Question

In which of the following molecules all the bonds are not equal?
(1) NF3
(2) CIF3
(3) BF3
(4) PF3​

Answer 1

ClF3 is the answer

due to trigonal bi pyramidal structure

so the bonds are not equal

Answer 2

Final answer: Option (2) $ClF_{3}$    

Given that: We are given, molecules

(1) $NF_{3}$

(2) $ClF_{3}$

(3) $BF_{3}$

(4) $PF_{3}$

To find: We have to find molecule in which all the bonds are not equal.

Explanation:

• In $ClF_{3}$ there are 3 bond pairs and 2 lone pairs around central atom $Cl$. They arrange themselves in a trigonal bipyramidal fashion due to $sp^{3}d$ hybridisation. In order to minimize 90° interactions involving lone pairs, the 2 lone pairs occupy two equatorial while 3 bond pairs occupy the remaining one equatorial position and the two axial position. As a result the molecule is T-shaped. Hence, all bonds in $ClF_{3}$ are not equal. So option (2)$ClF_{3}$ is correct.
• $NF_{3}$ exhibits a pyramidal structure due to $sp^{3}$ hybridisation. Where $N$ is surrounded by three $F$ atoms with a lone pair of electrons in the fourth orbital. So all bonds in $NF_{3}$ are equal. Hence, option (1) $NF_{3}$ is wrong answer.
• Molecular geometry of $BF_{3}$ is trigonal planar due to $sp^{2}$hybridization. So $BF_{3}$ have trigonal symmetric structure. Thus all bonds in $BF_{3}$ are equal. Hence, option (3) $BF_{3}$ is wrong answer.
• $PF_{3}$ exhibits a trigonal pyramidal geometry due to $sp^{3}$ hybridisation. Where central $P$ is surrounded by three $F$ atoms with a lone pair of electrons in the fourth orbital. So all bonds in $PF_{3}$ are equal. Hence, option (4) $PF_{3}$ is wrong answer.

The structures of $NF_{3}$, $ClF_{3}$, $BF_{3}$ and $PF_{3}$ are shown below figure.

To know more about the concept please go through the links

https://brainly.in/question/23705707

https://brainly.in/question/5246997

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