Saline hydrides:
A. are dehydrating agents
B. are formed by transition metals
C. are very poor reducing agents
D. are stable towards water and alcohol
Answer
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Hint: Saline hydride forms by s-block elements. Generally, hydrogen donates electrons during the formation of a compound. When hydrogen combines with the electron donor metals then hydrogen accepts electrons and a hydride forms.
Complete step-by-step answer :Saline hydrides are formed by group-I and II metals which are alkali and alkaline earth metals.
Alkali metals form MH type hydride and alkaline earth metal forms ${\text{M}}{{\text{H}}_{\text{2}}}$type hydride.
Saline hydride are dehydrating agents. They remove water molecules from compounds on reacting so option (A) is correct.
Saline hydrides are formed by alkali and alkaline earth metal, not by transition metal so, option (B) is incorrect.
Saline hydrides are a strong reducing agent. The adds hydrogen to the reacting species so, option (C) is incorrect.
Saline hydrides react with water and alcohol and release hydrogen gas. The reaction is shown as follows:
${\text{MH}}\,\,{\text{ + }}\,{\text{ROH}}\,\, \to {\text{R}}{{\text{O}}^ - }{{\text{M}}^ + }\,{\text{ + }}\,{{\text{H}}_{\text{2}}}$
${\text{M}}{{\text{H}}_2}\,\,{\text{ + }}\,{\text{ROH}}\,\, \to {\text{M(OR}}{{\text{)}}_2}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}$
${\text{MH}}\,\,{\text{ + }}\,{{\text{H}}_2}{\text{O}}\,\, \to {\text{MOH}}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}$
${\text{M}}{{\text{H}}_2}\,\,{\text{ + }}\,{{\text{H}}_2}{\text{O}}\,\, \to {\text{M(OH}}{{\text{)}}_2}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}$
Where, M is the alkali or alkaline earth metal.
So, saline hydrides are not stable for water and alcohol so, option (D) is incorrect.
So, saline hydrides are dehydrating agents.
Therefore, option (A) are dehydrating agents, is correct.
Note: Saline hydrides are also known as an ionic hydride. These are very reactive and volatile hydride. Saline and alkali or alkaline all mean basic. S-block elements are more electropositive than hydrogen. The anion of hydrogen is known as a hydride. The species which lose electrons is known as a reducing agent. The reducing agent itself gets oxidized.
Complete step-by-step answer :Saline hydrides are formed by group-I and II metals which are alkali and alkaline earth metals.
Alkali metals form MH type hydride and alkaline earth metal forms ${\text{M}}{{\text{H}}_{\text{2}}}$type hydride.
Saline hydride are dehydrating agents. They remove water molecules from compounds on reacting so option (A) is correct.
Saline hydrides are formed by alkali and alkaline earth metal, not by transition metal so, option (B) is incorrect.
Saline hydrides are a strong reducing agent. The adds hydrogen to the reacting species so, option (C) is incorrect.
Saline hydrides react with water and alcohol and release hydrogen gas. The reaction is shown as follows:
${\text{MH}}\,\,{\text{ + }}\,{\text{ROH}}\,\, \to {\text{R}}{{\text{O}}^ - }{{\text{M}}^ + }\,{\text{ + }}\,{{\text{H}}_{\text{2}}}$
${\text{M}}{{\text{H}}_2}\,\,{\text{ + }}\,{\text{ROH}}\,\, \to {\text{M(OR}}{{\text{)}}_2}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}$
${\text{MH}}\,\,{\text{ + }}\,{{\text{H}}_2}{\text{O}}\,\, \to {\text{MOH}}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}$
${\text{M}}{{\text{H}}_2}\,\,{\text{ + }}\,{{\text{H}}_2}{\text{O}}\,\, \to {\text{M(OH}}{{\text{)}}_2}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}$
Where, M is the alkali or alkaline earth metal.
So, saline hydrides are not stable for water and alcohol so, option (D) is incorrect.
So, saline hydrides are dehydrating agents.
Therefore, option (A) are dehydrating agents, is correct.
Note: Saline hydrides are also known as an ionic hydride. These are very reactive and volatile hydride. Saline and alkali or alkaline all mean basic. S-block elements are more electropositive than hydrogen. The anion of hydrogen is known as a hydride. The species which lose electrons is known as a reducing agent. The reducing agent itself gets oxidized.
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