Chemistry Test

Result

Chemistry Test - 4

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

Flint glass contains

A
borax

B
excess of sodium carbonate

C
hydrofluoric acid

D
lead oxide

Solution

Traditionally, flint glasses are lead glasses containing around 4-60% lead oxide.
Question 2
4 / -1

The reason for complex formation in d-block element is

A
small size

B
high positive charge density

C
presence of vacant d orbitals

D
all of these

Solution

The tendency of the transition metals to form complex ions is due to:

(a) Small size of the metals ion

(b) Large ionic or nuclear charge

(c) Available d- orbitals

(d) Low basicity of the metal ions

(e) Availability of various oxidation state

(f) High lewis acid nature of electron pair accepting tendency of the metal ion.
Question 3
4 / -1

The most important oxidation state of copper is

A
+ 2

B
+1

C
+2 and +3

D
+3

Solution

Cu(II) or the cupric ion is the most important oxidation state of copper. Cu(II) is the oxidation state of copper generally encountered in water. Cupric ions are coordinated with six water molecules in solution; the arrangement of the water molecules is distorted in that there are four molecules bound closely to the copper in a planar array while the other two are more loosely bound in polar position. Addition of ligands such as NH3 to the solution will successively displace only the four planar water molecules. Most cupric compounds and complexes are blue or green in color. They are frequently soluble in water.
Question 4
4 / -1

Which of the following co-ordination compounds would exhibit optical isomerism?

A
Pentaamminenitrocobalt (III) iodide

B
Diamminedinitroplatinum (II)

C
Trans-dicyanobis (ethylenediamine)

D
Tris-(ethylendiamine) cobalt (III) bromide

Solution

Tris (ethylenediamine) cobalt (III) bromide shows optical iso - merism.

Two optical isomers are

Two enantiomers (non - super - impossible mirror images)

The molecule is chiral on the whole and exhibits optical isomerism.
Question 5
4 / -1

Acid catalysed hydration of alkenes except ethene leads to the formation of

A
mixture of secondary and tertiary alcohols

B
mixture of primary and secondary alcohols

C
secondary or tertiary alcohol

D
primary alcohol

Solution

For alkenes other then ethene, addition follows Markowinkov`s rule.
Question 6
4 / -1

A nitro group can be reduced to primary amino group by

A
Sn/conc. HCl

B
Zn dust

C
Zn/NH4Cl

D
Zn/NaOH

Solution

A nitro (- NO2) group can be reduced to the corresponding amino (- NH2) group by treatment with a reducing metal such as tin or zinc in the presence of acid. Sodium bicarbonate is used to neutralize the acid from the first step. This two-step sequence is an important route to aniline.
Question 7
4 / -1

In the silver plating of copper, K[Ag(CN)₂] is used instead of AgNO₃. This is because

A
a thin layer of Ag is formed on Cu

B
more voltage is required

C
Ag⁺ ions are completely removed from solution

D
of less availability of Ag⁺ ions as Cu cannot displace Ag from [Ag(CN)₂]^- ion

Solution

In the silver plating of copper, K[Ag(CN)₂] is used instead of AgNO₃. This is because of less availability of Ag⁺ ions as Cu cannot displace Ag from [Ag(CN)₂]^-ion
Question 8
4 / -1

Which of the following conformations of butane is most unstable?

A
Anti-eclipsed

B
Skew eclipsed

C
Fully eclipsed

D
Eclipsed

Solution

Fully eclipsed conformations of butane are most unstable as the groups attached to the two carbon atoms are closest to each other.
Question 9
4 / -1

Which of the following compounds is purified by steam distillation?

A
Aniline

B
Benzoic acid

C
Petroleum

D
Naphathalene

Solution

Aniline is purified by steam distillation.
Question 10
4 / -1

The correct order of bond energies of the following bonds is:

(I) C(sp³)-C(sp³), (II) C(sp²)-C(sp²), (III) C(sp)-C(sp)

A
I < II < III

B
I > II > III

C
II > III > I

D
III > I > II

Solution

The sp hybridised orbital has 50% 's' character.

The sp² hybridised orbital has 33% 's' character.

The sp³ hybridised orbital has 25% 's' character.

Bond energy increases as the 's' character increases.

Hence, option (1) is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Chemistry Test - 4

Result

Chemistry Test - 4

Score

-

Rank

-

SHARING IS CARING

Question 1

borax

excess of sodium carbonate

hydrofluoric acid

lead oxide

Solution

Question 2

small size

high positive charge density

presence of vacant d orbitals

all of these

Solution

Question 3

+ 2

+1

+2 and +3

+3

Solution

Question 4

Pentaamminenitrocobalt (III) iodide

Diamminedinitroplatinum (II)

Trans-dicyanobis (ethylenediamine)

Tris-(ethylendiamine) cobalt (III) bromide

Solution

Question 5

mixture of secondary and tertiary alcohols

mixture of primary and secondary alcohols

secondary or tertiary alcohol

primary alcohol

Solution

Question 6

Sn/conc. HCl

Zn dust

Zn/NH4Cl

Zn/NaOH

Solution

Question 7

a thin layer of Ag is formed on Cu

more voltage is required

Ag+ ions are completely removed from solution

of less availability of Ag+ ions as Cu cannot displace Ag from [Ag(CN)2]- ion

Solution

Question 8

Anti-eclipsed

Skew eclipsed

Fully eclipsed

Eclipsed

Solution

Question 9

Aniline

Benzoic acid

Petroleum

Naphathalene

Solution

Question 10

I < II < III

I > II > III

II > III > I

III > I > II

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

Ag⁺ ions are completely removed from solution

of less availability of Ag⁺ ions as Cu cannot displace Ag from [Ag(CN)₂]^- ion