Question

Elements of group 14 used in semiconductors are:
(A) $C,Si,Ge$
(B) $Si,Ge,Sn$
(C) $Si,Ge$
(D) $B,Si,Ge$

Answer 1

Hint: The radii of gathering fourteen components are lesser than that of gathering thirteen components. This can be clarified by the expansion in the successful atomic charge and after which the expansion in the radii is less. This can be credited to the helpless protection of d and f orbitals, which increment the viable atomic charge, subsequently making the radii little.

Complete answer:The radii of group fourteen components are lesser than that of group thirteen components. This can be clarified by the expansion in the viable atomic charge. The expansion in the radii from $C$ to $Si$ is extensive, after which the increment in the radii is less. This can be ascribed to the helpless protection of d and f orbitals, which increment the viable atomic charge, accordingly making the radii little. The ionization energy of group fourteen components is more noteworthy than that of group 13 components. This can be ascribed to measure. Down the group, the Ionization Enthalpy diminishes. There is a sharp reduction from $C$ to $Si$, after which the lessening is ostensible. The request is as per the following, $C > Si > Ge > Pb > Sn$. Here $Pb$ has a more prominent Ionization Enthalpy than Sn because of incapable protection of d and f orbitals.Silicon is found in the premise of every single electronic gadget. It is a semiconductor. Germanium is another component utilized in semiconductors. Ultrapure types of Ge and Si are utilized to make interpreters and semiconductors.

Additional information:What are the compounds formed by group four elements?Hydrides : All the components of group fourteen structure hydrides. Carbon structures hydrides widely because of their capacity to catenate. The hydrides of carbon are arranged as beneath. Alkanes (paraffin): General equation: ${C_n}{H_{2n + 2}}$. Alkenes (olefins): General equation: ${C_n}{H_{2n}}$ . Alkynes (Acetylenes): General equation: ${C_n}{H_{2n - 2}}$.
Oxides of Group fourteen : Group 14 components structure oxides of the sort $MO$ and $M{O_2}$. Lead additionally structures an oxide $P{b_3}{O_4}$ which is a blended oxide of $PbO$ and $Pb{O_2}$ . Among the monoxides, $CO$ is nonpartisan, $GeO$ is fundamental while $SnO$ and $PbO$ are amphoteric. In $C{O_2}$, $C$ is $sp - $ hybridized. It is not the same as $Si{O_2}$ in which $Si$ is $s{p^3}$ hybridized. In $Si{O_2}$, every $O$ molecule is clung to two $Si$ bonds. This offers access to a three-dimensional structure for $Si{O_2}$. This additionally authenticates the high dissolving purpose of $Si{O_2}$.
Halides : They structure tetrahalides of the structure $M{X_4}$ . The focal iota is $s{p^3}$ hybridized and accepts a tetrahedral shape.

Hence the correct answer is option (C).

Note:Components beneath $C$ , have void d-orbitals, with which they can display back holding with the incandescent light$(p\Pi - d\Pi )$ . The radii of group fourteen components are lesser than that of group thirteen components. This can be clarified by the expansion in the successful atomic charge. The expansion in the radii from $C$ to $Si$ is significant, after which the expansion in the radii is less. This can be credited to the helpless protection of d and f orbitals, which increment the powerful atomic charge, accordingly making the radii little.