In which of the following cases C-C bond leng | vedclass

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(D) According to Bent's Rule,more electronegative substituents prefer to be attached to a hybrid orbital that has more $p-$character. As the number of

Vedclass · Accepted Answer

$CF_3-CF_3$

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(D) According to Bent's Rule,more electronegative substituents prefer to be attached to a hybrid orbital that has more $p-$character. As the number of fluorine atoms increases on the carbon atoms,the $p-$character in the $C-C$ bond increases. Greater $p-$character in the hybrid orbitals forming the $C-C$ bond leads to a longer bond length. In $CH_3-CF_3$,the $C-C$ bond is formed by $sp^3-sp^3$ hybridization. In $CF_3-CF_3$,the $C-C$ bond has the highest $p-$character due to the presence of the most electronegative fluorine atoms,resulting in the longest $C-C$ bond length.

List-$I$	List-$II$
$(a)$ $SF_6$	$(i)$ $sp^3d^2$
$(b)$ $PCl_5$	(ii) $sp^3d$
$(c)$ $XeF_4$	(iii) $sp^3d^3$
$(d)$ $IF_7$	(iv) $sp^3d^2$

$Column\, I$	$Column\, II$
$(A)\, XeF_6$	$(i)\, \text{Distorted octahedral}$
$(B)\, XeO_3$	$(ii)\, \text{Square planar}$
$(C)\, XeOF_4$	$(iii)\, \text{Pyramidal}$
$(D)\, XeF_4$	$(iv)\, \text{Square pyramidal}$

In which of the following cases $C-C$ bond length will be highest?

Similar Questions

Which compound does not possess linear geometry?

Match the following:
List-$I$ List-$II$
$(a)$ $SF_6$ $(i)$ $sp^3d^2$
$(b)$ $PCl_5$ (ii) $sp^3d$
$(c)$ $XeF_4$ (iii) $sp^3d^3$
$(d)$ $IF_7$ (iv) $sp^3d^2$

The correct answer is:

The correct order of hybridisations of the central atom in $NH_3$,$[PtCl_4]^{2-}$,$PCl_5$,and $BCl_3$ is:

Explain bond formation in $BCl_{3}$. It is a symmetrical trigonal molecule.

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