Question 513 Marks
Describe the general trends in the following properties of the elements in Groups $13$ and $14.$
Nature of halides.
Nature of halides.
Answer
View full question & answer→Groups 13 Nature of Halides: These elements react with halogens to from trihalides (except $TII _3$ ).
$2 E(s)+3 X_2(g) \rightarrow 2 EX_3(s)(X=F, Cl, Br, I)$
Halides of boron and Aluminum are electron deficient and act as Lewis acidic character of halides of boron decreases in the following order:
$BF_3<BCl_3<BBr_3<BI_3$
Groups 14
Nature of Halides: These elements can form halides of formula $MX _2$ and $MX _4( X = F , Cl , Br , I )$. Except carbon, all other members react directly with halogen under suitable conditions to make halides, most of the compounds $\left( MX _4\right)$ are covalent in nature.
The central metal atom in these halides undergoes $sp ^3$ hybridization and the molecule is tetrahedral in shape. Exceptions are $SnF _4$ and $PbF _1$, which are ionic in nature. $Pbl _4$ does not exist because Pb -I bond initially formed during the reaction does not release enough energy to unpair $6 s^2$ electrons and excite one of them to higher orbital to have four unpaird electrons around lead atom. Heavier members Ge to Pb are able to make halides of formula $MX _2$.
$2 E(s)+3 X_2(g) \rightarrow 2 EX_3(s)(X=F, Cl, Br, I)$
Halides of boron and Aluminum are electron deficient and act as Lewis acidic character of halides of boron decreases in the following order:
$BF_3<BCl_3<BBr_3<BI_3$
Groups 14
Nature of Halides: These elements can form halides of formula $MX _2$ and $MX _4( X = F , Cl , Br , I )$. Except carbon, all other members react directly with halogen under suitable conditions to make halides, most of the compounds $\left( MX _4\right)$ are covalent in nature.
The central metal atom in these halides undergoes $sp ^3$ hybridization and the molecule is tetrahedral in shape. Exceptions are $SnF _4$ and $PbF _1$, which are ionic in nature. $Pbl _4$ does not exist because Pb -I bond initially formed during the reaction does not release enough energy to unpair $6 s^2$ electrons and excite one of them to higher orbital to have four unpaird electrons around lead atom. Heavier members Ge to Pb are able to make halides of formula $MX _2$.