Question

(3) 3
(4) 7
Number of 120° bond angles present in BF, is
[NCERT Pg. 1
(1) 4
(2) 5​

Answer 1

Answer:

there are 4 120 degree bond angle in BF3 molecule

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Answer 2

Answer:

The number of 120° bond angles present in $BF_3$ is 3.

Explanation:

[Assuming the given molecule is $BF_3$]

• The Boron trifluoride, $BF_3$ is a trigonal planar molecule.
• Boron needs 3 hybridized orbitals to make a bond with 3 atoms of Fluorine, thus forming hybridized  $sp^2$ orbitals.
• The three hybridized $sp^2$ orbitals are usually arranged in a triangular shape.
• So, $BF_3$ forms a trigonal planar molecule
• In trigonal planar molecular geometry, three atoms are surrounded around an atom at the center.
• In $BF_3$ molecule, the 3 Flourine atoms surround around the boron atom at the center, making an equilateral triangle.
• $BF_3$ molecule has a symmetric charge distribution on the boron atom and is nonpolar.
• The F-B-F bond angle is 120° where all the atoms are in one plane.
• Therefore, all the three fluorine atoms are bonded with boron making 120° bond angle at each.

Hence, the number of 120° bond angles present in $BF_3$ is 3.

So, the correct answer is option 3.

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