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CBSE Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 4 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Coordination Compounds
Quiz 4
What is the electronic configuration of elements of
I
I
I
rd group
Report Question
0%
1
s
2
,
2
s
2
2
p
s
3
0%
1
s
2
,
2
s
2
2
p
6
,
3
s
2
3
p
1
0%
1
s
2
,
2
s
2
2
p
6
0%
1
s
2
,
2
s
2
2
p
6
,
3
s
1
Explanation
1
s
2
,
2
s
2
2
p
6
,
3
s
2
3
p
1
As it is p block element therefore the group is
12
+
1
=
13
(III A group)
The compounds
[
C
o
(
S
O
4
)
(
N
H
3
)
5
]
B
r
and
[
C
o
(
S
O
4
)
(
N
H
3
)
5
]
C
l
represents:
Report Question
0%
linkage isomerism
0%
ionisation isomerism
0%
coordination isomerism
0%
no isomerism
Explanation
The compounds
[
C
o
(
S
O
4
)
(
N
H
3
)
5
]
B
r
and
[
C
o
(
S
O
4
)
(
N
H
3
)
5
]
C
l
represent no isomerism as they have different molecular formulae.
Note: Isomers have the same molecular formula.
TEL is a compound used as
Report Question
0%
antibiotic
0%
antiseptic
0%
antiknocking
0%
None of these
Explanation
Anti knocking compounds are the chemicals which reduce knocking for improving the quality of gasoline. example:- Tetraethyl lead (TEL)
Hence, the correct option is
C
The solubilities of carbonates decrease down the magnesium group due to decrease in
Report Question
0%
lattice energies of solids
0%
hydration energies of carbon
0%
inter ionic attraction
0%
entropy of solution formation
Which of the following statements are correct?
I)
C
O
forms a complex with
C
u
C
l
in
c
o
n
.
H
C
l
II)
N
i
forms a volatile carbonyl compound with
C
O
III)
C
O
act as a lewis acid
(IV)
C
O
oxidizes
C
u
to
C
u
O
Report Question
0%
I and III
0%
I and II
0%
III and IV
0%
I and IV
Explanation
i)
Correct
. When
C
u
(
I
)
C
l
is dissolved in conc.
H
C
l
, it absorbs
C
O
forming an adduct
C
u
C
l
(
C
O
)
.
ii)
Correct
N
i
+
4
C
O
→
N
i
(
C
O
)
4
volatile
iii)
Correct
.
C
O
can act as a Lewis acid because
C
O
has a vacant
Π
∗
orbital (antibonding) that can accept electrons from an electron pair donor.
Ans. (A) and (B)
2.33
g of compound X(empirical formula
C
o
H
12
N
4
C
l
3
)
upon treatment with excess
A
g
N
O
3
solution produces
1.435
g of a white precipitate. The primary and secondary valences of cobalt in compound X, respectively, are.
[Given : Atomic mass:
C
o
=
59
,
C
l
=
35.5
,
A
g
=
108
].
Report Question
0%
3
,
6
0%
3
,
4
0%
2
,
4
0%
4
,
3
Explanation
Since
0.01
mole of
X
produces
0.01
mole of
A
g
C
l
,hence one
C
l
−
is out of coordination sphere or complex is
[
C
O
(
N
H
3
)
4
C
l
2
]
C
l
AND HENCE
C
O
(
3
+
)
,has
6
coordination number
Which can exist both as diastereoisomer and enantiomer?
Report Question
0%
[
P
t
(
e
n
)
3
]
4
+
0%
[
P
t
(
e
n
)
2
C
l
B
r
]
2
+
0%
[
R
u
(
N
H
3
)
4
C
l
2
]
0
0%
[
P
t
C
l
2
B
r
2
]
0
Explanation
It exists as lis trans which is diastereomer and its cis form is optically active and exits as enantiomer also,where as only exists as enantiomers.
Barium ions,
C
N
−
and
C
o
2
+
form an ionic complex. If that complex is supposed to be
75
%
ionised in water with vant Hoff factor 'i' equal to four, then the coordination number of
C
o
2
+
in the complex can be.
Report Question
0%
Six
0%
Five
0%
Four
0%
Six and Four both
Explanation
Complex formed is
B
q
2
[
C
O
(
C
N
)
6
]
in which oxidation of
C
O
is
(
2
+
)
and
C
N
of
C
O
is six.
When one mole of each of the following complex is treated with an excess of
A
g
N
O
3
.which will give maximum amount of
A
g
C
l
?
Report Question
0%
[
C
o
(
N
H
3
)
6
]
C
l
3
0%
[
C
o
(
N
H
3
)
5
C
l
]
C
l
2
0%
[
C
o
(
N
H
3
)
C
l
2
]
C
l
0%
[
C
o
(
N
H
3
)
3
C
l
3
]
Explanation
In coordination compounds, the counter ions can react to give another product but the complex ion will not split in a solution.
Here, when one mole of a complex given when reacted with
A
g
N
O
3
then the maximum counter ion compound will give the maximum amount of
A
g
C
l
.
Hence,
[
C
o
(
N
H
3
)
6
]
C
l
3
having a maximum number of 3 counter ions in the given options will produce more amount of
A
g
C
l
.
[
C
o
(
N
H
3
)
6
]
C
l
3
→
[
C
o
(
N
H
3
)
6
]
3
+
(
a
q
)
+
3
C
l
−
(
a
q
)
A
g
N
O
3
+
C
l
−
(
a
q
)
→
A
g
C
l
+
N
O
−
3
(
a
q
)
Hexa ammine nickel (II) hexa nitro cobaltate (III) can be written as:
Report Question
0%
[
N
i
(
N
H
3
)
6
C
o
(
N
O
2
)
6
]
0%
[
N
i
(
N
H
3
)
6
]
3
[
C
o
(
N
O
2
)
6
]
2
0%
[
N
i
(
N
H
3
)
6
]
[
C
o
(
N
O
2
)
6
]
2
0%
[
N
i
(
N
H
3
)
6
(
N
O
2
)
6
]
C
o
Explanation
[
N
i
(
N
H
3
)
6
]
3
[
C
O
(
N
O
2
)
6
]
2
≡
3
[
N
i
(
N
H
3
)
6
]
2
+
|
2
[
C
O
(
N
O
2
)
6
]
3
−
Since
3
N
i
(
2
+
)
balance with
2
C
o
(
3
+
)
Total no. of possible isomers for complex compound
[
C
u
(
N
H
3
)
4
]
[
P
t
C
l
4
]
are:
Report Question
0%
3
0%
2
0%
4
0%
none of these
Explanation
[
C
u
(
N
H
3
)
4
]
[
P
t
C
l
4
]
,
[
C
u
(
N
H
3
)
3
C
l
]
[
P
t
C
l
3
N
H
3
]
,
[
C
u
(
N
H
3
)
2
C
l
2
]
[
P
t
C
l
2
(
N
H
3
)
2
]
,
[
C
u
(
N
H
3
)
C
l
3
]
[
P
t
C
l
(
N
H
3
)
3
]
,
[
C
u
(
C
l
)
4
]
[
P
t
(
N
H
3
)
4
]
So five isomers are possible ,but third one do not exist due to unsatisfie charge on complex so only
4
exist.
Indicate the incorrect statement:
Report Question
0%
Number of hybrid orbitals formed is equal to no. of atomic involved
0%
2
p
x
and
2
p
y
orbitals of carbon can be hybridized to yield two new more stable orbitals
0%
Effective hybridisation is not possible with orbitals of widely different energies
0%
The concept of hybridization has a greater significance in the
V
B
theory of localised orbitals than
M
O
theory.
Explanation
2
p
x
&
2
p
y
orbitals of Carbon cannot be hybridized to yield
2
more stable orbitals. This is because hybridization takes place between orbitals of different atoms thus the statement of option
B
is incorrect. Hence answer is option
B
.
A violet colour compound is formed in detection of
S
in a compound:
Report Question
0%
N
a
4
[
F
e
(
C
N
)
5
N
O
S
]
0%
N
a
3
[
F
e
(
C
N
)
5
N
O
S
]
0%
N
a
2
[
F
e
(
C
N
)
5
N
O
S
]
0%
N
a
[
F
e
(
C
N
)
5
N
O
S
]
Explanation
N
a
2
S
+
N
a
2
[
F
e
(
C
N
)
5
N
O
]
→
N
a
4
[
F
e
(
C
N
)
5
N
O
S
]
(Violet Color Compond)
Which of the following complex has five unpaired electrons?
Report Question
0%
[
M
n
(
H
2
O
)
6
]
2
+
0%
[
M
n
(
C
N
)
6
]
3
−
0%
[
C
r
C
l
3
(
H
2
O
3
]
0%
[
A
g
(
N
H
3
)
2
]
+
Explanation
M
n
2
+
has
d
5
configuration, since
H
2
O
is weak ligand field, so it does not causes pairing of electrons and hence it has 5 unpaired electrons.
In which of the following pair of complexes, the experimental magnetic moment and the geometric shapes same?
Report Question
0%
K
[
M
n
O
4
]
and
K
2
[
N
i
C
l
4
]
0%
K
2
[
N
i
(
C
N
)
4
]
and
K
4
[
N
i
(
C
N
)
4
]
0%
K
2
[
N
i
(
C
N
)
4
]
and
K
2
[
N
i
(
N
H
3
)
2
C
l
2
]
0%
K
3
[
F
e
(
C
N
)
6
]
and
K
4
[
F
e
(
C
N
)
6
]
Explanation
K
2
[
N
i
(
C
N
)
4
]
⇒
N
i
+
2
⇒
3
d
8
4
s
0
With cyanide,
N
i
+
2
form square planar and
diamagnetic complex.
K
2
[
N
i
(
N
H
3
)
2
C
l
2
]
⇒
N
i
0
⇒
3
d
8
4
s
2
Ammonia is a strong field ligand, There won't be any lone pair. Diamagnetic and square planar
complex.
Option C is correct.
Which of the following will give maximum number of isomer?
Report Question
0%
[
C
o
(
p
y
)
3
(
N
H
3
)
3
]
3
+
0%
[
N
i
(
e
n
)
(
N
H
3
)
4
]
2
+
0%
[
F
e
(
C
2
O
4
)
(
e
n
)
2
]
2
−
0%
[
C
o
(
N
O
2
)
2
(
N
H
3
)
4
]
+
Explanation
[
C
o
(
N
O
2
)
2
(
N
H
3
)
4
]
+
shows the following isomerism:
1) Linkage isomerism through oxygen
[
−
O
N
O
]
and nitrogen
[
−
N
O
2
]
.
2) Geometrical isomerism: 2 cis isomers and 1 trans isomer
[
N
i
(
e
n
)
(
N
H
3
)
4
]
2
+
and
[
F
e
(
C
2
O
4
)
(
e
n
)
2
]
2
−
will show optical isomerism. But total isomers are 4. But it has 5 isomerism.
Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers:
Report Question
0%
K
[
C
r
(
H
2
O
)
2
(
C
2
O
4
)
2
]
0%
[
C
o
(
e
n
)
3
]
C
l
3
0%
[
C
o
(
N
H
3
)
5
(
N
O
2
)
]
(
N
O
3
)
2
0%
[
P
t
(
N
H
3
)
(
H
2
O
)
C
l
2
]
Explanation
A
- exhibits cis trans.
B
- Exhibits optical Isomerism.
C
- Exhibits linkage Isomerism
(
N
O
2
,
−
O
N
O
)
D
- Cis-Trans Isomerism.
Which of the following pairs of complexes whose aqueous solutions gives pale yellow and white precipitates respectively with
0.1
M
A
g
N
O
3
?
Report Question
0%
[
P
t
(
N
H
3
)
4
B
r
2
]
C
l
2
and
[
P
t
(
N
H
3
)
4
C
l
2
]
B
r
2
0%
[
C
o
(
N
H
3
)
5
N
O
3
]
B
r
and
[
C
o
(
N
H
3
)
5
B
r
]
N
O
3
0%
[
P
t
(
N
H
3
)
4
C
l
2
]
B
r
2
and
[
P
t
(
N
H
3
)
4
B
r
2
]
C
l
2
0%
[
C
o
(
N
H
3
)
5
N
O
3
]
C
l
and
[
C
o
(
N
H
3
)
5
C
l
]
N
O
3
Explanation
The complex having
B
r
and
C
l
out of coordination sphere only gives
A
g
B
r
(pale yellow ) and
A
g
C
l
(white ppt.) respectively upon reaction with
A
g
N
O
3
.
A six coordinate complex of formula
C
r
C
l
3
.
6
H
2
O
has green colour. A
0.1
M
solution of the complex when treated with excess of
A
g
N
O
3
gave
28.7
g
of white precipitate. The formula of the complex would be___________.
Report Question
0%
[
C
r
(
H
2
O
)
6
]
C
l
3
0%
[
C
r
(
H
2
O
)
5
C
l
]
C
l
2
.
H
2
O
0%
[
C
r
(
H
2
O
)
4
C
l
]
C
l
2
.
2
H
2
O
0%
[
C
r
(
H
2
O
)
3
C
l
3
]
.
3
H
2
O
Explanation
Since
0.1
M
complex gives
0.2
mole
A
g
c
l
means
2
C
l
−
are ionisable or out of coordination sphere so, complex is
[
C
r
(
H
2
O
)
5
C
l
]
C
l
2
.
H
2
O
Which of the following complexes exists in facial and meridional forms?
Report Question
0%
K
[
C
r
(
H
2
O
)
2
(
C
2
O
4
)
2
]
0%
[
C
o
(
N
H
)
3
(
N
O
2
)
3
]
0%
[
C
o
(
N
H
3
)
5
(
N
O
2
)
]
(
N
O
3
)
2
0%
[
P
t
(
N
H
3
)
(
H
2
O
)
C
l
2
]
The halogen form compound among themselves with the formula
A
A
′
,
A
A
′
3
,
A
A
′
5
and
A
A
′
7
where
A
is the heavier halogen. Which of the following pairs representing their structure and being polar and non-polar are correct ?
(A)
A
A
′
linear, polar
(B)
A
A
′
3
T
−
Shaped, polar
(C)
A
A
′
5
square pyramidal, polar
(D)
A
A
′
7
pentagonal bipyramidal, non-polar
Report Question
0%
(
A
)
and
(
C
)
are correct
0%
(
A
)
,
(
B
)
And
(
C
)
are correct
0%
(
B
)
,
(
C
)
and
(
D
)
are correct
0%
(
A
)
,
(
B
)
,
(
C
)
and
(
D
)
are correct
Explanation
A
A
′
linear,polar as both halogen atoms with different electronegativity has polarity. Eg-
I
−
C
l
A
A
′
5
square pyramidal and polar due to the high electronegativity of other atom. Eg-
I
F
5
Which of the following compounds would exhibit co-ordination isomerism?
Report Question
0%
[
C
r
(
H
2
O
)
5
]
C
l
2
0%
[
C
r
(
N
H
3
)
5
]
[
C
o
(
C
N
)
5
]
0%
[
C
r
(
e
n
)
2
]
N
O
2
0%
[
N
i
(
N
H
3
)
5
]
[
B
F
4
]
3
Explanation
Co-ordination isomerism is a form of structural isomerism in which the composition of the complexion varies. In a coordination isomer, the total ratio ligand to metal remains the same, but the ligand attached to specific metal ion change. In B option ratio of metal to the ligand is the same and they can exchange their ligands (
N
H
3
,
C
N
)
.
Several octahedral complexes are possible from combinations of
C
o
3
+
,
C
l
−
and
N
H
3
. Primary valence do not have chloride ion. The correct statement(s) regarding the octahedral coordination entities having the formula .
[
C
o
(
N
H
3
)
n
C
l
6
−
n
]
(
n
−
3
)
+
with
n
≥
3
, is/ are_____________.
Report Question
0%
At most six octahedral complexes are possible
0%
One of the complexes is homoleptic
0%
All the complexes are paramagnetic
0%
Some of the complexes dissociate in water to give
C
o
3
+
and
C
l
−
ions
Explanation
[
C
o
(
N
H
3
)
n
C
l
6
−
n
]
(
n
−
3
)
+
n
≥
3
For
n
=
3
,
n
=
4
,
n
=
5
,
n
=
6
For
n
=
6
⇒
[
C
o
(
N
H
3
)
6
]
3
+
=
is homoleptic due to all 6 same ligands.
C
o
3
+
⇒
[
A
r
]
3
d
6
with strong field ligand, it is diamagnetic.
For
n
=
5
⇒
[
C
o
(
N
H
3
)
5
C
l
]
2
+
.
C
o
2
+
⇒
[
A
r
]
3
d
7
with strong field ligand, it is paramagnetic.
For
n
=
4
⇒
[
C
o
(
N
H
3
)
4
C
l
2
]
+
.
C
o
+
⇒
[
A
r
]
3
d
8
with strong field ligand, it is paramagnetic.
For
n
=
3
⇒
[
C
o
(
N
H
3
)
3
C
l
3
]
.
C
o
⇒
[
A
r
]
4
s
2
3
d
7
with strong field ligand, it is paramagnetic.
Only
4
octahedral complex are possible.
One of them is homoleptic.
All the complexes are not paramagnetic.
No complex carries
C
o
3
+
and
C
l
−
both.
Option B is only correct.
Which among the following forms octahedral complex?
Report Question
0%
d
4
(low spin)
0%
d
8
(high spin)
0%
d
6
(low spin)
0%
All of these
Explanation
writing the respective configurations
Hence, all can form octahedral complexes.
What is the oxidation number of gold in the complex
[
A
u
C
l
4
]
1
−
?
Report Question
0%
+
4
0%
+
3
0%
+
2
0%
+
1
Explanation
Oxidation state of chlorine
C
l
is
−
1
.
Let the oxidation number of gold
A
u
be
x
.
∴
x
+
4
×
(
−
1
)
=
−
1
⟹
x
=
+
3
Thus oxidation number of gold is
+
3
.
Correct answer is option B.
Compound
[
C
o
(
C
N
)
2
(
e
n
)
2
]
C
l
can show:
Report Question
0%
only optical isomerism
0%
only geometrical and optical isomerism
0%
only structural isomerism
0%
structural, geometrical and optical isomerism
Explanation
Complex having
[
m
(
A
A
)
2
B
2
]
can show cis-trans and when it exits in Cis-, it shows optical isomerism also.
Which of the following statements is true about hybridisation?
Report Question
0%
The hybridised orbitals have different energies for each orbital.
0%
The number of hybrid orbitals is equal to the number of atomic orbitals that are hybridised.
0%
Hybrid orbitals form multiple bonds.
0%
The orbitals with different energies undergo hybridisation.
Explanation
In hybridization, the number of hybrid orbitals is equal to the number of atomic orbitals that are hybridized.
Let's take an example:-
C
H
4
Carbon electronic configuration is
2
S
2
2
P
2
in ground state, in excited state one electron of
s
orbital move into
p
orbital and thus we get
4
unpaired electrons in atomic orbital which will take part in bonding,
The hybridisation of carbon in
C
H
4
is
s
p
3
, means
4
hybrid orbitals. (one
s
orbital and three
p
orbitals)
So the correct option is [B]
If the value of C.F.S.E. for "Ni" is
Δ
0
then it is for Pd should be ?
Report Question
0%
1.1
Δ
0
0%
0.5
Δ
0
0%
1.5
Δ
0
0%
none of the above
Explanation
Ongoing from
3
d
to
4
d
series C.F.S.E increases by
50
%
Now, for
N
i
CFSE
=
△
o
⇒
For
P
t
=
△
o
+
50
100
△
o
=
1.5
△
o
Which one of the following complexes has largest value of
Δ
0
?
Report Question
0%
[
C
r
(
H
2
O
)
6
]
+
2
0%
[
C
r
(
H
2
O
)
6
]
+
3
0%
[
M
n
(
H
2
O
)
6
]
+
2
0%
[
M
n
(
H
2
O
)
6
]
+
3
Explanation
The value of
△
o
increases with increase in oxidation state of central metal, since maximum charges causes more left and hence orbital contracts lead to more
C
F
S
E
(
△
o
)
c
r
3
+
→
t
3
2
g
e
g
o
→
C
F
S
E
=
|
−
6
/
5
|
△
o
M
n
3
+
→
d
4
→
t
2
g
3
e
g
1
→
C
F
S
E
=
−
6
/
5
△
o
+
3
/
5
△
o
=
|
−
3
/
5
|
△
o
∴
△
o
=
c
r
3
+
>
M
n
3
+
According to Werner's theory of coordination compounds:
Report Question
0%
primary valency is ionisable
0%
secondary valency is ionisable
0%
primary and secondary valencies are ionisable
0%
neither primary nor secondary valency is ionisable
Explanation
Primary valency is ionisable according to Werner's theory of coordination compounds.
According to Werner's coordination theory, there are two kinds of valency, primary and secondary. The primary valency of a central metal ion is satisfied with anions.
For example, in
[
C
u
(
N
H
3
)
4
]
S
O
4
primary valency is 2 and secondary valency is 4.
Secondary valence refers to coordination number. Since copper is coordinated to 4 ammonia ligands, secondary valence is 4. Primary valence is satisfied by anions. Since sulphate ion has
−
2
charge, primary valence is 2.
Three arrangements are shown for the complex,
[
C
o
(
e
n
)
(
N
H
3
)
2
B
r
2
]
+
. Which one is the wrong statement?
Report Question
0%
I and II are geometrical isomers.
0%
II and III are optical isomers.
0%
I and III are optical isomers.
0%
II and III are geometrical isomers.
Explanation
I and II are geometrical isomers. I is cis as two
B
r
atoms are adjacent to each other. II is trans as two
B
r
atoms are opposite to each other.
II and III are geometrical isomers. II is trans as two
B
r
atoms are opposite to each other. III is cis as two
B
r
atoms are adjacent to each other.
I and III are optical isomers. They are non-superimposable mirror images of each other.
Copper sulphate dissolves in ammonia due to the formation of:
Report Question
0%
C
u
2
O
0%
[
C
u
(
N
H
3
)
4
]
S
O
4
0%
[
C
u
(
N
H
3
)
4
]
O
H
0%
[
C
u
(
H
2
O
)
4
]
S
O
4
Explanation
Copper sulphate dissolves in ammonia due to the formation of
[
C
u
(
N
H
3
)
4
]
S
O
4
.
C
u
S
O
4
+
4
N
H
3
→
[
C
u
(
N
H
3
)
4
]
S
O
4
[
C
u
(
N
H
3
)
4
]
S
O
4
contains complex cation
[
C
u
(
N
H
3
)
4
]
2
+
.
The correct labeling of different terms used in coordination compounds is:
Report Question
0%
(i) Central atom, (ii) Ionisation sphere, (iii) Coordination number, (iv) Ligands
0%
(i) Ligands, (ii) Coordination number, (iii) Valency, (iv) Ionisation sphere
0%
(i) Ionisation sphere, (ii) Ligands, (iii) Coordination number, (iv) Central atom
0%
(i) Ligands, (ii) Ionisation sphere, (iii) Coordination number, (iv) Central atom
Explanation
Ammonia is a ligand. It donates lone pair of electrons to Cobalt (central metal atom) and forms co-ordinate bond.
Co-ordination number is 6 as 6 ammonia ligand coordinate to central metal.
(ii) represents ionisation sphere ,
[
C
o
(
N
H
3
)
6
]
has
+
3
charge.
Hence ,option D is correct.
Among the following complexes (K-P) :
K
3
[
F
e
(
C
N
6
]
−
K
and
[
C
o
(
N
H
3
)
6
]
C
l
3
−
L
;
N
a
3
[
C
o
(
o
x
a
l
a
t
e
)
3
]
−
M
and
[
N
i
(
H
2
O
)
6
]
C
l
2
−
N
;
K
2
[
P
t
(
C
N
)
4
]
−
O
and
[
Z
n
(
H
2
O
)
6
]
(
N
O
3
)
2
−
P
;
the diamagnetic complexes are:
Report Question
0%
K,L,M,N
0%
K,M,O,P
0%
L,M,O,P
0%
all
Explanation
K
3
[
F
e
(
C
N
6
]
−
K
=
3
d
5
it is paramagnetic
[
C
o
(
N
H
3
)
6
]
C
l
3
−
L
=
3
d
6
is diamagnetic
N
a
3
[
C
o
(
o
x
a
l
a
t
e
)
3
]
−
M
=
3
d
6
is diamagnetic
[
N
i
(
H
2
O
)
6
]
C
l
2
−
N
=
3
d
8
is paramagnetic
K
2
[
P
t
(
C
N
)
4
]
−
O
=
d
3
it is diamagnetic
and
[
Z
n
(
H
2
O
)
6
]
(
N
O
3
)
2
−
P
;=
d
6
it is diamagnetic
Which one of the following has largest number of isomers?
(R = alkyl group,en = ethylenediamine)
Report Question
0%
[
R
u
(
N
H
3
)
4
C
l
2
]
+
0%
[
C
o
(
N
H
3
)
5
C
l
]
2
+
0%
[
I
r
(
P
R
3
)
2
H
(
C
O
)
]
2
+
0%
[
C
o
(
e
n
)
2
C
l
2
]
+
Explanation
[
C
o
(
e
n
)
2
C
l
2
]
+
show geometrical as well as optical isomerism.
[
R
u
(
N
H
3
)
4
C
l
2
]
+
and
[
I
r
(
P
R
3
)
2
H
(
C
O
)
]
2
+
shows geometrical isomerism only.
No isomerism is shown by
[
C
o
(
N
H
3
)
5
C
l
]
2
+
.
Both geometrical and optical isomerism are shown by:
Report Question
0%
[
C
o
(
e
n
)
2
C
l
2
]
+
0%
[
C
o
(
N
H
3
)
5
C
l
]
2
+
0%
[
C
o
(
N
H
3
)
4
C
l
2
]
+
0%
[
C
r
(
o
x
)
3
]
3
−
Explanation
Both geometrical and optical isomerism is shown by
[
C
o
(
e
n
)
2
C
l
2
]
+
.
No isomerism is shown by
[
C
o
(
N
H
3
)
5
C
l
]
2
+
.
[
C
o
(
N
H
3
)
4
C
l
2
]
+
shows geometrical isomerism.
[
C
r
(
o
x
)
3
]
3
−
shows optical isomerism.
According to Werner's coordination theory, there are _______ kinds of valency, _______ and ______. The primary valency of a central metal ion is satisfied with ________.
Report Question
0%
three, negative, positive, cations
0%
different, negative, positive, anions
0%
two, primary, secondary, anions
0%
two, saturated, unsaturated, cations
Explanation
According to Werner's coordination theory, there are two kinds of valency, primary and secondary. The primary valency of a central metal ion is satisfied with anions.
For example, in
[
C
u
(
N
H
3
)
4
]
S
O
4
primary valency is 2 and secondary valency is 4.
Secondary valence refers to coordination number. Since copper is coordinated to 4 ammonia ligands, secondary valence is 4. Primary valence is satisfied by anions. Since sulphate ion has
−
2
charge, primary valence is 2.
The compounds
[
C
o
(
S
O
4
)
(
N
H
3
)
5
]
B
r
and
[
C
o
(
S
O
4
)
(
N
H
3
)
5
]
C
l
represent:
Report Question
0%
linkage isomerism
0%
ionisation isomerism
0%
coordination isomerism
0%
no isomerism
Explanation
Isomers are compounds having the same chemical formula but a different structure.
The compounds
[
C
o
(
S
O
4
)
(
N
H
3
)
5
]
B
r
and
[
C
o
(
S
O
4
)
(
N
H
3
)
5
]
C
l
have different chemical formulas.
So they
represent no isomerism.
Which of the following shows maximum number of isomers?
Report Question
0%
[
C
o
(
N
H
3
)
4
C
l
2
]
0%
[
N
i
(
e
n
)
(
N
H
3
)
4
]
2
+
0%
[
N
i
(
C
2
O
4
)
(
e
n
)
2
]
2
−
0%
[
C
r
(
S
C
N
)
2
(
N
H
3
)
4
]
+
Explanation
[
C
r
(
S
C
N
)
2
(
N
H
3
)
4
]
+
shows maximum number of isomers. It shows geometrical (cis trans) isomerism as well as linkage isomerism.
S
C
N
−
is ambidentate ligand and can coordinate through either S or N atom.
Which of the following sets of examples and geometry of the compounds is not correct?
Report Question
0%
Octahedral -
[
C
o
(
N
H
3
)
6
]
3
+
,
[
F
e
(
C
N
)
6
]
3
−
0%
Square planar -
[
N
i
(
C
N
)
4
]
2
−
,
[
C
u
(
N
H
3
)
4
]
2
+
0%
Tetrahedral-
[
N
i
(
C
O
)
4
]
,
[
Z
n
C
l
4
]
2
−
0%
Trigonal bipyramidal -
[
F
e
(
N
H
3
)
6
]
2
+
,
[
C
u
C
l
4
]
2
−
Explanation
[
F
e
(
N
H
3
)
6
]
2
+
=
F
e
2
+
=
[
A
r
]
3
d
6
=
Octahedral.
Hybridisation is
=
s
p
3
d
2
.
[
C
u
C
l
4
]
2
+
=
C
u
2
+
=
[
C
u
]
3
d
9
=
Tetrahedral.
Hybridisation
=
s
p
3
These two do not have trigonal bipyramidal geometry.
Which of the following pairs of isomers is not correctly matched with its type of isomerism?
Report Question
0%
[
C
r
(
N
H
3
)
6
]
[
C
r
(
C
N
)
6
]
and
[
C
r
(
N
H
3
)
4
(
C
N
)
2
]
[
C
r
(
N
H
3
)
2
(
C
N
)
4
]
- Coordination isomerism
0%
C
o
[
(
N
H
3
)
5
N
O
2
]
C
l
2
and
[
C
o
(
N
H
3
)
5
O
N
O
]
C
l
2
- Linkage isomerism
0%
[
C
o
(
p
y
)
2
(
H
2
O
)
2
C
l
2
]
C
l
and
[
C
o
(
p
y
)
2
(
H
2
O
)
C
l
3
]
H
2
O
-Coordination isomerism
0%
[
P
t
(
N
H
3
)
4
B
r
2
]
C
l
2
and
[
P
t
(
N
H
3
)
4
C
l
2
]
B
r
2
-Ionisation isomerism
Explanation
Water molecules are exchanged inside and outside coordination sphere. Number of water molecules is different inside and outside coordination sphere. These are called hydrate isomers.
(i)
[
C
o
(
p
y
)
2
(
H
2
O
)
2
C
l
2
]
C
l
has two water molecules inside coordination sphere and zero water molecules outside coordination sphere.
(ii)
[
C
o
(
p
y
)
2
(
H
2
O
)
C
l
3
]
H
2
O
has one water molecule inside coordination sphere and one water molecule outside coordination sphere.
Match the column I with column II and mark the appropriate choice.
Column I
(Complex)
Column II
(Isomerism)
(A)
[
C
o
(
N
H
3
)
6
]
[
C
r
(
C
N
)
6
]
(i)
Geometrical isomerism
(B)
[
C
o
(
e
n
)
2
(
N
O
2
)
C
l
]
B
r
(ii)
Optical isomerism
(C)
[
P
t
(
e
n
)
2
C
l
2
]
(iii)
Coordination isomerism
(D)
[
C
r
(
C
N
)
2
(
N
H
3
)
4
]
2
+
(iv)
Linkage isomerism
Report Question
0%
A - (iv), B - (ii), C - (iii), D - (i)
0%
A - (ii), B - (iii), C - (i), D - (iv)
0%
A - (iii), B - (iv), C - (ii), D - (i)
0%
A - (i), B - (iii), C - (iv), D - (ii)
Explanation
[
C
O
(
N
H
3
)
6
]
[
C
r
(
C
N
)
6
]
⟺
[
C
O
(
C
N
)
6
]
[
C
r
(
N
H
3
)
6
]
(Co-ordinate Isomerism)
[
C
O
(
e
n
)
2
[
N
O
2
]
C
l
]
B
r
⇒
[
C
O
(
e
n
)
2
(
O
N
O
)
C
l
]
B
r
(Linkage Isomerism)
[
P
t
(
e
n
)
2
C
l
2
]
(Optical Isomerism)(Refer to image 01)
[
C
r
(
C
N
)
2
(
N
H
3
)
6
]
2
+
(Refer to image 02)
Geometric Isomerism (Refer to image 03)
Two isomers of a compound
[
C
o
(
N
H
3
)
3
C
l
3
]
(
M
A
3
B
3
type) are shown in the figures. The isomers can be classified as:
Report Question
0%
(i) fac-isomers (ii) mer-isomer
0%
(i) optical-isomer (ii) trans-isomer
0%
(i) mer-isomer (ii) fac-isomer
0%
(i) trans-isomer (ii) cis-isomer.
Explanation
In fac-isomer, three
C
l
atoms are present on one face of the octahedron. Similarly, 3 ammonia molecules are present on one face of the octahedron. All 3 same ligands have
90
o
bond angle between them.
In mer-isomer, there is a
90
o
;
90
o
;
180
o
bond angle between three same ligands.
Which of the following complexes exists as pair of enantiomers?
Report Question
0%
[
C
o
(
N
H
3
)
4
C
l
2
]
+
0%
[
C
r
(
e
n
)
3
]
3
+
0%
[
C
o
(
P
(
C
2
H
5
)
3
)
2
C
l
B
r
]
0%
t
r
a
n
s
−
[
C
o
(
e
n
)
2
C
l
2
]
+
Explanation
Optical isomerism is common in octahedral complexes of the formula
[
M
a
2
b
2
c
2
]
,
[
M
(
A
A
)
3
]
,
etc.
[
C
r
(
e
n
)
3
]
3
+
is octahedral complex of the formula
[
M
(
A
A
)
3
]
.
It shows optical isomerism and exists as a pair of enantiomers. Enantiomers are non superimposable mirror images of each other.
[
C
o
(
N
H
3
)
4
C
l
2
]
+
,
[
C
o
(
P
(
C
2
H
5
)
3
)
2
C
l
B
r
]
and
t
r
a
n
s
−
[
C
o
(
e
n
)
2
C
l
2
]
+
are optically inactive.
C
u
S
O
4
⋅
5
H
2
O
is blue is colour while
C
u
S
O
4
is colourless due to:
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0%
presence of strong field ligand in
C
u
S
O
4
⋅
5
H
2
O
0%
due to absence of water (ligand), d - d transitions are not possible in
C
u
S
O
4
0%
anhydrous undergoes d - d transitions due to crystal field splitting
0%
colour is lost due to unpaired electrons.
Explanation
In
C
u
S
O
4
⋅
5
H
2
O
, water acts as ligand and causes crystal filed splitting. This makes d - d transitions possible. On the other hand, in
C
u
S
O
4
, due to absence of ligand crystal filed splitting is not possible. Hence no colour is observed.
The correct energy level diagram for
[
C
o
(
C
N
)
6
]
3
−
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0%
0%
0%
0%
Explanation
Since cyanide ion is a strong field ligand, electrons pair up. All six electrons are present in lower
t
2
g
energy level.
The hybridisation involved in
[
C
o
(
C
2
O
4
)
3
]
3
−
is ?
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0%
s
p
3
d
2
0%
s
p
3
d
3
0%
d
s
p
3
0%
d
2
s
p
3
Explanation
[
C
o
(
C
2
O
4
)
3
]
3
−
=
C
O
3
+
=
[
A
r
]
3
d
6
=
Refer to image 01
C
2
O
2
−
4
=
Strong field ligand.
=
d
2
s
p
3
Hybridisation.
=
Octahedral.
(
C
2
O
4
)
−
2
=
Bidentate Ligand
[
P
t
(
N
H
3
)
4
]
[
C
u
C
l
4
]
and
[
C
u
(
N
H
3
)
4
]
[
P
t
C
l
4
]
are known as:
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0%
ionisation isomers
0%
coordination isomers
0%
linkage isomers
0%
polymerisation isomers.
Explanation
Different complex ions have the same molecular formula. Ligands are interchanged between the complex cation and complex anion. These type of complexes are called coordination isomers.
In
[
P
t
(
N
H
3
)
4
]
[
C
u
C
l
4
]
, ammonia ligands are attached to
P
t
metal and chloride ligands are attached to
C
u
metal.
In
[
C
u
(
N
H
3
)
4
]
[
P
t
C
l
4
]
, ammonia ligands are attached to
C
u
metal and chloride ligands are attached to
P
t
metal.
Energy profile diagram for an exothermic reaction,
A
1
→
B
2
→
C
3
→
D
, is given below.
The rate determining step of the reaction is :
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0%
A
⟶
B
0%
B
⟶
C
0%
C
⟶
D
0%
can not preditc
Explanation
In a chemical reaction, the overall rate of a reaction is determined by the slowest step, known as the rate determining step. The rate equation is simplified by using the approximation of rate determining step. Since, the slowest step in this reaction is when it has highest potential energy. Thus the rate determining step of reaction is
A
⟶
B
.
The synthetic steroid ethynylestardiol (1) is a compound used in the birth control pill. How many
s
p
3
hybridised carbon atoms are present in compound (1) ?
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0%
8
0%
9
0%
10
0%
11
0%
12
Explanation
s
p
3
hybridised carbon are those carbon having
4
single bond with
4
different groups.
(Refer to Image)
There are
12
s
p
3
hybridised carbon atoms.
0:0:1
1
2
3
4
5
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0
Answered
1
Not Answered
49
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Correct : 0
Incorrect : 0
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Practice Class 12 Engineering Chemistry Quiz Questions and Answers
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