Chemistry Test

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Chemistry Test - 19

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Weekly Quiz Competition

Question 1
3 / -1

What is the number of possible isomers for the complex [Pt(NH3)2Cl2]?

A
Two

B
Three

C
Four

D
One

Solution

Pt(II) is always square planar and so there are only two isomers the cis-(Cl)₂Pt(NH₃)₂ and trans isomers (Cl)(H₃N)Pt(NH₃)(Cl).

The cis isomer is the famous anticancer drug cis platin
Question 2
3 / -1

Solution of sodium metal in liquid ammonia is strongly reducing due to the presence of

A
sodium hydride

B
sodamide

C
sodium atoms

D
solvated electrons

Solution

The solutions of metal in liquid ammonia are strong reducing agents due to solvated electrons.
Question 3
3 / -1

The major product of nitration of benzoic acid is

A
3 - nitrobenzoic acid

B
4 - nitrobenzoic acid

C
2 - nitrobenzoic acid

D
2, 4 - dinitrobenzoic acid

Solution

Meta product is the major product in this reaction.
Question 4
3 / -1

The alkali metals are chemically similar because

A
for each element, the atomicity is one

B
they have the same electronic configuration in the highest energy level

C
their outermost electrons have the same principal quantum number

D
their outermost electrons have the same energy

Solution

Alkali metals show similarity in their chemical behaviour due to similar valence shell electronic configuration,
Question 5
3 / -1

The percentage s-character of the hybrid orbitals in methane, ethene and ethyne are respectively

A
25, 33, 50

B
25, 50, 75

C
50, 75, 100

D
10, 20, 40

Solution

The hybrid orbitals used for bonding in methane (sp³), ethane (sp²) and ethyne (sp) have 25, 33 and 50% S-character.
Question 6
3 / -1

Which of the following metals is present in vitamin B₁₂?

A
Aluminium

B
Zinc

C
Iron

D
Cobalt

Solution

Vitamin B - 12 is actually a group of compounds called cobalamins, which contains the metal ion - cobalt.
Question 7
3 / -1

A group, which deactivates the benzene ring towards electrophilic substitution and directs the incoming group principally to the O and P position, is

A
−NH₂

B
−Cl

C
−NO₂

D
−C₂H₅

Solution

Electron accepting groups which make the substitution difficult are known as deactivating groups. The group or substituent already present on the ring also decides the position of incoming group. Ortho and para directing groups are as follow.

- CH₃, C₂H₅ (- R), - NH₂, - OH, halogens, (Cl, Br, I)
Question 8
3 / -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 9
3 / -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 10
3 / -1

In single state, the hybridisation of carbon atom of carbene is

A
sp²

B
sp

C
sp³

D
None of these

Solution

In singlet state carbon atom is in sp2 hybridisation state. One of sp2 hybrid orbital contains unshared electron pair with opposite spins and two sp2 hybrid orbitals form two covalent bonds with planar geometry. Net magnetic moment in this state is zero.

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Chemistry Test - 19

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Chemistry Test - 19

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SHARING IS CARING

Question 1

Two

Three

Four

One

Solution

Question 2

sodium hydride

sodamide

sodium atoms

solvated electrons

Solution

Question 3

3 - nitrobenzoic acid

4 - nitrobenzoic acid

2 - nitrobenzoic acid

2, 4 - dinitrobenzoic acid

Solution

Question 4

for each element, the atomicity is one

they have the same electronic configuration in the highest energy level

their outermost electrons have the same principal quantum number

their outermost electrons have the same energy

Solution

Question 5

25, 33, 50

25, 50, 75

50, 75, 100

10, 20, 40

Solution

Question 6

Aluminium

Zinc

Iron

Cobalt

Solution

Question 7

−NH2

−Cl

−NO2

−C2H5

Solution

Question 8

small size

high positive charge density

presence of vacant d orbitals

all of these

Solution

Question 9

+ 2

+1

+2 and +3

+3

Solution

Question 10

sp2

sp

sp3

None of these

Solution

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−NH₂

−NO₂

−C₂H₅

sp²

sp³