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Byju's Answer
Standard XII
Chemistry
Valence Bond Theory
BF3 is planar...
Question
B
F
3
is planar and electron deficient compound. Hybridization and number of electrons around the central atom, respectively are :
A
s
p
2
and
8
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B
s
p
3
and
4
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C
s
p
3
and
6
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D
s
p
2
and
6
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Solution
The correct option is
D
s
p
2
and
6
Number of electrons around boron atom is
6
.
Hybridization of
B
is
s
p
2
.
Shape is trigonal planar.
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2
Similar questions
Q.
B
F
3
is planar and electron deficient compound. Hybridisation and number of electrons around the central atom, respectively are:
Q.
The hybridization of the central atoms of compounds
A
,
B
,
C
and
D
are
s
p
3
d
,
s
p
3
,
s
p
2
and
s
p
respectively. If the compounds
A
and
D
have the same shape like
I
−
3
and the compounds
B
and
C
have the same shape like water molecule, then the value of
P
+
Q
+
R
+
S
is:
(Given that
P
,
Q
,
R
and
S
are the number of lone pairs of electrons on central atoms of compounds
A
,
B
,
C
and
D
respectively.)
Q.
The geometry and the type of hybrid orbital present about the central atom in
B
F
3
are trigonal planar and
s
p
2
. If true enter 1, else enter 0.
Q.
The geometry and the hybridization present about the central atom in
B
F
3
is :
Q.
Consider two covalent compounds
A
L
n
1
and
B
L
n
2
. If central atom
(
A
)
of first compound has total of six valence electron pairs and central atom
(
B
)
of second compound contains total of five valence electron pairs and both compounds are planar and non-polar then calculate value of expression
(
n
1
–
n
2
)
2
.
[where
n
1
and
n
2
are number of monovalent surrounding atom
(
L
)
]
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