p-Block Elements — Page 10 | JEE Chemistry

Q91

The oxide which contains an odd electron at the nitrogen atom is :

A N2O

B NO2

C N2O3

D N2O5

Correct Answer

Option B

Solution

The oxide of nitrogen which contains odd electron is NO2

Q92

The number of bridged oxygen atoms present in compound B formed from the following reactions is Pb(NO3)2

\overset{{673\,K}}\longrightarrow

A + PbO + O2 A

\overset{{Dimerise}}\longrightarrow

B

A 0

B 1

C 2

D 3

Correct Answer

Option A

Solution

2 \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2} \stackrel{673 \mathrm{~K}}{\longrightarrow} 4 \mathrm{AO}_{2}+2 \mathrm{PbO}+\mathrm{O}_{2}

{}\begin{gathered}NO_{2}\\ \text{A}\end{gathered} \xrightarrow{\text{Dimerise} } \begin{gathered}N_{2}O_{4}\\ \text{B}\end{gathered}

Hence no bridged oxygen atom is present in N2O4

Q93

White precipitate of AgCl dissolves in aqueous ammonia solution due to formation of :

A [Ag(NH3)4]Cl2

B [Ag(Cl)2(NH3)2]

C [Ag(NH3)2]Cl

D [Ag(NH3)Cl]Cl

Correct Answer

Option C

Solution

\mathrm{AgCl} +2\mathrm{NH}_{3} \rightarrow \begin{gathered}\left[ \mathrm{Ag} \left( \mathrm{NH}_{3} \right)_{2} \right]^{+} \mathrm{Cl}^{-} \\ \text{Soluble} \end{gathered}

Q94

PCl5 is well known, but NCl5 is not. because,

A nitrogen is less reactive than phosphorus.

B nitrogen doesn't have d-orbitals in its valence shell.

C catenation tendency is weaker in nitrogen than phosphorus.

D size of phosphorus is larger than nitrogen.

Correct Answer

Option B

Solution

\mathrm{PCl}_{5}

is well known but

\mathrm{NCl}_{5}

is not because nitrogen does not have vacant d-orbitals in its valence shell.

So, nitrogen cannot expand its octet.

On the other hand, phosphorus has vacant

\mathrm{d}

-orbitals in its valence shell which enables it to expand its octet.

Q95

Identify the correct statement for B2H6 from those given below : (A) In B2H6, all B-H bonds are equivalent. (B) In B2H6, there are four 3-centre-2-electron bonds. (C) B2H6 is a Lewis acid. (D) B2H6 can be synthesized from both BF3 and NaBH4. (E) B2H6 is a planar molecule. Choose the most appropriate answer from the options given below :

A (A) and (E) only

B (B), (C) and (E) only

C (C) and (D) only

D (C) and (E) only

Correct Answer

Option C

Solution

It has two 3-centre-2-electron bonds and four 2-centre-2-electron bonds.

Hence, all B–H bonds are not equivalent.

It is an electron deficient compound as the octet of boron is incomplete.

Hence, it can behave as a Lewis acid.

It can be synthesized from both BF3 and NaBH4

2B{F_3} + 6NaH\overset{{450K}}\longrightarrow {B_2}{H_6} + 6NaF

2NaB{H_4} + {I_2} \to {B_2}{H_6} + 2NaI + {H_2}

It is a non-planar molecule. Hence, only Statements (C) and (D) are correct.

Q96

The most stable trihalide of nitrogen is :

A NF3

B NCl3

C NBr3

D NI3

Correct Answer

Option A

Solution

The stability of trihalides decreases down the group due to weakening of N – X bond and inability of N to accommodate large sized halogen atoms (Cl, Br, I) around it.

Q97

Which one of the following elemental forms is not present in the enamel of the teeth?

A Ca2+

B P3+

C F

-

D P5+

Correct Answer

Option B

Solution

$\mathrm{P}^{+3}$ is not present is enamel of teeth.

Th compound present is $\left[3 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \cdot \mathrm{CaF}_{2}\right]$ Which contains $\mathrm{Ca}^{+2}, \mathrm{P}^{+5}$ & $\mathrm{F}^{-}$

Q98

A

\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}^{573\,K}}

Red phosphorus

\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{under\,pressure}^{heat\,;\,803\,K}}

B Red phosphorus is obtained by heating "A" at 573 K, and can be converted to "B" by heating at 803 K under pressure. A and B, respectively, are

A

\beta

-black phosphorus and white phosphorus.

B white phosphorus and

\beta

-black phosphorus.

C

\alpha

-black phosphorus and white phosphorus.

D white phosphorus and

\alpha

-black phosphorus.

Correct Answer

Option D

Solution

Red phosphrous is obtained by heating white phosphorous at $573 \mathrm{~K}$ in inert atmosphere.

When red phosphorous is heated in a sealed tube at $803 \mathrm{~K}$ , we get $\alpha$ -black phosphorous.

$\beta$ -Black phosphorous is prepared by heating white phosphorous at $473 \mathrm{~K}$ under high pressure.

Q99

The geometry around boron in the product 'B' formed from the following reaction is

\begin{aligned} &\mathrm{BF}_{3}+\mathrm{NaH} \stackrel{450 \mathrm{~K}}{\rightarrow} \mathrm{A}+\mathrm{NaF} \\\\ &\mathrm{A}+\mathrm{NMe}_{3} \rightarrow \mathrm{B} \end{aligned}

A trigonal planar

B tetrahedral

C pyramidal

D square planar

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{BF}_3+\mathrm{NaH} \stackrel{450 \mathrm{~K}}{\longrightarrow} \underset{\text{diborane}}{\mathrm{B}_2 \mathrm{H}_6}+\mathrm{NaF} \\\\ & \mathrm{B}_2 \mathrm{H}_6+\mathrm{NMe}_3 \longrightarrow 2 \underset{\text{symmetrical cleavage}}{\left[\mathrm{BH}_3 \leftarrow \mathrm{NMe}_3\right]} \end{aligned}

Q100

The total number of acidic oxides from the following list is NO, N2O, B2O3, N2O5, CO, SO3, P4O10

A 3

B 4

C 5

D 6

Correct Answer

Option B

Solution

\mathrm{NO}, \mathrm{N}_{2} \mathrm{O}, \mathrm{CO}

- neutral oxides

\mathrm{B}_{2} \mathrm{O}_{3}, \mathrm{~N}_{2} \mathrm{O}_{5}, \mathrm{SO}_{3}, \mathrm{P}_{4} \mathrm{O}_{10}-\text{ acidic oxides }