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CBSE Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 11 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Coordination Compounds
Quiz 11
Hybridization of central atoms in the molecules
N
(
C
H
3
)
3
and
N
(
S
i
H
3
)
3
respectively are:
Report Question
0%
s
p
2
nad
s
p
2
0%
s
p
3
and
s
p
3
0%
s
p
2
and
s
p
3
0%
s
p
3
and
s
p
2
Hydrogen peroxide oxidises
[
F
e
(
C
N
)
6
]
4
−
to
[
F
e
(
C
N
)
6
]
3
−
in acidic medium but reduces
[
F
e
(
C
N
)
6
]
3
−
to
[
F
e
(
C
N
)
6
]
4
−
in alkaline medium. The other products formed are, respectively:
Report Question
0%
H
2
O
and
(
H
2
O
+
O
2
)
0%
H
2
O
and
(
H
2
O
+
O
H
−
)
0%
(
H
2
O
+
O
2
)
and
H
2
O
0%
None of these
Explanation
Which of the following a bidentate mono iron ligands?
Report Question
0%
Acetyl Acetoneate
0%
Oxalate ion
0%
Both of these
0%
None of the above
In quantitative analysis of second group in lab.,
H
2
S
gas is passed in acidic medium for ppt. When
C
u
+
2
and
C
d
+
2
react with
K
C
N
, then in which of the following condition, ppt will not be formed due to relative stability.
Report Question
0%
K
2
[
C
u
(
C
N
)
4
]
- More stable
K
2
[
C
d
(
C
N
)
4
]
−
Less stable
0%
K
2
[
C
u
(
C
N
)
4
]
−
Less stable
K
2
[
C
d
(
C
N
)
4
]
−
More stable
0%
K
j
[
C
u
(
C
N
)
4
]
- More stable
K
2
[
C
d
(
C
N
)
4
]
−
Less stable
0%
K
J
[
C
u
(
C
N
)
4
]
−
Less stable
K
J
[
C
d
(
C
N
)
4
]
−
More stable
Select the correct statement
Report Question
0%
(
C
H
3
)
3
ˉ
C
is pyramidal while
ˉ
C
(
C
N
)
3
is planar
0%
ˉ
C
(
C
N
)
3
is Pyramidal while
ˉ
C
(
C
H
3
)
3
is plannar
0%
both
ˉ
C
(
C
N
)
3
and
ˉ
C
(
C
H
3
)
3
are planar
0%
both
ˉ
C
(
C
N
)
3
and
ˉ
C
(
C
H
3
)
3
are Pyramidal
[
N
i
(
C
N
)
4
]
2
−
is isostructural with:
Report Question
0%
H
e
F
4
0%
S
i
F
4
0%
S
F
4
0%
[
P
t
(
N
H
)
4
]
C
l
2
The incorrect statement regarding
O
(
S
i
H
3
)
2
and
O
C
l
2
molecule is/are :
Report Question
0%
The strength of back bonding is more in
O
(
S
i
H
3
)
2
molecule than
O
C
l
2
molecule
0%
S
i
−
ˆ
O
−
S
i
bond angle in
O
(
S
i
H
3
)
2
is greater than
C
l
−
ˆ
O
−
C
l
bond angle in
O
C
l
2
0%
The nature of back bond in both molecules is
2
P
π
−
3
d
π
0%
Hybridisation of central O-atom in both molecules is same
Explanation
Option D is incorrect as hybridization of central
O
in both the molecules are different not same as follows:
Option D is correct answer
The conjugate base of
[
A
l
(
H
2
O
)
3
(
O
H
)
3
]
is
Report Question
0%
[
A
l
(
H
2
O
)
3
(
O
H
)
2
]
+
0%
[
A
l
(
H
2
O
)
3
(
O
H
)
2
]
−
0%
[
A
l
(
H
2
O
)
3
(
O
H
)
3
]
−
0%
[
A
l
(
H
2
O
)
2
(
O
H
)
4
]
−
The correct order of increasing s-character (in percentage) in the hybrid orbitals of following molecules and ions is
(I)
C
O
2
−
3
(II)
X
e
F
4
(III)
I
−
3
(IV)
N
C
l
3
(V)
B
e
C
l
2
Report Question
0%
I
I
<
I
I
I
<
I
V
<
I
<
V
0%
I
I
<
I
I
I
<
I
V
<
V
<
I
0%
I
I
I
<
I
I
<
I
<
V
<
I
V
0%
I
I
<
I
V
<
I
I
I
<
I
<
V
Identify the correct statement(s) regarding
[
C
r
(
e
n
)
2
]
+
and
[
C
o
(
C
2
O
4
)
2
(
N
H
3
)
2
]
complex ions:
(i) Both are chelate complexes
(ii) Both have octahedral geometry
(iii) Both are low spin complexes
(iv) Both contain metal-carbon bond i.e. , they are organometallic compounds
Report Question
0%
I,III,IV
0%
I,II
0%
II,III
0%
I,II,III
Which of the following is not correctly matched ?
Report Question
0%
[
F
e
(
C
N
)
6
]
3
−
d
2
s
p
3
, paramagnetic
0%
[
N
i
(
C
O
)
4
]
−
s
p
3
, diamagnetic
0%
[
F
e
(
e
n
)
3
]
+
3
−
s
p
3
d
2
, paramagnetic
0%
[
F
e
C
l
2
(
H
2
O
)
2
]
s
p
3
, paramagnetic
In
[
C
o
(
N
H
3
)
6
]
C
l
3
, the number of covalent bonds is
Report Question
0%
3
0%
6
0%
9
0%
18
which has the maximum conductivity in thier 0.1M solution?
Report Question
0%
A)
[
C
o
(
N
H
3
)
3
C
l
3
]
0%
B)
[
C
o
(
N
H
3
)
4
C
l
]
C
l
2
0%
C)
[
C
o
(
N
H
3
)
5
C
l
]
C
l
2
0%
D)
[
C
o
(
N
H
3
)
6
]
C
l
3
Explanation
The maximum conductivity is shown by that compound which have maximum ions on dissociation. Hence the compound
[
C
o
(
N
H
3
)
6
]
C
l
3
have the maximum number of ions
[
C
o
(
N
H
3
)
6
]
+
3
C
l
−
= 4 ions and hence the correct answer is D.
Which of these statements about
[
C
o
(
C
N
)
6
]
3
−
is true?
Report Question
0%
[
C
o
(
C
N
)
6
]
3
−
has zero unpaired electrons and will be in a low-spin configuration.
0%
[
C
o
(
C
N
)
6
]
3
−
has four unpaired electrons and will be in a high-spin configuration.
0%
[
C
o
(
C
N
)
6
]
3
−
has zero unpaired electrons and will be in a high-spin configuration.
0%
[
C
o
(
C
N
)
6
]
3
−
has four unpaired electrons and will be in a low-spin configuration.
Explanation
Oxidation no. of
C
o
is +3 and
electronic
configuration of
C
o
3
+
=
3
d
6
As
C
N
is a strong field ligand, so all electrons will be paired up and the complex will be low spin complex.
Total number of possible isomers of
[
C
u
(
N
H
3
)
4
]
[
P
t
C
l
4
]
is:
Report Question
0%
3
0%
4
0%
5
0%
6
Explanation
The total number of possible isomer of the
[
C
u
(
N
H
3
)
4
]
[
P
t
C
l
4
]
is 4.
The isomers can be formed by exchanging the ligands.
1.
[
C
u
(
N
H
3
)
3
C
l
]
[
P
t
(
C
l
)
3
N
H
3
]
2.
[
C
u
(
N
H
3
)
2
C
l
2
]
[
P
t
C
l
2
(
N
H
3
)
2
]
3.
[
C
u
(
N
H
3
)
C
l
3
]
[
P
t
C
l
(
N
H
3
)
3
]
4.
[
C
u
C
l
4
]
[
P
t
(
N
H
3
)
4
]
Hence, the correct option is B
Given the molecular formula of the hexa coordinated complexes (A)
C
o
C
l
3
.
6
N
H
3
(B)
C
o
C
l
3
.
5
N
H
3
(C)
C
o
C
l
3
.
4
N
H
3
. If the number of coordinated
N
H
3
molecules in
A
,
B
and
C
respectively are
6
,
5
and
4
, primary valency in
(
A
)
,
(
B
)
and
(
C
)
are
Report Question
0%
0
,
1
,
2
0%
3
,
2
,
1
0%
6
,
5
,
4
0%
3
,
3
,
3
Explanation
The complex can be written as follows.
A
−
[
C
o
(
N
H
3
)
6
]
C
l
3
B
−
[
C
o
(
N
H
3
)
5
C
l
]
C
l
2
C
−
[
C
o
(
N
H
3
)
4
C
l
4
]
C
l
Hence, number of primary valency are
3
,
2
and
1
respectively.
[
F
e
(
H
2
O
)
6
]
+
2
has a crystal field splitting energy value of
10
,
400
c
m
−
1
and a pairing energy value of
17
,
600
c
m
−
1
. Then, the compound is:
Report Question
0%
a low spin complex
0%
paramagnetic in nature
0%
diamagnetic in nature
0%
none of these
In the co-ordination compound
N
a
4
[
F
e
)
C
N
)
5
N
O
S
]
oxidation state of
F
e
is:
Report Question
0%
+1
0%
+2
0%
+3
0%
+4
Explanation
N
a
4
[
F
e
(
C
N
)
5
N
O
]
[
F
e
(
C
N
)
5
N
O
]
−
4
NO is a natural ligand thus it's
oxidation state is 'O'
Let oxidation state of Fe is 'x'
(
x
×
1
)
+
5
×
(
−
1
)
=
−
4
x
=
−
4
+
5
x
=
+
1
Oxidation state of Fe is +1 in above
compound.
Which isomer of
C
r
C
l
3
.
6
H
2
O
is dark green in colour and forms one mole of
A
g
C
l
with excess of
A
g
N
O
3
solution ?
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%
[
C
r
(
H
2
O
)
3
C
l
3
]
3
H
2
O
The complex ion having minimum wavelength of absorption in the visible region is:
Report Question
0%
[
C
o
(
N
H
3
)
6
]
3
+
0%
[
C
o
C
l
(
N
H
3
)
5
]
2
+
0%
C
i
s
−
[
C
o
C
l
2
(
N
H
3
)
4
]
+
0%
T
r
a
n
s
−
[
C
o
C
l
2
(
N
H
3
)
4
]
+
Select the incorrect statement.
Report Question
0%
Crystal field stabilisation energy of
d
2
configuration in an octahedral complex of a weak field ligand is
−
0.8
△
o
.
0%
[
M
n
F
6
]
4
−
is an example of a coordination compound having a weak field ligand and
d
5
configuration.
0%
Effective atomic number of Pt in
[
P
t
C
l
6
]
2
−
is 84.
0%
Geometrical isomerism is shown by
[
P
t
(
N
H
3
)
4
C
l
2
]
.
The number of unpaired electron in complex
[
C
r
(
N
O
2
)
3
(
N
H
3
)
3
]
is
Report Question
0%
2
0%
3
0%
1
0%
0
Nickel
(
Z
=
28
)
combines with a uni negative monodentate ligand
X
−
to form a paramagnetic complex,
[
N
i
X
4
]
2
−
. The number of unpaired electrons in the nickel and geometry of this complexion are, respectively:
Report Question
0%
One, tetrahedral
0%
Two, tetrahedral
0%
One, square planar
0%
None of these
The number of chloride ions which would be precipitated, when
C
r
C
l
3
.4
N
H
3
is treated with silver nitrate solution:
Report Question
0%
3
0%
2
0%
1
0%
0
Explanation
Chromium has coordination number six.
Therefore to write its formula, we will first add the maximum number of neutral elements inside the coordination sphere.
Hence, the formula is
[
C
r
(
N
H
3
)
4
C
l
2
]
C
l
.
As 1
C
l
−
ion is available for dissociation only one
C
l
−
ion will precipitate when it reacts with
A
g
N
O
3
.
Hence, the correct option is (C).
Both geometrical and optical isomerism was shown by
Report Question
0%
[Pt(NH
3
)
2
Cl
2
]
0%
[Pt(NH
3
)
4
Cl
2
]
0%
[Pt(en)
2
Cl
2
]
0%
[Pt(en)
3
Cl
2
]
The oxidation state of
C
r
in
[
C
r
(
N
H
3
)
4
C
l
2
]
+
is:
Report Question
0%
+2
0%
+3
0%
0
0%
+1
L
i
H
+
B
_
2
H
6
→
2
L
i
B
_
H
4
Find the change in hybridization of the underlined atom.
Report Question
0%
s
p
2
→
s
p
3
0%
s
p
3
→
s
p
2
0%
s
p
→
s
p
3
0%
None
For the following two complexes
A
:
[
N
i
C
l
6
]
4
−
and
[
N
i
C
l
4
]
2
−
the ratio of CFSE will be nearly:
Report Question
0%
3
2
0%
4
9
0%
9
4
0%
27
8
Explanation
[
N
i
C
l
6
]
4
−
→
octahedral complex (
t
2
g
=
6
,
e
g
=
4
)
[
N
i
C
l
4
]
2
−
→
Tetrahedral complex (
e
g
=
4
,
t
2
g
=
4
)
for octahedral CFSE =
−
0.4
×
6
+
0.6
×
2
=
−
1.2
for tetradidral CFSE
=
+
0.4
×
4
−
0.6
×
4
=
−
0.8
∴
C
F
S
E
(
1
)
C
F
S
E
(
2
)
=
−
1.2
−
0.8
=
3
2
Option A is correct.
Correct formula of potassium ferrocyanide is
Report Question
0%
K
4
[
F
e
(
C
N
)
6
]
0%
K
2
[
F
e
(
C
N
)
6
]
.
H
2
O
0%
K
3
[
F
e
(
C
N
)
6
]
0%
None of these
Number of isomers of
[
P
t
(
N
H
3
)
4
]
[
C
u
C
l
4
]
complex are
Report Question
0%
2
0%
3
0%
4
0%
5
Copper sulphate when dissolved in excess of
K
C
N
gives
Report Question
0%
C
u
(
C
N
)
2
0%
[
C
u
(
C
N
)
4
]
3
−
0%
[
C
u
(
C
N
)
4
]
2
−
0%
C
u
C
N
The number of moles of
B
a
S
O
4
precipitate obtained, when reagent
[
C
u
(
N
H
3
)
4
]
S
O
4
is treated with an excess of
B
a
C
l
2
:-
Report Question
0%
2
0%
0
0%
1
0%
None of these
Which of the following complex has no optical isomers, where
a
,
b
,
c
are monodentate ligands and
(
A
A
)
is symmetrical bidentate ligand:
Report Question
0%
[
M
a
2
b
2
]
0%
[
M
a
2
b
2
c
2
]
0%
[
M
(
A
A
)
2
a
2
]
0%
[
M
(
A
A
)
2
a
b
]
On adding
A
g
N
O
3
solution to a solution of
[
P
t
(
N
H
3
)
3
C
l
3
]
C
l
,
the percentage of total chloride ion precipitate is:
Report Question
0%
100
0%
75
0%
50
0%
25
The hybridization of Fe in
K
4
[
F
e
(
C
N
)
6
]
is:
Report Question
0%
s
p
3
0%
s
p
3
d
2
0%
d
2
s
p
3
0%
d
s
p
2
The species in which
(
n
−
1
)
d
x
2
−
y
2
orbital take part in hybridization?
Report Question
0%
P
C
l
5
0%
[
N
i
C
l
4
]
2
−
0%
[
C
r
(
N
H
3
)
6
]
3
+
0%
[
N
i
(
C
O
)
4
]
If the bidentate ligand
L
2
−
in
[
M
L
2
]
n
−
4
is replaced by neutral monodentate ligand
X
, the formula of the resulting complex ion is:
Report Question
0%
[
M
X
4
]
n
+
0%
[
M
X
2
]
n
+
0%
[
M
X
2
]
n
−
1
0%
[
M
X
4
]
2
+
Iron from
H
b
is stored in liver in the form of:
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0%
Ferric
0%
Ferritin
0%
Ferrous
0%
all of the above
According to Werner's theory state, which of the following statements are correct?
Ligands are connected to the metal ions by covalent bonds
Secondary valencies have directional properties
Secondary valencies are non-ionisable.
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0%
1, 2, and 3 are correct
0%
2 and 3 are correct
0%
1 and 2 are correct
0%
1 ,2 are correct,and 3 is not correct
Which of the following will give maximum number of isomers:-
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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
r
(
N
O
2
)
2
(
N
H
3
)
4
]
+
The hybridization of the complex
[
C
r
C
l
2
(
N
O
2
)
2
(
N
H
3
)
2
]
−
is:
Report Question
0%
s
p
3
d
2
0%
d
2
s
p
3
0%
s
p
3
d
0%
Cannot be predicted
Which one of the following high spin complexes has the largest
C
F
S
E
(crystal field stabilization energy)?
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
+
The number of electrons present in
e
g
set of orbital of triquatrifluoro cobalt
(
I
I
I
)
is:
Report Question
0%
2
0%
3
0%
4
0%
0
Which one is more stable in the following pairs of complexes? Give reasons.
Report Question
0%
[
C
o
(
N
O
2
)
6
]
4
−
and
[
C
o
(
N
O
2
)
6
]
3
−
0%
K
4
[
F
e
(
C
N
)
6
]
and
K
3
[
F
e
(
C
N
)
6
]
0%
[
C
o
(
H
2
O
)
6
]
2
+
and
[
C
o
(
N
H
3
)
6
]
2
+
0%
None of them
Number of chlorides satisfying the secondary valency in
C
o
C
l
3
.4
N
H
3
is:
Report Question
0%
1
0%
2
0%
3
0%
6
[
C
r
(
H
2
O
)
6
]
C
l
3
(atomic number of Cr =24) has a magnetic moment of 3.83 B.M. The correct distribution of 3d -electrons in the chromium present in the complex is:
Report Question
0%
3
d
1
x
y
,
3
d
1
y
z
,
3
d
1
z
x
0%
3
d
1
x
y
,
3
d
1
y
z
,
3
d
1
z
2
0%
3
d
1
x
2
−
y
2
,
3
d
1
z
2
,
3
d
1
z
x
0%
3
d
1
x
y
,
3
d
1
(
x
2
−
y
2
)
,
3
d
1
x
z
State the correct statement(s):
Report Question
0%
[
C
o
(
E
D
T
A
)
−
]
has two optical isomers
0%
[
C
o
(
N
H
3
)
3
(
N
O
2
)
]
2
+
show linkage isomerism
0%
For
[
P
t
(
N
H
3
)
B
r
C
l
(
N
O
2
)
p
y
]
, theoretically fifteen different geometrical isomers are possible
0%
[
C
r
(
H
2
O
)
4
C
l
2
]
C
l
2
.2
H
2
O
is an example of hydrate as well as ionisation isomerism
Which of the following complexes is/are optically inactive due to the presence of plane of symmetry?
Report Question
0%
0%
0%
0%
The number of possible isomers of an octahedral complex
[
C
o
(
C
2
O
4
)
2
(
N
H
3
)
2
]
is-
Report Question
0%
1
0%
2
0%
3
0%
4
Square planar complex that can not show cis-trans isomerism is:
Report Question
0%
[
C
u
(
N
H
3
)
4
]
2
+
0%
[
P
t
C
l
2
(
N
H
3
)
2
]
0%
[
P
t
(
N
H
3
)
(
H
2
O
)
(
N
O
2
)
(
C
l
)
]
0%
[
P
t
(
g
l
y
)
2
]
0:0:1
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Answered
1
Not Answered
49
Not Visited
Correct : 0
Incorrect : 0
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Practice Class 12 Engineering Chemistry Quiz Questions and Answers
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