Group 14

Given below are two statements :

Statement I : The second ionization enthalpy of B, Al and Ga is in the order of B > Al > Ga.

Statement II : The correct order in terms of first ionization enthalpy is Si < Ge < Pb < Sn.

In the light of the above statements, choose the correct answer from the options given below :

Given below are two statements :

Statement I : $\ce{F2O}<\ce{H2O}<\ce{Cl2O}$ is the correct trend in terms of bond angle.

Statement II : $\ce{SiF4}$, $\ce{SnF4}$ and $\ce{PbF4}$ are ionic in nature.

In the light of the above statements, choose the correct answer from the options given below :

The reaction of $\ce{Pb(NO3)2}$ and NaCl in water produces a precipitate that dissolves upon the addition of HCl of appropriate concentration. The dissolution of the precipitate is due to the formation of

Choose the correct statement(s) among the following.

A tin chloride Q undergoes the following reactions (not balanced)

$\mathbf Q + \ce{Cl-} \to \mathbf X$

$\mathbf Q + \ce{Me3N} \to \mathbf Y$

$\mathbf Q + \ce{CuCl2} \to \mathbf Z + \ce{CuCl}$

X is a monoanion having pyramidal geometry. Both Y and Z are neutral compounds. Choose the correct option(s)

The option(s) with only amphoteric oxides is(are)

Which one of the following is an oxide ?

Match the items in Column I with its main use listed in Column II :

Column I	Column II
(A) Silica gel	(i) Transistor
(B) Silicon	(ii) Ion-exchanger
(C) Silicone	(iii) Drying agent
(D) Silicate	(iv) Sealant

Assertion : Among the carbon allotropes, diamond is an insulator, whereas, graphite is a good conductor of electricity.

Reason : Hybridization of carbon in diamond and graphite are $sp^3$ and $sp^2$, respectively.

The gas evolved on heating $\ce{CaF2}$ and $\ce{SiO2}$ with concentrated $\ce{H2SO4}$, on hydrolysis gives a white gelatinous precipitate. The precipitate is :

The number and type of bonds in $\ce{C2^2-}$ ion in $\ce{CaC2}$ are :

With respect to graphite and diamond, which of the statement(s) given below is (are) correct?

STATEMENT-1: $\ce{Pb^{4+}}$ compounds are stronger oxidizing agents than $\ce{Sn^{4+}}$ compounds.

and

STATEMENT-2: The higher oxidation states for the group 14 elements are more stable for the heavier members of the group due to 'inert pair effect'.