The order H_2O

Question

(D) $1$. Boiling point: The boiling point of $H_2O$ is $100 \ ^\circ C$ $(373 \ K)$,while the boiling point of $H_2O_2$ is $150 \ ^\circ C$ $(423 \ K)

Vedclass · Accepted Answer

All of the above are correct

Vedclass · Answer

(D) $1$. Boiling point: The boiling point of $H_2O$ is $100 \ ^\circ C$ $(373 \ K)$,while the boiling point of $H_2O_2$ is $150 \ ^\circ C$ $(423 \ K)$. Thus,$H_2O $2$. Acidic nature: $H_2O_2$ is a stronger acid than $H_2O$ because the $O-H$ bond in $H_2O_2$ is more polar due to the electron-withdrawing effect of the second oxygen atom,and the resulting hydroperoxide ion $(HO_2^-)$ is more stable than the hydroxide ion $(OH^-)$. Thus,$H_2O $3$. Dipole moment: The dipole moment of $H_2O$ is $1.85 \ D$,while the dipole moment of $H_2O_2$ is $2.13 \ D$. Thus,$H_2O Since all statements are correct,the correct option is $D$.

Column-$I$	Column-$II$
$(i)$ Hydrogen peroxide	$(a)$ Used as air purifier in submarines
$(ii)$ Calcium sulphate hemihydrate	$(b)$ Used for water softening
$(iii)$ Potassium superoxide	$(c)$ Antiseptic
$(iv)$ Calgon	$(d)$ Plaster of Paris

List-$I$	List-$II$
$I$. $H_2O_2$	$P$. $Mg(HCO_3)_2 + Ca(OH)_2 \rightarrow$
$II$. $Mg(OH)_2$	$Q$. $BaO_2 + H_2SO_4 \rightarrow$
$III$. $BaCl_2$	$R$. $Ca(OH)_2 + MgCl_2 \rightarrow$
$IV$. $CaCO_3$	$S$. $BaO_2 + HCl \rightarrow$
	$T$. $Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow$

Column-$I$	Column-$II$
$A$. $H$	$1$. Used in the name of perhydrol.
$B$. $H_2$	$2$. Can be reduced to dihydrogen by $NaH$.
$C$. $H_2O$	$3$. Can be used in hydroformylation of olefin.
$D$. $H_2O_2$	$4$. Can be used in cutting and welding.

The order $H_2O < H_2O_2$ is correct for which of the following properties?

Similar Questions

Match the following columns:

Column-$I$ Column-$II$

$(i)$ Hydrogen peroxide $(a)$ Used as air purifier in submarines

$(ii)$ Calcium sulphate hemihydrate $(b)$ Used for water softening

$(iii)$ Potassium superoxide $(c)$ Antiseptic

$(iv)$ Calgon $(d)$ Plaster of Paris

Correct order of boiling point $(BP)$ is:

In which of the following reactions,hydrogen is liberated?
$(i)$ $Al_{(s)} + HCl_{(aq)} \rightarrow$
$(ii)$ $Al_{(s)} + NaOH_{(aq)} \rightarrow$
$(iii)$ $NaBH_4 + I_2 \rightarrow$

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