p-Block Elements — Page 17 | JEE Chemistry

Q161

The interhalogen compound formed from the reaction of bromine with excess of fluorine is a :

A hypohalite

B halate

C perhalate

D halite

Correct Answer

Option B

Solution

\mathrm{Br}_{2}+\underset{\text{ (Excess) }}{5 \mathrm{~F}_{2}} \longrightarrow 2 \mathrm{BrF}_{5}

If

\mathrm{BrF}_{5}

undergoes hydrolysis it will produce halide.

Q162

White phosphorus reacts with thionyl chloride to give :

A

\mathrm{PCl}_{5}, \mathrm{SO}_{2}

and

\mathrm{S}_{2} \mathrm{Cl}_{2}

B

\mathrm{PCl}_{3}, \mathrm{SO}_{2}

and

\mathrm{S}_{2} \mathrm{Cl}_{2}

C

\mathrm{PCl}_{3}, \mathrm{SO}_{2}

and

\mathrm{Cl}_{2}

D

\mathrm{PCl}_{5}, \mathrm{SO}_{2}

and

\mathrm{Cl}_{2}

Correct Answer

Option B

Solution

$\underset{\text{ White phosphorous }}{\mathrm{P}_{4}}+\underset{\text{ Thionyl chloride }}{8 \mathrm{SOCl}_{2}} \longrightarrow 4 \mathrm{PCl}_{3}+4 \mathrm{SO}_{2}+2 \mathrm{~S}_{2} \mathrm{Cl}_{2}$

Q163

Boric acid is solid, whereas

\mathrm{BF}_3

is gas at room temperature because of :

A Strong van der Waal's interaction in Boric acid

B Strong covalent bond in

\mathrm{BF}_{3}

C Strong hydrogen bond in Boric acid

D Strong ionic bond in Boric acid

Correct Answer

Option C

Solution

Boric acid is solid because of strong hydrogen bond in it.

Q164

One mole of

\mathrm{P}_{4}

reacts with 8 moles of

\mathrm{SOCl}_{2}

to give 4 moles of

\mathrm{A}, x

mole of

\mathrm{SO}_{2}

and 2 moles of

\mathrm{B} . \mathrm{A}, \mathrm{B}

and

x

respectively are :

A

\mathrm{POCl}_{3}, \mathrm{~S}_{2} \mathrm{Cl}_{2}

and 4

B

\mathrm{PCl}_{3}, \mathrm{~S}_{2} \mathrm{Cl}_{2}

and 4

C

\mathrm{POCl}_{3}, \mathrm{~S}_{2} \mathrm{Cl}_{2}

and 2

D

\mathrm{PCl}_{3}, \mathrm{~S}_{2} \mathrm{Cl}_{2}

and 2

Correct Answer

Option B

Solution

\mathrm{P}_4+8 \mathrm{SOCl}_2 \longrightarrow 4 \mathrm{PCl}_3+4 \mathrm{SO}_2+2 \mathrm{~S}_2 \mathrm{Cl}_2

Hence, $\mathrm{A}=\mathrm{PCl}_3, \mathrm{x}=4, \mathrm{~B}=\mathrm{S}_2 \mathrm{Cl}_2$

Q165

Which among the following is the most reactive?

A Br2

B I2

C ICl

D Cl2

Correct Answer

Option C

Solution

ICl

Order of reactivity of halogens

C{l_2} > B{r_2} > {I_2}

But, the interhalogen compounds are generally more reactive than halogens (except

{F_2}

), since the bond between two dissimilar electronegative elements is weaker than the bond between two similar atoms i.e,

X-X

Q166

The compound that is white in color is

A ammonium sulphide

B ammonium arsinomolybdate

C lead iodide

D lead sulphate

Correct Answer

Option D

Solution

Lead sulphate-white Ammonium sulphide-soluble Lead iodide-Bright yellow Ammonium arsinomolybdate-yellow

Q167

Among the oxides of nitrogen : N2O3, N2O4 and N2O5; the molecule(s) having nitrogen-nitrogen bond is / are :

A Only N2O5

B N2O3 and N2O5

C N2O4 and N2O5

D N2O3 and N2O4

Correct Answer

Option D

Solution

$\therefore$ N2O3 and N2O4 has N $-$ N bond.

Q168

Lithium aluminium hydride reacts with silicon tetrachloride to form :

A LiCl, AlH3 and SiH4

B LiCl, AlCl3 and SiH4

C LiH, AlCl3 and SiCl2

D LiH, AlH3 and SiH4

Correct Answer

Option B

Solution

LiAlH4 + SiCl4

\overset{\,}\longrightarrow

LiCl + AlCl3 + SiH4

Q169

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Both rhombic and monoclinic sulphur exist as

\mathrm{S}_8

while oxygen exists as

\mathrm{O}_2

. Reason (R) : Oxygen forms

p \pi-p \pi

multiple bonds with itself and other elements having small size and high electronegativity like

\mathrm{C}, \mathrm{N}

, which is not possible for sulphur. In the light of the above statements, choose the most appropriate answer from the options given below :

A Both (A) and (R) are correct but (R) is not the correct explanation of (A)

B (A) is correct but (R) is not correct

C (A) is not correct but (R) is correct

D Both

(\mathbf{A})

and

(\mathbf{R})

are correct and

(\mathbf{R})

is the correct explanation of

(\mathbf{A})

Correct Answer

Option D

Solution

Oxygen exists as

\mathrm{O}_2

Sulphur exists as

\mathrm{S}_8

Because of small size of oxygen, oxygen can form

2 \mathrm{p}_\pi-2 \mathrm{p}_\pi

bond.

$\Rightarrow$ Assertion (A) is correct.

$\Rightarrow$ Reason (R) is correct.

Reason (R) is correct explanation of Assertion (A).

Q170

The correct order of catenation is :

A C > Sn > Si

\approx

Ge

B Si > Sn > C > Ge

C C > Si > Ge

\approx

Sn

D Ge > Sn > Si > C

Correct Answer

Option C

Solution

The order of catenation property amongst 14th group elements is based on bond enthalpy values of identical atoms of the same element.

The decrasing order of catenation is C > Si > Ge $\approx$ Sn