Today's Test - Chemistry

The order of reactivity of phenyl magnesium bromide(PhMgBr) with the following compounds

III > II > I
II > I > III
I > III > II
I > II > III

Solution

The reactivity of the carbonyl group toward the nucleophilic addition reactions depend upon themagnitude of the positive charge on the carbonyl carbon atom (electronic factor) and also on the crowding around the carbonyl carbon atom in the transition state (steric factor). Both these factors predict the following order

(due to steric crowding).

The second order Bragg diffraction of X-rays with = 1.00 Å from a set of parallel planes in a metal occurs at an angle 60º. The distance between the scattering planes in the crystal is

0.575 Å
1.00 Å
2.00 Å
1.15 Å

Solution

Ketones react with Mg-Hg over water to give :

pinacolone
pinacols
alcohols
none of these

Solution

Among BMe₃, BCl₃, BF₃ and B₂H₆, which one will be the best Lewis acid ?

BCl₃
BMe₃
B₂H₆
BF₃

Solution

In BCl₃ and BF₃ due to –I effect of F and Cl these are better Lewis acids than BMe₃ and B₂H₆. Among BCl₃ and BF₃, the former is better Lewis acid as in it p?-p? back bonding is lesser than in BF₃.

The bond angle between two hybrid orbitals is 105°. The percentage of s-character of hybrid orbital is between

50 - 55%
9 - 12%
22 - 23%
11 - 12%

Solution

HCHO, HCOONH₄
HCOOH,HCOONH₄
HCOOH, HCONH₂
HCHO, HCONH₂

Solution

The test used for the distinction of HCOOH and CH₃COOH is treatment with :

sodium
sodium hydroxide
Tollen's reagent
alcohol

Solution

The test used for the distinction of HCOOH and CHCOOH₃ is treatment with Tollen’s reagent because formic acid can reduce Tollen’s reagent but acetic acid can not.

Which one is most reactive towards NS₁ reaction ?

C₆H₅CH(C₆H₅)Br
C₆H₅CH(CH₃)Br
C₆H₅C(CH₃)(C₆H₅)Br
C₆H₅CH₂

Solution

SN₁ reactions involve the formation of carbocations,order of stability of carbocationis 3° > 2° > 1° hence higher the stability of carbocation, more will be the reactivity of the parent alkyl halide. Moreover the tertiary carbocation formed from (c) is stabilized by two phenyl groups.

Set of isoelectronic species is

N₂, CO₂, CN^-,O^-
N₂, H₂S, CO
N₂ , CO, CN^- , O₂⁺²
Ca, Mg, Cl

Solution

The molecule which contains same number of electrons are called isoelectronic. eg. N₂ = CO = CN^- = O₂⁺² = 14e

Consider the following statements :
1.Atomic hydrogen is obtained by passing hydrogen through an electric arc.
2.Hydrogen gas will not reduce heated aluminium oxide.
3.Finely divided palladium adsorbs large volume of hydrogen gas
4.Pure nascent hydrogen is best obtained by reacting Na with C₂H₅OH Which of the above statements is/are correct ?

1 alone
2 alone
1, 2 and 3
2, 3 and 4

Solution

Pure hydrogen is evolved by reacting absolute alcohol and Na
C₂H₅OH+Na→C₂H₅ONa+½H₂
other statements are correct. See text.

Alum helps in purifying water by

forming Si complex with clay particles
sulphate part which combines with the dirt and remove sit
coagulaing the mud particles
making mud water soluble.

Solution

Alum furnishes Al³⁺ ions which bring about coagulation of negatively charged clay particles, bacteria etc.

The Ce (Z = 58) belongs to III rd group of periodic table.If it furnish one α particle to form an element ‘X’, then X belongs to

III^rd group
II^nd group
I^st group
zero group

Solution

₅₈Ce is lanthanide. Lanthanides are from 57 to 71 all present in III^rd group. Hence ₅₈Ce forms ₅₆X element on emission of one α particle with belongs to II^nd group.

Cyclohexene
Cyclohexadiene
Cyclogexatriene
None of these

Solution

An important reaction of acetone is auto condensation in presence of concentrated sulphuric acid to give the aromatic compound……

Mesitylene
Mesityl oxide
Trioxan
Phorone

Solution

Oxidation number of ‘N’ in N₃H (hydrazoic acid) is:

-\(\frac{1}{3}\)
+ 3
zero
– 3

Solution

Let oxidation number of N in N₃H =x
3x+ 1= 0 or 3x= – 1
∴ x= –\(\frac{1}{3}\)

In India at the occasion of marriages, the fire works used give green flame. Which one of the following radicals maybe present?

Na⁺
K⁺
Ba²⁺
Ca²⁺

The highest boiling point is expected for :

iso-octane
n-octane
2, 2, 3, 3-tetramethylbutane
n-butane

Solution

For isomeric alkanes,the one having longest straight chain has highest b.p. because of larger surface area.

n-Propyl bromide on treatment with ethanolic potassium hydroxide produces

propane
propene
propyne
propanol

Solution

Solution

Species acting as both Bronsted acid and base, is

(HSO4)^-
Na₂CO₃
NH₃
OH^-

Solution

(HSO₄)^- can accept and donate a proton
(HSO₄)^- + H⁺ ⇌ H₂SO₄ (acting as base)
(HSO₄)^- – H⁺ ⇌ SO₄^2-. (acting as acid)