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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)

(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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