
Select the incorrect statement among the following:
(1) ${\text{B}}{{\text{F}}_{\text{3}}}$and ${\text{BC}}{{\text{l}}_{\text{3}}}$ both have equal bond angles
(2) ${\text{Xe}}{{\text{O}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ has two $p\pi - p\pi $ bonds
(3) ${\text{S}}{{\text{F}}_{\text{6}}}$ has equal bond lengths
(4) ${\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$ is less polar than ${\text{CHC}}{{\text{l}}_{\text{3}}}$
Answer
577.5k+ views
Hint:. To know the bond angles in the given species, firstly their hybridizations should be considered. Polarity of any given compounds can be explained by their dipole moments which arise due to the presence of electronegative atoms attached to the carbon atoms.
Complete step by step answer:
In order to answer the above given type of question one has to check statements given in all the options.
1). The central atom in both ${\text{B}}{{\text{F}}_{\text{3}}}$ and ${\text{BC}}{{\text{l}}_{\text{3}}}$ is boron in which the number of lone pairs are zero and the number of bond pair of electrons are three as the available bonding atoms of fluorine as well as chlorine are three. So, the resulting hybridization is ${\text{s}}{{\text{p}}^{\text{3}}}$ and the bond angle in ${\text{s}}{{\text{p}}^{\text{3}}}$ hybridization is same in both the compounds which is ${120^ \circ }$.
2) The central atom in ${\text{Xe}}{{\text{O}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ is xenon. Fluorine is monovalent but Oxygen is a divalent atom so it forms double bonds with two of the lone pairs of electrons of xenon. This results in the formation of two $p\pi - p\pi $bonds.
3) In Sulphur hexafluoride the hybridization in central atom is ${\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$ which gives the octahedral arrangement of the orbitals in space. All the orbitals participate in bonding and the lengths of all six bonds are similar with 156.4 pikometer.
4) The direction of dipole in ${\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$ of C-H and C-Cl bonds in the horizontal array is towards chlorine and the direction in vertical bonds is also towards chlorine and the total moment would be the vector addition of both. But in chloroform ${\text{CHC}}{{\text{l}}_{\text{3}}}$ there are 3 C-Cl bonds, the dipoles of two C-Cl bonds arranged in opposite directions cancels out and the net dipole results from one C-Cl bond. This means the dipole moment is greater for dichloromethane than chloroform.
So, the correct answer is “Option D”.
Note: The total number of electrons available in Xenon to bond with two oxygen and two fluorine atoms would be 4 from 8 of its total valence electrons to form single sigma bonds.
Complete step by step answer:
In order to answer the above given type of question one has to check statements given in all the options.
1). The central atom in both ${\text{B}}{{\text{F}}_{\text{3}}}$ and ${\text{BC}}{{\text{l}}_{\text{3}}}$ is boron in which the number of lone pairs are zero and the number of bond pair of electrons are three as the available bonding atoms of fluorine as well as chlorine are three. So, the resulting hybridization is ${\text{s}}{{\text{p}}^{\text{3}}}$ and the bond angle in ${\text{s}}{{\text{p}}^{\text{3}}}$ hybridization is same in both the compounds which is ${120^ \circ }$.
2) The central atom in ${\text{Xe}}{{\text{O}}_{\text{2}}}{{\text{F}}_{\text{2}}}$ is xenon. Fluorine is monovalent but Oxygen is a divalent atom so it forms double bonds with two of the lone pairs of electrons of xenon. This results in the formation of two $p\pi - p\pi $bonds.
3) In Sulphur hexafluoride the hybridization in central atom is ${\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$ which gives the octahedral arrangement of the orbitals in space. All the orbitals participate in bonding and the lengths of all six bonds are similar with 156.4 pikometer.
4) The direction of dipole in ${\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$ of C-H and C-Cl bonds in the horizontal array is towards chlorine and the direction in vertical bonds is also towards chlorine and the total moment would be the vector addition of both. But in chloroform ${\text{CHC}}{{\text{l}}_{\text{3}}}$ there are 3 C-Cl bonds, the dipoles of two C-Cl bonds arranged in opposite directions cancels out and the net dipole results from one C-Cl bond. This means the dipole moment is greater for dichloromethane than chloroform.
So, the correct answer is “Option D”.
Note: The total number of electrons available in Xenon to bond with two oxygen and two fluorine atoms would be 4 from 8 of its total valence electrons to form single sigma bonds.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

10 examples of friction in our daily life

