Of the following statements which are true for $N O_{2}$ ?
$i)$ Nitrogen dioxide brown liquid turns pale on cooling and becomes colourless solid
ii) $N O_{2}$ undergoes dimerisation to form $N_{2} O_{4}$
iii) $N O_{2}$ is paramagnetic, $N_{2} O_{4}$ is diamagnetic

i and ii

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i and iii

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ii and iii

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All

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Solution

The correct option is D All
$N O_{2}$ - paramagnetic (odd no of $e^{-}$ )
$N_{2} O_{4}$ - dimagnetic (even no of $e^{-}$ )
$2 N O_{2} \to N_{2} O_{4}$ (dimerisation)
$2 N O_{2} \to N_{2} O_{4}$
Brown liquid colourless solid

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Similar questions

Q. Assertion :On cooling, the brown colour of nitrogen dioxide disappears. Reason: On cooling,

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undergoes dimerisation resulting in the pairing of odd electrons of

{N O}_{2}

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N₂O₄ (g) ⇌ 2NO₂ (g)

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Read the above assertion and reason and choose the corret option regarding it.

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Of the following statements which are true for NO2?i) Nitrogen dioxide brown liquid turns pale on cooling and becomes colourless solidii) NO2 undergoes dimerisation to form N2O4iii) NO2 is paramagnetic, N2O4 is diamagnetic

The correct option is D AllNO2- paramagnetic (odd no of e−)N2O4- dimagnetic (even no of e−)2NO2 → N2O4 (dimerisation)2NO2 → N2O4Brown liquid colourless solid

Of the following statements which are true for $N O_{2}$ ?
$i)$ Nitrogen dioxide brown liquid turns pale on cooling and becomes colourless solid
ii) $N O_{2}$ undergoes dimerisation to form $N_{2} O_{4}$
iii) $N O_{2}$ is paramagnetic, $N_{2} O_{4}$ is diamagnetic

The correct option is D All
$N O_{2}$ - paramagnetic (odd no of $e^{-}$ )
$N_{2} O_{4}$ - dimagnetic (even no of $e^{-}$ )
$2 N O_{2} \to N_{2} O_{4}$ (dimerisation)
$2 N O_{2} \to N_{2} O_{4}$
Brown liquid colourless solid