer from the options given below:

View answer and explanation

Correct option: D

Both Statement I and Statement II are correct

Option D (Both Statement I and Statement II are correct) is correct because Boron-10 is used in nuclear reactors due to its neutron absorption ability, and Aluminum has higher conductivity per weight than Copper.

50 Easy

50 of 163

Understanding +4 | -1

The order of Lewis acid nature

View answer and explanation

Correct option: B

B{F_3} < BC{l_3} < BB{r_3} < B{I_3}

Option B (

B{F_3} < BC{l_3} < BB{r_3} < B{I_3}

) is correct because the Lewis acidity increases down the group due to decreasing pπ-pπ back bonding.

51 Medium

51 of 163

Remembering +4 | -1

Which is not an alloy of aluminium?

View answer and explanation

Correct option: C

German silver

Option C (German silver) is correct because German silver is an alloy of copper, zinc, and nickel, not aluminum.

52 Easy

52 of 163

Understanding +4 | -1

Which is not correct?

View answer and explanation

Correct option: D

Al is ionic in all its compounds

Option D (Al is ionic in all its compounds) is correct because Aluminum can also form covalent compounds, such as

AlCl_3

, making the statement incorrect.

53 Medium

53 of 163

Understanding +4 | -1

Which of the following is correct statement?

View answer and explanation

Correct option: A

Aluminium dissolves in dilute mineral acids but is made passive by concentrated nitric acid

Option A (Aluminium dissolves in dilute mineral acids but is made passive by concentrated nitric acid) is correct because concentrated nitric acid forms a protective oxide layer on aluminum, preventing further reaction.

54 Medium

54 of 163

Understanding +4 | -1

In the following

\underset{(p)}{\left[{Al}\left({H}_2 {O}\right)_6\right]^{3+}}+\underset{(q)}{{HCO}_3^{-}} \Leftrightarrow \underset{(r)}{\left[{Al}\left({H}_2 {O}\right)_5 {OH}\right]^{2+}}+\underset{(s)}{{H}_2 {CO}_3}

species behaving as Brosnted-Lowry acids are

View answer and explanation

Correct option: A

(p), (s)

Option A (

(p), (s)

) is correct because in the forward reaction,

[{Al}({H}_2 {O})_6]^{3+}

donates a proton, and in the reverse reaction,

{H}_2 {CO}_3

donates a proton, thus acting as Bronsted-Lowry acids.

55 Medium

55 of 163

Analyzing +4 | -1

A compound

(A)

of boron reacts with

\mathrm{NMe}_3

to give an adduct

(B)

which on hydrolysis gives a compound

(C)

and hydrogen gas. Compound

(C)

is an acid. The adduct

(B)

View answer and explanation

Correct option: B

2 \mathrm{BH}_3 \cdot \mathrm{NMe}_3

Option D (

\mathrm{BH}_3 \cdot \mathrm{NMe}_3

) is incorrect. The correct answer should be

2 \mathrm{BH}_3 \cdot \mathrm{NMe}_3

because diborane (

B_2H_6

) reacts with two molecules of

NMe_3

56 Easy

56 of 163

Analyzing +4 | -1

Least basic among the following is

View answer and explanation

Correct option: C

B{\left( {OH} \right)_3}

Option C (

B{\left( {OH} \right)_3}

) is the least basic because boron's small size and high charge density make it electron-deficient, thus acting as a Lewis acid rather than a base.

57 Easy

57 of 163

Remembering +4 | -1

In aluminates coordination number of Al is:

View answer and explanation

Correct option: B

Option B (6) is correct because in aluminates, aluminum often exists in a complex state where it can coordinate with six ligands, such as water molecules or hydroxide ions.

58 Medium

58 of 163

Remembering +4 | -1

In Hall-Heroult process, the following is used for reducing

\mathrm{Al}_{2} \mathrm{O}_{3}

View answer and explanation

Correct option: C

Graphite

Option C (Graphite) is correct because graphite is used as the anode in the Hall-Héroult process to reduce aluminum oxide to aluminum.

59 Medium

59 of 163

Remembering +4 | -1

The element which exists in liquid state for a wide range of temperature and can be used for measuring high temperature is

View answer and explanation

Correct option: C

Ga

The correct answer is C (

Ga

) because Gallium (Ga) has a relatively low melting point and a high boiling point, allowing it to exist in the liquid state over a wide temperature range.

60 Easy

60 of 163

Remembering +4 | -1

Aluminium is not present in which of the following mineral?

View answer and explanation

Correct option: C

Fluorspar

Option C (Fluorspar) is correct because fluorspar is calcium fluoride (

\mathrm{Ca}{\mathrm{F}_2}

), which does not contain aluminum.

Revision notes and key takeaways

Review 13 TH GROUP ELEMENTS (P-BLOCK) in your textbook or class notes, then use this set to check your understanding.

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