- A. Sodium
- B. Iron
- C. Silver
- D. Lead
Sodium is extracted by an electrochemical process, namely electrolysis of molten sodium chloride (Downs process). Iron, silver and lead are generally extracted by chemical reduction or other metallurgical methods, not by electrolysis in their usual extraction routes.
- A. $\mathrm{In < Tl < Ga < Al}$
- B. $\mathrm{Al < Ga < In < Tl}$
- C. $\mathrm{Ga < In < Al < Tl}$
- D. $\mathrm{Tl < In < Ga < Al}$
The stability of the $+1$ oxidation state increases down group 13 due to the inert pair effect. Thus the order is $\mathrm{Tl > In > Ga > Al}$, i.e. $\mathrm{Tl < In < Ga < Al}$ in increasing stability. Hence option D is correct.
- A. $\mathrm{XeO_3}$
- B. $\mathrm{XeOF_4}$
- C. $\mathrm{XeO_2}$
- D. $\mathrm{XeO_2F_2}$
$\mathrm{XeF_6}$ on complete hydrolysis gives xenon trioxide. The overall reaction is: $$\mathrm{XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF}$$ So the correct product is $\mathrm{XeO_3}$. The other oxyfluorides are formed only in partial hydrolysis.
- A. $\mathrm{Co^{3+}}$ and $\mathrm{Fe^{2+}}$ have same number of unpaired electrons.
- B. Of all the known elements, Silver has the highest electrical conductivity at room temperature.
- C. To form a substitute alloy, the difference between the atomic radii of solvent and solute should be less than 15%.
- D. If the standard electrode potential ($E^\circ$) of a metal is large and negative, the metal is a powerful oxidising agent.
The incorrect statement is D. A large negative standard electrode potential means the metal is easily oxidised, so it acts as a powerful reducing agent, not an oxidising agent. A is correct because $\mathrm{Co^{3+}}$ is $d^6$ and low-spin octahedral, while $\mathrm{Fe^{2+}}$ is also $d^6$; in many common cases they are compared as having the same number of unpaired electrons. B is correct: silver has the highest electrical conductivity at room temperature. C is correct for substitutional alloys, where the atomic size difference should generally be less than 15%.
- B. 3
- C. 15
- D. 4
$[\mathrm{Pt(Py)(NH_3)(Br)(Cl)}]$ is a square planar complex of the type $\mathrm{MA_2BC}$ (all ligands different except the two positions occupied by none repeated? here effectively four different ligands around Pt). For square planar complexes with four different ligands, geometrical isomerism arises from different relative positions of pairs of ligands. The number of distinct arrangements is 3, corresponding to choosing which one is trans to a given ligand. Hence the correct answer is 3.
- A. Assertion is true but Reason is false.
- B. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
- C. Both Assertion and Reason are false.
- D. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
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Log in to view solution →- A. $\frac{1}{8}$ g
- B. $\frac{1}{2}$ g
- C. $\frac{1}{16}$ g
- D. $\frac{1}{4}$ g
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Log in to view solution →- A. $\mathrm{pH = 7 - \frac{1}{2}pK_a - \frac{1}{2}\log C}$
- B. $\mathrm{pH = 7 - \frac{1}{2}pK_a + \frac{1}{2}\log C}$
- C. $\mathrm{pH = 7 + \frac{1}{2}pK_a + \frac{1}{2}\log C}$
- D. $\mathrm{pH = 7 + \frac{1}{2}pK_a - \frac{1}{2}\log C}$
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Log in to view solution →- A. $0.086\ \mathrm{S\ cm^2\ mol^{-1}}$
- B. $2.88\ \mathrm{S\ cm^2\ mol^{-1}}$
- C. $28.8\ \mathrm{S\ cm^2\ mol^{-1}}$
- D. $11.52\ \mathrm{S\ cm^2\ mol^{-1}}$
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Log in to view solution →- A. The magnitude of the charge on the ion alone.
- B. Both magnitude and sign of the charge on the ion.
- C. The sign of charge on the ion alone.
- D. Size of the ion alone.
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Log in to view solution →- A. 1-chloropropane
- B. Ethyl chloride
- C. 2-chloro-2-methyl propane
- D. 2-chloropropane
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Log in to view solution →- A. Hexamethylene tetramine
- B. Formaldehyde
- C. Oxime
- D. Diacetone ammonia
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Log in to view solution →- A. $\mathrm{C_6H_5CHO}$
- B. $\mathrm{C_6H_5OH}$
- C. $\mathrm{C_6H_5NH_2}$
- D. $\mathrm{C_6H_5CH_2OH}$
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Log in to view solution →- A. Enantiomers
- B. Epimers
- C. Conformational isomers
- D. Anomers
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Log in to view solution →- A. enzymes
- B. antagonists
- C. molecular targets
- D. agonists
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Log in to view solution →- A. [Co(NH$_3$)$_4$Br$_2$]Cl and [Co(NH$_3$)$_4$ClBr]Br
- B. [Co(NH$_3$)$_5$NO$_2$] and [Co(NH$_3$)$_5$ONO]
- C. [Pt(NH$_3$)$_4$][PdCl$_4$] and [Pd(NH$_3$)$_4$][PtCl$_4$]
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