314408
\(\mathrm{BF}_{3}\) does not have proton but still acts as an acid and reacts with \(\mathrm{NH}_{3}\). Why is it so?
1 \(\mathrm{BF}_{3}\) is electron pair acceptor
2 \(\mathrm{BF}_{3}\) is electron pair donor
3 \(\mathrm{NH}_{3}\) is electron pair acceptor
4 \(\mathrm{NH}_{3}\) is electron pair donor
Explanation:
Electron pair acceptors are Lewis acids. \(\mathrm {\mathrm{BF}_{3}}\) is electron deficient and accepts electron pair as given: \(\mathrm {\mathrm{H}_{3} \mathrm{~N}: \rightarrow \mathrm{BF}_{3}}\).
CHXI07:EQUILIBRIUM
314409
Match Column I with Column II and choose the correct combination from the options given
1 A - R, S, B - P, C - S, D - Q , S
2 A - Q , S, B - Q , C - S, D - R , S
3 A - Q , S, B - R, C - R , S, D - P , R
4 A - R, B - P, C - R, S, D - Q , R
Explanation:
\(\mathrm{HSO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{O}^{\oplus}+\mathrm{SO}_{4}^{2-}\) (Bronsted acid) \(\mathrm{HSO}_{4}^{-}+\mathrm{H}^{\oplus} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}\) (Bronsted base) \(\mathrm{BF}_{3}\) is \(\mathrm{e}^{-}\)deficient and hence is a Lewis acid. \(\mathrm{H}_{3} \mathrm{~N}\) can donate lone pair of electrons and acts as Lewis base. \(\mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \stackrel{+}{\mathrm{N}} \mathrm{H}_{4}+\mathrm{OH}^{-}\)(Accepts \(\mathrm{H}^{+}\)and acts as Bronsted base) \(\mathrm{OH}^{-}\)acts as a Bronsted base by accepting \(\mathrm{H}^{+}\) ions. It can also donate lone pair of electrons and act as Lewis base.
CHXI07:EQUILIBRIUM
314410
Which of these is least likely to act as Lewis base?
1 \(\mathrm{F}^{-}\)
2 \(\mathrm{BF}_{3}\)
3 \(\mathrm{PF}_{3}\)
4 \(\mathrm{CO}\)
Explanation:
\({\rm{B}}{{\rm{F}}_{\rm{3}}}\) is Lewis acid (\({{\rm{e}}^{\rm{ - }}}\)pair acceptor)
CHXI07:EQUILIBRIUM
314429
The conjugate base of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in the following reaction is - \(\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HSO}_{4}^{-}\)
1 \(\mathrm{H}_{2} \mathrm{O}\)
2 \(\mathrm{H}_{3} \mathrm{O}^{+}\)
3 \(\mathrm{HSO}_{4}^{-}\)
4 \(\mathrm{SO}_{4}^{2-}\)
Explanation:
A pair of acid and its conjugate base differs by a proton only,i.e. \(\rm {\text { Acid } \rightleftharpoons \text { Conjugate base }+\mathrm{H}^{+}}\) \({\text{ For }}\mathop {{{\text{H}}_{\text{2}}}{\text{SO}}}\limits_{{\text{acid}}} {\text{,}}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \rightleftharpoons \mathop {{\text{HSO}}_{\text{4}}^{\text{ - }}{\text{ + }}{{\text{H}}^{\text{ + }}}}\limits_{{\text{ conjugate base }}} \)
314408
\(\mathrm{BF}_{3}\) does not have proton but still acts as an acid and reacts with \(\mathrm{NH}_{3}\). Why is it so?
1 \(\mathrm{BF}_{3}\) is electron pair acceptor
2 \(\mathrm{BF}_{3}\) is electron pair donor
3 \(\mathrm{NH}_{3}\) is electron pair acceptor
4 \(\mathrm{NH}_{3}\) is electron pair donor
Explanation:
Electron pair acceptors are Lewis acids. \(\mathrm {\mathrm{BF}_{3}}\) is electron deficient and accepts electron pair as given: \(\mathrm {\mathrm{H}_{3} \mathrm{~N}: \rightarrow \mathrm{BF}_{3}}\).
CHXI07:EQUILIBRIUM
314409
Match Column I with Column II and choose the correct combination from the options given
1 A - R, S, B - P, C - S, D - Q , S
2 A - Q , S, B - Q , C - S, D - R , S
3 A - Q , S, B - R, C - R , S, D - P , R
4 A - R, B - P, C - R, S, D - Q , R
Explanation:
\(\mathrm{HSO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{O}^{\oplus}+\mathrm{SO}_{4}^{2-}\) (Bronsted acid) \(\mathrm{HSO}_{4}^{-}+\mathrm{H}^{\oplus} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}\) (Bronsted base) \(\mathrm{BF}_{3}\) is \(\mathrm{e}^{-}\)deficient and hence is a Lewis acid. \(\mathrm{H}_{3} \mathrm{~N}\) can donate lone pair of electrons and acts as Lewis base. \(\mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \stackrel{+}{\mathrm{N}} \mathrm{H}_{4}+\mathrm{OH}^{-}\)(Accepts \(\mathrm{H}^{+}\)and acts as Bronsted base) \(\mathrm{OH}^{-}\)acts as a Bronsted base by accepting \(\mathrm{H}^{+}\) ions. It can also donate lone pair of electrons and act as Lewis base.
CHXI07:EQUILIBRIUM
314410
Which of these is least likely to act as Lewis base?
1 \(\mathrm{F}^{-}\)
2 \(\mathrm{BF}_{3}\)
3 \(\mathrm{PF}_{3}\)
4 \(\mathrm{CO}\)
Explanation:
\({\rm{B}}{{\rm{F}}_{\rm{3}}}\) is Lewis acid (\({{\rm{e}}^{\rm{ - }}}\)pair acceptor)
CHXI07:EQUILIBRIUM
314429
The conjugate base of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in the following reaction is - \(\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HSO}_{4}^{-}\)
1 \(\mathrm{H}_{2} \mathrm{O}\)
2 \(\mathrm{H}_{3} \mathrm{O}^{+}\)
3 \(\mathrm{HSO}_{4}^{-}\)
4 \(\mathrm{SO}_{4}^{2-}\)
Explanation:
A pair of acid and its conjugate base differs by a proton only,i.e. \(\rm {\text { Acid } \rightleftharpoons \text { Conjugate base }+\mathrm{H}^{+}}\) \({\text{ For }}\mathop {{{\text{H}}_{\text{2}}}{\text{SO}}}\limits_{{\text{acid}}} {\text{,}}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \rightleftharpoons \mathop {{\text{HSO}}_{\text{4}}^{\text{ - }}{\text{ + }}{{\text{H}}^{\text{ + }}}}\limits_{{\text{ conjugate base }}} \)
314408
\(\mathrm{BF}_{3}\) does not have proton but still acts as an acid and reacts with \(\mathrm{NH}_{3}\). Why is it so?
1 \(\mathrm{BF}_{3}\) is electron pair acceptor
2 \(\mathrm{BF}_{3}\) is electron pair donor
3 \(\mathrm{NH}_{3}\) is electron pair acceptor
4 \(\mathrm{NH}_{3}\) is electron pair donor
Explanation:
Electron pair acceptors are Lewis acids. \(\mathrm {\mathrm{BF}_{3}}\) is electron deficient and accepts electron pair as given: \(\mathrm {\mathrm{H}_{3} \mathrm{~N}: \rightarrow \mathrm{BF}_{3}}\).
CHXI07:EQUILIBRIUM
314409
Match Column I with Column II and choose the correct combination from the options given
1 A - R, S, B - P, C - S, D - Q , S
2 A - Q , S, B - Q , C - S, D - R , S
3 A - Q , S, B - R, C - R , S, D - P , R
4 A - R, B - P, C - R, S, D - Q , R
Explanation:
\(\mathrm{HSO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{O}^{\oplus}+\mathrm{SO}_{4}^{2-}\) (Bronsted acid) \(\mathrm{HSO}_{4}^{-}+\mathrm{H}^{\oplus} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}\) (Bronsted base) \(\mathrm{BF}_{3}\) is \(\mathrm{e}^{-}\)deficient and hence is a Lewis acid. \(\mathrm{H}_{3} \mathrm{~N}\) can donate lone pair of electrons and acts as Lewis base. \(\mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \stackrel{+}{\mathrm{N}} \mathrm{H}_{4}+\mathrm{OH}^{-}\)(Accepts \(\mathrm{H}^{+}\)and acts as Bronsted base) \(\mathrm{OH}^{-}\)acts as a Bronsted base by accepting \(\mathrm{H}^{+}\) ions. It can also donate lone pair of electrons and act as Lewis base.
CHXI07:EQUILIBRIUM
314410
Which of these is least likely to act as Lewis base?
1 \(\mathrm{F}^{-}\)
2 \(\mathrm{BF}_{3}\)
3 \(\mathrm{PF}_{3}\)
4 \(\mathrm{CO}\)
Explanation:
\({\rm{B}}{{\rm{F}}_{\rm{3}}}\) is Lewis acid (\({{\rm{e}}^{\rm{ - }}}\)pair acceptor)
CHXI07:EQUILIBRIUM
314429
The conjugate base of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in the following reaction is - \(\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HSO}_{4}^{-}\)
1 \(\mathrm{H}_{2} \mathrm{O}\)
2 \(\mathrm{H}_{3} \mathrm{O}^{+}\)
3 \(\mathrm{HSO}_{4}^{-}\)
4 \(\mathrm{SO}_{4}^{2-}\)
Explanation:
A pair of acid and its conjugate base differs by a proton only,i.e. \(\rm {\text { Acid } \rightleftharpoons \text { Conjugate base }+\mathrm{H}^{+}}\) \({\text{ For }}\mathop {{{\text{H}}_{\text{2}}}{\text{SO}}}\limits_{{\text{acid}}} {\text{,}}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \rightleftharpoons \mathop {{\text{HSO}}_{\text{4}}^{\text{ - }}{\text{ + }}{{\text{H}}^{\text{ + }}}}\limits_{{\text{ conjugate base }}} \)
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CHXI07:EQUILIBRIUM
314408
\(\mathrm{BF}_{3}\) does not have proton but still acts as an acid and reacts with \(\mathrm{NH}_{3}\). Why is it so?
1 \(\mathrm{BF}_{3}\) is electron pair acceptor
2 \(\mathrm{BF}_{3}\) is electron pair donor
3 \(\mathrm{NH}_{3}\) is electron pair acceptor
4 \(\mathrm{NH}_{3}\) is electron pair donor
Explanation:
Electron pair acceptors are Lewis acids. \(\mathrm {\mathrm{BF}_{3}}\) is electron deficient and accepts electron pair as given: \(\mathrm {\mathrm{H}_{3} \mathrm{~N}: \rightarrow \mathrm{BF}_{3}}\).
CHXI07:EQUILIBRIUM
314409
Match Column I with Column II and choose the correct combination from the options given
1 A - R, S, B - P, C - S, D - Q , S
2 A - Q , S, B - Q , C - S, D - R , S
3 A - Q , S, B - R, C - R , S, D - P , R
4 A - R, B - P, C - R, S, D - Q , R
Explanation:
\(\mathrm{HSO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{O}^{\oplus}+\mathrm{SO}_{4}^{2-}\) (Bronsted acid) \(\mathrm{HSO}_{4}^{-}+\mathrm{H}^{\oplus} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}\) (Bronsted base) \(\mathrm{BF}_{3}\) is \(\mathrm{e}^{-}\)deficient and hence is a Lewis acid. \(\mathrm{H}_{3} \mathrm{~N}\) can donate lone pair of electrons and acts as Lewis base. \(\mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \stackrel{+}{\mathrm{N}} \mathrm{H}_{4}+\mathrm{OH}^{-}\)(Accepts \(\mathrm{H}^{+}\)and acts as Bronsted base) \(\mathrm{OH}^{-}\)acts as a Bronsted base by accepting \(\mathrm{H}^{+}\) ions. It can also donate lone pair of electrons and act as Lewis base.
CHXI07:EQUILIBRIUM
314410
Which of these is least likely to act as Lewis base?
1 \(\mathrm{F}^{-}\)
2 \(\mathrm{BF}_{3}\)
3 \(\mathrm{PF}_{3}\)
4 \(\mathrm{CO}\)
Explanation:
\({\rm{B}}{{\rm{F}}_{\rm{3}}}\) is Lewis acid (\({{\rm{e}}^{\rm{ - }}}\)pair acceptor)
CHXI07:EQUILIBRIUM
314429
The conjugate base of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in the following reaction is - \(\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HSO}_{4}^{-}\)
1 \(\mathrm{H}_{2} \mathrm{O}\)
2 \(\mathrm{H}_{3} \mathrm{O}^{+}\)
3 \(\mathrm{HSO}_{4}^{-}\)
4 \(\mathrm{SO}_{4}^{2-}\)
Explanation:
A pair of acid and its conjugate base differs by a proton only,i.e. \(\rm {\text { Acid } \rightleftharpoons \text { Conjugate base }+\mathrm{H}^{+}}\) \({\text{ For }}\mathop {{{\text{H}}_{\text{2}}}{\text{SO}}}\limits_{{\text{acid}}} {\text{,}}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \rightleftharpoons \mathop {{\text{HSO}}_{\text{4}}^{\text{ - }}{\text{ + }}{{\text{H}}^{\text{ + }}}}\limits_{{\text{ conjugate base }}} \)
