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

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

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

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

13 TH GROUP ELEMENTS (P-BLOCK) MCQs for JEE CHEMISTRY — Easy

1 Easy

1 of 60

Remembering +4 | -1

{Al}_2\left({SO}_4\right)_3

is used in the following but not

View answer and explanation

Correct option: B

In plastic industry

Correct Answer + Core Reason: Option B (In leather industry) is the correct answer because

Al_2(SO_4)_3

is not primarily used in the leather industry.

Solution approach

Correct Answer + Core Reason: $Al_2(SO_4)_3$ is not primarily used in the leather industry.

Step-by-Step Solution:

Step 1: Recall the uses of $Al_2(SO_4)_3$ (Aluminum sulfate).

Step 2: Aluminum sulfate is used as a coagulant in water purification, a mordant in dyeing, and in paper sizing.

Step 3: While aluminum compounds can be used in tanning leather, aluminum sulfate itself is not the primary agent.

Step 4: Final Answer: In leather industry

Why other options are wrong:
- Option A: Aluminum sulfate is used as a coagulant in water treatment.
- Option C: Aluminum sulfate is used as a mordant in dyeing.
- Option D: Both options 2 and 3 are correct uses of aluminum sulfate.

Key Concepts:
- Aluminum Sulfate
- Uses of Chemical Compounds
- Coagulation
- Mordant

Memory Aid:
- Aluminum sulfate: Water, dye, paper.

Quick Tips:
- Remember common uses of chemical compounds.

2 Easy

2 of 60

Remembering +4 | -1

Which metal will give blue colour metaborate in borax bead test?

View answer and explanation

Correct option: A

{Cu}

Option A (

{Cu}

) is correct because copper gives a blue color in the borax bead test due to the formation of copper(II) metaborate.

Solution approach

Correct Answer + Core Reason: ${Cu}$ . Copper gives a blue colored copper metaborate in the borax bead test.

Step-by-Step Solution:

Step 1: Recall the borax bead test: A qualitative test used to identify certain metals based on the color they impart to a borax bead when heated.

Step 2: Consider the colors produced by each metal in the borax bead test:
- Copper (Cu): Blue
- Iron (Fe): Green or brown
- Chromium (Cr): Green
- Nickel (Ni): Brown

Step 3: Identify the metal that gives a blue color: Copper gives a blue color.

Step 4: Therefore, Copper is the correct answer.

Why other options are wrong:
- Option B: Iron gives a green or brown color.
- Option C: Chromium gives a green color.
- Option D: Nickel gives a brown color.

Key Concepts:
- Borax bead test
- Qualitative analysis
- Metal identification

Memory Aids:
- Copper gives a blue color in the borax bead test.

Quick Tips:
- Remember the characteristic colors produced by common metals in the borax bead test.

3 Easy

3 of 60

Remembering +4 | -1

Which of the following is called alum?

View answer and explanation

Correct option: B

\mathrm{N}{\mathrm{a}_2}\mathrm{S}{\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 B (

\mathrm{N}{\mathrm{a}_2}\mathrm{S}{\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 it matches the general formula of alum, which is a double sulfate salt with the formula

{M'_2}\mathrm{S}{\mathrm{O}_4} \cdot {M'''_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3} \cdot 24{\mathrm{H}_2}\mathrm{O}

.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{a}_2}\mathrm{S}{\mathrm{O}_4}\cdot\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3} \cdot 24{\mathrm{H}_2}\mathrm{O}$ is alum because it fits the general formula of alum.

Step-by-Step Solution:

Step 1: Recall the general formula for alum: ${M'_2}\mathrm{S}{\mathrm{O}_4} \cdot {M'''_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3} \cdot 24{\mathrm{H}_2}\mathrm{O}$ , where M' is a monovalent cation and M''' is a trivalent cation.

Step 2: Compare the given options with the general formula.

Step 3: Option B, $\mathrm{N}{\mathrm{a}_2}\mathrm{S}{\mathrm{O}_4}\cdot\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3} \cdot 24{\mathrm{H}_2}\mathrm{O}$ , matches the general formula, with Na as the monovalent cation and Al as the trivalent cation.

Step 4: Therefore, $\mathrm{N}{\mathrm{a}_2}\mathrm{S}{\mathrm{O}_4}\cdot\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3} \cdot 24{\mathrm{H}_2}\mathrm{O}$ is alum.

Why other options are wrong:
- Option A: $\mathrm{NaAl}{\mathrm{O}_2}$ is sodium meta-aluminate, not alum. Alums are double salts of sulfates.
- Option C: $\mathrm{KCl} \cdot \mathrm{MgC}{\mathrm{l}_2} \cdot 6{\mathrm{H}_2}\mathrm{O}$ is carnallite, a double salt of potassium and magnesium chlorides, not alum.
- Option D: $\mathrm{FeS}{\mathrm{O}_4} \cdot {\left( {\mathrm{N}{\mathrm{H}_4}} \right)_2}\mathrm{S}{\mathrm{O}_4} \cdot 6{\mathrm{H}_2}\mathrm{O}$ is Mohr's salt, a double salt of ferrous sulfate and ammonium sulfate, but it does not have the correct stoichiometry and trivalent cation to be considered an alum.

Key Concepts:
- Alums are double sulfate salts with a specific general formula.
- General formula of alum: ${M'_2}\mathrm{S}{\mathrm{O}_4} \cdot {M'''_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3} \cdot 24{\mathrm{H}_2}\mathrm{O}$

Memory Aids:
- Remember that alums contain a monovalent and a trivalent cation along with sulfate ions and water of crystallization.

Quick Tips:
- Recognize the general formula of alum to quickly identify it among other compounds.

4 Easy

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Understanding +4 | -1

Statement-I: Borazole is inorganic benzene.

Statement-II: Benzene and Borazole are isoelectronic and show structural similarity.

View answer and explanation

Correct option: A

Both I and II are correct

Option A (Both I and II are correct) is correct because borazole (

B_3N_3H_6

) is indeed known as inorganic benzene due to its structural similarity and isoelectronic nature with benzene (

C_6H_6

).

Solution approach

Correct Answer + Core Reason: Both I and II are correct. Borazole is isoelectronic and structurally similar to benzene.

Step-by-Step Solution:
Step 1: Define Borazole: Borazole has the formula $B_3N_3H_6$ .
Step 2: Define Benzene: Benzene has the formula $C_6H_6$ .
Step 3: Compare Structures: Both compounds have a cyclic structure with alternating atoms.
Step 4: Compare Electronic Structures: Benzene and borazole are isoelectronic, meaning they have the same number of electrons. Benzene has 42 electrons (6C + 6H = 6*6 + 6*1 = 42). Borazole has 42 electrons (3B + 3N + 6H = 3*5 + 3*7 + 6*1 = 15 + 21 + 6 = 42).
Step 5: Conclusion: Borazole is called inorganic benzene due to its structural similarity and isoelectronic nature with benzene.

Why other options are wrong:
- Option B (Both I and II are incorrect): Both statements are correct.
- Option C (I is correct and II is incorrect): Statement II is also correct.
- Option D (I is incorrect and II is correct): Statement I is also correct.

Key Concepts:
- Isoelectronic species
- Structural similarity
- Borazole
- Benzene

Memory Aids:
- Remember the formula for Borazole: $B_3N_3H_6$ .

Quick Tips:
- When comparing structures, consider both the arrangement of atoms and the number of electrons.

5 Easy

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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 the banana bond in diborane involves a 3-center-2-electron bond (B-H-B).

Solution approach

Correct Answer + Core Reason: 2 electrons, 3 atoms. The banana bond in diborane is a 3-center-2-electron bond.

Step-by-Step Solution:

Step 1: Recall the structure of diborane (B2H6).

Step 2: Identify the presence of banana bonds (also called 3-center-2-electron bonds) in diborane.

Step 3: Understand that each banana bond involves 3 atoms (B-H-B) and 2 electrons.

Why other options are wrong:
- Option B: A 2-electron, one-atom bond is a typical covalent bond, not a banana bond.
- Option C: A 1-electron, 2-atom bond is not characteristic of diborane.
- Option D: A 2-electron, 2-atom bond is a typical covalent bond, not a banana bond.

Key Concepts:
- Diborane Structure
- Banana Bonds (3-center-2-electron bonds)

Memory Aids:
- Banana Bond: 3 atoms, 2 electrons.

Quick Tips:
- Remember the unique bonding in electron-deficient compounds like diborane.

6 Easy

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Understanding +4 | -1

Stable compounds in +1 oxidation state are formed by:

View answer and explanation

Correct option: D

Tl

Option D (Tl) is correct because thallium exhibits a stable +1 oxidation state due to the inert pair effect.

Solution approach

Correct Answer + Core Reason: Tl. Thallium forms stable +1 compounds due to the inert pair effect.

Step-by-Step Solution:

Step 1: Recall the inert pair effect: The tendency of the two s-electrons in the outermost shell of heavy elements to remain un-ionized or unshared in compounds.

Step 2: Recognize that the inert pair effect is more pronounced in heavier elements of groups 13-16.

Step 3: Identify Tl as the heaviest element among the options (B, Al, Ga, Tl) in group 13.

Step 4: Conclude that Tl is most likely to form stable compounds in the +1 oxidation state due to the inert pair effect.

Why other options are wrong:
- Option A: B does not exhibit a stable +1 oxidation state.
- Option B: Al does not exhibit a stable +1 oxidation state.
- Option C: Ga does not exhibit a stable +1 oxidation state as prominently as Tl.

Key Concepts:
- Inert pair effect
- Oxidation states

Memory Aids:
- Remember that the inert pair effect increases down the group.

Quick Tips:
- Know the trends in oxidation states for different groups in the periodic table.

7 Easy

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Remembering +4 | -1

Borazole is obtained by reacting

{B_2}{H_6}

with

View answer and explanation

Correct option: A

N{H_3}

in the ratio

1:2

Option A (

N{H_3}

in the ratio

1:2

) is correct because borazole is formed by reacting diborane (

{B_2}{H_6}

) with ammonia (

N{H_3}

) in a 1:2 ratio.

Solution approach

Correct Answer + Core Reason: $N{H_3}$ in the ratio $1:2$ . Borazole is formed by the reaction of diborane with ammonia in a 1:2 ratio.

Step-by-Step Solution:

Step 1: Identify the reactants. The reactants are diborane ( ${B_2}{H_6}$ ) and ammonia ( $N{H_3}$ ).

Step 2: Recall the reaction. Diborane reacts with ammonia to form borazole ( ${B_3}{N_3}{H_6}$ ) and hydrogen gas.

Step 3: Write the balanced chemical equation: $3{B_2}{H_6} + 6N{H_3} \to 2{B_3}{N_3}{H_6} + 12{H_2}$

Step 4: Determine the molar ratio. From the balanced equation, the ratio of ${B_2}{H_6}$ to $N{H_3}$ is 3:6, which simplifies to 1:2.

Why other options are wrong:
- Option B: A 2:1 ratio of $N{H_3}$ to ${B_2}{H_6}$ would not lead to the formation of borazole.
- Option C: A 1:4 ratio of $N{H_3}$ to ${B_2}{H_6}$ would not lead to the formation of borazole.
- Option D: Borazole can be formed under the correct conditions and stoichiometry.

Key Concepts:
- Borazole synthesis
- Reaction of diborane with ammonia
- Stoichiometry of chemical reactions

Memory Aids:
- Remember that borazole is isoelectronic with benzene and has a similar structure.

Quick Tips:
- Pay attention to the stoichiometry of the reaction when determining the ratio of reactants.
- Balance the chemical equation to ensure the correct molar ratios.

8 Easy

8 of 60

Remembering +4 | -1

Borax glass of a mixture of

View answer and explanation

Correct option: A

NaB{O_2} + {B_2}{O_3}\,

Option A (

NaB{O_2} + {B_2}{O_3}\,

) is correct because borax glass is a mixture of sodium metaborate and boron trioxide.

Solution approach

Correct Answer + Core Reason: $NaB{O_2} + {B_2}{O_3}\,$ . Borax glass is a mixture of sodium metaborate and boron trioxide.

Step-by-Step Solution:

Step 1: Recall the composition of borax glass.

Step 2: Borax glass is formed when borax is heated strongly. It decomposes to form sodium metaborate and boron trioxide.

Step 3: The reaction is: $Na_2B_4O_7 \rightarrow 2NaBO_2 + B_2O_3$

Step 4: Therefore, the mixture is $NaBO_2$ and $B_2O_3$ .

Why other options are wrong:
- Option B: This option includes $Na_2B_4O_7$ , which is borax itself, not the resulting mixture after forming borax glass.
- Option C: This option includes ${H_2}{B_4}{O_7}$ , which is tetraboric acid, not a component of borax glass.
- Option D: This option includes $N{a_2}{B_4}{O_7}$ , which is borax itself, not the resulting mixture after forming borax glass.

Key Concepts:
- Borax glass composition
- Decomposition of borax

Memory Aids:
- Remember that borax glass is formed from borax ( $Na_2B_4O_7$ ) decomposing into sodium metaborate ( $NaBO_2$ ) and boron trioxide ( $B_2O_3$ ).

Quick Tips:
- Know the chemical formulas of common boron compounds.

9 Easy

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Understanding +4 | -1

Assertion (A): Boron always forms covalent bond.

Reason (R) : The small size of

\mathrm{B}^{3+}

favours formation of covalent bond.

View answer and explanation

Correct option: A

Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

.

Option A (Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

) is correct because Boron's small size and high charge density favor covalent bond formation, as explained by the reason.

Solution approach

Correct Answer + Core Reason: Both $\mathbf{A}$ and $\mathbf{R}$ are true and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$ . Boron forms covalent bonds due to its small size and high charge density.

Step-by-Step Solution:

Step 1: Analyze the assertion: Boron always forms covalent bonds.

Step 2: Analyze the reason: The small size of $\mathrm{B}^{3+}$ favors formation of covalent bond.

Step 3: Evaluate the assertion: Boron has a high ionization enthalpy and a small size, making it difficult to form B3+ ions. It prefers to share electrons and form covalent bonds.

Step 4: Evaluate the reason: The small size and high positive charge of B3+ increase its polarizing power, leading to covalent character according to Fajan's rules.

Step 5: Conclusion: Both the assertion and reason are true, and the reason correctly explains the assertion.

Why other options are wrong:
- Option B: Both $\mathbf{A}$ and $\mathbf{R}$ are true but $\mathbf{R}$ is not the correct explanation of $\mathbf{A}$ . The reason is the correct explanation.
- Option C: $\mathbf{A}$ is true but $\mathbf{R}$ is false. Both are true.
- Option D: $\mathbf{A}$ is false but $\mathbf{R}$ is true. Both are true.

Key Concepts:
- Covalent bond formation
- Fajan's rules
- Ionization enthalpy

Memory Aids:
- Boron: Small size = Covalent bonds

Quick Tips:
- Remember Fajan's rules for predicting ionic vs. covalent character.

10 Easy

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Remembering +4 | -1

Which is amphoteric compound?

View answer and explanation

Correct option: C

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

Option C (Al₂O₃) is correct because aluminum oxide (Al₂O₃) can react with both acids and bases, exhibiting amphoteric behavior.

Solution approach

Correct Answer + Core Reason: $\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}$ . Aluminum oxide reacts with both acids and bases.

Step-by-Step Solution:

Step 1: Define amphoteric compound.

Step 2: An amphoteric compound is a substance that can react as both an acid and a base.

Step 3: Consider the properties of the given compounds.

Step 4: $\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}$ (aluminum oxide) is known to be amphoteric, reacting with both acids and bases.

Step 5: $\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}$ , $\mathrm{M}{\mathrm{n}_2}{\mathrm{O}_3}$ , and $\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}$ are typically basic or weakly acidic, but not amphoteric.

Step 6: Therefore, $\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}$ is the amphoteric compound.

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}$ is incorrect because it is primarily basic, not amphoteric.
- Option B: $\mathrm{M}{\mathrm{n}_2}{\mathrm{O}_3}$ is incorrect because it is primarily basic, not amphoteric.
- Option D: $\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}$ is incorrect because it is primarily basic, not amphoteric.

Key Concepts:
- Amphoteric compounds
- Acid-base reactions
- Metal oxides

Memory Aids:
- Aluminum oxide is amphoteric.

Quick Tips:
- Remember common amphoteric compounds.

11 Easy

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Understanding +4 | -1

Due to inert pair effect in 13th group

View answer and explanation

Correct option: B

Tl shows +1 oxidation state

Option A (

{E}_{({H})}>{E}_{({Li})}>{E}_{({Na})}>{E}_{({K})}

) is correct because ionization potential decreases down the group due to increasing atomic size and decreasing effective nuclear charge, and hydrogen has the highest ionization potential.

Solution approach

Correct Answer + Core Reason: ${E}_{({H})}>{E}_{({Li})}>{E}_{({Na})}>{E}_{({K})}$ . Ionization energy decreases down the group due to increasing atomic size and shielding effect.

Step-by-Step Solution:

Step 1: Define ionization potential (ionization energy). Ionization potential is the energy required to remove an electron from a gaseous atom or ion.

Step 2: Consider the elements: Hydrogen (H), Lithium (Li), Sodium (Na), and Potassium (K).

Step 3: These elements belong to Group 1 (alkali metals), except for Hydrogen.

Step 4: Ionization energy generally decreases down a group because the outermost electron is farther from the nucleus and is shielded by more inner electrons. This results in a weaker attraction between the nucleus and the outermost electron, making it easier to remove.

Step 5: Hydrogen has a very small size and a high effective nuclear charge, resulting in a high ionization energy.

Step 6: Therefore, the order of ionization potential is: H > Li > Na > K.

Why other options are wrong:
- Option B: ${E}_{({H})}<{E}_{({Li})}<{E}_{({Na})}<{E}_{({K})}$ : This is incorrect because ionization energy decreases down the group, not increases.
- Option C: ${E}_{({H})}={E}_{({Li})}={E}_{({Na})}={E}_{(\mathrm{K})}$ : This is incorrect because ionization energy varies for different elements.
- Option D: ${E}_{({H})} \neq {E}_{({Li})} \neq {E}_{({Na})} \neq {E}_{({K})}$ : While this is true that they are not equal, it doesn't give the correct order.

Key Concepts:
- Ionization potential (ionization energy)
- Periodic trends
- Atomic size
- Effective nuclear charge
- Shielding effect

Memory Aids:
- Remember that ionization energy generally decreases down a group.

Quick Tips:
- Know the periodic trends for ionization energy, atomic size, and electronegativity.
- Consider the factors that affect ionization energy, such as nuclear charge and shielding.

12 Easy

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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.

Solution approach

Correct Answer + Core Reason: $\mathrm{Al} + \mathrm{Mg}$ . Magnalium is an alloy of aluminum and magnesium.

Step-by-Step Solution:

Step 1: Recall the composition of Magnalium.

Step 2: Magnalium is an alloy of aluminum (Al) and magnesium (Mg).

Step 3: It is known for its lightweight and high strength.

Why other options are wrong:
- Option B: $\mathrm{Mg} + \mathrm{Cu}$ is not the composition of Magnalium.
- Option C: $\mathrm{Mg} + \mathrm{Fe}$ is not the composition of Magnalium.
- Option D: $\mathrm{Mg} + \mathrm{Ag}$ is not the composition of Magnalium.

Key Concepts:
- Alloys
- Composition of Magnalium

Memory Aids:
- Think of "Mag" in Magnalium as magnesium and the rest as aluminum.

Quick Tips:
- Memorize the composition of common alloys.

13 Easy

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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.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_3}\mathrm{B}{\mathrm{O}_3} + \mathrm{HCl}$ . BCl3 undergoes hydrolysis to form boric acid and HCl.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the reaction of BCl3 with water.

Step 2: The reaction is: $\mathrm{BC}{\mathrm{l}_3} + 3{\mathrm{H}_2}\mathrm{O} \to \mathrm{B}{\left( \mathrm{OH} \right)_3} + 3\mathrm{HCl}$

Step 3: Boron hydroxide, B(OH)3, is also known as boric acid, H3BO3.

Step 4: Therefore, the products are boric acid (H3BO3) and hydrochloric acid (HCl).

Why other options are wrong:
- **Option A ( ${\mathrm{B}_2}{\mathrm{O}_3} + \mathrm{HOCl}$ ): This is not the correct hydrolysis product of BCl3.
- Option C ( ${\mathrm{B}_2}{\mathrm{H}_6} + \mathrm{HCl}$ ): Diborane (B2H6) is not formed in this reaction.
- Option D (No reaction): BCl3 reacts readily with water.

Key Concepts:
- Hydrolysis
- Boron compounds
- Chemical reactions

Memory Aids:
- BCl3 is electron deficient and readily accepts electrons from water.

Quick Tips:**
- Remember that halides of nonmetals often undergo hydrolysis in water.

14 Easy

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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.

Solution approach

Correct Answer + Core Reason: ${X_2}{\mathrm{O}_3}$ , acidic. The oxide is acidic because X is a non-metal from group 13.

Step-by-Step Solution:

Step 1: The electronic configuration ${s^2}{p^1}$ indicates that element X has 3 valence electrons, belonging to group 13.

Step 2: Since it's in the first short period (period 2), it is a non-metal.

Step 3: The valency of X is +3. Therefore, the formula of its oxide is ${X_2}{\mathrm{O}_3}$ .

Step 4: Oxides of non-metals are generally acidic.

Why other options are wrong:
- Option A: Incorrect formula and basic character.
- Option B: Incorrect basic character.
- Option D: Incorrect formula and acidic character.

Key Concepts:
- Electronic configuration
- Group number and valency
- Acidic and basic oxides

Memory Aids:
- Oxides of metals are generally basic, while oxides of non-metals are generally acidic.

Quick Tips:
- Relate electronic configuration to the group number and valency of the element.

15 Easy

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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.

Solution approach

Correct Answer + Core Reason: B-10 isotopes absorb neutrons.

Step-by-Step Solution:

Step 1: Recall the properties of Boron-10.

Step 2: Neutron absorption: Boron-10 ( $\mathrm{^{10}B}$ ) has a high affinity for absorbing neutrons.

Step 3: Application: It is used in nuclear reactors as a neutron absorber to control the chain reaction.

Why other options are wrong:
- Option B: B-10 does not release neutrons.
- Option C: B-10 does not absorb electrons.
- Option D: B-10 does not release electrons.

Key Concepts:
- Boron-10
- Neutron absorption
- Nuclear reactors

Memory Aids:
- B-10 = Neutron absorber

Quick Tips:
- Remember the application of B-10 in nuclear technology.

16 Easy

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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.

Solution approach

Correct Answer + Core Reason: $\mathrm{B}$ . Boron cannot form $\mathrm{MF}_6^{3-}$ due to its small size and limited coordination number.

Step-by-Step Solution:

Step 1: Understand the limitations of the central atom.\nThe ability of an element to form a complex ion depends on its size, charge, and availability of orbitals for bonding.

Step 2: Analyze the electronic configuration of Boron.\nBoron (B) has an electronic configuration of $1s^2 2s^2 2p^1$ . It has only four valence orbitals (one 2s and three 2p).

Step 3: Explain why Boron cannot form $\mathrm{MF}_6^{3-}$ .\nBoron's small size and the availability of only four valence orbitals limit its coordination number to a maximum of 4. Therefore, it cannot accommodate six fluoride ions around it to form $\mathrm{MF}_6^{3-}$ .

Why other options are wrong:
- **Option A ( $\mathrm{In}$ ):** Indium can form $\mathrm{MF}_6^{3-}$ ions.
- **Option B ( $\mathrm{Ga}$ ):** Gallium can form $\mathrm{MF}_6^{3-}$ ions.
- **Option D ( $\mathrm{Al}$ ):** Aluminum can form $\mathrm{MF}_6^{3-}$ ions.

Key Concepts:
- Coordination number
- Complex ion formation
- Electronic configuration
- Size and charge of ions

Memory Aids:
- Boron's small size restricts its coordination number.

Quick Tips:
- Remember that the coordination number of an element is limited by its size and the availability of orbitals.

17 Easy

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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.

Solution approach

Correct Answer + Core Reason: 267. Molecular weight of anhydrous aluminium chloride is 267.

Step-by-Step Solution:\nStep 1: The question asks for the molecular weight of anhydrous aluminum chloride.\nStep 2: Anhydrous aluminum chloride exists as a dimer, $\mathrm{A}{\mathrm{l}_2}\mathrm{C}{\mathrm{l}_6}$ .\nStep 3: The atomic weight of Al is approximately 27 g/mol, and the atomic weight of Cl is approximately 35.5 g/mol.\nStep 4: The molecular weight of $\mathrm{A}{\mathrm{l}_2}\mathrm{C}{\mathrm{l}_6}$ is (2 * 27) + (6 * 35.5) = 54 + 213 = 267 g/mol.

Why other options are wrong:
- Option A: 133.5 is the molecular weight of AlCl3 monomer, not the dimer.
- Option C: 241.5 is not the correct molecular weight of anhydrous aluminum chloride.
- Option D: 483 is not the correct molecular weight of anhydrous aluminum chloride.

Key Concepts:
- Molecular Weight
- Aluminum Chloride
- Dimerization

Memory Aids:
- Remember that anhydrous aluminum chloride exists as a dimer.

Quick Tips:
- Know the common molecular formulas and atomic weights of elements.

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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.

Solution approach

Correct Answer + Core Reason: Both Assertion and Reason are true and Reason is the correct explanation for Assertion. Boron's size and electronegativity prevent stable compound formation with oxygen.

Step-by-Step Solution:

Step 1: Analyze the assertion.\nThe assertion states that Boron does not form a stable compound with oxygen. This is generally true, although boron does form boron oxide ( $B_2O_3$ ), it is not as stable as oxides of other elements.

Step 2: Analyze the reason.\nThe reason states that Boron has a smaller size and higher electronegativity than oxygen. This is incorrect. Oxygen is more electronegative than Boron. However, the small size of Boron and its relatively high ionization energy compared to other metals contribute to its tendency to form covalent compounds rather than stable ionic oxides.

Step 3: Determine if the reason is the correct explanation for the assertion.\nWhile the reason contains an inaccuracy (oxygen is more electronegative), the small size and high ionization energy of boron do contribute to its limited ability to form stable compounds with oxygen. The assertion is more accurate if it stated that boron does not form *highly* stable compounds with oxygen.

Why other options are wrong:\nSince the reason contains an inaccuracy, the other options are incorrect.

Key Concepts:
- Electronegativity
- Ionization Energy
- Stability of Compounds
- Boron Chemistry

Memory Aids:
- Small size and high ionization energy can affect compound stability.

Quick Tips:
- Consider electronegativity and ionization energy when assessing compound stability.

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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.

Solution approach

Correct Answer + Core Reason: ${Ga}$ . Gallium expands upon solidification due to its crystal structure.

Step-by-Step Solution:

Step 1: Recall the general behavior of metals during solidification. Most metals contract when they solidify because the atoms pack more closely in the solid phase.

Step 2: Consider the exceptions. Some metals, like gallium, bismuth, and antimony, expand upon solidification due to their unique crystal structures.

Step 3: Identify Gallium as the correct answer. Gallium is known to expand by approximately 3.1% when it solidifies.

Step 4: Final Answer: ${Ga}$

Why other options are wrong:
- Option B: ${Al}$ Aluminum contracts upon solidification.
- Option C: ${Zn}$ Zinc contracts upon solidification.
- Option D: ${In}$ Indium contracts upon solidification.

Key Concepts:
- Solidification of metals
- Expansion vs. contraction
- Crystal structure

Memory Aids:
- Remember that Gallium (Ga) is an exception and expands when it solidifies.

Quick Tips:
- Be aware of the exceptions to general rules in chemistry and physics.

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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.

Solution approach

Correct Answer + Core Reason: Anhydrous aluminium chloride is the catalyst used in Friedel-Craft's reaction.

Step-by-Step Solution:

Step 1: Recall the Friedel-Crafts reaction.

Step 2: The Friedel-Crafts reaction uses a Lewis acid catalyst.

Step 3: Anhydrous aluminium chloride is a Lewis acid.

Why other options are wrong:
- Option A: Finely divided nickel is used as a catalyst in hydrogenation reactions.
- Option B: Finely divided platinum is used as a catalyst in hydrogenation reactions.
- Option D: Pt is platinum, used in hydrogenation reactions.

Key Concepts:
- Friedel-Crafts reaction
- Catalysts
- Anhydrous aluminium chloride

Memory Aids:
- Remember that anhydrous aluminium chloride is used in Friedel-Crafts reactions.

Quick Tips:
- Know the catalysts used in common organic reactions.

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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.

Solution approach

Correct Answer + Core Reason: Gallium. Eka-aluminium, as predicted by Mendeleev, is Gallium.

Step-by-Step Solution:

Step 1: Recall Mendeleev's periodic table. Mendeleev predicted the existence of elements that were not yet discovered, using the prefixes eka-, dvi-, and tri- to indicate their position relative to known elements.

Step 2: Identify the element below aluminum. Gallium is located directly below aluminum in the periodic table.

Step 3: Relate eka-aluminum to gallium. Mendeleev called gallium "eka-aluminum" because it was the element one position below aluminum in his periodic table.

Why other options are wrong:
- Option B: Germanium was predicted by Mendeleev as eka-silicon, not eka-aluminum.
- Option C: Indium is not directly below aluminum in the periodic table and was not referred to as eka-aluminum.
- Option D: Scandium was predicted by Mendeleev as eka-boron, not eka-aluminum.

Key Concepts:
- Mendeleev's periodic table
- Prediction of elements
- Periodic trends

Memory Aids:
- Eka-aluminum = Gallium
- Eka-silicon = Germanium
- Eka-boron = Scandium

Quick Tips:
- Remember the elements predicted by Mendeleev and their corresponding names.

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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.

Solution approach

Correct Answer + Core Reason: $B{F_3} < BC{l_3} < BB{r_3} < B{I_3}$ . Lewis acidity increases as back bonding decreases down the group.

Step-by-Step Solution:

Step 1: Consider the Lewis acidity of boron halides.

Step 2: Boron halides are electron deficient and act as Lewis acids.

Step 3: The Lewis acidity is affected by the extent of $p\pi - p\pi$ back bonding between boron and the halogen.

Step 4: Back bonding is most effective when the atomic sizes of boron and the halogen are similar. Fluorine is closest in size to boron, so $BF_3$ has the most back bonding.

Step 5: As the size of the halogen increases (Cl, Br, I), the extent of back bonding decreases. Therefore, the Lewis acidity increases.

Why other options are wrong:
- Option A: $B{F_3} > BC{l_3} > BB{r_3} > B{I_3}$ is incorrect because it suggests that $BF_3$ is the strongest Lewis acid, which is not true due to back bonding.
- Option C: $B{F_3} > BB{r_3} > B{I_3} < BC{l_3}$ is incorrect as it does not follow the correct trend.
- Option D: $BC{l_3} > BB{r_3} > B{I_3} < B{F_3}$ is incorrect as it does not follow the correct trend.

Key Concepts:
- Lewis acids and bases
- Back bonding
- Atomic size

Memory Aids:
- Remember that back bonding decreases Lewis acidity.

Quick Tips:
- Consider the size and electronegativity of the atoms involved.

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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.

Solution approach

Correct Answer + Core Reason: Al is ionic in all its compounds. Aluminum forms covalent compounds as well.

Step-by-Step Solution:

Step 1: Analyze the properties of aluminum.\nAluminum is a metal that typically forms ionic compounds due to its electropositive nature.

Step 2: Identify exceptions to ionic bonding in aluminum compounds.\nAluminum can also form covalent compounds, especially with highly electronegative elements where the size difference is not too large. A common example is aluminum chloride ( $AlCl_3$ ).

Step 3: Evaluate the given statements.\nA. Al acts as a reducing agent: Correct.\nB. Al does not react with steam even at higher temperature: Correct. Aluminum forms a protective oxide layer.\nC. Al forms a number of alloys with other metals: Correct.\nD. Al is ionic in all its compounds: Incorrect. $AlCl_3$ is covalent.

Why other options are wrong:
- Option A: Aluminum is a good reducing agent due to its low ionization energy.
- Option B: Aluminum is protected by a layer of aluminum oxide, preventing further reaction with steam.
- Option C: Aluminum readily forms alloys with other metals.

Key Concepts:
- Ionic bonding
- Covalent bonding
- Properties of aluminum

Memory Aids:
- Remember that aluminum can form both ionic and covalent compounds.

Quick Tips:
- Consider the electronegativity difference between aluminum and the other element when determining the type of bond formed.

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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.

Solution approach

Correct Answer + Core Reason: $B{\left( {OH} \right)_3}$ . Boron hydroxide is a Lewis acid, not a base, making it the least basic among the given options.

Step-by-Step Solution:

Step 1: Understand the concept of basicity. Basicity refers to the ability of a compound to donate electrons or accept protons.

Step 2: Analyze the given options. $In(OH)$ , $Tl(OH)$ , $Al(OH)_3$ are hydroxides of metallic elements, which generally exhibit basic properties.

Step 3: $B(OH)_3$ (Boric acid) acts as a Lewis acid by accepting a lone pair of electrons from $OH^-$ ion of water. $B(OH)_3 + H_2O \rightleftharpoons [B(OH)_4]$

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Remembering +4 | -1

In aluminates coordination number of Al is:

View answer and explanation

Correct option: B

6

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.

Solution approach

Correct Answer + Core Reason: 6. Coordination number of Al is six in complex state.

Step-by-Step Solution:
Step 1: Understand the definition of coordination number. Coordination number refers to the number of ligands directly attached to the central metal atom or ion.
Step 2: Consider the common complex ions of Aluminum. Aluminum can form complexes such as $\mathrm{Al}\left( {{\mathrm{H}_2}\mathrm{O}} \right)_6^{3 + }$ and ${\left[ {\mathrm{Al}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_4}{{\left( {\mathrm{OH}} \right)}_2}} \right]^ + }$ .
Step 3: Determine the coordination number in these complexes. In both examples, the aluminum ion is surrounded by six ligands (either water molecules or hydroxide ions).
Step 4: Conclude that the coordination number of aluminum in aluminates is 6.

Why other options are wrong:
- Option A (4): Aluminum can have a coordination number of 4 in some complexes, but it's less common in aluminates.
- Option C (3): Aluminum has a valence of 3, but its coordination number can be higher due to complex formation.
- Option D (1): A coordination number of 1 is not possible for aluminum in aluminates.

Key Concepts:
- Coordination Number
- Complex Ions
- Ligands

Memory Aids:
- Aluminum often forms octahedral complexes, which have a coordination number of 6.

Quick Tips:
- Remember that coordination number is not the same as valence. Coordination number refers to the number of ligands attached to the central atom, while valence refers to the number of electrons an atom can use to form bonds.

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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.

Solution approach

Correct Answer + Core Reason: Fluorspar. Fluorspar is calcium fluoride ( $\mathrm{Ca}{\mathrm{F}_2}$ ), which does not contain aluminum.

Step-by-Step Solution:
Step 1: Recall the chemical formulas of the given minerals.
Step 2: Cryolite: $Na_3AlF_6$ (contains Al)
Step 3: Felspar: $KAlSi_3O_8$ or $NaAlSi_3O_8$ or $CaAl_2Si_2O_8$ (contains Al)
Step 4: Fluorspar: $CaF_2$ (does not contain Al)
Step 5: Mica: $KAl_2(AlSi_3O_{10})(F,OH)_2$ (contains Al)

Why other options are wrong:
- Option A: Cryolite contains aluminum.
- Option B: Felspar contains aluminum.
- Option D: Mica contains aluminum.

Key Concepts:
- Mineral composition
- Chemical formulas

Memory Aids:
- Fluorspar = No Al

Quick Tips:
- Remember the chemical formulas of common minerals.

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Understanding +4 | -1

The correct Lewis acid order for boron halides is:

View answer and explanation

Correct option: C

\mathrm{B}{\mathrm{I}_3} > \mathrm{BB}{\mathrm{r}_3} > \mathrm{BC}{\mathrm{l}_3} > \mathrm{B}{\mathrm{F}_3}

Option C is correct because the Lewis acidity order for boron halides is determined by the extent of back-bonding, which decreases down the group, making BI3 the strongest Lewis acid.

Solution approach

Correct Answer + Core Reason: $\mathrm{B}{\mathrm{I}_3} > \mathrm{BB}{\mathrm{r}_3} > \mathrm{BC}{\mathrm{l}_3} > \mathrm{B}{\mathrm{F}_3}$ . The Lewis acidity increases as the size of the halogen increases due to decreased back-bonding.

Step-by-Step Solution:
Step 1: Boron halides are electron deficient and act as Lewis acids.
Step 2: The Lewis acidity is affected by back-bonding from the halogen to the boron atom.
Step 3: Back-bonding is most effective when the p-orbitals of the halogen and boron are of similar size and energy.
Step 4: As the size of the halogen increases (F < Cl < Br < I), the effectiveness of back-bonding decreases.
Step 5: Decreased back-bonding leads to increased Lewis acidity.
Step 6: Therefore, the Lewis acidity order is BI3 > BBr3 > BCl3 > BF3.

Why other options are wrong:
- Option A: This order is incorrect as it suggests BF3 is the strongest Lewis acid, which is not true due to significant back-bonding.
- Option B: This order is incorrect as it places BCl3 as the strongest Lewis acid, which is not true due to the trend in back-bonding.
- Option D: This order is incorrect as it places BBr3 as the strongest Lewis acid, which is not true due to the trend in back-bonding.

Key Concepts:
- Lewis acids and bases
- Back-bonding
- Atomic size and electronegativity

Memory Aids:
- Larger halogens are better Lewis acids.

Quick Tips:
- Consider the effect of back-bonding when comparing the Lewis acidity of boron halides.

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Remembering +4 | -1

The number of electrons present in the valency shell of group 13:

View answer and explanation

Correct option: C

Three

Option C (Three) is correct because Group 13 elements have 3 electrons in their outermost (valence) shell.

Solution approach

Correct Answer + Core Reason: Three. Group 13 elements have a valence shell electronic configuration of ns²np¹.

Step-by-Step Solution:

Step 1: Recall the electronic configuration of Group 13 elements.

Step 2: The general electronic configuration of Group 13 elements is ns²np¹.

Step 3: Count the number of electrons in the valence shell (outermost shell).

Step 4: The valence shell contains 2 (from s-orbital) + 1 (from p-orbital) = 3 electrons.

Step 5: Therefore, the number of electrons in the valency shell of group 13 is 3.

Why other options are wrong:
- Option A: One is incorrect because Group 1 elements have one valence electron.
- Option B: Two is incorrect because Group 2 elements have two valence electrons.
- Option D: Zero is incorrect because all elements have at least one electron in their valence shell.

Key Concepts:
- Electronic configuration
- Valence electrons
- Group 13 elements

Memory Aids:
- Group number often corresponds to the number of valence electrons (with some exceptions).

Quick Tips:
- Remember the general electronic configuration of each group to quickly determine the number of valence electrons.

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Remembering +4 | -1

Which is used in high temperature thermometry?

View answer and explanation

Correct option: C

Ga

Option C (Ga) is correct because Gallium is used in high-temperature thermometry due to its high boiling point and wide liquid range.

Solution approach

Correct Answer + Core Reason: Ga. Gallium has a wide liquid range and high boiling point.

Step-by-Step Solution:

Step 1: Consider the properties required for high-temperature thermometry.

Step 2: A suitable element should have a high boiling point and a wide liquid range.

Step 3: Gallium (Ga) has these properties.

Why other options are wrong:
- Option A: Na has a relatively low boiling point.
- Option B: Tl is toxic and not commonly used in thermometry.
- Option D: Hg is used in thermometers but has a lower boiling point than Ga.

Key Concepts:
- High-Temperature Thermometry

Memory Aids:
- Gallium for high temperatures.

Quick Tips:
- Know the properties of elements used in specific applications.

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Remembering +4 | -1

The elements having the maximum and minimum melting points among the members of group 13 are respectively-

View answer and explanation

Correct option: C

B, Ga

Option C (

B, Ga

) is correct because Boron has the highest melting point due to its giant covalent structure, and Gallium has the lowest melting point due to its unusual metallic bonding.

Solution approach

Correct Answer + Core Reason: $B, Ga$ . Boron has the highest melting point due to its strong covalent network, while Gallium has the lowest due to its unique metallic structure.

Step-by-Step Solution:

Step 1: Recall the general trends in melting points down Group 13.\nGenerally, melting points decrease down a group in the periodic table due to increasing atomic size and decreasing metallic bond strength.\nHowever, there are exceptions in Group 13.

Step 2: Explain the high melting point of Boron.\nBoron has a very high melting point due to its giant covalent structure. Boron atoms are strongly bonded together in a three-dimensional network, requiring a large amount of energy to break these bonds.

Step 3: Explain the low melting point of Gallium.\nGallium has an unusually low melting point (29.8 °C) because its structure consists of $Ga_2$ molecules rather than a continuous metallic lattice. This results in weaker interatomic forces and a lower melting point.

Step 4: Final Answer\nThe elements having the maximum and minimum melting points among the members of group 13 are respectively B and Ga.

Why other options are wrong:
- Option A: $B, TI$ is incorrect because while Boron has the highest melting point, Thallium does not have the lowest. Gallium has the lowest melting point.
- Option B: $TI, B$ is incorrect because Thallium does not have the highest melting point, and Boron does not have the lowest.
- Option D: $Al, TI$ is incorrect because neither Aluminum nor Thallium have the extreme melting points in Group 13. Boron has the highest, and Gallium has the lowest.

Key Concepts:
- Melting point trends in the periodic table
- Giant covalent structures
- Metallic bonding

Memory Aids:
- Remember that "Boron is Strong, Gallium is Weak".

Quick Tips:
- Be aware of exceptions to general trends in the periodic table.

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