sblock Elements 1 Question 4
4. The correct statements among (a) to (d) are:
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Saline hydrides produce $H_2$ gas when reacted with $H_2O$.
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Reaction of $LiAlH_4$ with $BF_3$ leads to $B_2H_6$.
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$PH_3$ and $CH_4$ are electron rich and electron precise hydrides, respectively.
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$HF$ 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)
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Answer:
Correct Answer: 4. (a)
Solution:
- The explanation of given statements are as follows :
The explanation of given statements are as follows :
- Saline or ionic hydrides produce $\mathrm{H}_2$ with $\mathrm{H}_2 \mathrm{O}$. $$ M \stackrel{\ominus}{\mathrm{H}}+\mathrm{H}_2 \mathrm{O} \quad \mathrm{H}_2+M \mathrm{OH} $$
Thus, statement (1) is correct.
- $3 \mathrm{LiAlH}_4+4 \mathrm{BF}_3$ Ether $2 \mathrm{~B}_2 \mathrm{H}_6+3 \mathrm{LiF}+3 \mathrm{AlF}_3$ (Diborane)
Thus, statement (2) is correct.
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$\mathrm{PH}_3$ and $\mathrm{CH}_4$ are covalent hydrides and in both of the hydrides, octet of $\mathrm{P}$ and $\mathrm{C}$ have been satisfied. But $\mathrm{P}$ in $\mathrm{PH}_3$ has one lone pair of electrons and $\mathrm{C}$ 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.
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$\mathrm{HF}$ and $\mathrm{CH}_4$ are called as molecular hydrides because of their discrete and sterically symmetrical structure.
Thus, statement (4) is also correct.
