Relative stability of the contributing structures is :
\((I)>(III)>(II)\)
\((I)>(II)>(III)\)
\((II)>(I)>(III)\)
\((III)>(II)>(I)\)
To determine the relative stability of the given contributing structures, we need to consider several factors, including the distribution of charges, the number of covalent bonds, and the octet rule. Let’s analyze each structure in detail:
Comparing these factors, we can deduce the relative stability as follows:
Hence, the correct order of stability is \((I) > (II) > (III)\).
Explanation:
1. Neutral structures are more stable than charged ones. Therefore, structure (I) is the most stable as it is neutral.
2. Among the charged structures, a positive charge (+) on a less electronegative atom (like carbon) is more stable than a positive charge on a more electronegative atom (like oxygen).
- Hence, structure (II), where C+ is present, is more stable than (III), where O+ is present.
Order: \( I > II > III.\)
Final Answer: Option (2).
In a resonance tube closed at one end. Resonance is obtained at lengths \( l_1 = 120 \, \text{cm} \) and \( l_2 = 200 \, \text{cm} \). If \( v_s = 340 \, \text{m/s} \), find the frequency of sound.
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :
