Download App

-p-Block elements - I — Chemistry STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceChemistry-p-Block elements - I1 Mark
MCQ
Boric acid is an acid because its molecule 
  • A
    contains replaceable $H^+$ ion
  • B
    gives up a proton
  • ✓
    accepts $OH^-$ from water releasing proton.
  • D
    combines with proton from water molecule.
✓

Answer

Correct option: C.
accepts $OH^-$ from water releasing proton.
c
Boric acid is an acid because its molecule accepts $\mathrm{OH}^{-}$ from water releasing proton

$\mathrm{B}(\mathrm{OH})_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow\left[\mathrm{B}(\mathrm{OH})_{4}\right]^{(-)}(\mathrm{aq})+\mathrm{H}^{(-)}(\mathrm{aq})$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from -p-Block elements - I→-p-Block elements - I — all question types→Chemistry STD 11 Science question papers→Rajasthan Board STD 11 Science question papers→Rajasthan Board question paper generator→

Similar questions

$AlCl_3$ on hydrolysis gives:
→
Which of the following resonating structures of ${N_2}O$ is the most contributing
→
In the purification of bauxite by Hall's process
→
The ethylene molecule have :
→
The enthalpy of fusion of water is $1.435\ kcal/mol.$ The molar entropy change for the melting of ice at $0\,^oC$ is
→
What is the work function of the metal if the light of wavelength $4000\, \mathop A\limits^o $ generates photoelectrons of velocity $6\times10^5\, ms^{-1}$ from it? ............. $\mathrm{eV}$ (Mass of electron $= 9\times10^{-31}\, kg$ Velocity of light $= 3\times10^8\, ms^{-1}$ Planck’s constant $= 6.626\times10^{-34}\, Js$ Charge of electron $= 1.6\times10^{-19}\, J\,eV^{-1}$)
→
The enthalpy of combustion at $25\,^o C $ of $H_2$, cyclohexane $(C_6H_{12})$  and cyclohexene $(C_6H_{10})$  are $-241, -3920 $ and $-800\ KJ / mole$  respectively. The heat of hydrogenation of cyclohexene is.....$KJ / mole$
→
benzene $+ ClC{H_2}C{H_2}C{H_3}\xrightarrow{{AlC{l_3}}}M\xrightarrow{{(i)\,{O_2}/\Delta ,\,\left( {ii} \right)\,\,{H_2}{O^ \oplus }\,}}N + $ phenol Here $M$ and $N$ are
→
Anhydrous $AlC{l_3}$ is obtained from
→
The state of hybridisation of $C_2, C_3, C_5$ and $C_6$ of the hydrocarbon,

$\begin{array}{*{20}{c}}
  {C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}\,\,\,\,\,\,\,} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {\mathop C\limits_7 {H_3} - \mathop C\limits_6  - \mathop C\limits_5 H = \mathop C\limits_4 H - \mathop C\limits_3 H - \mathop C\limits_2  \equiv \mathop C\limits_1 H} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} 
\end{array}$

is in the following sequence

→