Chemistry Test - 36

The arrangement of elements in the Modem Periodic Table is based on increasing atomic number in the horizontal rows.

In 1869, Russian chemist Dmitri Mendeleev created the framework that became the modern periodic table, leaving gaps for elements that were yet to be discovered. While arranging the elements according to their atomic weight, if he found that they did not fit into the group he would rearrange them.

Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halides due to resonance stabilization.

Due to resonance, \({C}\) - \({Cl}\) bond acquires partial double bond character and becomes shorter and stronger and cannot be easily replaced by nucleophiles.

Other reasons for the low reactivity of aryl halides are:

(1) Difference in the hybridization states of carbon atom in \({C}-{X}\) bond. In alkyl halides, the carbon atom is \(sp^{3}\) hybridized whereas, in aryl halides, it is \({sp}^{2}\) hybridized and hence, more electronegative. Therefore, \({C}-{X}\) bond in aryl halides is more difficult to break.

(2) Polarity of \({C}-{X}\) bond in aryl halides is lower than that in alkyl halides. Lower is the polarity, lower is the reactivity.

\(\begin{aligned} & \text { Hybridisation }=\frac{1}{2}\left[\left(\begin{array}{c}\text { No. of electrons } \\ \text { in valence } \\ \text { shell of atom }\end{array}\right)+\right. \\ & \left(\begin{array}{c}\text { No.of monovalent } \\ \text { atoms around it }\end{array}\right)-\left(\begin{array}{c}\text { Charge on } \\ \text { cation }\end{array}\right)+\left(\begin{array}{c}\text { Charge on } \\ \text { anion }\end{array}\right)\end{aligned}\)

(a) For \(\mathrm{AlH}_3\)

Hybridisation of \(\mathrm{Al}\) atom \(=\frac{1}{2}[3+3-0+0]=3=\mathrm{sp}^2\)

For \(\mathrm{AlH}_4\),

Hybridisation of \(\mathrm{A} 1\) atom \(=\frac{1}{2}[3+4-0+1]=4=\mathrm{sp}^3\)

(b) For \(\mathrm{H}_2 \mathrm{O}\),

Hybridisation of \(\mathrm{O}\) atom

\(=\frac{1}{2}[6+2-0+0]=4=\mathrm{sp}^3\)

For \(\mathrm{H}_3 \mathrm{O}^{+}\),

Hybridisation of \(\mathrm{O}\) atom

\(=\frac{1}{2}[6+3-1+0]=4=\mathrm{sp}^3\)

(c) For \(\mathrm{NH}_3\),

Hybridisation of \(\mathrm{N}\) atom

\(=\frac{1}{2}[5+3-0+0]=4=\mathrm{sp}^3\)

For \(\mathrm{NH}_4^{+}\),

Hybridisation of \(\mathrm{N}\) atom

\(=\frac{1}{2}[5+4-1+0]=4=\mathrm{sp}^3\)

Thus hybridisation changes only in option (a).

Given,

\(\nu=1368\) kHz \(=1368 \times 10^{3}\) Hz

As we know,

The speed of light, \(c=3 \times 10^8\) m/s

The wavelength is,

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

\(=\frac{3.00 \times 10^{8}}{1368 \times 10^{3}}\)

\(=219.3 \mathrm{~m}\)

In ionic solids the vacancies are produced due to absence of cations and anions in stoichiometric proportions resulting in defect called Schottky defect.

This defect arises when cation and anion are of similar sizes.

CsCl shows Schottky defect.

A reaction, \(\mathrm{A+B \rightarrow C+D+q}\) is found to have a positive entropy change. The reaction will bepossible at any temperature.

For a reaction to be spontaneous, \(\Delta \mathrm{G}\) should be negative \(\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}\)

According to the question, for the given reaction,

\(\Delta \mathrm{S}=\) positive

\(\Delta \mathrm{H}=\) negative (since heat is evolved)

That results in \(\Delta \mathrm{G}=\) negative

Therefore, the reaction is spontaneous at any temperature.

The compound having the least molecular mass should contain the minimum amount of sulphur or simply 1 atom of sulphur.

Let the molecular mass of the compound (in amu) be \(x\).

Mass of sulphur in the compond \(=8\%\) of the total molecular mass of \(x\)

\(=\frac{8 x}{100}\)

Since, the molecular mass of one mole of sulphur is 32, the compound must contain this amount of sulphur, i.e.,

\(\frac{8 x}{100}=32\)

which gives:

\(x=400\)

Hybridization of \(\mathrm{Fe}\) in \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) is \(d^{2} s p^{3}\).

In the given complex of \(\mathrm{Fe}^{3+}\), as the co-ordination number is ' 6 ', there are two possibilities. It can be \(d^{2} s p^{3}\) or \(s p^{3} d^{2}\). But the ligand attached with central metal ion, \(\mathrm{Fe}^{3+}\), is strong field ligand so it will promote inner orbital complex, \(d^{2} s p^{3}\).

Under ultra-violet light, three chlorine molecules add to benzene to produce benzene hexachloride, C₆H₆C_l6 which is also calledgammaxane.Benzene hexachloride is an isomer of hexachlorocyclohexane with a chemical formula C₆H₆C_l6.It is also known as Lindane or hexachloride.

Uses of Benzene hexachloride (C₆H₆C_l6):

• Benzene hexachloride is used as an insecticide on crops, in forestry, for seed treatment.
• It is used in the treatment of head and body lice.
• It is used in pharmaceuticals.
• It is used to treat scabies.
• It is used in shampoo.

All have \(\mathrm{d}_{10}\) configuration. Copper belongs to \(\mathrm{3 d}\) series, silver belongs to \(\mathrm{4 d}\) series an gold belongs to \(\mathrm{5 d}\) series.

• The size decreases from copper to silver, but as we move from silver to gold, the size remains almost constant due to the lanthanide contraction that occurs due to introduction of \(\mathrm{f}\)- orbital.
• \(\mathrm{f}\)- orbital have poor shielding and therefore, force of attraction between nucleus and valence electron increases and atomic size of the element decreases.
• Thus, the melting point decreases from copper to silver, but the melting point of gold becomes even higher than silver due to its higher density that happened due to contraction.

Therefore, the order will be:

\(\mathrm{C u>A u>A g}\)