3
You visited us
3
times! Enjoying our articles?
Unlock Full Access!
Byju's Answer
Standard X
Chemistry
sp Hybridisation
Total number ...
Question
Total number of geometrical isomers for the complex
[
R
h
C
l
(
C
O
)
(
P
P
h
3
)
(
N
H
3
)
]
is
Open in App
Solution
[
R
h
C
l
(
C
o
)
(
P
P
h
3
)
(
N
H
3
)
]
has
d
s
p
2
hybridisation and square plannar shape. It has total 3 geometrical isomer.
Suggest Corrections
0
Similar questions
Q.
Total number of geometrical isomers for the complex
[
R
h
C
l
(
C
O
)
(
P
P
h
3
)
(
N
H
3
)
]
is
Q.
The possible number of geometrical isomers for the complex
[
P
t
(
N
H
3
)
(
N
H
2
O
H
)
(
N
O
2
)
(
p
y
)
]
N
O
2
are:
Q.
The number of geometrical isomers of the complex
[
C
o
(
N
O
2
)
3
(
N
H
3
)
3
]
is
Q.
For the given complex
[
C
o
C
l
2
(
e
n
)
(
N
H
3
)
2
]
+
, the number of geometrical isomers, the number of optical isomers and total number of isomers of all type possible respectively are:
Q.
The number of geometrical isomers for octahedral
[
C
o
C
l
4
(
N
H
3
)
2
]
−
complex is _____
View More
Join BYJU'S Learning Program
Grade/Exam
1st Grade
2nd Grade
3rd Grade
4th Grade
5th Grade
6th grade
7th grade
8th Grade
9th Grade
10th Grade
11th Grade
12th Grade
Submit
Related Videos
sp Hybridisation
CHEMISTRY
Watch in App
Explore more
sp Hybridisation
Standard X Chemistry
Join BYJU'S Learning Program
Grade/Exam
1st Grade
2nd Grade
3rd Grade
4th Grade
5th Grade
6th grade
7th grade
8th Grade
9th Grade
10th Grade
11th Grade
12th Grade
Submit
Solve
Textbooks
Question Papers
Install app