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Identify the correct answer.
$CO_3^{2-}$ has three equivalent canonical forms (one C=O, two C–O⁻ per form, rotated).
The work done during reversible isothermal expansion of one mole of hydrogen gas at $25^\circ C$ from pressure of 20 atmosphere to 10 atmosphere is:
(Given $R = 2.0\ cal\ K^{-1}\ mol^{-1}$)
Work done by gas: $W = nRT \ln(P_1/P_2) = 1\cdot 2\cdot 298\cdot \ln 2 \approx 413.14$ cal (expansion → positive).
Consider the following reaction in a sealed vessel at equilibrium with concentrations of $N_2 = 3.0\times 10^{-3}\,M$, $O_2 = 4.2\times 10^{-3}\,M$ and $NO = 2.8\times 10^{-3}\,M$.
$2\,NO_{(g)} \rightleftharpoons N_2{}_{(g)} + O_2{}_{(g)}$
If $0.1\,\text{mol L}^{-1}$ of $NO_{(g)}$ is taken in a closed vessel, what will be degree of dissociation ($\alpha$) of $NO_{(g)}$ at equilibrium?
$K_c \approx 1.607$; solving $x^2/(0.1-2x)^2 = K_c$ gives $x \approx 0.0359$, so $\alpha = 2x/0.1 \approx 0.717$.
For the given reaction: Ph–CH=CH–Ph $\xrightarrow{KMnO_4/H^+}$ 'P' (major product). 'P' is:
Hot acidic KMnO₄ cleaves the C=C oxidatively; each side becomes a –COOH on the phenyl ring → benzoic acid.
The pair of lanthanoid ions which are diamagnetic is
$Ce^{4+}$ is $f^0$ and $Yb^{2+}$ is $f^{14}$ — both have no unpaired electrons.
The products A and B obtained in the following reactions, respectively, are:
$3\,ROH + PCl_3 \to 3\,RCl + A$
$ROH + PCl_5 \to RCl + HCl + B$
PCl₃ + 3ROH → 3RCl + H₃PO₃; PCl₅ + ROH → RCl + HCl + POCl₃.
Given below are certain cations. Using inorganic qualitative analysis, arrange them in increasing group number from 0 to VI.
A. $Al^{3+}$ B. $Cu^{2+}$ C. $Ba^{2+}$ D. $Co^{2+}$ E. $Mg^{2+}$
Choose the correct answer from the options given below:
Cu²⁺ Gr II → Al³⁺ Gr III → Co²⁺ Gr IV → Ba²⁺ Gr V → Mg²⁺ Gr VI.
Given below are two statements:
Statement I: $[\text{Co(NH}_3)_6]^{3+}$ is a homoleptic complex whereas $[\text{Co(NH}_3)_4\text{Cl}_2]^+$ is a heteroleptic complex.
Statement II: Complex $[\text{Co(NH}_3)_6]^{3+}$ has only one kind of ligands but $[\text{Co(NH}_3)_4\text{Cl}_2]^+$ has more than one kind of ligands.
In the light of the above statements, choose the correct answer from the options given below:
Homoleptic = one kind of ligand; heteroleptic = more than one. Both statements hold.
A compound X contains 32% of A, 20% of B and remaining percentage of C. Then, the empirical formula of X is:
(Given atomic masses of $A = 64$; $B = 40$; $C = 32$ u)
Moles: $A = 32/64 = 0.5$; $B = 20/40 = 0.5$; $C = 48/32 = 1.5$. Ratio $1:1:3$ → $ABC_3$.
The plot of osmotic pressure ($\Pi$) vs concentration ($\text{mol L}^{-1}$) for a solution gives a straight line with slope $25.73\ \text{L bar mol}^{-1}$. The temperature at which the osmotic pressure measurement is done is:
(Use $R = 0.083\ \text{L bar mol}^{-1}\text{K}^{-1}$)
$\Pi = MRT$, slope $= RT = 25.73$, $T = 25.73/0.083 = 310$ K $= 37^\circ$ C.
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