The dipole moment depends on the electronegativity difference and the molecular geometry. Here’s the analysis:
Thus, the decreasing order of dipole moments is:
\[ {H}_2{O} > {NH}_3 > {H}_2{S} > {NF}_3 \]Final Answer:
\( {H}_2{O} > {NH}_3 > {H}_2{S} > {NF}_3 \)
Observe the following reactions:
\( AB(g) + 25 H_2O(l) \rightarrow AB(H_2S{O_4}) \quad \Delta H = x \, {kJ/mol}^{-1} \)
\( AB(g) + 50 H_2O(l) \rightarrow AB(H_2SO_4) \quad \Delta H = y \, {kJ/mol}^{-1} \)
The enthalpy of dilution, \( \Delta H_{dil} \) in kJ/mol\(^{-1}\), is:
Identify the sets of species having the same bond order:
(i) \( {F}_2, {O}_2^{2-} \)
(ii) CO, NO\(^{+}\)
(iii) \(N_2\), \(O_2\)
(iv) \(H_2\), \(B_2\)
The correct option is: