Question

Consider the following statements.
I. In $SiM{e}_{2}C{l}_2$, $\angle Cl-Si-Cl$ is smaller than $\angle Me-Si-Me$.
II. $\angle Me-O-Me$ of $M{e}_{2}O$ is smaller than $\angle F-O-F$ of $F_{2}O$.
III. $C-Cl$ bond length of $C{H}_{3}Cl$ is shorter than $C-Cl$ bond length of $C{F}_{3}Cl$.
IV. $C-F$ bond length of $C{H}_{3}F$ is greater than $C-F$ bond length of $C{H}_2{F}_2$.

• A. TFTF
• B. TFFT
• C. FTTF
• D. FTFT

Answer 1

The correct option is B TFFT

According to the Bent's rule, more electronegative atom occupies orbital having less $s$ character.
I. In $SiM{e}_{2}C{l}_2$, $Si-Cl$ bond has more $p$ character which results in lesser bond angle than $Si-Me$ bond having more $s$ charater.
II. Bent' rule suggests that a hybrid orbital having more $s$ character will be oriented toward less electropositive lone pair whereas orbital having more $p$ character will be directed toward more electronegative $F$ atom in $F_{2}O$. Hence, $F_{2}O$ has larger bond angle.
III. $C-Cl$ bond of $C{H}_{3}Cl$ has more p character than $C-Cl$ bond of $C{F}_{3}Cl$ according to Bent's rule. Hence the former bond is longer than later.
IV. $C-F$ bond of $C{H}_{3}F$ has more p character than $C-F$ bond of $C{H}_2{F}_2$ since in the later, p character is shared by two $F$ atoms. Hence former bond length is longer than the later.