$STATEMENT-1$: Alkali metals dissolve in liquid ammonia to give a blue solution because
$STATEMENT-2$: Alkali metals in liquid ammonia give solvated species of the type $[M(NH_3)_n]^{+}$ ($M=$ alkali metals).
$STATEMENT-2$: Alkali metals in liquid ammonia give solvated species of the type $[M(NH_3)_n]^{+}$ ($M=$ alkali metals).
- A$Statement-1$ is True,$Statement-2$ is True; $Statement-2$ is a correct explanation for $Statement-1$
- B$Statement-1$ is True,$Statement-2$ is True; $Statement-2$ is $NOT$ a correct explanation for $Statement-1$
- C$Statement-1$ is True,$Statement-2$ is False
- D$Statement-1$ is False,$Statement-2$ is True
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Similar Questions
The correct order of degree of hydration of $M^{+}$ ions of alkali metals is
Medium
View SolutionThe total number of elements among $S$,$P$,$Zn$,$Al$,$Sn$,$Pb$ which produce $H_2$ gas with $NaOH$ is
Medium
View SolutionComment on each of the following observations:
$(a)$ The mobilities of the alkali metal ions in aqueous solution are
$Li^{+} < Na^{+} < K^{+} < Rb^{+} < Cs^{+}$
$(b)$ Lithium is the only alkali metal to form a nitride directly.
$(c)$ $E^{\circ}$ for $M^{2+}_{(aq)} + 2e^{-} \longrightarrow M_{(s)}$ (where $M = Ca, Sr$ or $Ba$) is nearly constant.
$(a)$ The mobilities of the alkali metal ions in aqueous solution are
$Li^{+} < Na^{+} < K^{+} < Rb^{+} < Cs^{+}$
$(b)$ Lithium is the only alkali metal to form a nitride directly.
$(c)$ $E^{\circ}$ for $M^{2+}_{(aq)} + 2e^{-} \longrightarrow M_{(s)}$ (where $M = Ca, Sr$ or $Ba$) is nearly constant.
Medium
View SolutionFill in the blanks:
$(1)$ ......... method is used to prepare sodium carbonate.
$(2)$ Sodium carbonate exists as .........
$(3)$ Sodium chloride is commonly known as .........
$(4)$ Sodium chloride melts at ......... temperature.
$(1)$ ......... method is used to prepare sodium carbonate.
$(2)$ Sodium carbonate exists as .........
$(3)$ Sodium chloride is commonly known as .........
$(4)$ Sodium chloride melts at ......... temperature.
Difficult
View SolutionAmong the hydroxides and carbonates of $IA$ and $IIA$ group elements respectively,on moving down the group,the correct solubility order is:
Difficult
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