Chemistry Test - 49

Such a large difference in electronegativity shows that Caesium is electropositive while Fluorine is electronegative in nature. So, Caesium can donate an electron to Caesium which can accept it. So, they form ionic bond or electrovalent bond.

Ba < Ca < Se < S < Ar represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se and Ar.

Ionisation energy increases along a period from left to right and decreases down a group. The position of given elements in the periodic table is as:

Liquefied petroleum gas (LPG) mainly consists of propane and butane. Ethyl Mercaptan (CH₃CH₂SH) is an organic sulphur compound. It has a distinct smell that is used to detect the leakage of otherwise odourless LPG cooking gas.

All the elements except beryllium (Be) combine with hydrogen upon heating to form their hydrides, \(MH _{2}\).

\(BeH _{2}\), however, can be prepared by the reaction of \(BeCl _{2}\) with \(LiAlH _{4}\).

\(2 BeCl _{2}+ LiAlH _{4} \rightarrow 2 BeH _{2}+ LiCl + AlCl _{3}\)

Given,

Wavelength \(=5800 Å=5800 \times 10^{-10} \mathrm{~m}\)

As we know,

The speed of light,

\(c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\)

(a) Wavenumber is given as,

\(\bar{\nu}=\frac{1}{\lambda}\)

\(=\frac{1}{5800 \times 10^{-10} }\)

\(=1.724 \times 10^{6} \mathrm{~m}^{-1}\)

\(=1.724 \times 10^{4} \mathrm{~cm}^{-1}\)

(b) The frequency is given as,

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

\(=\frac{3 \times 10^{8} }{5800 \times 10^{-10}}\)

\(=5.172 \times 10^{14}\) Hz

\(\because d=\frac{Z \times H}{N_{A} \times V}\)

\(z=\frac{d \times N_{A} \times V}{M}\)

a = 5Å = 5 × 10^-8 cm

\(\therefore V=\left(5 \times 10^{-9}\right)^{3} \mathrm{cm}^{2}\)

Molecular mass, \(M=56+16=72\) g mol \(^{-1}\)

\(z=\frac{4 \times 6.023 \times 10^{23} \times\left(5 \times 10^{-8}\right)^{3}}{72}\)

\(z=4.18 \approx 4\)

Four molecules of ferrous oxide, FeO per unit cell, means that four Fe²⁺ and four O^2- ions are present in each unit cell. So, FeO forms a CCP lattice.

\(2 K(s)+F_{2}(g) \rightarrow 2 K^{+} F^{-}(s)\) is a type of redox reaction.

• The oxidation number of \(K\) increases from 0 to \(+1\) and the oxidation number of \(F_{2}\) decreases from 0 to \(-1 .\)
• Therefore, \(K\) is oxidized and \(F_{2}\) is reduced.
• Thus, it is a redox reaction.

\(\frac{E_{H_{+}}}{H_{2}}=\frac{E_{H^{+}}^{\circ}}{H_{2}}-\frac{0.059}{2} \log \frac{P_{H_{2}}}{\left[\left(H^{+}\right)\right]^{2}}\)

\(\frac{E_{H+}}{H_{2}}=-\frac{059}{2} P_{H_{2}}+.059 \log \left[H^{+}\right]\)

If \(\left[H^{+}\right]\) is increased by a factor of \(n\) then,

\(\frac{E_{H+}}{H_{2}}=\frac{0.059}{2} P_{H_{2}}+0.059 \log \left[\left(H^{+}\right)\right]+0.059 \log n\)

Thus, \(\Delta E=.059 \log n\)

Thus, \(\log n=\frac{\Delta E}{ .059}=\frac{.0282}{ .059}=.4780 \approx \log 3\)

So, \(n=3\)

In \(\mathrm{sp}, \mathrm{sp}^2\) and \(\mathrm{sp}^3\) hybridization of \(\mathrm{C}\), the number of hybrid orbitals formed are 2,3 and 4 respectively and the number of pure orbitals are 2,1 and 0 respectively. In this compound, \(2 \mathrm{~sp}\) hybridized, \(4 \mathrm{~sp}^2\) hybridized and \(1 \mathrm{~sp}^3\) hybridized \(\mathrm{C}\) atom is present. The number of pure orbitals in these atoms are \(4+4+\) \(0=8\)

Also there are \(6 \mathrm{~H}\) atoms which contribute 6 pure orbitals. Hence,the total number of pure orbitals are \(8+6=14\)

The number of hybrid orbitals are \(4+12+4+0=20\)

Hence, the ratio of pure orbitals to hybrid orbitals is \(14: 20\) or \(7: 10\)