MCQ
$H -$ bonding is maximum in
- A${C_6}{H_5}OH$
- ✓${C_6}{H_5}OH$
- C$C{H_3}C{H_2}OH$
- D$C{H_3}COC{H_3}$
✓
Answer
Correct option: B.
${C_6}{H_5}OH$
(b) Due to greater resonance stabilization.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The electronegativity of the following elements increases in the order
During an electrolysis of conc. $H_2SO_4$ , perdisulphuric acid $(H_2S_2O_8)$ and $O_2$ form in equimolar amount. The amount of $H_2$ that will form simultaneously will be $(2{H_2}S{O_4} \to {H_2}{S_2}{O_8} + 2{H^ + } + 2{e^ - })$
→The solubility of $PbC{l_2}$ is
→Benzoic acid on treatment with hydrazoic acid in the presence of concentrated sulphuric acid gives:
Which of the following formula does not correctly represent the bonding capacity of the atom involved
A litre of buffer solution contains $0.1$ mole of each of $NH _3$ and $NH _4 Cl$. On the addition of $0.02\,mole$ of $HCl$ by dissolving gaseous $HCl$, the $pH$ of the solution is found to be $......\times 10^{-3}($ Nearest integer)
→$\left[\right.$ Given : $pK _{ b }\left( NH _3\right)=4.745$
$\log 2=0.301$
$\log 3=0.477$
$T =298\,K ]$
The formation of $PH_4^+$ is difficult compared to $NH_4^+$ because
→$A$ and $B$ are :-
→Match List$-I$ with List$-II$
→| List$-I$ | List$-II$ | ||
| $(a)$ | Sucrose | $(i)$ |
$\beta$ -D-Galactose and $\beta$ -D-Glucose |
| $(b)$ | Lactose | $(ii)$ | $\alpha$ -D-Glucose and $\beta$ -D-Fructose |
| $(c)$ | Maltose | $(iii)$ | $\alpha$ -D-Glucose and $\alpha$ -D-Glucose |
Choose the correct answer from the options given below:
.......... $(\%)$ percent of glucose solution is isotonic with $1.5\%\, (w/V)$ urea solution.
→