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11. THE P-BLOCK ELEMENTS

189635 Aluminium chloride exists as dimer, $A l_{2} C l_{6}$ in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives

1

[A l (O H)_{6}]^{3 -} + 3 H C l

2

[A l (H_{2} O)_{6}]^{3 +} + 3 H C l

3

A l^{3 +} + 3 C l^{-}

4

A l_{2} O_{3} + 6 H C l

Explanation:

Correct option is B)
Aluminium chloride exists as dimer $(A l_{2} C l_{6})$ in a solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives
${[A l (O H)_{6}]}^{3 -} + 3 H C l$
$A l_{2} C l_{6} + 12 H_{2} O \to 2 {[A l (O H)_{6}]}^{3 -} + 12 H C l$
Thus aluminum chloride is hydrolyzed to form the complex ion containing s hydroxide groups.

11. THE P-BLOCK ELEMENTS

189636 The hardest substance amongst the following is

1

B e_{2} C

2 Graphite

3 Titanium

4

B_{4} C

Explanation:

(d) $B_{4} C$ is the hardest substance along with diamond.

11. THE P-BLOCK ELEMENTS

189619 The type of hybridisation of boron in diborane is

1

s p

- hybridisation

2

s p^{2} -

hybridisation

3

s p^{3} -

hybridisation

4

s p^{3} d^{2} -

hybridisation

Explanation:

Boron has three valence electrons, so it is supposed to make $3$ bonds in a molecule with hybridization, $s p^{2}$ as only s and two $p$ orbitals are used in hybridization and last $p$ orbitalis vacant.
But diborane, $B_{2} H_{6}$ contains two electrons each, three centred bonds. Each Boron atom is in a link with four hydrogen atoms. This makes tetrahedral geometry.
Hence, each Boron atom is $s p^{3}$ - hybridized.

11. THE P-BLOCK ELEMENTS

189620 In the reaction $B_{2} O_{3} + C + C l_{2} \to A + C O .$ The $A$ is

1

B C l_{3}

2

B C l_{2}

3

B_{2} C l_{2}

4

C C l_{2}

Explanation:

(a) $B_{2} O_{3} + 3 C + 3 C l_{2} \to 2 B C l_{3} + 3 C O$
$B C l_{3}$ is obtained by passing chlorine over the heated mixture of $B_{2} O_{3}$ and powdered charcoal.

11. THE P-BLOCK ELEMENTS

189635 Aluminium chloride exists as dimer, $A l_{2} C l_{6}$ in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives

1

[A l (O H)_{6}]^{3 -} + 3 H C l

2

[A l (H_{2} O)_{6}]^{3 +} + 3 H C l

3

A l^{3 +} + 3 C l^{-}

4

A l_{2} O_{3} + 6 H C l

Explanation:

Correct option is B)
Aluminium chloride exists as dimer $(A l_{2} C l_{6})$ in a solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives
${[A l (O H)_{6}]}^{3 -} + 3 H C l$
$A l_{2} C l_{6} + 12 H_{2} O \to 2 {[A l (O H)_{6}]}^{3 -} + 12 H C l$
Thus aluminum chloride is hydrolyzed to form the complex ion containing s hydroxide groups.

11. THE P-BLOCK ELEMENTS

189636 The hardest substance amongst the following is

1

B e_{2} C

2 Graphite

3 Titanium

4

B_{4} C

Explanation:

(d) $B_{4} C$ is the hardest substance along with diamond.

11. THE P-BLOCK ELEMENTS

189619 The type of hybridisation of boron in diborane is

1

s p

- hybridisation

2

s p^{2} -

hybridisation

3

s p^{3} -

hybridisation

4

s p^{3} d^{2} -

hybridisation

Explanation:

Boron has three valence electrons, so it is supposed to make $3$ bonds in a molecule with hybridization, $s p^{2}$ as only s and two $p$ orbitals are used in hybridization and last $p$ orbitalis vacant.
But diborane, $B_{2} H_{6}$ contains two electrons each, three centred bonds. Each Boron atom is in a link with four hydrogen atoms. This makes tetrahedral geometry.
Hence, each Boron atom is $s p^{3}$ - hybridized.

11. THE P-BLOCK ELEMENTS

189620 In the reaction $B_{2} O_{3} + C + C l_{2} \to A + C O .$ The $A$ is

1

B C l_{3}

2

B C l_{2}

3

B_{2} C l_{2}

4

C C l_{2}

Explanation:

(a) $B_{2} O_{3} + 3 C + 3 C l_{2} \to 2 B C l_{3} + 3 C O$
$B C l_{3}$ is obtained by passing chlorine over the heated mixture of $B_{2} O_{3}$ and powdered charcoal.

11. THE P-BLOCK ELEMENTS

189635 Aluminium chloride exists as dimer, $A l_{2} C l_{6}$ in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives

1

[A l (O H)_{6}]^{3 -} + 3 H C l

2

[A l (H_{2} O)_{6}]^{3 +} + 3 H C l

3

A l^{3 +} + 3 C l^{-}

4

A l_{2} O_{3} + 6 H C l

Explanation:

Correct option is B)
Aluminium chloride exists as dimer $(A l_{2} C l_{6})$ in a solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives
${[A l (O H)_{6}]}^{3 -} + 3 H C l$
$A l_{2} C l_{6} + 12 H_{2} O \to 2 {[A l (O H)_{6}]}^{3 -} + 12 H C l$
Thus aluminum chloride is hydrolyzed to form the complex ion containing s hydroxide groups.

11. THE P-BLOCK ELEMENTS

189636 The hardest substance amongst the following is

1

B e_{2} C

2 Graphite

3 Titanium

4

B_{4} C

Explanation:

(d) $B_{4} C$ is the hardest substance along with diamond.

11. THE P-BLOCK ELEMENTS

189619 The type of hybridisation of boron in diborane is

1

s p

- hybridisation

2

s p^{2} -

hybridisation

3

s p^{3} -

hybridisation

4

s p^{3} d^{2} -

hybridisation

Explanation:

Boron has three valence electrons, so it is supposed to make $3$ bonds in a molecule with hybridization, $s p^{2}$ as only s and two $p$ orbitals are used in hybridization and last $p$ orbitalis vacant.
But diborane, $B_{2} H_{6}$ contains two electrons each, three centred bonds. Each Boron atom is in a link with four hydrogen atoms. This makes tetrahedral geometry.
Hence, each Boron atom is $s p^{3}$ - hybridized.

11. THE P-BLOCK ELEMENTS

189620 In the reaction $B_{2} O_{3} + C + C l_{2} \to A + C O .$ The $A$ is

1

B C l_{3}

2

B C l_{2}

3

B_{2} C l_{2}

4

C C l_{2}

Explanation:

(a) $B_{2} O_{3} + 3 C + 3 C l_{2} \to 2 B C l_{3} + 3 C O$
$B C l_{3}$ is obtained by passing chlorine over the heated mixture of $B_{2} O_{3}$ and powdered charcoal.

11. THE P-BLOCK ELEMENTS

189635 Aluminium chloride exists as dimer, $A l_{2} C l_{6}$ in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives

1

[A l (O H)_{6}]^{3 -} + 3 H C l

2

[A l (H_{2} O)_{6}]^{3 +} + 3 H C l

3

A l^{3 +} + 3 C l^{-}

4

A l_{2} O_{3} + 6 H C l

Explanation:

Correct option is B)
Aluminium chloride exists as dimer $(A l_{2} C l_{6})$ in a solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives
${[A l (O H)_{6}]}^{3 -} + 3 H C l$
$A l_{2} C l_{6} + 12 H_{2} O \to 2 {[A l (O H)_{6}]}^{3 -} + 12 H C l$
Thus aluminum chloride is hydrolyzed to form the complex ion containing s hydroxide groups.

11. THE P-BLOCK ELEMENTS

189636 The hardest substance amongst the following is

1

B e_{2} C

2 Graphite

3 Titanium

4

B_{4} C

Explanation:

(d) $B_{4} C$ is the hardest substance along with diamond.

11. THE P-BLOCK ELEMENTS

189619 The type of hybridisation of boron in diborane is

1

s p

- hybridisation

2

s p^{2} -

hybridisation

3

s p^{3} -

hybridisation

4

s p^{3} d^{2} -

hybridisation

Explanation:

Boron has three valence electrons, so it is supposed to make $3$ bonds in a molecule with hybridization, $s p^{2}$ as only s and two $p$ orbitals are used in hybridization and last $p$ orbitalis vacant.
But diborane, $B_{2} H_{6}$ contains two electrons each, three centred bonds. Each Boron atom is in a link with four hydrogen atoms. This makes tetrahedral geometry.
Hence, each Boron atom is $s p^{3}$ - hybridized.

11. THE P-BLOCK ELEMENTS

189620 In the reaction $B_{2} O_{3} + C + C l_{2} \to A + C O .$ The $A$ is

1

B C l_{3}

2

B C l_{2}

3

B_{2} C l_{2}

4

C C l_{2}

Explanation:

(a) $B_{2} O_{3} + 3 C + 3 C l_{2} \to 2 B C l_{3} + 3 C O$
$B C l_{3}$ is obtained by passing chlorine over the heated mixture of $B_{2} O_{3}$ and powdered charcoal.

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