106
Q:

Boron cannot form which one of the following anions?

(1) BO2-

(2) BF63-

(3) BH4-

(4) B(OH)4-

A) Option 1 B) Option 2
C) Option 3 D) Option 4

Answer:   B) Option 2



Explanation:

Due to absence of low lying vacant d orbital in B, sp3d2hybridization is not possible hence BF63-will not formed. 

Subject: Chemistry
Exam Prep: AIEEE
Q:

The reduction potential of hydrogen half-cell will be negative if :
(1) PH2= 2 atm and [H+] = 2.0 M
(2) PH2= 1 atm and [H+] = 2.0 M
(3) PH2= 1 atm and [H+] = 1.0 M
(4) PH2= 2 atm and [H+] = 1.0 M

A) Option 1 B) Option 2
C) Option 3 D) Option 4
 
Answer & Explanation Answer: D) Option 4

Explanation:

 H+-e-12H2   

Apply Nernst equation  

 

E=0-0.0591logPH212H+  

 

 E=-0.0591log2121  

 

Therefore E is Negative

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26 15294
Q:

Trichloroacetaldehyde was subject to Cannizzaro's reaction by using NaOH. The mixture of the products contains sodium trichloroacetate and another compound. The other compound is :

A) Chloroform B) 2, 2, 2-Trichloroethanol
C) Trichloromethanol D) 2, 2, 2-Trichloropropanol
 
Answer & Explanation Answer: B) 2, 2, 2-Trichloroethanol

Explanation:

Cl|Cl-C-CHO|ClCannizzaro's ReactionNaOH Cl|Cl-C-COONa|Cl+ Cl|Cl-C-CH2 |21Cl-OH

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7 5220
Q:

Silver Mirror test is given by which one of the following compounds?

1. Benzophenone

2. Acetaldehyde

3. Acetone

4. Formaldehyde

A) Only 1 B) Only 2
C) 1 and 3 D) 2 and 4
 
Answer & Explanation Answer: D) 2 and 4

Explanation:

Both Formaldehyde and Acetaldehyde will give this test

 HCHO[Ag(NH3)2+]    Silver mirrorAg + Organic compound

 CH3-CHO [Ag(NH3)2+] AgSilver mirror+ Organic compound 

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66 64681
Q:

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm3 at 27°C is to a volume of 100 dm3

 

A) 42.3 J/ mole / K B) 38.3 J/ mole / K
C) 35.8 J/ mole / K D) 32.3 J/ mole / K
 
Answer & Explanation Answer: B) 38.3 J/ mole / K

Explanation:

s=nRlnv2v1 

=2.303 nR logv2v1 = 2.303×2×8.314×log10010 = 38.3 J mol-1k-1

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43 16196
Q:

A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions is at 680 nm, the other is at :

A) 518 nm B) 1035 nm
C) 325 nm D) 743 nm
 
Answer & Explanation Answer: D) 743 nm

Explanation:

1λ=1λ1+1λ2

1355=1680+1λ2

1λ2=680-355680×355

λ2=743nm.

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12 9303
Q:

In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is

1) A2B5

2) A3B2

3) A2B3

4) A3B4

A) Option 1 B) Option 2
C) Option 3 D) Option 4
 
Answer & Explanation Answer: A) Option 1

Explanation:

 ZA  = 88  ; ZB = 52

So formula of compound is AB52 

i.e., A2B5.

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13 9147
Q:

The degree of dissociation (α) of a weak electrolyte, AxBy is related to van't Hoff factor (i) by the expression:

1. α=x+y+1i-1

2.α=x+y-1i-1

3. α=i-1x+y+1

4. α=i-1x+y-1

A) 1 is correct B) 2 is correct
C) 3 is correct D) 4 is correct
 
Answer & Explanation Answer: D) 4 is correct

Explanation:

Van't Hoff factor (i) = Observed Colligative PropertyNormal Colligative Property

 

AxBy1-αxA+yxα+yB-xyα  

 

Total Moles = 1-α+xα+yα = 1-αx+y-1

 

i=1+αx+y-11

 

α=i-1x+y-1

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6 7266
Q:

Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of :

A) An acetylenic triple bond B) Two ethylenic double bonds
C) A vinyl group D) An isopropyl group
 
Answer & Explanation Answer: C) A vinyl group

Explanation:

C=}CH2OzonolysisHCHO

Vinylic group.

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3 4817