MCQ
The heat of neutralisation will be highest in
- A$N{H_4}OH$ and $C{H_3}COOH$
- B$N{H_4}OH$ and $HCl$
- C$KOH$ and $C{H_3}COOH$
- ✓$KOH$ and $HCl$
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$\mathop C\limits_6 {H_3} - \mathop C\limits_5 H = \mathop C\limits_4 H - \mathop C\limits_3 {H_2} - \mathop C\limits_2 \equiv \mathop C\limits_1 H$
The state of hybridization of carbons $1, 3$ and $5$ are in the following sequence
$(a)\ Pb^{+2} > Pb^{+4} , Tl^{+1} < Tl^{+3}$
$(b)\ Bi^{+3} < Sb^{+3 }, Sn^{+2} < Sn^{+4}$
$(c)\ Pb^{+2} > Pb^{+4} , Bi^{+3} > Bi^{+5}$
$(d)\ Tl^{+3} < In^{+3} , Sn^{+2} > Sn^{+4}$
$(e)\ Sn^{+24} < Pb^{+2} , Sn^{+4} > Pb^{+4}$
$(f)\ Sn^{+2} < Pb^{+2} , Sn^{+4} < Pb^{+4}$