Chemistry Test - 54

F < O < Nis the correct order of the atomic radii of the elements oxygen, fluorine, and nitrogen.

Oxygen (8), fluorine (9), and nitrogen (7) belong to the same period of the periodic table, in the order nitrogen, oxygen, and fluorine. Now in a period, on moving from left to right the atomic radius of the elements decreases. Therefore, the atomic radius of nitrogen is the largest.

The numbering of carbon chain is done from the carbon which is having two chlorine atoms. 

The parent alkane containing two carbon atoms is called as ethane and therefore, the IUPAC name of this compound is 1,1-Dichloroethane.

Let vapour pressure of \(A=P_A{ }^{\circ}\)

Vapour pressure of \(B=P_B{ }^{\circ}\)

In first solution,

Mole fraction of \(\mathrm{A}\left(\mathrm{X}_{\mathrm{A}}\right)=\frac{1}{1+2}=\frac{1}{3}\)

Mole fraction of \(\mathrm{B}\left(\mathrm{X}_{\mathrm{B}}\right)=\frac{2}{1+2}=\frac{2}{3}\)

According to Raoult's law,

Total vapour pressure \(=250=\mathrm{P}_{\mathrm{A}}{ }^{\circ} \mathrm{X}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}{ }^0 \mathrm{X}_{\mathrm{B}}\)

\(250=\frac{1}{3} \mathrm{P}_{\mathrm{A}}{ }^{\circ}+\frac{2}{3} \mathrm{P}_{\mathrm{B}}{ }^{\circ}\)

In second solution

Mole fraction of \(\mathrm{A}\left(\mathrm{X}_{\mathrm{A}}\right)=\frac{2}{2+2}=\frac{2}{4}=\frac{1}{2}\)

Mole fraction of \(\mathrm{B}\left(\mathrm{X}_{\mathrm{B}}\right)=\frac{2}{4}=\frac{1}{2}\)

\(\therefore\) Total vapour pressure \(=300=\mathrm{P}_{\mathrm{A}}{ }^{\infty} \mathrm{X}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}{ }^{\circ} \mathrm{X}_{\mathrm{B}}\)

\(300=\frac{1}{2} \mathrm{P}_{\mathrm{A}}{ }^{\circ}+\frac{1}{2} \mathrm{P}_{\mathrm{B}}{ }^{\circ}\)

Multiplying equation (i) by \(\frac{1}{2}\) and equation (ii) by \(\frac{1}{3}\)

\(\frac{1}{6} \mathrm{P}_{\mathrm{A}}^{\circ}+\frac{2}{6} \mathrm{P}_{\mathrm{B}}{ }^{\circ}=125 \)

\(\frac{\frac{1}{6} \mathrm{P}_{\mathrm{A}}{ }^{\circ}+\frac{1}{6} \mathrm{P}_{\mathrm{B}}{ }^{\circ}=100}{\frac{1}{6} \mathrm{P}_{\mathrm{B}}{ }^{\circ}=25}\)

\(\mathrm{P}_{\mathrm{B}}{ }^{\circ}=25 \times 6=150 \mathrm{mmH} g\) On substituting value of \(\mathrm{P}_{\mathrm{B}}{ }^{\circ}\) in equation (ii)

we get

\(300=\mathrm{P}_{\mathrm{A}}^{\circ} \times \frac{1}{2}+150 \times \frac{1}{2} \)

\(\mathrm{P}_{\mathrm{A}}^{\circ}=450 \mathrm{mmH} g\)

The compounds formed by the Boron family are both ionic and covalent.

Ionic compound formation’s tendency increases from Boron to thallium. Boron can only form covalent compounds, whereas aluminium can form both covalent as well as ionic compounds. Gallium forms mainly ionic compounds.

We know that:

Density, \(\rho=\frac{Z \times M}{a^{3} \times N_{0}}\)

where,

\(Z:\) no. of atoms in the bcc unit cell \(=2\)

\({M}:\) molar mass of \({CsBr}=133+80=213\)

\({a}:\) edge length of unit cell \(=436.6 {pm}\)

\(=436.6 \times 10^{-10} {~cm}\)

Therefore,

Density, \(\rho=\frac{2 \times 213}{\left(436.6 \times 10^{-10}\right)^{3} \times 6.02 \times 10^{23}}\)

\(=8.50 {~g} / {cm}^{3}\)

For a unit cell \(=\frac{8.50}{2}=4.25 {~g} / {cm}^{3}\)

In an ideal mono-atomic gas, the internal energy is contributed only due to the translational kinetic energy.

For example, if we take \(He\) or \(Ne\), the atoms are not bound to each other and hence they do not vibrate. Rotational energy is neglected since their atomic moment of inertia is minimal and higher energy excitation electronically also is not possible (except at very high temperatures).Thus, the internal energy changes in an ideal monoatomic gas are purely translational and there are \(3\) degrees of freedom. In other words, translation can take place in \(3D\) space i.e. \(x, y,\) and \(z\) directions.

Phenol on hydrogenation with Ni at 433K gives cyclohexanol.

The hydrogenation reaction is the addition of hydrogen to compounds.

Raney Ni, which is a finely divided Nickel powder is used in the hydrogenation process because it adsorbs the hydrogen molecules and provides high catalytic efficiency.

When phenol is treated with Ni at 433K, it converts to Cyclohexanol.

It is a highly selective reaction.

The reaction of reduction of phenol is given as:

When phenol is reacted with zinc dust it forms benzene.

Zinc oxidized itself and become ZnO and reduces phenol to benzene.

The electronic configuration of \(\mathrm{La}(\mathrm{Z}=57)\) is \([\mathrm{Xe}] 5 \mathrm{~d}^{1} 6 \mathrm{~s}^{2}\).

Therefore, further addition of electrons occurs in a lower energy 4 f-orbital till it is exactly half filled at \(\operatorname{Eu}(\mathrm{Z}=63)\).

Thus, the electronic configuration of Eu is \([\mathrm{Xe}] 4 \mathrm{f}^{7} 6 \mathrm{~s}^{2}\).

Therefore, addition of next electron doesn't occur in a more stable exactly halffilled \(4 \mathrm{f}^{7}\) shell but occur in a little higher energy \(5 \mathrm{~d}\)-orbital.

Thus, the electronic configuration of \(\mathrm{Gd}(\mathrm{Z}=64)\) is \([\mathrm{Xe}] 4 \mathrm{f}^{7} 5 \mathrm{~d}^{1} 6 \mathrm{~s}^{2}\).

\({A}_{2} {X}_{3} \rightarrow 2 {~A}_{3}++3 {X}^{-2}\)

\({Ksp}=\left[{A}^{3+}\right]^{2}\left[{X}^{2-}\right]^{3}\)

Let solubility of \({A}_{2} {X}^{3}\) be \({S}\)

\(\therefore {Ksp}=[2 {~S}]^{2}[3 {~S}]^{3}=108 {~S}^{5}\)

\(\therefore1.1 \times 10^{-23}=108 \mathrm{~S}^{5}\)

\(\therefore S 5=1.1 \times 10^{-23} / 108=1 \times 10^{-25}\)

\(\therefore S=1.0 \times 10^{-5} {~mol} / {L}\)