sblock Elements 1 Question 4
4. The correct statements among (a) to (d) are:
-
Saline hydrides produce $H_2$ gas when reacted with $H_2O$.
-
Reaction of $LiAlH_4$ with $BF_3$ leads to $B_2H_6$.
-
$PH_3$ and $CH_4$ are electron rich and electron precise hydrides, respectively.
-
$HF$ and $CH_4$ are called as molecular hydrides.
(a) (1),(2),(3) and (4)
(b) (1),(2) and (3) only
(c) (3) and (4) only
(d) (1),(3) and (4) only
(2019 Main, 8 April II)
Show Answer
Answer:
Correct Answer: 4. (a)
Solution:
The explanation of given statements are as follows :
- Saline or ionic hydrides produce $\mathrm{H}_2$ with $\mathrm{H}_2 \mathrm{O}$.
$$ \stackrel{\oplus}{M} \stackrel{\ominus}{\mathrm{H}}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_2 \uparrow+\mathrm{MOH} $$
Thus, statement ( 1 ) is correct.
- $3 \mathrm{LiAlH}_4+4 \mathrm{BF}_3 \xrightarrow{\text { Ether }} \underset{\text { (Diboranc) }}{2 \mathrm{~B}_2 \mathrm{H}_6}+3 \mathrm{LiF}+3 \mathrm{AlF}_3$
Thus, statement (2) is correct.
- $\mathrm{PH}_3$ and $\mathrm{CH}_4$ are covalent hydrides and in both of the hydrides, octet of P and C have been satisfied. But P in $\mathrm{PH}_3$ has one lone pair of electrons and C in $\mathrm{CH}_4$ does not have so $\mathrm{PH}_3$ (group 15) and $\mathrm{CH}_4$ (group 14) are electron rich and electron precise hydrides, respectively.
Thus, statement ( 3 ) is correct.
- HF and $\mathrm{CH}_4$ are called as molecular hydrides because of their discrete and sterically symmetrical structure.
Thus, statement (4) is also correct.
BETA
