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

13 th group elements (p-block) practice · Hard

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1 Hard

1 of 9Question permalink

Understanding +4 | -1

Which of the following elements does not exist in aqueous medium, as species like tetrahedral

\left[M(\mathrm{OH})_4\right]^{-}

and octahedral

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

View answer and explanation

Correct option: A

Boron

Option A (Boron) is correct because boron, due to its small size and high charge density, does not form stable octahedral complexes in aqueous solution.

Solution approach

Correct Answer + Core Reason: Boron. Boron does not exist in aqueous medium as tetrahedral $[M(OH)_4]^-$ and octahedral $[M(H_2O)_6]^{3+}$ species due to its small size and high charge density.

Step-by-Step Solution:

Step 1: Consider the properties of Boron.\nBoron is a small atom with high charge density. It has a limited ability to expand its coordination number beyond 4.

Step 2: Consider the properties of other elements.\nThallium, Gallium, and Indium are larger and can form octahedral complexes in aqueous solution.

Step 3: Explain why Boron is different.\nBoron's small size and high charge density make it difficult to accommodate six water molecules in an octahedral arrangement. It prefers to form tetrahedral $[B(OH)_4]^-$ species.

Why other options are wrong:
- Option B: Thallium can form octahedral complexes like $[Tl(H_2O)_6]^{3+}$ .
- Option C: Gallium can form octahedral complexes like $[Ga(H_2O)_6]^{3+}$ .
- Option D: Indium can form octahedral complexes like $[In(H_2O)_6]^{3+}$ .

Key Concepts:
- Coordination complexes
- Size and charge density of ions
- Hydration of ions

Memory Aids:
- Boron is too small to form octahedral complexes easily.

Quick Tips:
- Remember the size trends in the group 13 elements.

2 Hard

2 of 9Question permalink

Application +4 | -1

The number of neutrons present in the more abundant isotope of boron is

x.

Amorphous boron upon heating with air forms a product, in which the oxidation state of boron is

y.

The value of

x+y

View answer and explanation

Correct option: D

Option D (9) is correct because the more abundant isotope of boron is

^{11}B

(x=6), and boron reacts with air to form

B_2O_3

where boron's oxidation state is +3 (y=3), so x+y = 6+3 = 9.

Solution approach

Correct Answer + Core Reason: 9. The value of x+y is 9, where x is the number of neutrons in the more abundant isotope of boron and y is the oxidation state of boron in its oxide.

Step-by-Step Solution:

Step 1: Determine the more abundant isotope of Boron.\nBoron has two isotopes: $^{10}B$ (approximately 20%) and $^{11}B$ (approximately 80%). The more abundant isotope is $^{11}B$ .

Step 2: Calculate the number of neutrons in $^{11}B$ .\nNumber of neutrons = Mass number - Atomic number = 11 - 5 = 6. Therefore, x = 6.

Step 3: Determine the oxidation state of Boron in its oxide.\nBoron reacts with air to form $B_2O_3$ . The oxidation state of oxygen is -2. Let the oxidation state of Boron be y.
$2y + 3(-2) = 0$
$2y - 6 = 0$
$2y = 6$
$y = +3$

Step 4: Calculate x + y.\nx + y = 6 + 3 = 9

Why other options are wrong:
- Option A: 4 is incorrect. The correct value is 9.
- Option B: 6 is incorrect. The correct value is 9.
- Option C: 3 is incorrect. The correct value is 9.

Key Concepts:
- Isotopes
- Number of neutrons
- Oxidation states

Memory Aids:
- Remember the isotopes of Boron.

Quick Tips:
- Remember the formula for calculating the number of neutrons.

3 Hard

3 of 9Question permalink

Understanding +4 | -1

Diborane

({B_2}{H_6})

reacts independently with

{O_2}

and

{H_2}O

to produce, respectively.

View answer and explanation

Correct option: A

{B_2}{O_3}

and

{H_3}B{O_3}

Option A (

B_2O_3

and

H_3BO_3

) is correct because diborane reacts with oxygen to form boron oxide and with water to form boric acid.

Solution approach

Correct Answer + Core Reason: ${B_2}{O_3}$ and ${H_3}B{O_3}$ . Diborane reacts with oxygen to form boron oxide and with water to form boric acid.

Step-by-Step Solution:

Step 1: Recall the reaction of diborane with oxygen: $B_2H_6 + 3O_2 \rightarrow B_2O_3 + 3H_2O$

Step 2: Recall the reaction of diborane with water: $B_2H_6 + 6H_2O \rightarrow 2H_3BO_3 + 6H_2$

Step 3: Conclusion: Diborane reacts with $O_2$ to produce $B_2O_3$ and with $H_2O$ to produce $H_3BO_3$ .

Why other options are wrong:
- **Option B ( ${B_2}{O_3}$ and ${[B{H_4}]^ - }$ ):** Diborane does not form $[BH_4]^-$ with water.
- **Option C ( ${H_3}B{O_3}$ and ${B_2}{O_3}$ ): The order is reversed. Diborane reacts with oxygen to form boron oxide and with water to form boric acid.
- Option D ( $HB{O_2}$ and ${H_3}B{O_3}$ ):** While $HBO_2$ can be formed under certain conditions, $B_2O_3$ is the primary product with oxygen.

Key Concepts:
- Diborane chemistry
- Reactions with oxygen and water

Memory Aids:
- Remember the products of diborane reactions with oxygen and water.

Quick Tips:
- Know the common reactions of diborane.

4 Hard

4 of 9Question permalink

Understanding +4 | -1

Assertion (A) :

{AlCl}_3

forms dimer

{Al}_2 {Cl}_6

but is dissolves in

{H}_2 {O}

forming

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

and

{Cl}^{-}

ions.

Reason (R) : Aqueous solution of

{AlCl}_3

is acidic due to hydrolysis.

View answer and explanation

Correct option: B

Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is not the correct explanation of

\mathbf{A}

Option B (Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is not the correct explanation of

\mathbf{A}

) is correct because both the assertion and reason are true, but the reason does not explain why

AlCl_3

forms a dimer.

Solution approach

Correct Answer + Core Reason: Both $\mathbf{A}$ and $\mathbf{R}$ are true but $\mathbf{R}$ is not the correct explanation of $\mathbf{A}$ . $AlCl_3$ forms a dimer to achieve stability by completing its octet, and its aqueous solution is acidic due to hydrolysis, but these are separate phenomena.

Step-by-Step Solution:

Step 1: Analyze Assertion (A).
$AlCl_3$ exists as a dimer ( $Al_2Cl_6$ ) in the vapor phase and in non-polar solvents. In water, it forms $[Al(H_2O)_6]^{3+}$ and $Cl^-$ ions. The assertion is true.

Step 2: Analyze Reason (R).\nThe aqueous solution of $AlCl_3$ is acidic due to the hydrolysis of $Al^{3+}$ ions.
$[Al(H_2O)_6]^{3+}(aq) + H_2O(l) \rightleftharpoons [Al(H_2O)_5OH]^{2+}(aq) + H_3O^+(aq)$ \nThe reason is also true.

Step 3: Determine the relationship between A and R.\nThe formation of the dimer is due to the electron deficiency of Al and the need to complete its octet. The acidity of the aqueous solution is due to the hydrolysis of the $Al^{3+}$ ion. The reason does not explain the assertion.

Why other options are wrong:
- Option A: Reason is not the correct explanation for the assertion.
- Option C: Reason is true.
- Option D: Assertion is true.

Key Concepts:
- Dimerization of $AlCl_3$
- Hydrolysis of $Al^{3+}$ ions
- Acidity of aqueous solutions

Memory Aids:
- $AlCl_3$ dimerizes to complete its octet.
- Hydrolysis of metal ions leads to acidic solutions.

Quick Tips:
- Understand the difference between the reasons for dimerization and acidity.

5 Hard

5 of 9Question permalink

Understanding +4 | -1

A compound

X

, of boron reacts with

\mathrm{NH}_3

on heating to give another compound

Y

, which is called inorganic benzene. The compound

X

can be prepared by treating

\mathrm{BF}_3

with lithium aluminium hydride. The compounds

X

and

Y

are represented by the formulae

View answer and explanation

Correct option: A

\mathrm{B}_2 \mathrm{H}_6, \mathrm{~B}_3 \mathrm{~N}_3 \mathrm{H}_6

Option A (

\mathrm{B}_2 \mathrm{H}_6, \mathrm{~B}_3 \mathrm{~N}_3 \mathrm{H}_6

) is correct because diborane (

B_2H_6

) reacts with ammonia to give borazine (

B_3N_3H_6

Solution approach

Correct Answer + Core Reason: $\mathrm{B}_2 \mathrm{H}_6, \mathrm{~B}_3 \mathrm{~N}_3 \mathrm{H}_6$ . Diborane reacts with ammonia to form borazine, also known as inorganic benzene.

Step-by-Step Solution:

Step 1: Identify compound X: $BF_3$ reacts with lithium aluminum hydride ( $LiAlH_4$ ) to form diborane ( $B_2H_6$ ).

Step 2: Identify compound Y: Diborane ( $B_2H_6$ ) reacts with ammonia ( $NH_3$ ) to form borazine ( $B_3N_3H_6$ ).

Step 3: Conclusion: X is $B_2H_6$ and Y is $B_3N_3H_6$ .

Why other options are wrong:
- **Option B ( $\mathrm{B}_2 \mathrm{O}_3, \mathrm{~B}_3 \mathrm{~N}_3 \mathrm{H}_6$ ):** $B_2O_3$ does not react with ammonia to form borazine.
- **Option C ( $\mathrm{BF}_3, \mathrm{~B}_3 \mathrm{~N}_3 \mathrm{H}_6$ ):** $BF_3$ is not the direct precursor to borazine in this context.
- **Option D ( $\mathrm{B}_3 \mathrm{~N}_3 \mathrm{H}_6, \mathrm{~B}_2 \mathrm{H}_6$ ): The order is reversed. Diborane is the reactant, and borazine is the product.

Key Concepts:
- Borazine formation
- Diborane synthesis

Memory Aids:**
- Remember the reaction sequence: $BF_3 \rightarrow B_2H_6 \rightarrow B_3N_3H_6$ .

Quick Tips:
- Know the synthesis of diborane and borazine.

6 Hard

6 of 9Question permalink

Analyzing +4 | -1

Assertion (A) : In aqueous solution

{AlCl}_3

forms dimer

{Al}_2 {Cl}_6

.
Reason (R) : The high enthalpy of hydration is sufficient to break the covalent dimer into

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

and

{Cl}^{-}

ions.

View answer and explanation

Correct option: D

\mathbf{A}

is false but

\mathbf{R}

is true.

Option D (

\mathbf{A}

is false but

\mathbf{R}

is true) is correct because

AlCl_3

does not form a dimer in aqueous solution; it forms hydrated ions, and the high enthalpy of hydration is indeed sufficient to break any potential covalent dimer.

Solution approach

Correct Answer + Core Reason: $\mathbf{A}$ is false but $\mathbf{R}$ is true. $AlCl_3$ exists as hydrated ions in aqueous solution, and the high hydration enthalpy breaks any potential dimer.

Step-by-Step Solution:

Step 1: Analyze Assertion (A): In aqueous solution ${AlCl}_3$ forms dimer ${Al}_2 {Cl}_6$ . This is false. In aqueous solution, $AlCl_3$ exists as hydrated ions, $[\mathrm{Al}(\mathrm{H_2O})_6]^{3+}$ and $Cl^-$ .

Step 2: Analyze Reason (R): The high enthalpy of hydration is sufficient to break the covalent dimer into $\left[M\left({H}_2 {O}\right)_6\right]^{3+}$ and ${Cl}^{-}$ ions. This is true. The high hydration enthalpy of $Al^{3+}$ ions favors the formation of hydrated ions over the dimer.

Step 3: Conclusion: Assertion (A) is false, but Reason (R) is true.

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

Key Concepts:
- Hydration of ions
- Enthalpy of hydration
- Structure of aluminium chloride in aqueous solution

Memory Aids:
- Remember that $AlCl_3$ forms hydrated ions in water.

Quick Tips:
- Consider the hydration enthalpy when predicting the behavior of ionic compounds in water.

7 Hard

7 of 9Question permalink

Understanding +4 | -1

The structure of diborane

\left(B_2 H_6\right)

contains '

x

' number of

2 {c}-2 {e}

bonds and '

{y}

' number of

3 c-2 e^{-}

bonds and then calculate the value of

x^2-y^3

View answer and explanation

Correct option: A

Option A (8) is correct because diborane has 4

2c-2e

bonds and 2

3c-2e

bonds, so

x=4

and

y=2

, and

x^2 - y^3 = 4^2 - 2^3 = 16 - 8 = 8

Solution approach

Correct Answer + Core Reason: 8. Diborane has 4 terminal B-H bonds (2c-2e) and 2 bridging B-H-B bonds (3c-2e). Therefore, $x^2 - y^3 = 4^2 - 2^3 = 8$ .

Step-by-Step Solution:

Step 1: Understand the structure of diborane ( $B_2H_6$ ).

Step 2: Identify the types of bonds in diborane: There are four terminal B-H bonds (2-center-2-electron bonds) and two bridging B-H-B bonds (3-center-2-electron bonds).

Step 3: Assign values to x and y: x = number of 2c-2e bonds = 4, y = number of 3c-2e bonds = 2.

Step 4: Calculate $x^2 - y^3$ : $x^2 - y^3 = 4^2 - 2^3 = 16 - 8 = 8$ .

Step 5: Conclusion: The value of $x^2 - y^3$ is 8.

Why other options are wrong:
- Option B (9): Incorrect calculation.
- Option C (10): Incorrect calculation.
- Option D (11): Incorrect calculation.

Key Concepts:
- Structure of diborane
- 2-center-2-electron bonds
- 3-center-2-electron bonds

Memory Aids:
- Visualize the structure of diborane to count the different types of bonds.

Quick Tips:
- Know the structure of diborane and the types of bonds present.

8 Hard

8 of 9Question permalink

Understanding +4 | -1

Borazine, also known as inorganic benzene, can be prepared by the reaction of 3-equivalents of "

X

" with 6-equivalents of "

Y

". "

X

" and "

Y

", respectively are:

View answer and explanation

Correct option: B

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

and

\mathrm{NH}_3

Option B (

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

and

\mathrm{NH}_3

) is correct because borazine is formed by the reaction of diborane with ammonia.

Solution approach

Correct Answer + Core Reason: $\mathrm{B}_2 \mathrm{H}_6$ and $\mathrm{NH}_3$ . Borazine is prepared by the reaction of diborane with ammonia.

Step-by-Step Solution:

Step 1: Recall the reaction for the formation of borazine: $3B_2H_6 + 6NH_3 \rightarrow 2B_3N_3H_6 + 12H_2$

Step 2: Identify the reactants: Diborane ( $B_2H_6$ ) and ammonia ( $NH_3$ ).

Step 3: Conclusion: Borazine is formed by the reaction of diborane with ammonia.

Why other options are wrong:
- **Option A ( $\mathrm{B}(\mathrm{OH})_3$ and $\mathrm{NH}_3$ ): Boric acid and ammonia do not directly form borazine.
- Option C ( $\mathrm{B}_2 \mathrm{H}_6$ and $\mathrm{HN}_3$ ):** Hydrazine ( $HN_3$ ) is not used in the formation of borazine.
- **Option D ( $\mathrm{NH}_3$ and $\mathrm{B}_2 \mathrm{O}_3$ ): Boron oxide and ammonia do not directly form borazine.

Key Concepts:
- Borazine formation
- Reaction of diborane with ammonia

Memory Aids:
- Remember the reaction of diborane with ammonia to form borazine.

Quick Tips:**
- Know the synthesis of borazine.

9 Hard

9 of 9Question permalink

Understanding +4 | -1

Consider the following statements a. Boric acid is a weak monobasic protonic acid b. Orthoboric acid on heating above 370 K forms metaboric acid c. Diborane on hydrolysis forms boric acid. The correct statements are

View answer and explanation

Correct option: B

b and c only

Option B (b and c only) is correct because orthoboric acid forms metaboric acid upon heating and diborane hydrolyzes to form boric acid, while boric acid is a weak monobasic Lewis acid, not a protonic acid.

Solution approach

Correct Answer + Core Reason: b and c only. Boric acid is a Lewis acid, not a protonic acid. Orthoboric acid forms metaboric acid on heating, and diborane hydrolyzes to boric acid.

Step-by-Step Solution:

Step 1: Analyze statement a: Boric acid ( $H_3BO_3$ ) is a weak monobasic protonic acid. Boric acid acts as a Lewis acid by accepting a hydroxide ion from water, not by donating a proton. Thus, it is not a protonic acid.

Step 2: Analyze statement b: Orthoboric acid on heating above 370 K forms metaboric acid. This is correct. The reaction is:
$H_3BO_3 \xrightarrow{\Delta} HBO_2 + H_2O$

Step 3: Analyze statement c: Diborane on hydrolysis forms boric acid. This is correct. The reaction is:
$B_2H_6 + 6H_2O \rightarrow 2H_3BO_3 + 6H_2$

Step 4: Conclusion: Statements b and c are correct, while statement a is incorrect.

Why other options are wrong:
- Option A: a and b only: Statement a is incorrect.
- Option C: a and c only: Statement a is incorrect.
- Option D: a, b and c: Statement a is incorrect.

Key Concepts:
- Properties of boric acid
- Lewis acids and protonic acids
- Hydrolysis of diborane
- Thermal decomposition of boric acid

Memory Aids:
- Boric acid is a Lewis acid, not a protonic acid.

Quick Tips:
- Remember the difference between Lewis acids and Bronsted-Lowry acids.

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13 th group elements (p-block) practice · Hard

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