Chemistry Test - 4

The coordination number of octahedral geometry is 6. Whose radius ratio lies between 0.732 to 0.414.

Hence,option A is correct.

We had earlier seen that XeF₂ has a linear shape. The 3 lone pairs of electrons of central Xe atom lie on the plane perpendicular to the axial Xe ^– F bonds. The molecule has a trigonal bipyramidal geometry

In this molecule, if one oxygen atom takes the place of a lone pair, we get the geometry of XeOF₂. Do note that Xenon has a double bond with the oxygen atom. Let us look at the ground state and excited state:

Let us look at the geometry:

Now try to deduce the shape and geometry of XeO₂F₂:

A logical leap is to assume that the structure would be similar to replacing two of the three lone pairs with two oxygen atoms in the XeF₂ structure. It is a good guess!

As for the geometry, XeF₂, XeOF₂ and XeO₂F₂ all have the same trigonal bipyramidal geometry. XeF₂ has a linear shape. XeOF₂ is T - Shaped. XeO₂F₂ is see-saw shaped.

Earlier we had seen that the geometry of XeF₄ has a square bipyramidal geometry and square planar shape:

In the above geometry, if we replace a lone pair with an Oxygen atom such that there is a double bond between Xe and O, then we get the geometry of XeOF₄. The four F atoms line in a plane perpendicular to the Xe – O axial sigma bond. This geometry is still the same as XeF_4, but the shape will be square pyramidal!

So, options b is the required answer.

Energy in to break \(\mathrm{N} \equiv \mathrm{N}\) and \(3 \mathrm{H}-\mathrm{H}=945+(3 \times 436)=2253 \mathrm{kJ}\)

Energy out to form \(6 \mathrm{N}-\mathrm{H}=6 \times 391=2346 \mathrm{kJ}\)

\(\Delta H\) for reaction (which involves formation of two \(\mathrm{mol}\) of \(\left.\mathrm{NH}_{3}\right)=2253-2346=-93 \mathrm{kJ}\)

Enthalpy of formation of ammonia \(=1 / 2 x-93=-46.5 \mathrm{kJ} \mathrm{mol}^{-1}\)

Hence the correct answer is option (a).

The given element is having 14 nutrons and 27 mass number means number of electrons or protons are (z−n) 27−14=13 each.

Distribution of electrons according to 2n² (n = principal quantum number) is 2(K),8(L),3(M). Hence, the valence shell of this element is M.

Hence the correct answer is option (c).

The transition metal with more unpaired electrons corresponds to the higher magnetic moment.

\(\mathrm{V}^{4+} \rightarrow 4 \mathrm{s}^{0} 3 \mathrm{d}^{1}, 1\) unpaired electron;

\(\mathrm{Mn}^{4+} \rightarrow 4 \mathrm{s}^{0} 3 \mathrm{d}^{3}, 3\) unpaired electron;

\(\mathrm{Fe}^{3+} \rightarrow 4 \mathrm{s}^{0} 3 \mathrm{d}^{5}, 5\) unpaired electron and

\(\mathrm{Ni}^{2+} \rightarrow 4 \mathrm{s}^{0} 3 \mathrm{d}^{8}, 2\) unpaired electron.

Hence, option B is correct.

For walking 1 km the man needs 150KJ of energy. To walk 5 kms he needs 150×5=750 KJ energy. If only 30% of energy is available for muscular work, the total energy needed will be

750/30×100=2500KJ1 mole of glucose produces 3000 KJ. Then for producing 2500 KJ, 5/6 mole of glucose =5/6×180=150 gms of glucose will be needed (assuming the molecular weight of glucose=180 gms)

Hence the correct answer is option (d).

(i) \(\mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} \Delta \mathrm{H}=-40\)

(ii) \(\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O} \Delta \mathrm{H}=-10\)

(iii) \(\mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\)

(i) +2 (ii) - (iii)

\(\mathrm{C}+2 \mathrm{H}_{2} \rightarrow \mathrm{CH}_{4}\)

\(\Delta \mathrm{H}=-40-20+20\)

\(=-40 \mathrm{k} \mathrm{cal}\)

Hence the correct answer is option (A).

The species which by definition has zero standard molar enthalpy of formation at 298 K is\(\mathrm{Cl}_{2}(\mathrm{g}) . \mathrm{Cl}_{2}\) is gas at room temperature.

\(\mathrm{Br}_{2}\) is in liquid state at this temperature.

Hence, option B is correct.