Chemistry Test - 48

Fluorine is the most reactive non-metal.

Fluorine is the most reactive non-metal because it is the most electronegative of all of the non-metal elements of the periodic table. Due to its strong electro negativity & small size, Fluorine has a strong tendency to accept electrons from other atoms or ions. As a result, it oxidises all other substances.

The reaction given is a Sandmeyer's Reaction.

• The Sandmeyer reaction is used to prepare aryl halides from aryl diazonium salts. 
• It is named after the Swiss chemist Traugott Sandmeyer.
• The reaction is a method for substitution of an aromatic amino group via preparation of its diazonium salt followed by its displacement with a nucleophile, catalyzed by copper (I) salts. 

The reaction takes place as follows:

\(\mathrm{x}(\mathrm{g}) \rightleftharpoons \mathrm{y}(\mathrm{g})+\mathrm{z}(\mathrm{g}) \mathrm{k}_{\mathrm{p}_1}=3\)

Initial moles \(\mathrm{n}\) - -

at equilibrium \(n-\) \(\alpha\)n \(\alpha\)n \(\alpha\)n

\(k_{p_1}=\frac{\left(\frac{\alpha}{1+\alpha} \times p_1\right)^2}{\frac{1-\alpha}{1+\alpha} p_1} \)

\(3=\frac{\alpha^2 \times p_1}{1-\alpha^2} \)

\(A(g) \rightleftharpoons 2 B(g) k_{p_2}=1\)

Initial mole \(\mathrm{n}-\)

at equilibrium \(\mathrm{x}-\) an \(2 \alpha \mathrm{p}_{\text {total }}=\mathrm{p}_2\)

\(\mathrm{k}_{\mathrm{p}_2}=\frac{\left(\frac{2 \alpha}{1+\alpha} \times \mathrm{p}_2\right)^2}{\frac{1-\alpha}{1+\alpha} \times \mathrm{p}_2} \)

\(1=\frac{4 \alpha^2 \times \mathrm{p}_2}{1-\alpha^2} \)

\(\frac{\mathrm{k}_{\mathrm{p}_1}}{\mathrm{k}_{\mathrm{p}_2}}=\frac{\mathrm{p}_1}{4 \mathrm{p}_2} \)

\(\frac{3}{1}=\frac{\mathrm{p}_1}{4 \mathrm{p}_2} \quad \therefore \mathrm{p}_1: \mathrm{p}_2=12: 1 \)

\(\mathrm{x}=12\)

All the oxides and hydroxides of Boron family are acidic, basic and amphoteric in nature.

On moving top to bottom of the group, there is a change from acidic to amphoteric and then to basic character, of oxides and hydroxides of group \(13\) elements. Only boric acid is soluble in water while the other hydroxides are insoluble in water.

The amorphous solids are the solids which have short range order and are isotropic in nature which means that the physical properties are same along the crystal length.

Thus the option (D) is incorrect as amorphous solids are isotropic nature.

From the expression of heat \(\mathrm{(q)}\),

\(\mathrm{q}=\mathrm{m} \mathrm{c} \Delta \mathrm{T}\)

Where,

\(\mathrm{c}=\) molar heat capacity

\(\mathrm{m}=\) mass of substance

\(\Delta \mathrm{T}=\) change in temperature

Given,

\(\mathrm{m}=60 \mathrm{~g}\)

\(\mathrm{c}=24 \mathrm{~J} \mathrm{\textrm {mol } ^ { - 1 } \mathrm { K } ^ { - 1 }}\)

\(\Delta \mathrm{T}=(55-35)^{\circ} \mathrm{C}\)

\(\Delta \mathrm{T}=(328-308) \mathrm{K}=20 \mathrm{~K}\)

Substituting the values in the expression of heat:

\(\mathrm{q} =\left(\frac{60}{27} \mathrm{~mol}\right)\left(24 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}\right)(20 \mathrm{~K})\)

\(\mathrm{q} =1066.7 \mathrm{~J}\)

\(\mathrm{q} =1.07 \mathrm{~kJ}\)

Given,

Wavelength, \(\lambda=300 \mathrm{~nm}=300 \times 10^{-9}\mathrm{~m}\)

Planck's constant, \(h=6.626 \times 10^{-34} \mathrm{Js}\)

The speed of light, \(=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}\)

Energy of one photon, \(E=h \nu\)

\(=\frac{h c} {\lambda}\)

\(=\frac{6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s} \times 3.0 \times 10^{8} \mathrm{~m} \mathrm{~s}}{300 \times 10^{-1} \mathrm{~m}}\)

\(=6.626 \times 10^{-19} \mathrm{~J}\)

As we know,

Avogadro's number \(= 6.022 × 10 \mathrm{~mol}^{-1}\)

The energy of one mole of photons,

\(=6.626 \times 10^{-19} \mathrm{~J} \times 6.022 \times 10^{23} \mathrm{~mol}^{-1}\)

\(=3.99 \times 10^{5} \mathrm{~J} \mathrm{~mol}\)

The minimum energy needed to removeone mole of electrons from sodium\(=(3.99-1.68) \times 10^{5} \mathrm{~J} \mathrm{~mol}^{-1}\)

\(=2.31 \times 10^{5} \mathrm{~J} \mathrm{~mol}^{-1}\)

The minimum energy for one electron\(=\frac{2.31 \times 10^{5} \mathrm{~J} \mathrm{~mol}^{-1}}{6.022 \times 10^{23} \text { electrons } \mathrm{mol}^{-1}}\)

\(=3.84 \times 10^{-19} \mathrm{~J}\)

In nitroprusside ion, the iron and NO exist as Fe(II) and \(NO^{+}\) rather than Fe(III) and NO. These forms can be distinguished by measuring the solid-state magnetic moment.

Nitroprusside ion is \(\left[\mathrm{Fe}(\mathrm{CN})_{5} \mathrm{NO}\right]^{2-}\). If the central atom iron is present here in \(\mathrm{Fe}^{2+}\) form, its effective atomic number will be \(26-2+(6 \times 2)=36\) and the distribution of electrons in valence orbitals (hybridised and unhybridized) of the \(\mathrm{Fe}^{2+}\) will be

It has no unpaired electron. So this anionic complex is diamagnetic. If the nitroprusside ion has \(\mathrm{Fe}^{3+}\) and \(\mathrm{NO}\), the electronic distribution will be such that it will have one unpaired electron i.e. the complex will be paramagnetic.

Thus, magnetic moment measurement establishes that in nitroprusside ion, the Fe and NO exist as F(II) and \(NO^{+}\) rather than Fe(III) and NO.

All the given options are correctly matched.

Aldehyde has \(C _{n} H _{2} O\) general formula. Example: Ethanal \(- C _{2} H _{4} O , n =2\)

Carboxylic acid has \(C _{n} H _{2 n} O _{2}\) general formula. Example: Propionic acid \(- C _{3} H _{6} O _{2}, n =3\)

Cycloalkane has \(C _{ n } H _{2 n }\) general formula. Example: Cyclobutane \(- C _{4} H _{8}, n =4\)

Alkane also has \(C _{n} H _{2 n +2}\) general formula. Example: Ethane \(- C _{2} H _{6}, n =2\)

Transition metals do not show any basic character and from cations complexes in higher oxidation states.

• In the higher oxidation states, the transition metals show acidic nature and form anionic complexes.
• In higher oxidation states metal oxides are more covalent in nature as the polarizing power of the metal ion proportional to its oxidation states. So, their oxides are acidic in nature.