Question

All members of group 14 when heated in oxygen form oxides. Which of the following is the correct trend of oxides?
A. Dioxides $C{{O}_{2}}$, $Si{{O}_{2}}$ and $Ge{{O}_{2}}$ are acidic while $Sn{{O}_{2}}$ and $Pb{{O}_{2}}$ are amphoteric
B. CO, GeO, SnO and PbO are amphoteric.
C. Monoxides react with hemoglobin to form toxic compounds.
D. All oxides burn with blue flame.

Answer 1

Hint: 14 group elements are also known by carbon group elements and it is the second group in the p-block of the periodic table. The members of this group are carbon, silicon, germanium, tin, lead and flavourium. General electronic configuration of group 14 elements is shown by $n{{s}^{2}}n{{p}^{2}}$.

Complete Step by step solution: General oxidation states exhibited by the group 14 elements are +2 and +4. Covalent radii of group 14 elements is lesser as compared to 13 group elements. This is due to the increase in the effective nuclear charge but the ionization energy of group 14 elements is greater than group 13 elements which can be explained on the basis of. Group 14 elements are less electropositive in nature. Group 14 elements form two types of oxides i.e. MO and $M{{O}_{2}}$. Lead also form an oxide shown by $P{{b}_{3}}{{O}_{4}}$ which is a mixed oxide of PbO and $Pb{{O}_{2}}$. Among the monoxides, CO is neutral, GeO is basic while SnO and PbO are amphoteric. From the above discussion we can say that option A is correct i.e. Dioxides $C{{O}_{2}}$, $Si{{O}_{2}}$ and $Ge{{O}_{2}}$ are acidic while $Sn{{O}_{2}}$ and $Pb{{O}_{2}}$ are amphoteric it can be explained as:
$C{{O}_{2}}$, $Si{{O}_{2}}$ and $Ge{{O}_{2}}$ are acidic in nature so they react only with bases and form salts.
$C{{O}_{2}}+2NaOH\to N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O$
$Si{{O}_{2}}+2NaOH\to N{{a}_{2}}Si{{O}_{3}}+{{H}_{2}}O$
$Ge{{O}_{2}}+2NaOH\to N{{a}_{2}}Ge{{O}_{3}}+{{H}_{2}}O$
$Sn{{O}_{2}}$ and $Pb{{O}_{2}}$ are amphoteric in nature which means they can react with both acids and bases
$Sn{{O}_{2}}+2NaOH\to N{{a}_{2}}Sn{{O}_{3}}+{{H}_{2}}O$
$Sn{{O}_{2}}+4HCl\to SnC{{l}_{4}}+2{{H}_{2}}O$
$Pb{{O}_{2}}+2NaOH\to N{{a}_{2}}Pb{{O}_{3}}+{{H}_{2}}O$
$Pb{{O}_{2}}+4HCl\to PbC{{l}_{4}}+2{{H}_{2}}O$

Hence option A is correct.

Note: Other than oxides group 14 elements can also form halides and hydrides. Out of all elements carbon forms hydrides extensively due to their ability to catenate i.e. linking power with other atoms. Generally group 14 elements form tetra halides in which the central atom is $s{{p}^{3}}$ hybridized and assumes a tetrahedral shape.