Discuss the consequences of high enthalpy of $H-H$ bond in terms of chemical reactivity of dihydrogen.
(N/A) The bond dissociation enthalpy of the $H-H$ bond is very high $(435.88 \, kJ \, mol^{-1})$.
This high value indicates that the $H-H$ bond is very stable and strong.
Consequently,dihydrogen is relatively inert at room temperature.
It requires high temperatures or the presence of a catalyst to react with other elements.
Additionally,the high ionization enthalpy $(1312 \, kJ \, mol^{-1})$ of hydrogen makes it difficult to lose an electron to form $H^+$ ions,which is why it primarily forms covalent bonds rather than ionic ones.
This high value indicates that the $H-H$ bond is very stable and strong.
Consequently,dihydrogen is relatively inert at room temperature.
It requires high temperatures or the presence of a catalyst to react with other elements.
Additionally,the high ionization enthalpy $(1312 \, kJ \, mol^{-1})$ of hydrogen makes it difficult to lose an electron to form $H^+$ ions,which is why it primarily forms covalent bonds rather than ionic ones.
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Similar Questions
Which of the following is a hydride?
Easy
View SolutionIdentify the correct statements about the hydrides.
$(I)$ In saline hydrides,the oxidation state of hydrogen is $-1$.
$(II)$ $LaH_{2.87}$ is an example of an interstitial hydride.
$(III)$ $NH_3$ and $H_2O$ have the tendency to form hydrogen bonds.
$(IV)$ Electrolysis of molten sodium hydride liberates $H_2$ gas at the anode.
$(I)$ In saline hydrides,the oxidation state of hydrogen is $-1$.
$(II)$ $LaH_{2.87}$ is an example of an interstitial hydride.
$(III)$ $NH_3$ and $H_2O$ have the tendency to form hydrogen bonds.
$(IV)$ Electrolysis of molten sodium hydride liberates $H_2$ gas at the anode.
Which of the following reactions can produce hydrogen gas?
Difficult
View SolutionWhich of the following reactions give $H_2$ as one of the products? (Reactions are not balanced)
$I) NaBH_4 + I_2 \longrightarrow$
$II) B_2H_6 + N(CH_3)_3 \longrightarrow$
$III) Al + NaOH + H_2O \longrightarrow$
$IV) BF_3 + NaH \longrightarrow$
$I) NaBH_4 + I_2 \longrightarrow$
$II) B_2H_6 + N(CH_3)_3 \longrightarrow$
$III) Al + NaOH + H_2O \longrightarrow$
$IV) BF_3 + NaH \longrightarrow$
Complete the following reactions:
$(i) \, H_{2(g)} + M_m O_{o(s)} \xrightarrow{\Delta }$
$(ii) \, CO_{(g)} + H_{2(g)} \xrightarrow[catalyst]{\Delta }$
$(iii) \, C_3 H_{8(g)} + 3 H_2 O_{(g)} \xrightarrow[catalyst]{\Delta }$
$(iv) \, Zn_{(s)} + NaOH_{(aq)} \xrightarrow{heat}$
$(i) \, H_{2(g)} + M_m O_{o(s)} \xrightarrow{\Delta }$
$(ii) \, CO_{(g)} + H_{2(g)} \xrightarrow[catalyst]{\Delta }$
$(iii) \, C_3 H_{8(g)} + 3 H_2 O_{(g)} \xrightarrow[catalyst]{\Delta }$
$(iv) \, Zn_{(s)} + NaOH_{(aq)} \xrightarrow{heat}$
Medium
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