Processing math: 100%
MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 13 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Coordination Compounds
Quiz 13
N
C
l
3
and
B
C
l
3
are mixed in a container. Which the following is true for hybridisation of
N
and
B
the product formed by bonding of
N
C
l
3
and
B
C
l
3
?
Report Question
0%
Hybridisation of
N
changes but for
B
it remains same
0%
Hybridisation of
B
changes but for
N
it remains same
0%
Hybridisation of both
N
and
B
change
0%
Hybridisation of neither
N
nor
B
changes
The number of
C
l
−
combined by secondary valency in
[
F
e
(
N
H
3
)
4
C
l
2
]
C
l
is ______
Report Question
0%
1
0%
2
0%
3
0%
0
Which of the following complexes is not correctly matched with hybridization of its central metal?
Report Question
0%
[
C
o
F
6
]
3
−
;
d
2
s
p
3
0%
[
N
i
(
C
O
)
4
]
;
s
p
3
0%
[
C
r
(
H
2
O
)
6
]
3
+
;
d
2
s
p
3
0%
[
P
t
(
N
H
3
)
4
]
2
+
;
d
s
p
2
Which compound to use for treatment of tumor?
Report Question
0%
C
i
s
[
P
d
C
l
2
(
N
H
3
)
2
]
0%
T
r
a
n
s
[
P
d
C
l
2
(
N
H
3
)
2
]
0%
C
i
s
[
P
t
C
l
2
(
N
H
3
)
2
]
0%
T
r
a
n
s
[
P
t
C
l
2
(
N
H
3
)
2
]
Explanation
The given compound is cis plating which is used in the treatment of cancer or tumor.
Which of the following represent E isomer?
Report Question
0%
0%
0%
0%
Which of the following complexes form facial and meridional isomers?
Report Question
0%
[
C
r
(
H
2
O
)
5
C
l
]
+
2
0%
[
C
r
(
N
H
3
)
5
B
r
]
S
O
4
0%
[
C
r
(
N
H
3
)
3
(
N
O
2
)
3
]
0%
[
C
r
(
o
x
)
3
]
3
−
Tetravalency of carbon is possible in the case of
Report Question
0%
s
p
3
-hybridisation
0%
s
p
2
-hybridisation
0%
s
p
-hybridisation
0%
all of these
The given compounds are______isomer of each other.
Report Question
0%
Positional
0%
Chain
0%
Geometrical
0%
Functional
Which of the following statement is incorrect?
Report Question
0%
The tendency to attract bonded pair of the electron in case of hybrid orbitals follows the order:
s
p
>
s
p
2
>
s
p
3
0%
Alkali metals generally have negative value of electron gain enthalpy
0%
C
s
+
(
g
)
releases more energy upon gain of an electron than
C
l
(
g
)
0%
The electronegativity values for
2
p-series elements is less than that for
3
p-series elements on account of small size and high inter electronic configuration
The compound
P
t
C
l
2
⋅
2
N
H
3
does not react with
A
g
N
O
3
. This compound can be represented as?
Report Question
0%
[
P
t
(
N
H
3
)
2
C
l
]
C
l
0%
[
P
t
(
N
H
3
)
2
C
l
2
]
0%
[
P
t
(
N
H
3
)
2
]
C
l
2
0%
both B and C
Which of the following represent
s
p
−
s
p
−
s
p
2
−
s
p
−
s
p
2
−
s
p
3
hybridization from left to right order?
Report Question
0%
C
H
2
=
C
H
−
C
=
H
|
C
−
C
≡
N
0%
C
H
≡
C
−
C
H
=
C
=
C
H
−
C
H
3
0%
N
≡
C
−
H
|
C
−
C
|
H
−
C
H
−
C
H
3
0%
Explanation
Carbon having single bond is
s
p
3
hybridised, carbon with double bond is
s
p
2
hybridised and the carbon having triple bond is
s
p
hybridised.
In option B
C
H
is
s
p
hybridised,
C
is
s
p
hybridised,
C
H
is
s
p
2
hybridised,
C
is
s
p
hybridised,
C
H
is
s
p
2
hybridised and
C
H
3
is
s
p
3
hybridised.
In
C
u
S
O
4
.5
H
2
O
how many molecules of water are indirectly connected to Cu?
Report Question
0%
5
0%
4
0%
2
0%
1
Explanation
Solution:- (D)
1
In
C
u
S
O
4
.5
H
2
O
, Four water molecules form coordinate bond with
C
u
2
+
ion while one water molecule is associated with
H
- bond
Which among the following is used in the treatment of cancer?
Report Question
0%
cis-
[
P
t
(
e
n
)
2
C
l
2
]
0%
cis-
[
P
t
C
l
2
(
N
H
3
)
2
]
0%
trans-
[
P
t
(
e
n
)
2
C
l
2
]
0%
trans-
[
P
t
(
N
H
3
)
2
C
l
2
]
Explanation
c
i
s
−
[
P
t
C
l
2
(
N
H
3
)
2
]
also known as cisplatin is used as an anticancer drug in the chemotherapy process of cancer treatment.
Hence, option B is correct.
When two or more complexes are combined, it is
Report Question
0%
Pattern
0%
Theme
0%
Complex
0%
All of these
The compound that inhibits the growth of tumors is:
Report Question
0%
c
i
s
−
[
P
d
(
C
l
)
2
(
N
H
3
)
2
]
0%
c
i
s
−
[
P
t
(
C
l
)
2
(
N
H
3
)
2
]
0%
t
r
a
n
s
−
[
P
t
(
C
l
)
2
(
N
H
3
)
2
]
0%
t
r
a
n
s
−
[
P
d
(
C
l
)
2
(
N
H
3
)
2
]
Explanation
Solution:- (B) cis-
[
P
t
(
C
l
)
2
(
N
H
3
)
2
]
c
i
s
−
[
P
t
(
C
l
)
2
(
N
H
3
)
2
]
is used in chemotherapy to inhibits the growth of tumors.
The complex ion that will lose its crystal field stabilization energy upon oxidation of its metal to
+
3
state is:
Report Question
0%
[
F
e
(
p
h
e
n
)
3
]
2
+
0%
[
Z
n
(
p
h
e
n
)
3
]
2
+
0%
[
N
i
(
p
h
e
n
)
3
]
2
+
0%
[
C
o
(
p
h
e
n
)
3
]
2
+
Explanation
By oxidation of
F
e
2
+
into
F
e
3
+
, the CFSE value decrease.
(2)
[
Z
n
(
p
h
e
n
)
3
]
2
−
e
−
→
[
Z
n
(
p
h
e
n
)
3
]
3
+
Z
n
2
+
:
3
d
10
Z
n
3
+
:
3
d
9
C.F.S.E
=
0
C.F.S.E=
−
0.6
Δ
o
By oxidation of
Z
n
2
+
into
Z
n
3
+
, the CFSE value increase.
(3)
[
N
i
(
p
h
e
n
)
3
]
2
+
−
e
−
→
[
N
i
(
p
h
e
n
)
3
]
3
+
N
i
2
+
:
3
d
8
N
i
3
+
:
3
d
7
C.F.S.E
=
−
1.2
Δ
0
C.F.S.E
=
−
1.8
Δ
0
By oxidation of
N
i
2
+
into
N
i
3
+
, the CFSE value increase.
(4)
[
C
o
(
p
h
e
n
)
3
]
2
+
−
e
−
→
[
C
o
(
p
h
e
n
)
3
]
3
+
C
o
2
+
:
3
d
7
C
o
3
+
:
3
d
6
C
.
F
.
S
.
E
=
−
1.8
Δ
0
C
.
F
.
S
.
E
=
−
2.4
Δ
0
By oxidation of
C
o
2
+
into
C
o
3
+
, the CFSE value increase.
Complete removal of both the axial ligands (along the z-axis) from an octahedral complex leads to which of the following splitting patterns? (relative orbital energies not on scale).
Report Question
0%
0%
0%
0%
Explanation
If both ligands present along z-axis removed from octahedral field and converted into a square planar field, then the energy level of orbitals is shown.
The degenerate orbitals of
[
C
r
(
H
2
O
)
6
]
3
+
are:
Report Question
0%
d
y
z
and
d
z
2
0%
d
z
2
and
x
x
z
0%
d
x
z
and
d
y
z
0%
d
x
2
−
y
2
and
d
x
y
Explanation
Degenerate orbitals of
[
C
r
(
H
2
O
)
6
]
3
+
is shown in the diagram.
Hence according to the options given, degenerate orbitals are
d
x
z
and
d
y
z
.
So, C option is correct.
Which complex among the following gives a white precipitate on treatment with an aqueous solution of barium chloride?
Report Question
0%
[
P
t
(
N
H
3
)
4
B
r
2
]
C
l
2
0%
[
C
o
(
N
H
3
)
5
S
O
4
]
N
O
2
0%
[
C
o
(
N
H
3
)
5
N
O
2
]
S
O
4
0%
[
P
t
(
N
H
3
)
4
C
l
2
]
B
r
2
Explanation
White precipitate on treatment with barium chloride is due to the formation of
B
a
S
O
4
and for this sulphate ion should be present in ionisable part.
[
C
o
(
N
H
3
)
5
N
O
2
]
S
O
4
+
B
a
C
l
2
(
a
q
)
→
[
C
o
(
N
H
3
)
5
N
O
2
]
C
l
2
+
B
a
S
O
4
(
s
)
↓
B
a
S
O
4
is a white precipitate.
The other compounds do not give white ppt. due to the absence of
S
O
2
−
4
ions.
Hence, option
C
is correct.
A molecule
(
X
)
has (i) four sigma bonds formed by the overlap of
s
p
2
and
s
orbitals; (ii) one sigma bond formed by
s
p
2
and
s
p
2
- orbitals and (iii) one
π
−
b
o
n
d
formed by
p
z
and
p
z
orbitals. Which of the following is
X
?
Report Question
0%
C
2
H
6
0%
C
2
H
5
C
l
0%
C
2
H
2
C
l
2
0%
C
2
H
4
The complexes
[
C
o
N
O
2
(
N
H
3
)
5
]
C
l
2
and
[
C
o
(
O
N
O
)
(
N
H
3
)
5
]
C
l
2
are not the examples of.
Report Question
0%
Geometrical isomers
0%
Optical isomers
0%
Coordination isomers
0%
Linkage isomers
The total number of possible coordination isomer for the given compound
[
P
t
(
N
H
3
)
4
C
l
2
]
[
P
t
C
l
4
]
is?
Report Question
0%
2
0%
4
0%
5
0%
3
The octahedral complex
C
o
S
O
4
C
l
5
N
H
3
exists in two isomeric forms
X
and
Y
. Isomers
X
reacts with
A
g
N
O
3
to give a white precipitate, but does not react with
B
a
C
l
2
. Isomer
Y
gives white precipitate with
B
a
C
l
2
but does not react with
A
g
N
O
3
.
Isomers
X
and
Y
are:
Report Question
0%
ionization isomers
0%
linkage isomers
0%
coordinating isomers
0%
solvent isomers
Explanation
[
C
o
(
N
H
3
)
5
S
O
4
]
C
l
(X)
A
g
N
O
3
→
A
g
C
l
white ppt.
↓
[
C
o
(
N
H
3
)
5
C
l
]
S
O
4
(Y)
B
a
C
l
2
→
B
a
S
O
4
white ppt.
↓
[
X
]
and
[
Y
]
are ionization isomers.
Brown ring complex known as:
Report Question
0%
K
4
[
F
e
(
C
N
)
6
]
0%
[
F
e
(
N
O
)
(
H
2
O
)
5
]
S
O
4
0%
[
F
e
(
N
O
)
2
)
(
H
2
O
)
4
]
B
r
2
0%
[
F
e
F
6
]
3
−
Explanation
6
F
e
S
O
4
+
3
H
2
S
O
4
+
2
H
N
O
3
→
3
F
e
2
(
S
O
4
)
3
+
4
H
2
O
+
2
N
O
F
e
S
O
4
+
N
O
+
5
H
2
O
→
[
F
e
(
N
O
)
(
H
2
O
)
5
]
S
O
4
Brown ring complex
Option B is correct.
The complexes
[
C
r
(
N
H
3
)
6
]
[
C
o
(
C
N
)
6
]
and
[
C
o
(
N
H
3
)
6
]
[
C
r
(
C
N
)
6
]
are not the example of?
Report Question
0%
Geometrical isomers
0%
Optical isomers
0%
Coordination isomers
0%
Linkage isomers
Select the correct statement(s) :
Report Question
0%
B
F
3
>
M
e
3
B
0%
H
C
≡
C
−
H
(
s
p
c
a
r
b
o
n
)
<
H
2
C
=
C
H
−
H
(
s
p
2
c
a
r
b
o
n
)
<
H
3
C
−
C
H
2
−
H
(
s
p
3
c
a
r
b
o
n
)
0%
B
F
3
<
M
e
3
B
0%
H
C
≡
C
−
H
(
s
p
c
a
r
b
o
n
)
>
H
2
C
=
C
H
−
H
(
s
p
2
c
a
r
b
o
n
)
>
H
3
C
−
C
H
2
−
H
(
s
p
3
c
a
r
b
o
n
)
VSEPR theory suggest a small departure from
109
0
28
′
for
N
H
3
molecule. The H-N-H bond angle has been found to be
107
0
what kind of hybrid orbital will be occupied by the non -bonding pair of electron?
Report Question
0%
s
p
3
0%
s
p
3.4
0%
s
p
2.13
0%
s
p
2
Select the correct statement for
P
4
O
10
Report Question
0%
It has four
s
p
3
-hybridized phosphorous atoms
0%
It has higher
s
%
charcter in
P
−
O
bond than the
P
4
O
6
0%
It has cage -like structure
0%
It has
p
π
−
d
π
bonding
Explanation
From the above diagrams, we can conclude that :
All the four P atoms in P
₄O₁₀ are sp
³
hybridized (4 sigma bonds).
It has a cage-like structure.
The percentage s character in phosphorus pentoxide is higher than that in phosphorus trioxide due to the presence of double bond between P and O.
The bond between P atom and one of the O atoms of each PO
₄ unit in
P
₄O₁₀
is a double bond and involves p
π-dπ bonding.
Hence, all the above options are correct.
Select the correct options for following statement(s) :
Report Question
0%
s
p
3
hybrid orbitals are at
90
0
to one another
0%
s
p
3
d
2
adjacent hybrid orbitals are at
90
0
to one another
0%
s
p
2
hybrid orbitals are at
120
0
to one another
0%
Bond order of
N
−
O
bond in
N
O
−
3
is
1
1
3
.
Choose the correct statement regarding the complexes X and Y:
(I) Both complexes have the same six geometrical isomer.
(II) Both complexes have nearly the same conductivity in
H
2
O
.
(III) In both complexes,
N
H
3
is involved only to secondary valency.
(IV) In both complexes, electronic configuration of
t
2
g
is same.
Report Question
0%
I, II and III
0%
I, II, III and IV
0%
II, III and IV
0%
I, III and IV
Which of the following sets of molecule(s) is/are having a linear shape but different hybridization?
Report Question
0%
I
−
3
and
C
O
2
0%
H
g
C
l
2
and
I
C
l
−
2
0%
X
e
F
2
and
C
2
H
2
0%
X
e
F
2
and
O
C
l
−
Explanation
A)
I
−
3
and
C
O
2
are linear in shape.
I
−
3
undergoes
s
p
3
hybridization and
C
O
2
undergoes
s
p
hybridization.
B)
H
g
C
l
2
and
I
C
l
−
2
are linear in shape.
H
g
C
l
2
undergoes
s
p
hybridization and
I
C
l
−
2
undergoes
s
p
3
d
hybridization .
C)
X
e
F
2
and
C
2
H
2
are linear in shape.
X
e
F
2
undergoes
s
p
3
d
2
and
C
2
H
2
undergoes
s
p
hybridization.
D)
X
e
F
2
and
O
C
l
−
are linear in shape.
X
e
F
2
undergoes
s
p
3
d
2
and
O
C
l
−
undergoes
s
p
3
hybridization .
Hence, options A, B, C, and D all are correct.
Which of the following sets of the molecule(s) is/are having a V-shape but different hybridization?
Report Question
0%
S
n
C
l
2
and
H
2
O
0%
S
O
2
and
N
O
+
2
0%
B
F
−
2
and
S
C
l
2
0%
O
F
2
and
S
C
l
2
In V.B.T., the idea of hybridization was required to explain which of the following facts:
Report Question
0%
The equivalence of the bonds in most of the com-pounds
0%
The stereochemistry of the molecules
0%
The better overlapping of the orbitals
0%
None of these
Match the complex species given in Column-I with the possible isomerism given in Column-II and assign the correct code.
Column-I(Complex species)
Column-II(Isomerism)
(a)
[
C
o
(
N
H
3
)
4
C
l
2
]
+
(p) Optical
(b) cis-
[
C
o
(
e
n
)
2
C
l
2
]
+
(q) Ionization
(c)
[
C
o
(
N
H
3
)
5
(
N
O
2
)
]
C
l
2
(r) Coordination
(d)
[
C
o
(
N
H
3
)
6
]
[
C
r
(
C
N
)
6
]
(s) Geometrical
(t) Linkage
Report Question
0%
a(p) b(q) c(s) d(t)
0%
a(s) b(r) c(q) d(p)
0%
a(s) b(p) c(q,t) d(r)
0%
a(s) b(p) c(q) d(r)
Complexes A and B are respectively.
Report Question
0%
[
C
r
(
N
H
3
)
4
C
l
2
]
B
r
,
[
C
r
(
N
H
3
)
4
B
r
]
C
l
2
0%
[
C
r
(
N
H
3
)
4
B
r
]
C
l
2
,
[
C
r
(
N
H
3
)
4
C
l
2
]
B
r
0%
[
C
r
(
N
H
3
)
4
C
l
B
r
]
C
l
,
[
C
r
(
N
H
3
)
4
C
l
2
]
B
r
0%
[
C
r
(
N
H
3
)
4
C
l
2
]
B
r
,
[
C
r
(
N
H
3
)
4
C
l
B
r
]
C
l
Complex A is?
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
2
]
C
l
⋅
2
H
2
O
0%
None of these
One mole of the complex compound
C
o
(
N
H
3
)
5
C
l
3
gives
3
moles of ions on dissolution in water. One mole of the same complex reacts with two moles of
A
g
N
O
3
solution to yield two moles of
A
g
C
l
(
s
)
. The structure of the complex is?
Report Question
0%
[
C
o
(
N
H
3
)
3
C
l
3
]
⋅
2
N
H
3
0%
[
C
o
(
N
H
3
)
4
C
l
2
]
⋅
2
N
H
3
0%
[
C
o
(
N
H
3
)
4
C
l
]
C
l
2
⋅
N
H
3
0%
[
C
o
(
N
H
3
)
5
C
l
]
C
l
2
Explanation
Total
3
moles of ions are generated on dissolution in water, that is, one complex ion and two simple ions.
One mole of complex reacts with two moles of
A
g
N
O
3
; thus, it must have
2
Cl atoms bonded by ionic linkage. The coordination number of Co is
6
.
[
C
o
(
N
H
3
)
5
C
l
]
C
l
2
→
[
C
o
(
N
H
3
)
5
C
l
]
2
+
+
2
C
l
−
[
C
o
(
N
H
3
)
5
C
l
]
C
l
2
+
2
A
g
N
O
3
→
[
C
o
(
N
H
3
)
5
C
l
]
(
N
O
3
)
2
+
2
A
g
C
l
(
p
p
t
.
)
Coordination compounds have great importance in biological systems. In this context, which of the following statements is incorrect?
Report Question
0%
Cyanobalamin is
B
12
and contains cobalt.
0%
Hemoglobin is the red pigment of blood and contains irons.
0%
Chlorophylls are green pigments in plants and contains calcium.
0%
Carboxypeptidase-A is an enzyme and contains zinc.
Explanation
The correct statement should be chlorophylls are green pigments in plants and contain magnesium.
The correct order regarding electronegativity of hybrid orbitals of carbon is:
Report Question
0%
s
p
<
s
p
2
<
s
p
3
0%
s
p
<
s
p
2
>
s
p
3
0%
s
p
>
s
p
2
<
s
p
3
0%
s
p
>
s
p
2
>
s
p
3
Explanation
Electronegativity increases as s-character in hybrid orbital increases. Percentage of s character in
s
p
,
s
p
2
and
s
p
3
are 50%, 33.33% and 25% respectively.
So, the correct option is D.
Carbon dioxide is isostructural with:
Report Question
0%
H
g
C
l
2
0%
S
n
C
l
2
0%
C
2
H
2
0%
N
O
2
2
F
e
2
(
S
O
4
)
3
+
3
K
4
[
F
e
(
C
N
)
6
]
→
X
+
6
K
2
S
O
4
What is the name of X? if-
Report Question
0%
represents brown ring complex
0%
represents sodium nitroprusside
0%
represents Prussian blue
0%
represents chromyl chloride
What is the formula of X if
Report Question
0%
represents
C
r
O
2
−
4
0%
represent
C
r
O
5
0%
represent
C
r
2
O
2
−
7
0%
represents
C
r
O
2
C
l
2
Explanation
Option D is correct
Here, X represents
C
r
O
2
C
l
2
.
N
a
2
S
+
N
a
2
[
F
e
(
C
N
)
5
N
O
]
→
X
(purple color)
What is the formula of X?
Report Question
0%
F
e
4
[
F
e
(
C
N
)
6
]
0%
[
F
e
(
H
2
O
)
5
N
O
]
S
O
4
0%
[
F
e
(
H
2
O
)
5
N
O
]
0%
N
a
4
[
F
e
(
C
N
)
5
(
N
O
S
)
]
A solution containing
2.675
g of
C
o
C
l
3
⋅
6
N
H
3
(molar mass
=
267.5
g
m
o
l
−
1
) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of
A
g
N
O
3
to give
4.78
g of
A
g
C
l
(molar mass
=
143.5
g
m
o
l
−
1
). The formula of the complex is?
Report Question
0%
[
C
o
(
N
H
3
)
6
]
C
l
3
0%
[
C
o
C
l
2
(
N
H
3
)
4
]
C
l
0%
[
C
o
C
l
3
(
N
H
3
)
3
]
0%
[
C
o
C
l
(
N
H
3
)
5
]
C
l
2
Explanation
A
g
C
l
obtained is
4.78
g, that is,
4.78
143.5
moles.
Thus, in
2.675
267.5
=
10
−
2
moles of complex,
4.78
143.5
≈
1
30
moles of
C
l
−
ions were isonizable.
That is, in
10
−
2
moles of complex,
1
30
moles
C
l
−
are produced.
∴
1
mole of complex
→
1
30
×
10
2
moles of
C
l
−
=
100
30
=
3.3
mols.
Thus, the formula of complex should be
[
C
o
(
N
H
3
)
6
]
C
l
3
.
Complex manure is that which supplies:
Report Question
0%
S
,
K
and
N
0%
N
,
K
and
P
0%
S
and
N
0%
S
,
N
and
P
Explanation
Complex manures is that which supplies N, K and P that is nitrogen, potassium and phosphorus.
Nitrogen is good at making the leaves grow.
Phosphorus improves fruit and/or flower production as well as root growth.
Potassium is great for overall plant health.
Match each coordination compound in List-I with an appropriate pair of characteristics from List-II and select the correct answer using the code given in the lists.
{
e
n
=
H
2
N
C
H
2
C
H
2
N
H
2
'atomic numbers;
T
i
=
22
;
C
r
=
24
;
C
o
=
27
P
t
=
78
}
List-I
List-II
(P)
[
C
r
(
N
H
3
)
4
C
l
2
]
C
l
(
1
)
Paramagnetic and exhibits ionization isomerism
(Q)
[
T
i
(
H
2
O
)
5
C
l
]
(
N
O
3
)
2
(
2
)
Dimagnetic and exhibits cis-trans isomerism
(R)
[
P
t
(
e
n
)
(
N
H
3
)
C
l
]
N
O
3
(
3
)
Paramagnetic and exhibits cis-trans isomerism
(S)
[
C
o
(
N
H
3
)
4
(
N
O
3
)
2
]
N
O
3
(
4
)
Dimagnetic and exhibits ionization isomerism
Report Question
0%
P-
4
, Q-
2
, R-
3
, S-
1
0%
P-
3
, Q-
1
, R-
4
, S-
2
0%
P-
2
, Q-
1
, R-
3
, S-
4
0%
P-
1
, Q-
3
, R-
4
, S-
2
Which one of the following has the largest number of isomers?
Report Question
0%
[
I
r
(
P
R
3
)
2
H
(
C
O
)
]
2
+
0%
[
C
o
(
N
H
3
)
5
C
l
]
2
+
0%
[
R
u
(
N
H
3
)
4
C
l
2
]
+
0%
[
C
o
(
e
n
)
2
C
l
2
]
+
(
R
=
alkyl group,
e
n
=
ethylenediamine)
What is the formula of brown ppt.?
Report Question
0%
C
u
2
I
2
0%
C
u
2
I
2
+
I
−
3
0%
C
u
I
2
0%
C
u
S
O
4
Which of the following complex is formed when A reacts with
K
4
[
(
F
e
(
C
N
)
6
]
?
Report Question
0%
Prussian blue
0%
Turnbull's blue
0%
Brown ring complex
0%
Sodium nitroprusside
What is the formula of brown ppt?
Report Question
0%
F
e
(
O
H
)
3
0%
F
e
(
O
H
)
2
0%
F
e
C
l
3
0%
None of these
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
1
Not Answered
49
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 12 Engineering Chemistry Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
