
The calculated covalent radius of fluorine from $C - F$ bond length is least in case of:
A. $C{H_3}F$
B. \[C{H_2}{F_2}\]
C. $CH{F_3}$
D. $C{F_4}$
Answer
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Hint:The carbon and fluorine bond is polar in nature because fluorine is very electronegative in nature and makes the bond polar. The comparison of bond length is done by the bent’s rule. On applying bent’s rule, $C{F_4}$ will have the least bond length.
Complete answer:
Bent’s rule states that a central atom bonded to multiple groups will hybridise so that orbitals with more $s$ character are directed towards electropositive groups, while orbitals with more $p$ character will be directed towards groups that are more electronegative. As, the elements which have $p$ character have the highest covalent radius.
Bonds which are formed between elements of different electronegativities will be polar and the electron density in such bonds will be shifted towards the more electronegative element. Applying this to the given molecules, fluorine is more electronegative than carbon, the electron density in the \[C - F\] bonds will be towards fluorine. The energy of those electrons will depend heavily on the hybrid orbitals that carbon contributes to these bonds because of the increased electron density near the fluorine. By increasing the amount of p character in those hybrid orbitals, the energy of those electrons can be increased because p orbitals are higher in energy than s orbitals.Carbon tetrafluoride has the least bond length among the given options.By applying the bent’s rule in the above options we will get the bond length order as:
$C{F_4} < CH{F_3} < C{H_2}{F_2} < C{H_3}F$
So, the option D is correct.
Note:
As fluorine is highly electronegative so the electron density is concentrated around the fluorine, leaving the carbon relatively electron poor. This contributes ionic character to the bond through partial charges. The partial charges on the fluorine and carbon are attractive, they give the bond the unusual bond strength of the carbon–fluorine bond. So, the carbon fluorine bond is one of the strongest bonds in organic chemistry.
Complete answer:
Bent’s rule states that a central atom bonded to multiple groups will hybridise so that orbitals with more $s$ character are directed towards electropositive groups, while orbitals with more $p$ character will be directed towards groups that are more electronegative. As, the elements which have $p$ character have the highest covalent radius.
Bonds which are formed between elements of different electronegativities will be polar and the electron density in such bonds will be shifted towards the more electronegative element. Applying this to the given molecules, fluorine is more electronegative than carbon, the electron density in the \[C - F\] bonds will be towards fluorine. The energy of those electrons will depend heavily on the hybrid orbitals that carbon contributes to these bonds because of the increased electron density near the fluorine. By increasing the amount of p character in those hybrid orbitals, the energy of those electrons can be increased because p orbitals are higher in energy than s orbitals.Carbon tetrafluoride has the least bond length among the given options.By applying the bent’s rule in the above options we will get the bond length order as:
$C{F_4} < CH{F_3} < C{H_2}{F_2} < C{H_3}F$
So, the option D is correct.
Note:
As fluorine is highly electronegative so the electron density is concentrated around the fluorine, leaving the carbon relatively electron poor. This contributes ionic character to the bond through partial charges. The partial charges on the fluorine and carbon are attractive, they give the bond the unusual bond strength of the carbon–fluorine bond. So, the carbon fluorine bond is one of the strongest bonds in organic chemistry.
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