MCQ
Which has weakest bond
  • A
    Diamond
  • Neon (Solid)
  • C
    $KCl$
  • D
    Ice

Answer

Correct option: B.
Neon (Solid)
b
Diamond containg a strong covalent bonds.

In potassium chloride $( K C l )$ is containing an ionic bond.

In ice molecule containing hydrogen bond.

therefore, decreasing order of strong bond is as follows:-

Ionic $\,>\,$ Covalent $\,>\,$ Hydrogen $\,>\,$ Dipole-Dipole $\,>\,$ VanderWaal

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

Similar questions

Which one of the following pairs of solution is not an acidic buffer?
The number of unpaired electrons in $F{e^{3 + }}(Z = 26)$ are
Secondary butyl chloride will undergo alkaline hydrolysis in the polar solvent by the mechanism:
Metal which emits photo-electrons very easily is:
An ideal gas undergoes a reversible isothermal expansion from state $I$ to state $II$ followed by a reversible adiabatic expansion from state $II$ to state $III$. The correct plot($s$) representing the changes from state $I$ to state $III$ is(are)

( $p$ : pressure, $V$ : volume, $T$ : temperature, $H$ : enthalpy, $S$ : entropy)

If the atomic weight of an element is $ 23 $ imes that of the lightest element and it has $11$  protons, then it contains
$KMn{O_4}$ reacts with ferrous ammonium sulphate according to the equation

$MnO_4^ - + 5F{e^{2 + }} + 8{H^ + } \to M{n^{2 + }} + 5F{e^{3 + }} + 4{H_2}O$,

here $10\, ml$ of $0.1\, M$ $KMn{O_4}$ is equivalent to

Aluminium chloride exists as a dimer, $Al_2Cl_6,$ in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives:
Find the equilibrium constant for equilibrium
$HCO{O^ - } + {H_2}O \rightleftharpoons HCOOH + O{H^ - }$
In a solution of $0.1\, M\, HCOONa$. $K_a(HCOOH) = 1.8 \times 10^{-4}$
The value of $\log\,K$ for the reaction $A \leftrightharpoons B$ at $298\,K$ is (Nearest integer)

Given: $\Delta H ^0=-54.07\,kJ\,mol ^{-1}$

$\Delta S ^{\circ}=10\,JK ^{-1}\,mol ^{-1}$

(Take $2.303 \times 8.314 \times 298=5705$ )