Chemistry Test - 22

Westron has formula C₂H₂Cl₄. It is also called 1,1,2,2−tetrachloroethane.

Westron is an organic compound in which there is a total of 8 elements. There are three elements that in different ratios form the Westron compound. There are carbon atoms, hydrogen atoms, and chlorine atoms are present. There are 2 atoms of carbon element, 2 atoms of hydrogen element, and 4 atoms of chlorine element.

Silver Nitratechemical compound is also known as Lunar Caustic.

• Silver Nitrate is an inorganic compound also known as Lunar Caustic with the formula of\(\mathrm{AgNO}_{3}\).
• It is used in photography. It is also less light-sensitive than halides.
• It was once called lunar caustic, since the ancient alchemists, who linked silver with the moon, called silver luna.
• In solid silver nitrate, in a trigonal planar structure, silver ions are three-coordinated.
• Silver nitrate can be treated with nitric acid by reacting with silver, such as silver bullion or silver foil, resulting in silver nitrate, water, and nitrogen oxides. The by-products of the reaction depend on the nitric acid concentration used.

\(3 \mathrm{Ag}+4 \mathrm{HNO}_{3}\) (cold and diluted) \(\rightarrow 3 \mathrm{AgNO}_{3}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{NO}\)

\(\mathrm{Ag}+2 \mathrm{HNO}_{3}\) (hot and concentrated)\(\mathrm{AgNO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{NO}_{2}\)

If we place an element with atomic number 115 in the periodic table, it will come under the 15^th group. So, it will be a p-block element.

\(P\) atom \(=\) number of bond pairs \(+\) number of lone pairs \(= 3 + 1 = 4\)

\(1(s) + 3(p) → 4(sp^3)\)

We know that, Percentage of p-character \(=\frac{\text{No. of p -orbitals}}{\text{Total no. of orbitals}} × 100\)

Thus, percentage of p-character \(=\frac{3}{4} \times 100=75 \%\)

Therefore, percentage of \(p\)-character in the orbitals forming \(P-P\) bonds in \({P}_{4}\) is \(75 \%\).

\(\mathrm{R}-\mathrm{CH}=\mathrm{CH}_{2} \stackrel{\mathrm{B}_{2} \mathrm{H}_{6}}{\longrightarrow}\left(\mathrm{R}-\mathrm{CH}_{2}-\mathrm{CH}_{2}\right)_{3}-\mathrm{B} \stackrel{\mathrm{OH}^{-}}{\longrightarrow} \mathrm{R}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{OH}\)

Hydroboration of alkanes followed by hydrolysis in basic medium yield alcohols and not the alkanes.

Given,

One gram of charcoal adsorbs \(400 \mathrm{~mL}\) of \(0.5 \mathrm{M}\) acetic acid to form a monolayer, and the molarity of acetic acid reduces to \(0.49 \mathrm{M}\).

\(400 \mathrm{~mL}\) of \((0.5-0.49=0.01) \mathrm{M}\) acetic acid contains \(0.4 \times\) \(0.01=0.2\) moles of acetic acid or \(4 \times 10^{-3} \times 6.023 \times 10^{23}=2.409 \times 10^{21}\) molecules of acetic acid

Total surface area of \(1 \mathrm{~g}\) of charcoal is \(3.01 \times 10^{2} \mathrm{~m}^{2}\).

Thus, the surface area occupied by 1 molecule of acetic acid \(=\frac{3.01 \times 10^{2} \mathrm{~m}^{2}}{2.409 \times 10^{21}}\)

\(=1.2 \times 10^{-19} \mathrm{~m}^{2}\)

Let the oxidation number of a desired atom be \(x\).

We know that:

Oxidation number of \(H=N a=+1, C N=-1, O=-2, M g=+2,N H_{4}=+1\)

(A): \(\mathrm{P}\) in \(\mathrm{NaH}_{2} \mathrm{PO}_{4}: 1+2 \times 1+x+4 \times(-2)=0\)

\(x-5=0\) or \(x=+5\)

(B): Ni in \(\left[N i(C N)_{6}\right]^{4-}: x+6 \times(-1)=-4\)

\(x-6=-4\) 

or, \(x=+2\)

(C): \(\mathrm{P}\) in \(\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}: 2 \times 2+2 \times x+7 \times(-2)=0\)

\(2 x-10=0\) 

or, \(x=+5\)

(D): \(\mathrm{Cr}\) in \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}: 2 \times 1+2 \times x+7 \times(-2)=0\) 

\(2 x-12=0\) 

or, \(x=+6\)

Given:-

Time \((t)=10 \min\)

\(=10 \times 60\)

\(=600 {~s}\)

Current passed \(({i})=0.75 {~A}\)

From Faraday's law of electrolysis, \({Q}={n F}\)

Whereas, \(n\) is the no. of moles of electrons used

\(\Rightarrow {i} \times {t}={n} {F}(\because {q}={I} \times {t})\)

\(\Rightarrow {n}=\frac{{i} \times {t}}{{F}}\)

\(\Rightarrow {n}=\frac{0.75 \times 600}{96500}=\frac{4.5}{965} {~mol}\)

Now, \({H}_{2} {O} \longrightarrow 4 {H}^{+}+{O}_{2}+4 {e}^{-}\)

Amount of \({O}_{2}\) released at STP when \(4\) mole of electrons are used \(=22400 {~mL}\)

Amount of \({O}_{2}\) released at STP when \(\frac{4.5}{965}\) mole of electrons are used \(=\frac{22400}{4}\) \(× (\frac{4.5}{965})\)

\(=\) \(26.11 {~mL}\)

Hence the volume of oxygen liberated at the anode (at STP) will be \(26.11 mL\).

The ease of nucleophilic substitution is depended upon the nature of leaving the group. When the leaving tendency of a group in a compound is high, then the compound is more reactive towards nucleophilic substitution.  The nucleophilic acyl substitution is completed in two-step as shown below.

The reactivity of the compound may be explained on the basicity of the leaving group. A weaker base is a better leaving group. The basicity order is as: \(\mathrm{Cl}^{-}<\mathrm{RCOO}^{-}<\mathrm{RO}^{-}<\mathrm{NH}_{2}^{-}\)

Thus, the order of leaving tendency is \(\mathrm{Cl}^{-}>\mathrm{RCOO}^{-}>\mathrm{RO}^{-}>\mathrm{NH}_{2}^{-}\) and therefore, the order of reactivity of acyl compounds is as: Acyl chloride \(>\) Acid anhydride \(>\) Ester \(>\) Amide