Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions?
(N/A) hydrogen atom has only one electron,which is one electron less than the stable inert gas configuration of helium $(1s^2)$.
To attain the stable electronic configuration of helium,each hydrogen atom shares its single electron with another hydrogen atom.
This sharing of electrons results in the formation of a stable diatomic molecule $(H_2)$.
To attain the stable electronic configuration of helium,each hydrogen atom shares its single electron with another hydrogen atom.
This sharing of electrons results in the formation of a stable diatomic molecule $(H_2)$.
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Similar Questions
The most abundant element in the universe is
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.
The product$(s)$ of the reaction $NaBH_4 + I_2 \longrightarrow$ 'Products' is/are
Fill in the blanks:
$(1)$ In the crystal structure of ice, the oxygen atom is at a distance of $276 \ pm$ from four hydrogen atoms.
$(2)$ The relative proportion of tritium to protium is $1: 10^{18}$.
$(1)$ In the crystal structure of ice, the oxygen atom is at a distance of $276 \ pm$ from four hydrogen atoms.
$(2)$ The relative proportion of tritium to protium is $1: 10^{18}$.
Easy
View SolutionWhich of the following sets are correctly matched?
$(i)$ $B_2H_6$ - electron deficient hydride
$(ii)$ $NH_3$ - electron precise hydride
$(iii)$ $H_2O$ - electron rich hydride
$(i)$ $B_2H_6$ - electron deficient hydride
$(ii)$ $NH_3$ - electron precise hydride
$(iii)$ $H_2O$ - electron rich hydride
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