E represents an element belonging to carbon family- E-2X2 - EX4 X-F- Cl- Br- I

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Preparation and Properties of Beryllium Hydride

E represents ...

Question

E represents an element belonging to carbon family. $E + 2 X_{2} \to E X_{4} (X = F, C l, B r, I)$

Stability of $E X_{4}$ decreases down the $14^{t h}$ group

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$P b I_{4}$ does not exist

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Ge and Pb form $E X_{2}$ as well

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All are correct

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Solution

The correct option is D
All are correct

Stability of $E X_{4}$ decreases down the $14^{t h}$ group because of the inert pair effect. Lead (IV) iodide is not known yet probably because of the oxidizing nature of $P b^{4 +}$ and the reducing nature of $I^{-}$ . Ge and Pb definitely form $E X_{2}$ .

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Similar questions

E represents an element belonging to carbon family. $E + 2 X_{2} \to E X_{4} (X = F, C l, B r, I)$

Q. The correct order of polarizability is

I^{-}, B r^{-}, C l^{-}, F^{-}

Q. Standard reduction potentials of the half reactions are given below:

F_{2 (g)} + 2 e^{-} \to 2 F_{(a q)}^{-}; E^{o} = + 2.85 V

{C l}_{2 (g)} + 2 e^{-} \to 2 {C l}_{(a q)}^{-}; E^{o} = + 1.36 V

{B r}_{2 (l)} + 2 e^{-} \to 2 {B r}_{(a q)}^{-}; E^{o} = + 1.06 V

I_{2 (g)} + e^{-} \to 2 I_{(a q)}^{-}; E^{o} = + 0.53 V

The strongest oxidising and reducing agents respectively are:

Q. The increasing order of the atomic radii of the following elements is :

(a) C (b) O (c) F (d) Cl (e) Br

Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy ) is
(a) F > Cl > Br > I (b) F < Cl < Br < I (c) F < Cl > Br > I (d) F < Cl < Br < I

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E represents an element belonging to carbon family. E+2X2→EX4(X=F,Cl,Br,I)

The correct option is D All are correct Stability of EX4 decreases down the 14th group because of the inert pair effect. Lead (IV) iodide is not known yet probably because of the oxidizing nature of Pb4+ and the reducing nature of I−. Ge and Pb definitely form EX2.

E represents an element belonging to carbon family. $E + 2 X_{2} \to E X_{4} (X = F, C l, B r, I)$

The correct option is D
All are correct

Stability of $E X_{4}$ decreases down the $14^{t h}$ group because of the inert pair effect. Lead (IV) iodide is not known yet probably because of the oxidizing nature of $P b^{4 +}$ and the reducing nature of $I^{-}$ . Ge and Pb definitely form $E X_{2}$ .