Chemistry Test - 12

Methane is the simplest member of the alkane family and indeed the simplest of organic compounds, as all other compounds are derived by altering this compound.

Argyrol is a silver solution and is used as an eye lotion. Silver bromide, gelatin and glass plate are used in the making photographic plates.

Photographic plates and films are prepared by coating an emulsion of the light sensitive silver bromide in gelatin over glass plates or celluloid films.

When steam is passed over red hot iron, ferric oxide and hydrogen is formed.

\(3 Fe +4 H _{2} O \longrightarrow \underset{Ferric\; oxide} {Fe _{3} O _{4}}+\underset{Dihydrogen}{4 H _{2} }\)

The iron is oxidized to ferric oxide whereas the water is reduced to hydrogen.

The precipitate will form only

When,

lonic product \(>\) Solubility product

\({CaF}_{2}\) dissociates as follows

\({CaF}_{2} \rightleftharpoons {Ca}^{2+}+2 {~F}^{-}\)

Thus,

Ionic product \(=\left[10^{-2}\right]\left[10^{-3}\right]^{2}\)

\(=10^{-2} \times 10^{-6}=10^{-8}\)

This value is greater than \(K_{s p}\) ,

i.e.,\(1.7 \times 10^{-10},\)

So, precipitatewill form.

\(\mathrm{V}=3 \mathrm{~d}^3 4 \mathrm{~s}^2 ; \mathrm{V}^{2+}=3 \mathrm{~d}^3=3 \text { unpaired electrons } \)

\(\mathrm{Cr}=3 \mathrm{~d}^5 4 \mathrm{~s}^1 ; \mathrm{Cr}^{2+}=3 \mathrm{~d}^4=4 \text { unpaired electrons } \)

\(\mathrm{Mn}=3 \mathrm{~d}^5 4 \mathrm{~s}^2 ; \mathrm{Mn}^{2+}=3 \mathrm{~d}^5=5 \text { unpaired electrons } \)

\(\mathrm{Fe}=3 \mathrm{~d}^6 4 \mathrm{~s}^2 ; \mathrm{Fe}^{2+}=3 \mathrm{~d}^6=4 \text { unpaired electrons }\)

Hence the correct order of paramagnetic behaviour

\(\mathrm{V}^{2+}<\mathrm{Cr}^{2+}=\mathrm{Fe}^{2+}<\mathrm{Mn}^{2+}\)

(b) For the same oxidation state, the ionic radii generally decreases as the atomic number increases in a particular transition series, hence the order is

\(\mathrm{Mn}^{2+}>\mathrm{Fe}^{2+}>\mathrm{Co}^{2+}>\mathrm{Ni}^{2+}\)

(c) Larger size, least hydrated more stable in aqueous solution. As we move across the period \(\left(\mathrm{Sc}^{3+} \rightarrow \mathrm{Cr}^{3+} \rightarrow\right.\) \(\mathrm{Fe}^{3+} \rightarrow \mathrm{Co}^{3+}\) ), the ionic size usually decreases. \(\mathrm{Sc}^{3+}\) with the large size as least hydrated and hence more stable.

\(\text { (d) } \mathrm{Sc}-(+2),(+3) \)

\(\mathrm{Ti}-(+2),(+3),(+4) \)

\(\mathrm{Cr}-(+2),(+3),(+4),(+5),(+6) \)

\(\mathrm{Mn}-(+2),(+3),(+4),(+5),(+6),(+7) \)

\(\text { i.e. } \mathrm{Sc}<\mathrm{Ti}<\mathrm{Cr}<\mathrm{Mn}\)

(A) \(\mathrm{N H_{4} N O_{2}}\) is Ammonium Nitrite. The thermal decomposition occurs as follows-

\(\mathrm{NH}_{4} \mathrm{NO}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{N}_{2}+2 \mathrm{H}_{2} \mathrm{O}\)

The reaction produces Nitrogen gas and water vapour.

(B) \(\mathrm{N a N_{3}}\) is sodium azide. The thermal decomposition occurs as follows-

\(2 \mathrm{N a N_{3}} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{N a}+3 \mathrm{N_{2}}\)

The reaction produces Sodium metal and Nitrogen gas. Very pure nitrogen is obtained by this method.

(C) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is Ammonium dichromate. These are red coloured crystals. The thermal decomposition occurs as follows-

\(\left(\mathrm{NH}_{4}\right){ }_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \stackrel{\Delta}{\longrightarrow} \mathrm{Cr}_{2} \mathrm{O}_{3}+\mathrm{N}_{2}+4 \mathrm{H}_{2} \mathrm{O}\)

Ammonium dichromate is thermodynamically unstable and hence it decomposes into dark green chromium (III) oxide and Nitrogen gas. The reaction occurs with flashes of light.

We can see that, \(\mathrm{N}_{2}\) is released in all the reactions.

As we move from top to bottom the acidic nature of hydrides of the group increases as the electronegativity of atom decreases.

So, the order of acidity of hydrides of the Vth group is:

\(\mathrm{H}_{2} \mathrm{O}<\mathrm{H}_{2} \mathrm{~S}<\mathrm{H}_{2} \mathrm{Se}<\mathrm{H}_{2} \mathrm{Te}\)

Strongest acid among the given is \(\mathrm{H}_{2} \mathrm{Te}\).

Bohr's model can only clearly explain hydrogen or hydrogen-like atoms, it fails when applied to larger and heavier atoms like iron, gold, mercury, etc.

In the Bohr model of the atom, electrons travel in defined circular orbits around the nucleus. The orbits are labeled by an integer, the quantum number n. Electrons can jump from one orbit to another by emitting or absorbing energy.

Nitriles can be converted to 1° amines by reaction with LiAlH₄.

Mechanism

(i) Hydride nucleophile attacks the electrophilic carbon in nitrile to form imine anion.

(ii) Hydrolysis of amine derivative.

2-hydroxy-2-methyl propionitrile on acidic hydrolysis gives

2-hydroxy-2-methylpropionic acid.

Mechanism

The last step of dehydration takes place in presence of H₂SO₄ to form 2-methylacrylic acid.

2-methylacrylic acid is the product (B).

Hence, correct answer is (c).