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Tardigrade
Question
Chemistry
Borazine B 3 N 3 H 6(A) is converted into disubstituted product B 3 N 3 H 4 X2(B). Number of isomers of B would be
Q. Borazine
B
3
N
3
H
6
(
A
)
is converted into disubstituted product
B
3
N
3
H
4
X
2
(
B
)
. Number of isomers of
B
would be
103
224
The p-Block Elements
Report Error
A
2
B
4
C
6
D
No isomer, exists as single product
Solution:
Correct answer is (b) 4
