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Byju's Answer
Standard XII
Chemistry
Freidel Craft Alkylation Acylation
The number of...
Question
The number of geometrical isomers that can be shown by a square planar complex of the type [M(abcd)] are:
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Solution
The three possible geometrical isomers are as follow
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Similar questions
Q.
For the square planar complex,
[
M
(
a
)
(
b
)
(
c
)
(
d
)
]
(Where M
=
central metal and a, b, c and d are monodentate ligands), the number of possible geometrical isomers are:
Q.
For the square planar complex [Mabcd] where M is the central atom and a,b,c,d are monodentate ligands, the number of possible geometrical isomers are:
Q.
The number of geometric isomers that can exist for square planar
[
P
t
(
C
l
)
(
p
y
)
(
N
H
3
)
(
N
H
2
O
H
)
]
+
is:
Q.
The number of geometrical isomers that can exist for square planar
[
P
t
(
C
l
)
(
p
y
)
(
N
H
3
)
(
N
H
2
O
H
)
]
+
is (py
=
pyridine.)
Q.
The number of geometrical isomer(s) possible for
[
A
u
C
l
2
B
r
2
]
(square) planar is/are :
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