- A$\Delta H_{mix} = 0$
- B$\Delta U_{mix} = 0$
- C$\Delta P = P_{obs} -P_{calculated \,by \,Raoult's\, law} = 0$
- ✓$\Delta G_{mix} = 0$
Since the enthalpy of mixing (solution) is zero, the change in Gibbs energy on mixing is determined solely by the entropy of mixing ( $\Delta S _{\text {solution }}$ ).
So the $\Delta G$ is not zero.
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$(a)$ Octahedral $Co(III)$ complexes with strong field ligands have very high magnetic moments
$(b)$ When $\Delta_{0}< P$, the $d-$electron configuration of $Co(III)$ in an octahedral complex is $t_{\text {eg }}^{4} e_{g}^{2}$
$(c)$ Wavelength of light absorbed by $\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}$ is lower than that of $\left[\mathrm{CoF}_{6}\right]^{3-}$
$(d)$ If the $\Delta_{0}$ for an octahedral complex of $\mathrm{Co}(\mathrm{III})$ is $18,000 \;\mathrm{cm}^{-1},$ the $\Delta_{\mathrm{t}}$ for its tetrahedral complex with the same ligand will be $16,000\;\mathrm{cm}^{-1}$
$(A)$ $\mathrm{Ni}(\mathrm{CO})_{4}$
$(B)$ $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$
$(C)$ $\mathrm{Na}_{2}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]$
$(D)$ $\mathrm{PdCl}_{2}\left(\mathrm{PPh}_{3}\right)_{2}$
$A$ || $B$
($F = 96,500\;C\;mo{l^{ - 1}}; \,\, R = 8.314\;J{K^{ - 1}}mo{l^{ - 1}})$