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
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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.
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