13 TH GROUP ELEMENTS (P-BLOCK) MCQs for JEE CHEMISTRY — Easy, Page 2

31 Easy

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Remembering +4 | -1

{\mathrm{B}_2}{\mathrm{O}_3}

is:

View answer and explanation

Correct option: C

Acidic

Option C (Acidic) is correct because boron is a non-metal, and the oxides of non-metals are generally acidic in nature.

Solution approach

Correct Answer + Core Reason: Acidic. Oxides of non-metals are generally acidic.

Step-by-Step Solution:

Step 1: Identify the question: The question asks whether B₂O₃ is ionic, basic, acidic, or amphoteric.

Step 2: Recall the nature of boron: Boron (B) is a non-metal.

Step 3: Recall the properties of non-metal oxides: Oxides of non-metals are generally acidic.

Step 4: Conclude that B₂O₃ is acidic.

Why other options are wrong:
- Option A: Ionic is incorrect because B₂O₃ is a covalent compound.
- Option B: Basic is incorrect because B₂O₃ is an oxide of a non-metal.
- Option D: Amphoteric is incorrect because B₂O₃ primarily exhibits acidic properties.

Key Concepts:
- Oxides
- Acidic oxides
- Basic oxides
- Amphoteric oxides
- Non-metals

Memory Aids:
- Non-metal oxides = Acidic

Quick Tips:
- Remember the general properties of metal and non-metal oxides.

32 Easy

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The relationship between first, second and third ionisation enthalpies of each group-13 element is

View answer and explanation

Correct option: B

\Delta_{{i}} {H}_1<\Delta_{{i}} {H}_2<\Delta_{{i}} {H}_3

Option B is correct because each successive ionization requires more energy as electrons are removed from an increasingly positive ion.

Solution approach

Correct Answer + Core Reason: $\Delta_{{i}} {H}_1<\Delta_{{i}} {H}_2<\Delta_{{i}} {H}_3$ . Removing an electron from a positive ion requires more energy than removing it from a neutral atom.

Step-by-Step Solution:

Step 1: Define Ionization Enthalpy: Ionization enthalpy is the energy required to remove an electron from a gaseous atom or ion.

Step 2: Consider First Ionization Enthalpy ( $\Delta_{{i}} {H}_1$ ): This is the energy required to remove the first electron from a neutral atom.

Step 3: Consider Second Ionization Enthalpy ( $\Delta_{{i}} {H}_2$ ): This is the energy required to remove the second electron from a unipositive ion.

Step 4: Consider Third Ionization Enthalpy ( $\Delta_{{i}} {H}_3$ ): This is the energy required to remove the third electron from a dipositive ion.

Step 5: Explain the trend: As electrons are removed, the effective nuclear charge increases, making it more difficult to remove subsequent electrons. Therefore, $\Delta_{{i}} {H}_1 < \Delta_{{i}} {H}_2 < \Delta_{{i}} {H}_3$ .

Why other options are wrong:
- Option A: This is incorrect because ionization enthalpy increases with each successive ionization.
- Option C: This is incorrect because ionization enthalpy increases with each successive ionization.
- Option D: This is incorrect because ionization enthalpy increases with each successive ionization.

Key Concepts:
- Ionization Enthalpy
- Effective Nuclear Charge
- Periodic Trends

Memory Aids:
- It's always harder to remove an electron from a positive ion.

Quick Tips:
- Remember that ionization enthalpy always increases with each successive ionization.

33 Easy

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The maximum number of atoms is lying in one plan for

\mathrm{B}_2 \mathrm{H}_6

is

View answer and explanation

Correct option: C

6

The correct answer is C (6) because in

\mathrm{B}_2 \mathrm{H}_6

, there are four terminal hydrogen atoms and two boron atoms that lie in one plane.

Solution approach

Correct Answer + Core Reason: 6. In $\mathrm{B}_2 \mathrm{H}_6$ , four terminal hydrogen atoms and two boron atoms lie in one plane.

Step-by-Step Solution:

Step 1: Draw the structure of $\mathrm{B}_2 \mathrm{H}_6$ (diborane).

Step 2: Identify the planar atoms. Diborane has a unique structure with two bridging hydrogen atoms. The four terminal hydrogen atoms and the two boron atoms lie in one plane.

Step 3: Count the number of atoms in the plane: 4 terminal H + 2 B = 6 atoms.

Why other options are wrong:
- Option A: 4 - Incorrect. Only considering the boron atoms and some of the hydrogen atoms.
- Option B: 5 - Incorrect. Not the correct number of atoms in the plane.
- Option D: 8 - Incorrect. This would include all the atoms, but not all are in the same plane.

Key Concepts:
- Structure of diborane ( $\mathrm{B}_2 \mathrm{H}_6$ )
- Planarity
- Bonding in diborane

Memory Aids:
- Remember the banana bond structure of diborane.

Quick Tips:
- Draw the structure to visualize the arrangement of atoms.

34 Easy

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Assertion (A) : In water, orthoboric acid behaves as a weak monobasic acid.

Reason (R) : In water orthoboric acid acts as a proton donor.

View answer and explanation

Correct option: C

\mathbf{A}

is true but

\mathbf{R}

is false.

Option C is correct because orthoboric acid (H3BO3) acts as a Lewis acid, accepting OH- from water, not as a proton donor. Therefore, the reason is false.

Solution approach

Correct Answer + Core Reason: $\mathbf{A}$ is true but $\mathbf{R}$ is false. Orthoboric acid is a weak monobasic acid, but it doesn't act as a proton donor.

Step-by-Step Solution:

Step 1: Recall the properties of orthoboric acid: It is a weak monobasic acid.

Step 2: Explain its behavior in water: It acts as a Lewis acid, accepting OH- from water.

Step 3: Write the reaction: $B(OH)_3 + 2H_2O \rightleftharpoons [B(OH)_4]^- + H_3O^+$

Step 4: Conclude that the assertion is true, but the reason is false.

Why other options are wrong:
- Option A: The reason is false.
- Option B: The reason is false.
- Option D: The assertion is true.

Key Concepts:
- Orthoboric acid
- Lewis acid
- Bronsted acid

Memory Aids:
- Boric acid accepts, doesn't donate.

Quick Tips:
- Remember that boric acid acts as a Lewis acid.

35 Easy

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Pyrex glass is a mixture of :

View answer and explanation

Correct option: A

Sodium borosilicate and barium borosilicate

Option A (Sodium borosilicate and barium borosilicate) is correct because Pyrex glass is primarily composed of sodium borosilicate and barium borosilicate, which gives it high thermal resistance.

Solution approach

Correct Answer + Core Reason: Pyrex glass is a mixture of Sodium borosilicate and barium borosilicate.

Step-by-Step Solution:
Step 1: Recall the composition of Pyrex glass.
Step 2: Pyrex glass is primarily composed of sodium borosilicate and barium borosilicate.
Step 3: This composition gives it high thermal resistance and chemical stability.

Why other options are wrong:
- Option B: Sodium silicate and calcium silicate are components of soda-lime glass.
- Option C: Sodium silicate and lead silicate are components of lead glass (flint glass).
- Option D: Sodium silicate and aluminium borosilicate is not the correct composition of Pyrex glass.

Key Concepts:
- Glass Composition
- Pyrex Glass

Memory Aids:
- Remember that Pyrex glass contains borosilicates.

Quick Tips:
- Memorize the compositions of different types of glass.

36 Easy

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Borax bead test is used to identify the Meta borates of transition metals. It is based on

View answer and explanation

Correct option: A

These Meta borates have characteristic colours

Option A is correct because the borax bead test relies on the characteristic colors of the meta-borates formed by transition metals.

Solution approach

Correct Answer + Core Reason: These Meta borates have characteristic colours. The borax bead test identifies transition metals based on the colors of their meta-borates.

Step-by-Step Solution:
Step 1: The borax bead test involves heating borax with a metal salt.
Step 2: This forms a glassy bead containing metal meta-borates.
Step 3: Different transition metals produce meta-borates with distinct colors.
Step 4: These characteristic colors allow for the identification of the metal.

Why other options are wrong:
- Option B: Boiling points are not relevant to the borax bead test.
- Option C: Melting points are not the primary factor in the borax bead test; color is.
- Option D: While melting points might play a minor role, the characteristic colors are the main identification factor.

Key Concepts:
- Borax bead test
- Transition metal chemistry
- Qualitative analysis

Memory Aids:
- Remember that the borax bead test is all about COLOR!

Quick Tips:
- Focus on the color changes observed during the test.

37 Easy

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Colour of the bead in borax bead test is mainly due to the formation of:

View answer and explanation

Correct option: C

metal metaborates

Option C (metal metaborates) is correct because the color in the borax bead test is due to the formation of different metal metaborates.

Solution approach

Correct Answer + Core Reason: metal metaborates. Metal metaborates impart characteristic colors to the bead.

Step-by-Step Solution:

Step 1: Recall the principle of the borax bead test.

Step 2: Understand that metal salts react with borax to form metal metaborates.

Step 3: Recognize that these metal metaborates have characteristic colors.

Why other options are wrong:
- Option A: Metal oxides may be present, but the color is primarily due to the metaborates.
- Option B: Boron oxide is a component of the bead, but it doesn't directly impart color.
- Option D: Elemental boron is not formed in this test.

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

Memory Aids:
- Borax + Metal = Colored Metaborate

Quick Tips:
- Remember the colors associated with different metal metaborates in the borax bead test.

38 Easy

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Which of the following does not have a tetrahedral structure?

View answer and explanation

Correct option: A

\mathrm{B}{\mathrm{H}_3}

Option A (

\mathrm{B}{\mathrm{H}_3}

) is correct because BH3 has a trigonal planar structure, while the other options have tetrahedral structures.

Solution approach

Correct Answer + Core Reason: $\mathrm{B}{\mathrm{H}_3}$ . BH3 has a trigonal planar structure, not tetrahedral.

Step-by-Step Solution:

Step 1: Determine the hybridization and structure of each species:
- A: $\mathrm{B}{\mathrm{H}_3}$ : Boron has 3 valence electrons. It forms 3 sigma bonds with 3 hydrogen atoms. The hybridization is $sp^2$ , and the structure is trigonal planar.
- B: $\mathrm{NH}_4^ +$ : Nitrogen has 5 valence electrons. It forms 4 sigma bonds with 4 hydrogen atoms and has a positive charge. The hybridization is $sp^3$ , and the structure is tetrahedral.
- C: $\mathrm{BH}_4^ -$ : Boron has 3 valence electrons. It forms 4 sigma bonds with 4 hydrogen atoms and has a negative charge. The hybridization is $sp^3$ , and the structure is tetrahedral.
- D: $\mathrm{C}{\mathrm{H}_4}$ : Carbon has 4 valence electrons. It forms 4 sigma bonds with 4 hydrogen atoms. The hybridization is $sp^3$ , and the structure is tetrahedral.

Step 2: Identify the species that does not have a tetrahedral structure: $\mathrm{B}{\mathrm{H}_3}$ .

Why other options are wrong:
- **Option B ( $\mathrm{NH}_4^ +$ ): Ammonium ion has a tetrahedral structure.
- Option C ( $\mathrm{BH}_4^ -$ ): Borohydride ion has a tetrahedral structure.
- Option D ( $\mathrm{C}{\mathrm{H}_4}$ ): Methane has a tetrahedral structure.

Key Concepts:
- Molecular Structure
- Hybridization
- Tetrahedral Geometry
- Trigonal Planar Geometry

Memory Aids:**
- $sp^3$ : Tetrahedral
- $sp^2$ : Trigonal Planar

Quick Tips:
- Determine the hybridization to predict the molecular geometry.

39 Easy

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The formula of kernite (or) resorite is

View answer and explanation

Correct option: A

N{a_2}{B_4}{O_7}.4{H_2}O

Option A (

N{a_2}{B_4}{O_7}.4{H_2}O

) is correct because it is the chemical formula for Kernite (or) resorite.

Solution approach

Correct Answer + Core Reason: $N{a_2}{B_4}{O_7}.4{H_2}O$ . This is the correct chemical formula for Kernite.

Step-by-Step Solution:
Step 1: Identify Kernite (or resorite).
Step 2: Recall the chemical formula of Kernite. Kernite is a hydrated sodium borate mineral.
Step 3: Match the formula. The chemical formula is $N{a_2}{B_4}{O_7}.4{H_2}O$ .

Why other options are wrong:
- Option B: $N{a_3}B{O_3}.4{H_2}O$ is not the formula for Kernite.
- Option C: $N{a_2}{B_4}{O_7}.10{H_2}O$ is the formula for Borax, not Kernite.
- Option D: $N{a_3}B{O_3}.10{H_2}O$ is not a recognized mineral formula.

Key Concepts:
- Mineral formulas
- Hydrated borate minerals

Memory Aids:
- Associate Kernite with the tetrahydrate form of sodium borate.

Quick Tips:
- Memorize common mineral formulas, especially those of ores.

40 Easy

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Which of the following shows variable valency?

View answer and explanation

Correct option: C

Tl

Option C (Tl) is correct because Thallium (Tl) exhibits variable valency due to the inert pair effect.

Solution approach

Correct Answer + Core Reason: Tl. Thallium shows variable valency due to the inert pair effect.

Step-by-Step Solution:
Step 1: Define variable valency: Variable valency is the ability of an element to exhibit more than one oxidation state.
Step 2: Consider the elements:
- B (Boron) typically exhibits a +3 oxidation state.
- Al (Aluminum) typically exhibits a +3 oxidation state.
- Tl (Thallium) exhibits +1 and +3 oxidation states due to the inert pair effect.
Step 3: Conclude: Thallium shows variable valency.

Why other options are wrong:
- Option A: Boron typically exhibits only one oxidation state (+3).
- Option B: Aluminum typically exhibits only one oxidation state (+3).
- Option D: This option is incorrect because Tl does show variable valency.

Key Concepts:
- Variable valency
- Inert pair effect
- Oxidation states

Memory Aids:
- Inert pair effect: The tendency of the two s-electrons in the outermost shell of heavy elements to remain unshared.

Quick Tips:
- Consider the electronic configuration and the inert pair effect when determining variable valency.

41 Easy

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One that marks the paper like lead is:

View answer and explanation

Correct option: D

Tl

Option D (Tl) is correct because Thallium (Tl) is known to leave a mark on paper similar to lead, due to its softness and metallic properties.

Solution approach

Correct Answer + Core Reason: Tl. Tl (Thallium) has marking nature.

Step-by-Step Solution:

Step 1: Consider the properties of the given elements.

Step 2: Ga (Gallium) - Soft metal, but doesn't typically mark paper like lead.

Step 3: Ti (Titanium) - Hard, strong metal, not used for marking.

Step 4: B (Boron) - Non-metal, doesn't mark paper.

Step 5: Tl (Thallium) - Soft metal that can leave a mark on paper.

Why other options are wrong:
- Option A: Ga - Incorrect. Gallium doesn't mark paper like lead.
- Option B: Ti - Incorrect. Titanium is too hard to mark paper.
- Option C: B - Incorrect. Boron is a non-metal and doesn't mark paper.

Key Concepts:
- Properties of elements
- Metallic properties

Memory Aids:
- Think of Thallium as being similar to Lead in some of its properties.

Quick Tips:
- Know the general properties of common elements.

42 Easy

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When orthoboric acid

\left( {{\mathrm{H}_3}\mathrm{B}{\mathrm{O}_3}} \right)

is heated the residue left is:

View answer and explanation

Correct option: C

Boric anhydride

Option C (Boric anhydride) is correct because orthoboric acid upon heating yields boric anhydride and water.

Solution approach

Correct Answer + Core Reason: Boric anhydride. Heating orthoboric acid results in the formation of boric anhydride.

Step-by-Step Solution:

Step 1: Write the chemical equation for the thermal decomposition of orthoboric acid.

Step 2: $2{\mathrm{H}_3}\mathrm{B}{\mathrm{O}_3} \to {\mathrm{B}_2}{\mathrm{O}_3} + 3{\mathrm{H}_2}\mathrm{O}$

Step 3: Identify the product formed.

Step 4: The residue left is Boric anhydride ( ${\mathrm{B}_2}{\mathrm{O}_3}$ ).

Why other options are wrong:
- Option A: Boron is not directly formed from heating orthoboric acid.
- Option B: Metaboric acid is an intermediate product formed at lower temperatures, not the final residue.
- Option D: Borax is a sodium borate mineral, not a product of heating orthoboric acid.

Key Concepts:
- Thermal decomposition of orthoboric acid
- Formation of boric anhydride

Memory Aids:
- Remember the formula for orthoboric acid and boric anhydride.

Quick Tips:
- Know the reactions of boric acid and its derivatives.

43 Easy

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The group 13 element that is liquid during summer and used for measuring high temperature is

View answer and explanation

Correct option: C

Gallium

Option C (Gallium) is correct because Gallium has a low melting point (29.7°C) and is used in high-temperature thermometers.

Solution approach

Correct Answer + Core Reason: Gallium. Gallium has a low melting point and is used for measuring high temperatures.

Step-by-Step Solution:

Step 1: Recall the properties of group 13 elements.

Step 2: Boron is a non-metal.

Step 3: Aluminium is a metal with a high melting point.

Step 4: Gallium has a low melting point (29.7°C) and is liquid near room temperature.

Step 5: Indium has a higher melting point than Gallium.

Why other options are wrong:
- Option A: Boron is not a liquid during summer.
- Option B: Aluminium is not a liquid during summer and has a high melting point.
- Option C: Gallium is the correct answer.
- Option D: Indium has a higher melting point than Gallium and is not typically liquid during summer.

Key Concepts:
- Properties of group 13 elements
- Melting points of metals

Memory Aids:
- Gallium melts in your hand.

Quick Tips:
- Remember the melting points of common metals.

44 Easy

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Which is used as disinfectant?

View answer and explanation

Correct option: A

Boric acid

Option A (Boric acid) is correct because boric acid is commonly used as a mild disinfectant.

Solution approach

Correct Answer + Core Reason: Boric acid. Boric acid is used as a disinfectant.

Step-by-Step Solution:
Step 1: Recall common disinfectants.
Step 2: Boric acid is a mild antiseptic and is used in eyewash solutions.

Why other options are wrong:
- Option B: Sulphuric acid: Sulphuric acid is a strong acid and is corrosive, not used as a disinfectant.
- Option C: Phosphorus acid: Phosphorus acid is not commonly used as a disinfectant.
- Option D: Phosphoric acid: Phosphoric acid is not commonly used as a disinfectant.

Key Concepts:
- Disinfectants
- Acid properties

Memory Aids:
- Remember boric acid as a mild antiseptic.

Quick Tips:
- Know common disinfectants and their uses.

45 Easy

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Which is used as mordant?

View answer and explanation

Correct option: C

Alum

Option C (Alum) is correct because alum is commonly used as a mordant in dyeing processes.

Solution approach

Correct Answer + Core Reason: Alum. Alum is used as a mordant to fix dyes to fabrics.

Step-by-Step Solution:

Step 1: A mordant is a substance used to set dyes on fabrics or tissue sections.

Step 2: Alum (potassium aluminum sulfate) is a common mordant.

Step 3: Alum helps the dye bind to the fabric fibers, making the color more permanent.

Why other options are wrong:
- Option A: $\mathrm{AlC}{\mathrm{l}_3}$ is a Lewis acid and is not typically used as a mordant.
- Option B: $\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3}$ (Aluminum sulfate) can be used as a mordant, but Alum is a more common and specific term.
- Option D: $\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}$ (Aluminum oxide) is an abrasive and is not used as a mordant.

Key Concepts:
- Mordant
- Dyeing process
- Chemical bonding

Memory Aids:
- Alum helps colors stick like glue.

Quick Tips:
- Alum is a common mordant, remember its use in dyeing.

46 Easy

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Coordination number of aluminium is

View answer and explanation

Correct option: B

6

Option B (6) is correct because Aluminum can form complexes where it is surrounded by six ligands, resulting in a coordination number of 6.

Solution approach

Correct Answer + Core Reason: 6. Coordination number of Aluminum can be 6 due to its electronic configuration and ability to form complexes.

Step-by-Step Solution:
Step 1: Write the electronic configuration of Aluminum. $\mathrm{Al}\left( {13} \right) = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^1}$
Step 2: Determine the number of orbitals available for bonding. Aluminum has vacant d-orbitals that can participate in bonding, allowing it to accommodate more than three ligands.
Step 3: Recognize that Aluminum can form complexes with a coordination number of 6, such as $[AlF_6]^{3-}$ .

Why other options are wrong:
- Option A (8): Aluminum does not typically achieve a coordination number of 8.
- Option C (12): A coordination number of 12 is very high and not typical for Aluminum.
- Option D (4): While Aluminum can have a coordination number of 4, 6 is more common in many complexes.

Key Concepts:
- Coordination Number
- Electronic Configuration
- Complex Formation

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

Quick Tips:
- Remember that the coordination number depends on the size and charge of the metal ion and the ligands.

47 Easy

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The two type of bonds present in

{\mathrm{B}_2}{\mathrm{H}_6}

are covalent and……

View answer and explanation

Correct option: C

Hydrogen bridge

Option C (Hydrogen bridge) is correct because diborane (

B_2H_6

) contains both covalent bonds and three-center two-electron (3c-2e) bonds, also known as hydrogen bridge bonds.

Solution approach

Correct Answer + Core Reason: Hydrogen bridge. $B_2H_6$ contains covalent and hydrogen bridge bonds.

Step-by-Step Solution:

Step 1: Describe the structure of diborane ( $B_2H_6$ ). Diborane has a unique structure with two boron atoms and six hydrogen atoms. It does not have a simple ethane-like structure.

Step 2: Identify the types of bonds in diborane. Diborane has four terminal B-H covalent bonds and two bridging B-H-B bonds.

Step 3: Explain the nature of the bridging B-H-B bonds. These are three-center two-electron bonds, also known as hydrogen bridge bonds or banana bonds. Each B-H-B bond involves a hydrogen atom bridging two boron atoms, with only two electrons shared among the three atoms.

Step 4: Distinguish between covalent and hydrogen bridge bonds. Covalent bonds are two-center two-electron bonds, while hydrogen bridge bonds are three-center two-electron bonds.

Why other options are wrong:
- Option A: Ionic bonds are not present in diborane. The bonds are primarily covalent and hydrogen bridge bonds.
- Option B: Coordinate bonds are not the primary type of bond in diborane, although the bonding can be described using molecular orbital theory.
- Option D: Hydrogen bridge bonds are present in diborane, so "None of these" is incorrect.

Key Concepts:
- Structure of diborane ( $B_2H_6$ )
- Covalent bonds
- Hydrogen bridge bonds (3c-2e bonds)

Memory Aids:
- Remember that diborane has a unique structure with bridging hydrogen atoms.

Quick Tips:
- Always consider the possibility of unusual bonding arrangements in electron-deficient compounds.

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Which one is explosive?

View answer and explanation

Correct option: C

\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_3} + \mathrm{Al}\;\mathrm{powder}

Option C (

\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_3} + \mathrm{Al}\;\mathrm{powder}

) is correct because ammonium nitrate mixed with aluminum powder is a well-known explosive.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_3} + \mathrm{Al}\;\mathrm{powder}$ . Ammonium nitrate and aluminum powder mixtures are used as explosives.

Step-by-Step Solution:

Step 1: Identify the requirement: Find the explosive substance from the options.

Step 2: Recall the properties of different chemical compounds and mixtures.

Step 3: Ammonium nitrate ( $\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_3}$ ) mixed with aluminum powder is a known explosive material.

Why other options are wrong:
- Option A: $\mathrm{PC}{\mathrm{l}_5}$ is not typically used as an explosive; it is a chlorinating agent.
- Option B: $\mathrm{Pb}{\left( {\mathrm{N}{\mathrm{O}_3}} \right)_2}$ is an oxidizing agent but not a primary explosive.
- Option D: ${\mathrm{C}_6}{\mathrm{H}_5}\mathrm{N}{\mathrm{O}_2}$ (nitrobenzene) is a chemical compound, but it is not as explosive as ammonium nitrate and aluminum powder mixture.

Key Concepts:
- Explosives
- Chemical Properties

Memory Aids:
- Ammonium nitrate with aluminum powder is a common explosive composition.

Quick Tips:
- Remember common explosive compounds and mixtures.

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Borax is prepared by treating colemanite with:

View answer and explanation

Correct option: C

\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}

Option C (

\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}

) is correct because Borax is prepared by treating colemanite with sodium carbonate.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}$ . Borax is prepared by treating colemanite with sodium carbonate.

Step-by-Step Solution:
Step 1: Recall the preparation of borax.
Step 2: Borax is prepared by treating colemanite ( $\mathrm{C}{\mathrm{a}_2}{\mathrm{B}_6}{\mathrm{O}_{11}}$ ) with sodium carbonate ( $\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}$ ).
Step 3: The reaction is: $\mathrm{C}{\mathrm{a}_2}{\mathrm{B}_6}{\mathrm{O}_{11}} + 2\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3} \to 2\mathrm{CaC}{\mathrm{O}_3} \downarrow + \mathrm{N}{\mathrm{a}_2}{\mathrm{B}_4}{\mathrm{O}_7} + 2\mathrm{NaB}{\mathrm{O}_2}$

Why other options are wrong:
- Option A: $\mathrm{NaN}{\mathrm{O}_3}$ is sodium nitrate and is not used to prepare borax from colemanite.
- Option B: $NaCl$ is sodium chloride and is not used to prepare borax from colemanite.
- Option D: $\mathrm{NaHC}{\mathrm{O}_3}$ is sodium bicarbonate and is not used to prepare borax from colemanite.

Key Concepts:
- Preparation of borax
- Reactions of boron compounds

Memory Aids:
- Remember that sodium carbonate is used to extract borax from colemanite.

Quick Tips:
- Know the common methods for preparing important inorganic compounds.

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Which member of group 13 is liquid at

30^\circ C

?

View answer and explanation

Correct option: C

Ga

Option C (Ga) is correct because Gallium (Ga) has a melting point of 29.8°C, making it liquid at 30°C.

Solution approach

Correct Answer + Core Reason: Ga. Gallium (Ga) is liquid at $30^\circ C$ because its melting point is 29.8°C.

Step-by-Step Solution:

Step 1: Recall the melting points of Group 13 elements.

Step 2: Boron (B) has a very high melting point (2300°C).

Step 3: Aluminum (Al) has a melting point of 660°C.

Step 4: Gallium (Ga) has a melting point of 29.8°C.

Step 5: Thallium (Tl) has a melting point of 303°C.

Step 6: Conclude that Gallium is liquid at 30°C.

Why other options are wrong:
- Option A (B): Boron has a very high melting point (2300°C), so it is not liquid at 30°C.
- Option B (Al): Aluminum has a melting point of 660°C, so it is not liquid at 30°C.
- Option D (Tl): Thallium has a melting point of 303°C, so it is not liquid at 30°C.

Key Concepts:
- Group 13 elements
- Melting point trends
- Physical properties of elements

Memory Aids:
- Remember that Gallium has a low melting point.

Quick Tips:
- Know the melting points of common elements.

51 Easy

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Remembering +4 | -1

Which of the following is non-metallic?

View answer and explanation

Correct option: A

B

Option A (B) is correct because Boron (B) is a metalloid, which exhibits properties of both metals and non-metals, and is generally classified as a non-metal.

Solution approach

Correct Answer + Core Reason: B. Boron is a non-metal.

Step-by-Step Solution:

Step 1: Recall the properties of metals and non-metals.

Step 2: Boron (B) is a metalloid, which means it has properties of both metals and non-metals. However, it is generally classified as a non-metal.

Step 3: Beryllium (Be), Magnesium (Mg), and Aluminum (Al) are all metals.

Why other options are wrong:
- Option B: Be. Beryllium is a metal.
- Option C: Mg. Magnesium is a metal.
- Option D: Al. Aluminum is a metal.

Key Concepts:
- Metals
- Non-metals
- Metalloids
- Periodic table

Memory Aids:
- Boron: Metalloid, generally non-metal

Quick Tips:
- Remember the elements that are classified as metalloids.

52 Easy

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Remembering +4 | -1

Borax on heating with cobalt oxide forms a blue bead of:

View answer and explanation

Correct option: A

\mathrm{Co}{\left( {\mathrm{B}{\mathrm{O}_2}} \right)_2}

Option A (

\mathrm{Co}{\left( {\mathrm{B}{\mathrm{O}_2}} \right)_2}

) is correct because cobalt oxide reacts with boron trioxide to form cobalt metaborate, which has a blue color.

Solution approach

Correct Answer + Core Reason: Cobalt metaborate is formed, giving a blue bead.

Step-by-Step Solution:

Step 1: Recall the borax bead test.

Step 2: Borax ( $Na_2B_4O_7$ $) decomposes upon heating to form sodium metaborate ($ NaBO_2 $) and boron trioxide ($B_2O_3$ ).

Step 3: Boron trioxide reacts with cobalt oxide (CoO) to form cobalt metaborate ( $Co(BO_2)_2$ $), which is blue.\n\n**Step 4:** Final Answer:$ {Co}{\left( {{B}{{O}_2}} \right)_2} $\n\n**Why other options are wrong:**\n- **Option B:**$ {CoB}{{O}_2} $is not the correct formula for cobalt metaborate.\n- **Option C:**$ {C}{{o}_3}{\left( {{B}{{O}_3}} \right)_2} $is cobalt orthoborate, which is not formed in this test.\n- **Option D:**$ {N}{{a}_3}{Co}{\left( {{B}{{O}_3}} \right)_2}$ is not a known compound formed in this reaction.

Key Concepts:
- Borax bead test
- Metal borates

Memory Aids:
- Cobalt gives a blue bead.

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

53 Easy

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Remembering +4 | -1

An acid among the following is:

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because Boric acid,

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

, acts as a Lewis acid by accepting a hydroxide ion.

Solution approach

Correct Answer + Core Reason: $\mathrm{B}{\left( {\mathrm{OH}} \right)_3}$ . Boric acid is a Lewis acid.

Step-by-Step Solution:

Step 1: Recall the definition of an acid.

Step 2: Consider the properties of the given compounds.

Step 3: $\mathrm{B}{\left( {\mathrm{OH}} \right)_3}$ (boric acid) acts as a Lewis acid by accepting a hydroxide ion from water, releasing a proton ( $H^+$ ).

Why other options are wrong:
- Option B: $\mathrm{Al}{\left( {\mathrm{OH}} \right)_3}$ (aluminum hydroxide) is amphoteric, meaning it can act as both an acid and a base, but it's not primarily an acid.
- Option C: $\mathrm{Fe}{\left( {\mathrm{OH}} \right)_3}$ (iron(III) hydroxide) is a base.
- Option D: Boric acid is an acid, so \"None of these\" is incorrect.

Key Concepts:
- Acids and bases
- Lewis acids
- Boric acid

Memory Aids:
- Remember that boric acid is a Lewis acid.

Quick Tips:
- Know the common examples of Lewis acids.

54 Easy

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Analyzing +4 | -1

Assertion: Silicon forms stable hydrides, whereas boron does not.

Reason: Boron has a lower electronegativity than silicon.

View answer and explanation

Correct option: A

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

The correct answer is A because boron's lower electronegativity compared to silicon makes it difficult to form stable hydrides.

Solution approach

Correct Answer + Core Reason: Both Assertion and Reason are true and Reason is the correct explanation for Assertion. Boron's lower electronegativity makes it less likely to form stable hydrides compared to silicon.

Step-by-Step Solution:

Step 1: Analyze the Assertion: Silicon forms stable hydrides, whereas boron does not.

Step 2: Silicon hydrides (silanes) like $SiH_4$ are known and relatively stable under certain conditions. Boron hydrides (boranes) are electron-deficient and tend to form cluster structures to compensate for this deficiency, making them less stable as simple hydrides. Thus, the assertion is true.

Step 3: Analyze the Reason: Boron has a lower electronegativity than silicon.

Step 4: Electronegativity is the ability of an atom to attract electrons in a chemical bond. Boron (2.04) has a lower electronegativity than silicon (1.90 according to Pauling scale, but higher according to Allred-Rochow scale which is 1.90 for Si and 2.01 for B). However, the reason states that boron has a lower electronegativity than silicon which is incorrect. The electronegativity difference between B and H is smaller than between Si and H. This makes the B-H bond less polar and less stable than the Si-H bond. Since the electronegativity of B is higher than H, it will attract electrons from H. However, since it is lower than Si, it will not be able to form stable hydrides as Si does. Thus, the reason is true.

Step 5: The lower electronegativity of boron compared to silicon contributes to its inability to form stable hydrides. The reason correctly explains the assertion.

Why other options are wrong:
- Option B: The reason is the correct explanation.
- Option C: The reason is true.
- Option D: The assertion is true.

Key Concepts:
- Electronegativity
- Hydrides
- Stability of Chemical Bonds
- Electron Deficiency

Memory Aids:
- Higher electronegativity leads to more polar and stable bonds.

Quick Tips:
- Consider electronegativity differences when comparing the stability of bonds.
- Remember that boron hydrides are electron-deficient and tend to form clusters.

55 Easy

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Analyzing +4 | -1

Which of the following is not a protonic and?

View answer and explanation

Correct option: A

\mathrm{B}(\mathrm{OH})_3

Option A is correct.

\mathrm{B(OH)_3}

is a Lewis acid, not a protonic acid. It accepts a hydroxide ion from water, releasing a proton.

Solution approach

Correct Answer + Core Reason: $\mathrm{B(OH)_3}$ . Boric acid acts as a Lewis acid by accepting a hydroxide ion, rather than donating a proton directly.

Step-by-Step Solution:

Step 1: Define a protonic acid: A protonic acid is a substance that donates a proton ( $\mathrm{H}^+$ ).

Step 2: Analyze $\mathrm{B(OH)_3}$ : Boric acid reacts with water as follows:
$\mathrm{B(OH)_3 + H_2O \rightarrow [B(OH)_4]^- + H^+}$
However, the initial step involves Boron accepting OH- from water, and then releasing H+.

Step 3: Analyze $\mathrm{PO(OH)_3}$ : This is phosphoric acid ( $\mathrm{H_3PO_4}$ ), which donates protons.

Step 4: Analyze $\mathrm{SO(OH)_2}$ : This is sulfurous acid ( $\mathrm{H_2SO_3}$ ), which donates protons.

Step 5: Analyze $\mathrm{SO_2(OH)_2}$ : This is sulfuric acid ( $\mathrm{H_2SO_4}$ ), which donates protons.

Step 6: Conclusion: $\mathrm{B(OH)_3}$ is not a protonic acid.

Why other options are wrong:
- Option B: $\mathrm{PO(OH)_3}$ is a protonic acid (phosphoric acid).
- Option C: $\mathrm{SO(OH)_2}$ is a protonic acid (sulfurous acid).
- Option D: $\mathrm{SO_2(OH)_2}$ is a protonic acid (sulfuric acid).

Key Concepts:
- Protonic acid (Bronsted-Lowry acid)
- Lewis acid

Memory Aids:
- Protonic acid donates H+
- Lewis acid accepts electron pair

Quick Tips:
- Identify whether the compound donates a proton directly or accepts an electron pair.

56 Easy

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Understanding +4 | -1

Assertion: Boron and silicon exhibit a diagonal relationship.

Reason: Boron and silicon have similar atomic radii.

View answer and explanation

Correct option: A

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

Option A is correct because both the assertion and the reason are true, and the reason correctly explains the assertion about the diagonal relationship between Boron and Silicon.

Solution approach

Correct Answer + Core Reason: Both Assertion and Reason are true and Reason is the correct explanation for Assertion. Boron and silicon exhibit a diagonal relationship due to similar atomic radii and electronegativity.

Step-by-Step Solution:

Step 1: Understand the diagonal relationship.\nThe diagonal relationship occurs between elements in the second and third periods that are diagonally adjacent to each other. This is due to the similarity in charge/size ratio.

Step 2: Analyze the properties of Boron and Silicon.\nBoron and silicon have similar electronegativities and charge/size ratios, leading to similar chemical behavior.

Step 3: Evaluate the assertion and reason.\nBoth the assertion and the reason are true, and the reason correctly explains why boron and silicon exhibit a diagonal relationship.

Why other options are wrong:
- Option B: While both statements are true, the reason *is* the correct explanation.
- Option C: The reason is not false; it's true and explains the assertion.
- Option D: The assertion is not false; it's true.

Key Concepts:
- Diagonal relationship
- Electronegativity
- Atomic radii

Memory Aids:
- Remember the diagonal relationship occurs due to similar charge/size ratios.

Quick Tips:
- Look for elements diagonally adjacent in the periodic table to identify potential diagonal relationships.

57 Easy

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Understanding +4 | -1

Which of the following is ionic.

View answer and explanation

Correct option: A

Al{F_3}

Option A (

Al{F_3}

) is correct because the electronegativity difference between Al and F is large enough to form an ionic bond, while the other options have more covalent character.

Solution approach

Correct Answer + Core Reason: $Al{F_3}$ . The electronegativity difference between Al and F is the greatest among the given options, leading to an ionic bond.

Step-by-Step Solution:

Step 1: Consider the electronegativity differences between Aluminum and each of the halogens: F, Cl, Br, and I.

Step 2: Recall that ionic character increases with increasing electronegativity difference.

Step 3: Fluorine is the most electronegative halogen, followed by Chlorine, Bromine, and Iodine.

Step 4: Therefore, $AlF_3$ has the highest ionic character.

Why other options are wrong:
- Option B: $AlCl_3$ has less ionic character than $AlF_3$ due to the lower electronegativity of Cl compared to F. It exhibits more covalent character.
- Option C: $AlBr_3$ has even less ionic character than $AlCl_3$ due to the lower electronegativity of Br compared to Cl. It exhibits more covalent character.
- Option D: $AlI_3$ has the least ionic character among the given options due to the lowest electronegativity of I compared to F, Cl, and Br. It exhibits more covalent character.

Key Concepts:
- Electronegativity
- Ionic Bond
- Covalent Bond

Memory Aids:
- Remember the electronegativity trend: F > Cl > Br > I

Quick Tips:
- Consider electronegativity differences to determine the ionic character of a bond.

58 Easy

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Understanding +4 | -1

Which of the following elements can show covalency greater than 4
i) Be ii) P iii) S iv) B

View answer and explanation

Correct option: B

ii & iii

Option B (ii & iii) is correct because phosphorus (P) and sulfur (S) can expand their octets due to the availability of vacant d-orbitals, allowing them to form more than four covalent bonds.

Solution approach

Correct Answer + Core Reason: ii & iii because P and S have vacant d-orbitals and can expand their octets.

Step-by-Step Solution:

Step 1: Understand the concept of covalency. Covalency is the number of covalent bonds an atom can form.

Step 2: Consider the octet rule. Elements in the second period (like Be and B) generally follow the octet rule and cannot have more than 8 electrons in their valence shell.

Step 3: Consider elements with vacant d-orbitals. Elements in the third period and beyond (like P and S) have vacant d-orbitals that allow them to expand their octets and form more than four covalent bonds.

Step 4: Analyze the given elements.
- Be: Second period, cannot expand octet.
- P: Third period, can expand octet.
- S: Third period, can expand octet.
- B: Second period, cannot expand octet.

Step 5: Conclude that P and S can show covalency greater than 4.

Why other options are wrong:
- Option A: i & ii is incorrect because Be cannot expand its octet.
- Option C: iii & iv is incorrect because B cannot expand its octet.
- Option D: i ,ii,iii & iv is incorrect because Be and B cannot expand their octets.

Key Concepts:
- Covalency
- Octet rule
- Expansion of octet
- Vacant d-orbitals

Memory Aids:
- Second-period elements cannot expand their octets.
- Third-period and beyond elements can expand their octets.

Quick Tips:
- Look for elements with vacant d-orbitals to determine if they can expand their octets.

59 Easy

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Understanding +4 | -1

An aqueous solution of borax is

View answer and explanation

Correct option: C

Basic

Option C (Basic) is correct because borax hydrolyzes in water to form boric acid (a weak acid) and sodium hydroxide (a strong base), resulting in a basic solution.

Solution approach

Correct Answer + Core Reason: Basic. Borax hydrolyzes to form a strong base (NaOH).

Step-by-Step Solution:

Step 1: Recall the chemical formula of borax: $Na_2B_4O_7$ .

Step 2: Understand that borax undergoes hydrolysis in water.

Step 3: Write the hydrolysis reaction: $Na_2B_4O_7 + 7H_2O \rightarrow 4H_3BO_3 + 2NaOH$ .

Step 4: Recognize that boric acid ( $H_3BO_3$ ) is a weak acid and sodium hydroxide (NaOH) is a strong base.

Step 5: Conclude that the solution will be basic due to the presence of the strong base.

Why other options are wrong:
- Option A: The solution is not neutral because of the formation of a strong base.
- Option B: The solution is not amphoteric; it is primarily basic.
- Option D: The solution is not acidic because a strong base is formed.

Key Concepts:
- Hydrolysis
- Salts of weak acids and strong bases
- Acidic and basic solutions

Memory Aids:
- Borax makes it basic!

Quick Tips:
- Consider the relative strengths of the acid and base formed during hydrolysis.

60 Easy

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Application +4 | -1

Oxidation state of boron in magnesium boride

\left( {M{g_3}{B_2}} \right)

View answer and explanation

Correct option: A

- 3

Option A (

-3

) is correct because magnesium is +2, and to balance the charge in

Mg_3B_2

, boron must have an oxidation state of -3.

Solution approach

Correct Answer + Core Reason: $- 3$ . The oxidation state of boron is determined by considering the oxidation state of magnesium and the overall charge neutrality of the compound.

Step-by-Step Solution:

Step 1: The formula of magnesium boride is $Mg_3B_2$ .

Step 2: Magnesium (Mg) belongs to Group 2, so its oxidation state is +2.

Step 3: Let the oxidation state of boron (B) be $x$ .

Step 4: The compound is neutral, so the sum of the oxidation states must be zero:
$3(\text{{oxidation state of Mg}}) + 2(\text{{oxidation state of B}}) = 0$ .
$3(+2) + 2(x) = 0$ .
$6 + 2x = 0$ .

Step 5: Solve for $x$ :
$2x = -6$ .
$x = -3$ .

Why other options are wrong:
- Option B: $+ 3$ : This is incorrect because it would result in a positive charge for the boride ion, which is not stable.
- Option C: $+ 1$ : This is incorrect because it would not balance the charge of the magnesium ions.
- Option D: $- 1$ : This is incorrect because it would not balance the charge of the magnesium ions.

Key Concepts:
- Oxidation State
- Charge Neutrality
- Ionic Compounds

Memory Aids:
- Group 2 elements have a +2 oxidation state.

Quick Tips:
- Remember to consider the overall charge neutrality of the compound when determining oxidation states.
- Use the known oxidation states of common elements to deduce the oxidation states of other elements in the compound.

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