JEE Advanced Mix Test 13

Result

JEE Advanced Mix Test 13

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Weekly Quiz Competition

Question 1
4 / -1

The hybridisation and geometry of the anion of ICl₃ in liquid phase is:

A
sp³ and bent

B
sp³d² and square planar

C
sp³d and see–saw

D
sp³ and tetrahedral

Solution
Question 2
4 / -1

Select the incorrect statement about the XeF₆.

A
It has capped octahedron or distorted octahedral structure with one lone pair.

B
In various non-aqueous solvents it forms a tetramer Xe₄F₂₄.

C
Solid Xenon hexafluoride is polymorphic.

D
None of these

Solution

In tetramer, four square pyramidal XeF₅⁺ ions are joined to two similar ions by means of two bridging F^- ions. The
Xe-F distances are 1.84 Å on the square pyramidal units and 2.23 Å and 2.60 Å in the bridging groups. The solid has various crystalline forms, of which three are tetrameric and a fourth has both hexamers and tetramers.
Question 3
4 / -1

Based on the Valence Shell Electron Pair Repulsion (VSEPR) model, what is the geometry around the sulphur, carbon and nitrogen in the thiourea – S , S – dioxide, O₂SC(NH₂)₂? (consider the Lewis structure with zero formal charges on all atoms).

A
Trigonal planar, T–shaped, trigonal planar respectively.

B
T-shaped, trigonal planar, trigonal planar respectively.

C
Trigonal planar, trigonal planar, trigonal pyramidal respectively.

D
Trigonal planar, trigonal planar, trigonal planar respectively.

Solution
Question 4
4 / -1

The following flow diagram represents the manufacturing of sodium carbonate

Which of the following options describes the reagents, products and reaction conditions (given in parentheses)?

A

B

C

D

Solution
Question 5
4 / -1

When boron is fused with potassium hydroxide which pair of species are formed ?

A

B

C

D

Solution
Question 6
4 / -1

A
borazine and hydrogen

B
boron and hydrogen

C
boron nitride and hydrogen

D
None of these

Solution
Question 7
4 / -1

Orthosilicate ions,(SiO₄^4-) undergo polycondensation to forms pyrosilicate, [O₃Si – O – SiO₃]^6-, in presence of :

A
acid

B
water

C
both (A) and (B)

D
none of these

Solution
Question 8
4 / -1

Select the correct statement(s) with respect to the pπ–dπ dative bond.

A
In (Ph₃Si)₂O , the Si—O—Si group is nearly linear.

B
Silanols such as (CH₃)₃ SiOH are stronger protonic acids then their carbon analogs.

C
Trisilysl phosphine (H₃Si)₃ P is planar.

D
All are correct

Solution

(A) Steric repulsions of bulkier groups and pπ–dπ dative bonding favour for a linear Si–O–Si group.

(B) Due to stabillization of the conjugate base anion by O(pπ) → Si(dπ) bonding motion.

(C) It is pyramidal because pπ–dπ bonding is not effective due to the bigger size of phosphorus atom.

(D) Most effective pπ–dπ overlapping due to small size of chlorine.
Question 9
4 / -1

Which of the following statements is/are correct ?

A
Out of trimethylamine and trimethylphosphine, trimethylamine has higher dipole moment ?

B
Out of (SiH₃)2O and (CH3)2O, (SiH₃)₂O is more basic.

C
C–C bond length (in pm) in C₂ molecule is greater than O–O bond length in O₂ molecule.

D
All are correct

Solution

(A) Nitrogen is more electronegative than phosphorus.

So, dipole moment of trimethylamine is greater than trimethy phosphine.

In trisilyl ether the lone pair of electron on oxygen atom is less easily available for donation because of pπ-dπ delocalisation due to presence of the vacant d-orbital with Si. This however is not possible with carbon in CH₃–O–CH₃ due to the absence of d-orbital making it more basic.
(C) Bond order of C₂
and O₂ are same i.e., 2. In C₂ molecules both bonds are π-bonds whereas, there is one σ and one π-bond in O₂ molecule
C₂ = 131 pm ; O₂ = 121 pm.
Question 10
4 / -1

Highly pure dilute solution of sodium in liquid ammonia :

A
on evaporation yield metals

B
exhibits electrical conductivity

C
acts as powerful reducing agent

D
All are correct

Solution

Blue colour of the solution is due to ammoniated electrons and good conductor of electricity because of both ammoniated cations and ammoniated electrons.

(change of oxidation states from 0 to – 1).