Chemistry Test - 3

Helium accounts for up to 23 percent by mass of the Universe and the Sun. It is the most abundant element in the universe after hydrogen; Helium is rare in the atmosphere because its atoms are small and travel fast enough to escape the pull of Earth’s gravitational field. Do note that all the other noble gases occur in the atmosphere. There is no primordial Helium. In other words, there was no Helium from before the formation of Earth.

That leaves us with the question – where does the Helium on Earth come from? Helium comes into existence as α particles from radioactive decay and tends to accumulate in natural gas deposits.

Hence the correct answer is option (b).

100g solution contains \(99 \mathrm{g} \mathrm{H}_{2} \mathrm{SO}_{4}\)

Moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}=\frac{99}{98},\) Volume of solution \(=\frac{100}{1.9} \mathrm{mL}\)

Therefore, Molarity \(=\frac{99}{98} \times \frac{1.9}{100} \times 1000=19.19 \mathrm{M}\)

Hence, the correct option is C.

Li only forms normal oxide Li₂O because Li⁺ is small in size and stabilized by an even smaller oxide ion (O^2-).

Hence the correct answer is option (c).

The magnetic moment increases with increasing number of unpaired electrons.

Ti²⁺: [Ar] 3d²; 2 unpaired electrons

V²⁺ : [Ar] 3d³; 3 unpaired electrons

Cr²⁺: [Ar] 3d⁴; 4 unpaired electrons

Mn²⁺: [Ar] 3d⁵; 5 unpaired electrons

Hence, the correct order of magnetic moments is Ti²⁺ < V²⁺ < Cr²⁺ < Mn²⁺.

Hence the correct answer is option (a).

The formation of gaseous $Pt{O}_2$ at $1200°C$ or higher is an endothermic reaction which can be used for chemical transport. The $Pt{O}_2$ diffuses at lower temperatures where it may deposit well-formed crystals of platinum metal.

Van Arkel method is used for the purification of certain metals makes uses of an exothermic reaction between metal and iodine to form a gaseous iodide, for example,$Cr{I}_2$. Since the formation of metal, iodide is exothermic, and so the metal is redeposited at a higher temperature.

Oxidation state of Co in the given complex is +3.

Co in ground state: 3d⁷4s²

So, Co⁺³: 3d⁶4s⁰

There will beno unpaired electrons in this coordination compound because NH₃ provides a strongligand field and is able to pair up electrons of the given compound forming d²sp³ hybridisation.Hence, the given complex is diamagnetic.

Hence the correct answer is option (b).

Optically inactive compounds, ie, the compounds which have no asymmetric or chiral carbon atom, do not rotate the plane polarised light. The structures of the given compounds is

Thus, 4-hydroxy heptane, due to the absence of asymmetric carbon atom does not rotate the plane polarised light.

Hence the correct answer is option (c).

[Co(NH₃)₃Cl₃] exists in two forms (facial and meridonial).

Both of these forms are achiral. Hence, [Co(NH₃)₃Cl ₃] does not show optical isomerism.

Hence the correct answer is option (b).

The conversion reaction of rhombic sulphur to monoclinic sulphur is given by,

\(\mathrm{S}_{\text {rhombic }} \rightarrow \mathrm{S}_{\text {monoclinic }}\)

Hence, \(\Delta \mathrm{H}_{\text {conversion }}=\Delta \mathrm{H}_{\text {combustion of } \mathrm{S}(\text { rhombic })}-\Delta \mathrm{H}_{\text {combustion of } \mathrm{S}(\text { monoclinic })}\)

\(\Delta \mathrm{H}_{\text {conversion }}=[(-70960)-(-71030)]=70 \mathrm{cal} .\)

Hence the correct answer is option (c).