The reaction of $\text{XeF}_6$ with a limited amount of quartz gives compound A. Then on reaction with an equivalent amount of $\text{XeO}_3$, A gives B. The products A and B are, respectively.
a.
$\text{XeOF}_2$ and $\text{XeO}_2\text{F}_2$
b.
$\text{XeOF}_4$ and $\text{XeO}_2\text{F}_2$
c.
$\text{XeOF}_2$ and $\text{XeOF}_4$
d.
$\text{XeO}_2\text{F}_2$ and $\text{XeOF}_4$
Step-by-Step AI Explanation
Verified Method
Reaction of $\text{XeF}_6$ with quartz (a limited amount of $\text{SiO}_2$) yields xenon oxytetrafluoride:
$$2\text{XeF}_6 + \text{SiO}_2 \longrightarrow 2\text{XeOF}_4\text{ (A)} + \text{SiF}_4$$
Subsequent reaction of $\text{XeOF}_4$ with an equivalent amount of $\text{XeO}_3$ yields xenon dioxydifluoride:
$$\text{XeOF}_4 + \text{XeO}_3 \longrightarrow 2\text{XeO}_2\text{F}_2\text{ (B)}$$
Thus, A is $\text{XeOF}_4$ and B is $\text{XeO}_2\text{F}_2$.
Q1mcq
For the given reaction,
$$[{}^*\mathrm{Co(L)}_n]^{2+} + [\mathrm{Co(L)}_n]^{3+} \longrightarrow [{}^*\mathrm{Co(L)}_n]^{3+} + [\mathrm{Co(L)}_n]^{2+}$$
the correct statement with respect to the rate of electron transfer process is
$o$-phen = $o$-phenanthroline; ${}^*\mathrm{Co}$ is labeled atom.
a.
fast electron transfer; $\mathrm{L = NH_3}$; $\mathrm{n = 6}$
b.
slow electron transfer; $\mathrm{L} = o\text{-phen}$; $\mathrm{n = 3}$
c.
very slow electron transfer; $\mathrm{L = NH_3}$; $\mathrm{n = 6}$
d.
very slow electron transfer; $\mathrm{L} = o\text{-phen}$; $\mathrm{n = 3}$
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Q2mcq
The geometry/shape of the $\text{Fe}_4$ core of the cluster $[\text{Fe}_4\text{C}(\text{CO})_{12}]^{2-}$ is
a.
tetrahedron
b.
square pyramid
c.
butterfly
d.
trigonal bipyramid
Step-by-Step AI Explanation
Verified Method
Using Wade-Mingos rules for the cluster $[\text{Fe}_4\text{C}(\text{CO})_{12}]^{2-}$:
Total valence electrons (TVE) = $4 \times 8\text{ (Fe)} + 4\text{ (C)} + 12 \times 2\text{ (CO)} + 2\text{ (charge)} = 32 + 4 + 24 + 2 = 62$.
For a tetranuclear cluster:
- $n = 4$
- TVE $= 60$ corresponds to a tetrahedron (closo, 6 skeletal pairs).
- TVE $= 62$ corresponds to an arachno structure with 4 vertices, which is a butterfly (an edge-opened tetrahedron).
Q2mcq
Correct statement(s) among the following with respect to ionization energy (IE) is/are
(i) $(\mathrm{IE}_1 + \mathrm{IE}_2 + \mathrm{IE}_3)$ for indium is more than that of aluminium
(ii) $\mathrm{IE}_1$ of scandium is higher than that of cobalt
(iii) $\mathrm{IE}_1$ of gallium is lower than that of selenium
(iv) $\mathrm{IE}_1$ of nitrogen is greater than that of oxygen
a.
(ii) and (iv)
b.
(iii) and (iv)
c.
(i) and (iv)
d.
(ii) and (iii)
Step-by-Step AI Explanation
Verified Method
Statement (iii): Along a period, ionization energy generally increases; Ga (Group 13) has a significantly lower first ionization energy ($579\text{ kJ/mol}$) than Se (Group 16, $941\text{ kJ/mol}$). Thus, statement (iii) is correct.
Statement (iv): Nitrogen has a stable, exactly half-filled $2p^3$ subshell, so its $\mathrm{IE}_1$ ($1402\text{ kJ/mol}$) is greater than that of oxygen ($1314\text{ kJ/mol}$), which has a $2p^4$ configuration where electron-electron pairing repulsion lowers $\mathrm{IE}_1$. Thus, statement (iv) is correct.
Statements (i) and (ii) are incorrect, making (iii) and (iv) the correct pair.
Q3mcq
The energies of interaction for (i) ion pair, (ii) ion-dipole, and (iii) dipole-dipole interactions are inversely proportional to
a.
$r$, $r^2$ and $r^3$ respectively
b.
$r^2$, $r$ and $r^3$ respectively
c.
$r$, $r^2$ and $r^6$ respectively
d.
$r^2$, $r$ and $r^6$ respectively
Step-by-Step AI Explanation
Verified Method
Based on classical electrostatics:
(i) Ion–ion (ion pair) interaction energy is governed by Coulomb's law and is proportional to $1/r$.
(ii) Ion–dipole interaction energy is proportional to $1/r^2$.
(iii) Stationary dipole–dipole interaction energy is proportional to $1/r^3$.
Therefore, the interaction energies are inversely proportional to $r$, $r^2$, and $r^3$, respectively.
Q3mcq
Correct statement for ‘Inductively Coupled Plasma Atomic Emission Spectroscopy’ is
a.
It is unsuitable for all non-metals
b.
Simultaneous determination of only metals is possible
c.
Induction coil stabilizes plasma
d.
Oxide formation lowers atomization of metals.
Step-by-Step AI Explanation
Verified Method
In Inductively Coupled Plasma Atomic Emission Spectroscopy (ICP-AES), the radiofrequency (RF) induction coil generates a time-varying magnetic field that couples energy into the argon gas, sustaining and stabilizing the high-temperature plasma. The other options are incorrect: ICP can determine many non-metals (e.g., P, S, halogens), simultaneous multielement analysis includes metalloids/non-metals, and the extremely high plasma temperature ($6000–10000\text{ K}$) effectively prevents oxide formation and ensures complete atomization.
Q4mcq
During the binding of $\mathrm{O}_2$ to myoglobin (consider ‘heme’ in xy-plane), the molecular orbital of $\mathrm{O}_2$ and atomic orbital of $\mathrm{Fe}$ involved in the formation of the $\sigma$-bond is
a.
$\pi^*$ and $\mathrm{d}_{z^2}$
b.
$\pi^*$ and $\mathrm{d}_{xz}$
c.
$\pi$ and $\mathrm{d}_{xz}$
d.
$\pi$ and $\mathrm{d}_{z^2}$
Step-by-Step AI Explanation
Verified Method
With the heme porphyrin ring situated in the $xy$-plane, the axial coordination sites lie along the $z$-axis. The vacant or partially occupied iron atomic orbital oriented along this axis is $\mathrm{d}_{z^2}$. In the bent $\text{Fe}-\text{O}_2$ complex, $\mathrm{O}_2$ coordinates in an end-on fashion using its singly occupied $\pi^*$ antibonding molecular orbital to donate electron density into/form a $\sigma$-type bond with the $\mathrm{Fe}(\mathrm{d}_{z^2})$ orbital. Thus, the $\sigma$-bond is formed between the $\pi^*$ orbital of $\mathrm{O}_2$ and the $\mathrm{d}_{z^2}$ orbital of iron.
Q4mcq
The correct electronic configuration of frontier MO's of $Mn(\eta^5\text{-C}_5\text{Me}_5)_2$ is
a.
$e_{2g}^2 a_{1g}^1 e_{1g}^2$
b.
$e_{2g}^4 a_{1g}^1$
c.
$e_{2g}^3 a_{1g}^2$
d.
$a_{1g}^2 e_{2g}^3$
Step-by-Step AI Explanation
Verified Method
Decamethylmanganocene, $\mathrm{Mn}(\eta^5\text{-C}_5\text{Me}_5)_2$, is a low-spin $d^5$ 17-electron metallocene because the permethylated rings ($ ext{Cp}^*$) provide a stronger ligand field than unsubstituted $\text{Cp}$. In metallocenes ($D_{5d}/D_{5h}$ symmetry), the metal $d$-orbitals split into $a_{1g} (d_{z^2})$, $e_{2g} (d_{x^2-y^2}, d_{xy})$, and $e_{1g}^* (d_{xz}, d_{yz})$. Due to ring methylation and steric/electronic effects in $\text{MnCp}^*_2$, the ground-state low-spin configuration is $^2E_{2g}$, corresponding to the electronic configuration $a_{1g}^2 e_{2g}^3$ (often written as $e_{2g}^3 a_{1g}^2$).
Q5mcq
The pair of compounds in which both members show LMCT band in their electronic spectra is
a.
$[\mathrm{FeCl}_4]^{2-}$ and $[\mathrm{Fe(bpy)}_3]^{2+}$
b.
$[\mathrm{FeBr}_4]^{2-}$ and $[\mathrm{TcO}_4]^-$
c.
$[\mathrm{ReO}_4]^-$ and $[\mathrm{Ru(bpy)}_3]^{2+}$
d.
$[\mathrm{Fe(phen)}_3]^{2-}$ and $[\mathrm{FeCl}_4]^{2-}$
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Q5mcq
The correct match for Column-A with Column-B is
\begin{tabular}{ll}
\textbf{Column-A} & \textbf{Column-B} \\
(A) $f \not\to f$ transition & (i) $\text{Tb}^{3+}$ \\
(B) ${}^5D_0 \to {}^7F_n$ emission & (ii) $\text{Lu}^{3+}$ \\
(C) ${}^5D_4 \to {}^7F_n$ emission & (iii) $\text{Sm}^{3+}$ \\
(D) Overlapping J levels & (iv) $\text{Eu}^{3+}$
\end{tabular}
(A) $\text{Lu}^{3+}$ has a completely filled $4f^{14}$ shell, so $f \to f$ transitions are not possible ($f \not\to f$).
(B) $\text{Eu}^{3+}$ ($4f^6$) shows characteristic luminescence corresponding to ${}^5D_0 \to {}^7F_n$ transitions (red emission).
(C) $\text{Tb}^{3+}$ ($4f^8$) shows characteristic luminescence corresponding to ${}^5D_4 \to {}^7F_n$ transitions (green emission).
(D) $\text{Sm}^{3+}$ and $\text{Eu}^{3+}$ have low-lying excited $J$-levels where the energy gap is comparable to $k_B T$, leading to populated/overlapping $J$ levels at room temperature.
Therefore, the correct match is (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii).
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On heating $[\text{Mo}(\text{N}_2)_2(\text{PMe}_2\text{Ph})_4]$ and $[\text{ReCl}(\text{N}_2)(\text{PMe}_2\text{Ph})_4]$, the products formed are respectively
a.
$[(\eta^6\text{-PhPMe}_2)\text{Mo}(\text{PMe}_2\text{Ph})_3]$ and $[(\eta^6\text{-PhPMe}_2)\text{Re}(\text{PMe}_2\text{Ph})_3]\text{Cl}$
b.
$[(\text{PMe}_2\text{Ph})_4\text{Mo}(\mu\text{-N}_2)_2\text{Mo}(\text{PMe}_2\text{Ph})_4]$ and $[(\text{PMe}_2\text{Ph})_4\text{Re}(\mu\text{-Cl})_2\text{Re}(\text{PMe}_2\text{Ph})_4]$
c.
$[(\eta^6\text{-PhPMe}_2)\text{Mo}(\text{PMe}_2\text{Ph})_3]$ and $[(\text{PMe}_2\text{Ph})_4\text{Re}(\mu\text{-Cl})_2\text{Re}(\text{PMe}_2\text{Ph})_4]$
d.
$[(\text{PMe}_2\text{Ph})_4\text{Mo}(\mu\text{-N}_2)_2\text{Mo}(\text{PMe}_2\text{Ph})_4]$ and $[(\eta^6\text{-PhPMe}_2)\text{Re}(\text{PMe}_2\text{Ph})_3]\text{Cl}$
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Consider the following statements for Allred-Rochow electronegativity ($\chi_{\text{AR}}$):
(I) $\chi_{\text{AR}}$ is directly proportional to $Z_{\text{eff}}$.
(II) $\chi_{\text{AR}}$ is inversely proportional to $Z_{\text{eff}}$.
(III) $\chi_{\text{AR}}$ is inversely proportional to $r(\text{covalent})$
(IV) $\chi_{\text{AR}}$ is inversely proportional to $r^2(\text{covalent})$
The correct statements are
a.
(I) and (IV)
b.
(II) and (IV)
c.
(I) and (III)
d.
(II) and (III)
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Among the following complexes, the one showing the highest rate of substitution of a $\text{CO}$ ligand on heating with one equivalent of $\text{PPh}_3$ in decalin, is
In the following sequence of reactions the correct (P), (Q) and (R) are respectively
$$\underset{\text{in dil. AcOH}}{\textbf{(P)}} \xrightarrow[\text{thiourea}]{\text{solid}} \text{N}_2\text{ gas}\!\uparrow + \text{ Solution} \xrightarrow{\text{dil. }\text{FeCl}_3} \underset{\text{blood red}}{\textbf{(Q)}} \text{ soln.} \xrightarrow{\text{dil. }\text{HgCl}_2} \underset{\text{Colourless}}{\textbf{(R)}}$$
The value of magnetic moment will be independent of temperature for ($\text{acac} = \text{acetylacetonato}$; $\text{OAc} = \text{acetate}$; $o\text{-phen} = o\text{-phenanthroline}$; $\text{Pz} = \text{pyrazolyl}$)
$\text{I}_2$ is violet in the solid as well as in gas phase. However in acetone or ethanol, it turns brown. Choose the correct statement(s) for this colour change:
(A) Dissociation of $\text{I}_2$ in atomic state.
(B) Interaction of low-lying $\sigma^*$-orbital of iodine with lone pair of $\text{O}$ (solvent)
(C) Formation of a charge-transfer complex.
a.
(A) only
b.
(B) only
c.
(A) and (B) only
d.
(B) and (C) only
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The infrared (IR) spectrum of the product (A) for the following reaction,
shows three STRONG bands at 1986, 1935 and $1601\text{ cm}^{-1}$. The correct structure of 'A' is
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Thermometric titration gives best results when
(A) $|\Delta\text{H}|$ is high
(B) $\Delta\text{G} < 0$
(C) Heat of mixing of titrant with titrand is high
(D) Titrant is used as dilute solution
The answer is
a.
A and B
b.
B and C
c.
C and D
d.
A and D
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The products (A) and (B) for the given reaction
$$[\text{Co}(\text{NH}_3)_5\text{Cl}]^{2+} + [\text{Cr}(\text{OH}_2)_6]^{2+} + 5\text{H}_3\text{O}^+ \longrightarrow (\text{A}) + (\text{B})$$
are, respectively
In the catalytic cycle of Cytochrome P450, the generation of $[(\text{porphyrin})^{+\bullet} \text{Fe}^{\text{IV}}(\text{O})]$ from $[(\text{porphyrin})\text{Fe}^{\text{III}}(\text{OOH})]$ involves
a.
one electron oxidation of $[(\text{porphyrin})\text{Fe}^{\text{III}}(\text{OOH})]$
b.
formation of the intermediate $[(\text{porphyrin})\text{Fe}^{\text{IV}}(\text{OH})]$
c.
homolytic $\text{O}-\text{O}$ cleavage of $[(\text{porphyrin})\text{Fe}^{\text{IV}}(\text{OOH})]$
d.
heterolytic $\text{O}-\text{O}$ cleavage of $[(\text{porphyrin})\text{Fe}^{\text{III}}(\text{OOH})]$
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The number of expected electronic transitions in $[\text{Cr}(\text{en})_3]^{3+}$ and $\text{trans-}[\text{Cr}(\text{en})_2\text{F}_2]^+$ at $4\text{ K}$ is, respectively ($\text{en}=\text{ethylenediamine}$)
a.
$3$ and $3$
b.
$3$ and $4$
c.
$3$ and $5$
d.
$3$ and $6$
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Choose the correct statement(s) for the group 15 halides
(I) $\text{AsCl}_3$ can form complexes of type $[\text{AsCl}_4]^-$ in the presence of a chloride source.
(II) $\text{PF}_3$ acts as a strong $\sigma$-donor ligand towards d-metals
(III) In the solid state, $\text{SbF}_5$ has a trigonal bipyramidal structure
(IV) The reaction of $\text{SbF}_5$ with anhdrous $\text{HF}$ generates $[\text{H}_2\text{F}]^+$ ions
a.
(I) and (IV)
b.
(II), (III) and (IV)
c.
(I), (III) and (IV)
d.
(II) and (III)
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The number of peaks with relative intensities observed in the $^1\text{H}$ NMR spectra of $[(\text{Cp})_2\text{Fe}(\text{CO})_2]$ at $+30^\circ\text{C}$ and $-80\ ^\circ\text{C}$ in diethyl ether are respectively,
a.
two peaks ($1 : 1$) and four peaks ($5 : 2 : 2 : 1$)
b.
One peak and two peaks ($1 : 1$)
c.
two peaks ($1 : 1$) at both the temperatures
d.
one peak and four peaks ($5 : 2 : 2 : 1$)
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The major product formed in the following reaction sequene is
$\text{Ns} = \text{2-Nitrobenzenesulfonyl}$
$\text{DEAD} = \text{Diethyl azodicarboxylate}$
a.
(a)
b.
(b)
c.
(c)
d.
(d)
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For the reaction, $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \longrightarrow 2\text{NH}_3(\text{g})$, the thermodynamic quantity which depends upon the pressure at which equilibrium is arrived at is (superscript 'o' represents standard state)
a.
$\Delta\text{G}^0$
b.
$\Delta\text{H}^0$
c.
$\text{K}_\text{p}$
d.
ratio of mole fractions, $\frac{x_{\text{NH}_3}^2}{x_{\text{N}_2}x_{\text{H}_2}^3}$, at equilibrium
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The major product formed in the following reaction is
$\text{TBSOTf} = \text{t-Butyldimethylsilyl triflate}$
$\text{TBAF} = \text{tetra-n-butylammonium fluoride}$
a.
(a)
b.
(b)
c.
(c)
d.
(d)
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The following is the character TABLE for the $\text{C}_{2\text{v}}$ point group
There are two functions $f_1$ and $f_2$ belonging to $A_2$ and $B_1$ representations, respectively. The correct option for the product of two functions $f_1$ and $f_2$ and the integral $\int f_1 f_2 \text{d}\tau$ is
a.
The product belongs to $\text{A}_2$ representation and the integral is non-zero
b.
The product belongs to $\text{B}_2$ representation and the integral is zero
c.
The product belongs to $\text{A}_1$ representation and the integral is zero
d.
The product belongs to $\text{B}_1$ representation and the integral is non-zero.
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For a reaction, raising the temperature from $200\text{K}$ to $300\text{K}$ results in the increase of rate constant by a factor of 2. The activation energy for this reaction in $\text{kJ mol}^{-1}$ is closest to ($\ln 2 = 0.69$)
a.
7.0
b.
3.5
c.
14.0
d.
0.83
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The y-intercept of the least square fitted straight line to the data in the table is closest to
$$\begin{array}{|c|c|c|c|c|} \hline x & -2.01 & -0.98 & 1.01 & 1.99 \\ \hline y & -4.01 & -1.98 & 2.01 & 3.99 \\ \hline \end{array}$$
a.
$-0.2$
b.
$-0.1$
c.
$0$
d.
$0.2$
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The equilibrium dissociation eneryg of a diatomic molecule is $4.75\text{ eV}$ and its stretching frequency corresponds to $0.5\text{ eV}$. The minimum energy required to dissociate the mmolecule in $\text{eV}$ is
a.
$4.75$
b.
$4.25$
c.
$4.50$
d.
$5.00$
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In the given transformation, the intermediate (A) and the major product (B) are
$$\underset{(2\text{ equiv.})}{\text{PhCHO}} + \text{N}_2\text{CHCO}_2\text{Et} \xrightarrow{\text{CuCl}} [\textbf{A}] \longrightarrow [\textbf{B}]$$
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A monatomic perfect gas undergoes expansion from $(p_1, V_1)$ to $(p_2, V_2)$ under isothermal or adiabatic conditions. The pressure of the gas will fall more rapidly under adibatic conditions because
a.
$p \propto \frac{1}{V}$
b.
$p \propto \frac{1}{V^{7/5}}$
c.
$p \propto \frac{1}{V^{3/2}}$
d.
$p \propto \frac{1}{V^{5/3}}$
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The y-intercept obtained from the plot of viscosity of a series of polymer solutions against the concenration is $0.05$. The proportionally constant $\text{K}$ and exponent $\text{a}$ for this polymer-solvent pair are $5\times 10^{-5}$ and $0.5$, respectively. The molar mass of the polymer in $\text{g mol}^{-1}$ is
a.
$10^9$
b.
$10^5$
c.
$10^6$
d.
$10^7$
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The expectation value of $p^2$ of a particle in a cubic box of side $\ell$, having the wavefunction
$$\psi_{n_x, n_y, n_z}(x, y, z) = \left(\frac{2}{\ell}\right)^{3/2} \sin \frac{2\pi x}{\ell} \sin \frac{3\pi y}{\ell} \sin \frac{2\pi z}{\ell},\text{ is}$$
a.
$\frac{17h^2}{4\ell^2}$
b.
$\frac{7h^2}{4\ell^2}$
c.
$\frac{3h^2}{\ell^2}$
d.
$\frac{13h^2}{4\ell^2}$
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The $E^0(M^+/M)$ of the cell, $\text{SHE} \parallel \text{MX} \mid \text{M}$ can be obtained from the plot of ($E_{\text{cell}}$ is the cell potential and $m$ is the molality of ideal dilute solution of $\text{MX}$)
a.
$E_{\text{cell}}$ against $(T \log m)$
b.
$E_{\text{cell}}$ against $\left(\frac{1}{T}\log m\right)$
c.
$E_{\text{cell}}$ against $(T\sqrt{m})$
d.
$E_{\text{cell}}$ against $\left(\frac{\sqrt{m}}{T}\right)$
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Molar conductivities of an electrolyte are are $100$ and $50\text{ mS cm}^2\text{ mol}^{-1}$ when concentrations are $4$ and $9\text{ mM}$, respectively. The limiting molar conductivity of this electrolyte in $\text{mS cm}^2\text{ mol}^{-1}$ is closest to
a.
150
b.
200
c.
250
d.
300
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Consider a particle on a ring that is perturbed by interacting with an applied electric field (E) with the perturbation being $H' = \mu E \cos\phi$, where $\mu$ is the dipole moment. The energy levels correct upto first order are
A gas is known to satisfy Langmuir isotherm when adsorbed on a certain metal surface. If the fractional coverage of the gas is $0.5$ when the gas pressure is $1.0\text{ Pa}$, the fractional coverage at $3.0\text{ Pa}$ would be closest to
a.
0.67
b.
0.75
c.
0.80
d.
1.0
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The degree of polymerization at $t = 10\text{ h}$ of a polymer formed by a stepwise process with polymerization rate constant of $3\times 10^{-2}\text{ M}^{-1}\text{s}^{-1}$ and an initial monomer concentration of $50\text{ mM}$ is
a.
55
b.
65
c.
550
d.
505
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Substance 'A' is exposed to $600\text{ nm}$, $100\text{ W}$ light source for $6626\text{ s}$, with $50\%$ of the incident light being absorbed. 'A' decomposes according to the reaction $\text{A} \rightarrow 2\text{B}$ . As a result of irradiation, $0.2\text{ mol}$ of B is produced. The quantum yield of the reaction is closest to
a.
$6\times 10^6$
b.
$6\times 10^{-4}$
c.
$6\times 10^{-2}$
d.
6
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The energy separation of $^{12}\text{C}^{16}\text{O}$ rotational energy levels between $J'' = 3$ and $J'' = 9$ is $24\text{ cm}^{-1}$. The rotational constant of $^{13}\text{C}^{16}\text{O}$ in $\text{cm}^{-1}$ is closest to
a.
2.98
b.
0.88
c.
1.90
d.
2.08
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The concentration of (C) of a compound undergoing decomposition as a function of time (t) is given below
$$\begin{array}{|c|c|c|c|c|c|} \hline t / \text{min} & 0 & 1 & 2 & 4 & 6 \\ \hline C / M & 1.00 & 0.80 & 0.67 & 0.50 & 0.40 \\ \hline \end{array}$$
The order of reaction is
a.
0
b.
1
c.
2
d.
3
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For a model system of three non-interacting electrons confined in a two dimensional square box of length L, the ground state energy in units of $\left(\frac{h^2}{8mL^2}\right)$ is
a.
14
b.
6
c.
4
d.
9
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The entropy in terms of internal enregy $\{U-U(0)\}$, and canonical partition function (Q) is given by $S = \frac{U-U(0)}{T} + k\ln Q$. Assuming that the atoms are distinguishable, the corresponding expression for a monatomic perfect gas is
[Here $n, m, N_A, k$ and $h$ represent number of moles, mass of atom, Avogadro constant, Boltzmann constant and Planck constant, respectively, $U(0)$ is internal energy at $T = 0$]
The vibrational transition energies of a diatomic molecule corresponding to $v = 1 \leftarrow v = 0$ and $v = 2 \leftarrow v = 1$ are $2143.1\text{ cm}^{-1}$ and $2116.1\text{ cm}^{-1}$, respectively. The anharmonic constant $(\omega_e x_e)$ of the molecule in $\text{cm}^{-1}$ is
a.
27
b.
13.5
c.
10
d.
54
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The excess molar entropy of mixing of liquid A with liquid B is $-R\ln 2$. The experimentally observed change in entropy upon mixing $1.0\text{ mol}$ of liquid A with $1.0\text{ mol}$ of liquid B is
a.
$-2R\ln 2$
b.
$+4R\ln 2$
c.
0
d.
$+R\ln 2$
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An aqueous solution contains $0.02\text{ mol kg}^{-1}\text{ NaCl}$ and $0.03\text{ mol kg}^{-1}\text{Ca(NO}_3)_2$. The logarithm of the mean ionic activity coefficient $(\log \gamma_\pm)$ of this solution at $25^\circ\text{C}$ is
a.
$-\sqrt{0.095}$
b.
$\sqrt{0.154}$
c.
$-\sqrt{0.033}$
d.
$-\sqrt{0.11}$
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The change in chemical potential (in J) of one mole of an ideal gas, when it is compressed isothermally at $300\text{K}$ from $1.0\text{ atm}$ to $2.0\text{ atm}$, is closest to $(\ln 2 = 0.69)$
a.
1225
b.
1725
c.
2425
d.
2725
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