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Help me solve this - Hydrogen - JEE Main

The correct statements among (a) to (d) are :

(a) saline hydrides produce $H_{2}$ gas when reacted with $H_{2}O.$

(b) reaction of $LiAlH_{4}$ with $BF_{3}$ leads to $B_{2}H_{6}.$

(c) $PH_{3}$ and $CH_{4}$ are electron -rich respectively.

(d) $HF$ and $CH_{4}$ are called as molecular hydrides.

Option 1)
(a),(b),(c) and (d).
Option 2)
(c) and (d) only.
Option 3)
(a),(c) and (d) only.
Option 4)
(a),(b) and (c) only.

Answers (1)

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V Vakul

(a) Saline hydride like $NaH,LiH$ react with water to form a base & $H_{2}$ gas $MH_{3}+H_{2}O(an)\rightarrow MOH(an)+H_{2}(g)$

(b) $3LiAlH_{4}+4Bf_{3}\rightarrow 3Lif+3Alf_{3}+2B_{2}H_{6}$

(c) Electron-Rich Hydride - it is a type of hydrides in which the number of $e^{-s}$ present are more than the number of $e^{-}$ required for making octet for ex- $PH_{3},NH_{3}$

Electron precise hydride -is the type of hydrides in which the number of $e^{s}$ presents is equal to the no.of $e^{-s}$ required (octet or duplet) $e_{n}-CH_{4}$

(d) $CH_{4},HF,H_{2}O,NH_{3}$ are covalent or molecular hydrides .

Option 1)
(a),(b),(c) and (d).
Option 2)

(c) and (d) only.

Option 3)

(a),(c) and (d) only.

Option 4)

(a),(b) and (c) only.

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