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

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

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31 Easy
31 of 121
Remembering +4 | -1
Hardest element of III A group of gp.13 is:
View answer and explanation
Correct option: A
B
Option A (B) is correct because Boron is a non-metal and exists as a giant covalent structure, making it the hardest element in Group 13.
32 Easy
32 of 121
Understanding +4 | -1
Aluminium becomes passive in nitric acid because it:
View answer and explanation
Correct option: B
Forms a thin film of oxide
Option B (Forms a thin film of oxide) is correct because aluminum forms a passive layer of aluminum oxide (Al2O3Al_2O_3) on its surface when exposed to nitric acid, protecting it from further reaction.
33 Medium
33 of 121
Application +4 | -1
Colemanite is an important mineral of boron. It is represented Ca2 BxO11⋅6H2O\mathrm{Ca}_2 \mathrm{~B}_x \mathrm{O}_{11} \cdot 6 \mathrm{H}_2 \mathrm{O}. What is the value of xx ?
View answer and explanation
Correct option: A
6
Option A (6) is correct because the formula of Colemanite is Ca2B6O11⋅5H2OCa_2B_6O_{11} \cdot 5H_2O, so x = 6.
34 Easy
34 of 121
Remembering +4 | -1
In diborane, the banana bond is formed between
View answer and explanation
Correct option: A
2 electrons, 3 atoms
Option A (2 electrons, 3 atoms) is correct because in diborane, the banana bond involves 2 electrons shared between 3 atoms (B-H-B).
35 Hard
35 of 121
Understanding +4 | -1
Diborane (B2H6)({B_2}{H_6}) reacts independently with O2{O_2} and H2O{H_2}O to produce, respectively.
View answer and explanation
Correct option: A
B2O3{B_2}{O_3} and H3BO3{H_3}B{O_3}
Option A (B2O3B_2O_3 and H3BO3H_3BO_3) is correct because diborane reacts with oxygen to form boron oxide and with water to form boric acid.
36 Easy
36 of 121
Understanding +4 | -1
Boron compound behave as Lewis acids because of their
View answer and explanation
Correct option: C
Electron deficient nature
Option C (Electron deficient nature) is correct because boron compounds act as Lewis acids due to their electron-deficient nature, having an incomplete octet.
37 Medium
37 of 121
Analyzing +4 | -1
BX3+NH3⟶ R.T. BX3⋅NH3+\mathrm{BX}_3+\mathrm{NH}_3 \stackrel{\text { R.T. }}{\longrightarrow} \mathrm{BX}_3 \cdot \mathrm{NH}_3+ Heat of adduct formation (ΔH).(\Delta \mathrm{H}). The numerical value of ΔH\Delta \mathrm{H} is found to be maximum for:
View answer and explanation
Correct option: D
BI3\mathrm{BI}_3
The correct answer is D (BI3) because BI3 has the least pi-backbonding, leading to the strongest Lewis acid character and thus the maximum heat of adduct formation.
38 Easy
38 of 121
Remembering +4 | -1
Which is used as antiseptic for eyes
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Correct option: A
aq.H3BO3aq.{H_3}B{O_3}
Option A (aq.H3BO3aq.{H_3}B{O_3}) is correct because boric acid in aqueous solution is commonly used as a mild antiseptic, especially for eyes.
39 Easy
39 of 121
Remembering +4 | -1
Aluminium appears like gold when it is mixed with:
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Correct option: A
90% Cu
Option A is correct because a mixture of 90% Copper and 10% Aluminum creates an alloy that resembles gold in appearance.
40 Easy
40 of 121
Understanding +4 | -1
Due to inert pair effect in 13th group
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Correct option: B
Tl shows +1 oxidation state
Option B (Tl shows +1 oxidation state) is correct because thallium (Tl) exhibits a stable +1 oxidation state due to the inert pair effect.
41 Medium
41 of 121
Analyzing +4 | -1
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) In TlI3Tl{I_3}, isomorphous to CsI3Cs{I_3}, the metal is present in + 1 oxidation state.
Reason (R) TlTl metal has fourteen f−f -electrons in the electronic configuration.
In the light of the above statements, choose the most appropriate answer from the options given below.
View answer and explanation
Correct option: B
Both A and R are correct and R is the correct explanation of A.
Option A is correct because Tl in TlI3 exists as Tl+ and I3- due to the inert pair effect, and the presence of 14 f-electrons contributes to this effect.
42 Medium
42 of 121
Analyzing +4 | -1
Assertion (A): HBF4\mathrm{HBF}_4 is a very strong acid.

Reason (R):B(OH)3(\mathbf{R}): \mathrm{B}(\mathrm{OH})_3 behaves as an acid in the following reaction :
B(OH)3+4HF⟶H3O++[BF4]−+2H2O.\mathrm{B}(\mathrm{OH})_3+4 \mathrm{HF} \longrightarrow \mathrm{H}_3 \mathrm{O}^{+}+\left[\mathrm{BF}_4 \right]^{-}+2 \mathrm{H}_2 \mathrm{O}.
View answer and explanation
Correct option: C
A\mathbf{A} is true but R\mathbf{R} is false.
The correct answer is C (A is true but R is false) because HBF4 is a strong acid, but the reaction provided demonstrates B(OH)3 acting as a Lewis acid, not a Bronsted-Lowry acid.
43 Medium
43 of 121
Understanding +4 | -1
Which is electron deficient compound?
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Correct option: B
B2H6{\mathrm{B}_2}{\mathrm{H}_6}
Option B (B2H6{\mathrm{B}_2}{\mathrm{H}_6}) is correct because Diborane (B2H6{\mathrm{B}_2}{\mathrm{H}_6}) is an electron-deficient compound with bridging hydrogen atoms.
44 Easy
44 of 121
Analyzing +4 | -1
Assertion: Boron is a good conductor of electricity.

Reason: Boron has a low ionization energy and a high electron affinity.
View answer and explanation
Correct option: D
Assertion is false, but Reason is true
Option D is correct because Boron is a semiconductor, not a good conductor, and the reason provided is true concerning ionization energy and electron affinity.
45 Easy
45 of 121
Understanding +4 | -1
The ratio of Fe2O3\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3} and Al, in thermite is
View answer and explanation
Correct option: C
3:13:1
Option C (3:1) is correct because the thermite reaction involves a mixture of iron(III) oxide (Fe2O3Fe_2O_3) and aluminum (Al) in a 3:1 ratio by mass.
46 Easy
46 of 121
Remembering +4 | -1
Which of the following is most abundant in the earth crust?
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Correct option: D
Al
Option D (Al) is correct because Aluminum is the most abundant metal and the third most abundant element in the Earth's crust.
47 Easy
47 of 121
Remembering +4 | -1
Alums are used for
View answer and explanation
Correct option: D
All of these
Option D is correct because Alums are used in tanning leather, coagulating blood, and purifying water due to their ability to precipitate impurities.
48 Medium
48 of 121
Analyzing +4 | -1
Which of the following statement(s) is/are incorrect ?

(i) Trichlorides on hydrolysis in water form tetrahedral [M(OH)4]−\left[\mathrm{M}(\mathrm{OH})_4\right]^{-}species.

(ii) Hybridisation state of metal in tetrahedral species is sp3\mathrm{sp}^3.

(iii) Aluminium chloride in acidified aqueous solution forms [Al(OH)4]−\left[\mathrm{Al}(\mathrm{OH})_4\right]^{-}ion.
View answer and explanation
Correct option: C
(iii) only
Option C ((iii) only) is correct because Aluminium chloride in acidified aqueous solution forms [Al(H2O)6]3+\left[\mathrm{Al}(\mathrm{H}_2\mathrm{O})_6\right]^{3+} ion, not [Al(OH)4]−\left[\mathrm{Al}(\mathrm{OH})_4\right]^{-} ion.
49 Medium
49 of 121
Remembering +4 | -1
The 13th group element which has the least melting point is
View answer and explanation
Correct option: C
Ga
Option C is correct because Gallium (Ga) has the lowest melting point among the group 13 elements listed.
50 Hard
50 of 121
Understanding +4 | -1
Assertion (A) : AlCl3{AlCl}_3 forms dimer Al2Cl6{Al}_2 {Cl}_6 but is dissolves in H2O{H}_2 {O} forming [Al(H2O)6]3+\left[{Al}\left({H}_2 {O}\right)_6\right]^{3+} and Cl−{Cl}^{-}ions.

Reason (R) : Aqueous solution of AlCl3{AlCl}_3 is acidic due to hydrolysis.
View answer and explanation
Correct option: B
Both A\mathbf{A} and R\mathbf{R} are true but R\mathbf{R} is not the correct explanation of A\mathbf{A}.
Option B (Both A\mathbf{A} and R\mathbf{R} are true but R\mathbf{R} is not the correct explanation of A\mathbf{A}) is correct because both the assertion and reason are true, but the reason does not explain why AlCl3AlCl_3 forms a dimer.
51 Easy
51 of 121
Application +4 | -1
A non-metallic element of group 13 , used in making bullet proof vests is extremely hard solid of black colour. It can exist in many allotropic forms and has unusually high melting point. Its trifluoride acts as
View answer and explanation
Correct option: B
Lewis acid
Option B (Lewis acid) is correct because boron trifluoride (BF3BF_3) is electron deficient and can accept a pair of electrons.
52 Easy
52 of 121
Analyzing +4 | -1
Which one of the following compounds, is not a protonic acid?
View answer and explanation
Correct option: C
B(OH)3\mathrm{B}{\left( {\mathrm{OH}} \right)_3}
The correct answer is C (B(OH)3) because it acts as a Lewis acid by accepting OH- ions from water, rather than donating protons (H+).
53 Medium
53 of 121
Analyzing +4 | -1
Which of the following ions do not exist?
i.   B3+ii.   Al+3iii.   Ga3+iv.   Tl3+\begin{array}{cccccccccccccccccccc} {{\text{i}}.\,\,\,{B^{3 + }}}&{{\text{ii}}.\,\,\,A{l^{ + 3}}}&{{\text{iii}}.\,\,\,G{a^{3 + }}}&{{\text{iv}}.\,\,\,T{l^{3 + }}} \end{array}
View answer and explanation
Correct option: D
i,iv
Option D (i,iv) is correct because B3+B^{3+} is highly unlikely due to the high ionization energy required to remove 3 electrons from Boron, and Tl3+Tl^{3+} is less stable than Tl+Tl^+ due to the inert pair effect.
54 Medium
54 of 121
Remembering +4 | -1
Boric acid is prepared from borax by the action of:
View answer and explanation
Correct option: A
Hydrochloric acid
Option A (Hydrochloric acid) is correct because boric acid is prepared from borax by reacting it with hydrochloric acid.
55 Medium
55 of 121
Remembering +4 | -1
In Gold Schmidt reaction, certain metallic oxides are reduced to the metallic state by-heating with:
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Correct option: B
Metallic aluminium
Option B (Metallic aluminium) is correct because in the Goldschmidt reaction, metallic oxides are reduced by heating with metallic aluminum.
56 Medium
56 of 121
Analyzing +4 | -1
Assertion (A) : B2Cl4,GaCl2,GaS{B}_2 {Cl}_4, {GaCl}_2, {GaS} are diamagnetic.

Reason (R) : They show +2 oxidation state.
View answer and explanation
Correct option: B
Both A\mathbf{A} and R\mathbf{R} are true but R\mathbf{R} is not the correct explanation of A\mathbf{A}.
Option B is correct because both the assertion and the reason are true, but the reason is not the correct explanation for the assertion. The diamagnetic nature arises from the pairing of electrons, not solely the +2 oxidation state.
57 Easy
57 of 121
Remembering +4 | -1
Which of the following is non-metallic?
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Correct option: A
B
Option A is correct because Boron (B) is a metalloid, exhibiting properties intermediate between metals and nonmetals, and is generally considered non-metallic.
58 Medium
58 of 121
Remembering +4 | -1
Alum is not used:
View answer and explanation
Correct option: B
As an insecticide
Option B (As an insecticide) is correct because alum is not used as an insecticide; it is used as a mordant, in water purification, and in leather tanning.
59 Easy
59 of 121
Application +4 | -1
Oxidation state of boron in magnesium boride (Mg3B2)\left( {M{g_3}{B_2}} \right)
View answer and explanation
Correct option: A
−3- 3
Option A (−3-3) is correct because magnesium is +2 and the overall compound is neutral, leading to a -3 oxidation state for boron.
60 Hard
60 of 121
Understanding +4 | -1
A compound XX, of boron reacts with NH3\mathrm{NH}_3 on heating to give another compound YY, which is called inorganic benzene. The compound XX can be prepared by treating BF3\mathrm{BF}_3 with lithium aluminium hydride. The compounds XX and YY are represented by the formulae
View answer and explanation
Correct option: A
B2H6, B3 N3H6\mathrm{B}_2 \mathrm{H}_6, \mathrm{~B}_3 \mathrm{~N}_3 \mathrm{H}_6
Option A (B2H6, B3 N3H6\mathrm{B}_2 \mathrm{H}_6, \mathrm{~B}_3 \mathrm{~N}_3 \mathrm{H}_6) is correct because diborane (B2H6B_2H_6) reacts with ammonia to give borazine (B3N3H6B_3N_3H_6).
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