CBSE 12th Standard Chemistry Subject The p-Block Elements Ncert Exemplar 2 Mark Questions With Solution 2021
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CBSE 12th Standard Chemistry Subject The p-Block Elements Ncert Exemplar 2 Mark Questions With Solution 2021
12th Standard CBSE
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Reg.No. :
Chemistry
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In the ring test of NO3- ion, Fe2+ ion reduces nitrate ion to nitric oxide, which combines with Fe2+(aq) ion to form brown complex. Write the reactions involved in the formation of brown ring.
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Explain why the acidity oxoacids of chlorine increases in the order given below:
HCIO < HCIO2 < HCIO3 -
P4O6 reacts with water according to equation,
P4O6 + 6H2O \(\rightarrow\) 4H3PO3 . Calculate the volume of 0.1M NaOH solution required to neutralise the acid formed by dissolving 1.1g of P4O6 in H2O. -
White phosphorus reacts with chlorine and the product hydrolyses in the presence of water. Calculate the mass of HCl obtained by the hydrolysis of the product formed by the reaction of 62g of white phosphorus with chlorine in the presence of water.
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Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state.
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CBSE 12th Standard Chemistry Subject The p-Block Elements Ncert Exemplar 2 Mark Questions With Solution 2021 Answer Keys
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\({ NO }_{ 3 }^{ - }\left( aq \right) +3{ Fe }^{ 2+ }\left( aq \right) +4{ H }^{ + }\left( aq \right) \longrightarrow NO\left( g \right) +3{ Fe }^{ 3+ }\left( aq \right) +3{ H }_{ 2 }O\left( l \right) \)
\(\\ Nitrate\quad ion\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad Nitric\quad oxide\)
\({ \left[ Fe{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ 2+ }+NO\left( g \right) \longrightarrow { \left[ Fe{ \left( { H }_{ 2 }O \right) }_{ 5 }NO \right] }^{ 2+ }+{ H }_{ 2 }O\)
\(\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad Pentaaquanitrosonium\quad iron\left( I \right) \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left( Brown\quad complex \right) \) -
Reason: All these acids on losing a proton give their corresponding conjugate bases, i.e.,\(Cl{ O }^{ - },\ Cl{ O }_{ 2 }^{ - },\ Cl{ O }_{ 3 }^{ - }\ and\ Cl{ O }_{ 4 }^{ - }.\) Their structures are :
Since oxygen is more electronegative than chlorine, therefore, the dispersal of the -ve charge present on oxygen atom (singly bonded to CI)increases as the number of oxygen atoms attached by a double bond to chlorine increases due to \(d\pi -p\pi \) back bonding. In other words, stability of the conjugate bases increases in the order: \({ ClO }^{ - }<{ ClO }_{ 2 }^{ - }<{ ClO }_{ 3 }^{ - }<{ ClO }_{ 4 }^{ - }.\)
Thus, due to increase in stability of the conjugate base, acidic strength increases in the same order :\( \ (HClO\) ) -
The chemical equations for the reactions involved are :
Now 220 g of \({ P }_{ 4 }{ O }_{ 6 }\) require NaOH for neutralization = 320 g
\(\therefore \) 1.1 g of \({ P }_{ 4 }{ O }_{ 6 }\) will require NaOH = \(\frac { 320 }{ 220 } \times 1.1=1.6g\)
Now 1000 mL of 0.1 MHCl contain NaOH = 40 X 0.1 = 4 g
In other words, 4 g of NaOH are present in 0.1 M NaOH = 1000 mL
\(\therefore \) 1.6 g of NaOH will be present in 0.1 M NaOH = \(\frac { 1000 }{ 4 } \times 1.6=400\quad mL=0.4L\) -
The chemical equations for the reactions involved are :
Now 124 g of white P give HCl = 12 X 36.5 g
\(\therefore \) 62 g of white P will give HCl \(=\frac { 12\times 36.5 }{ 124 } \times 62=219g\) -
Three oxoacids of nitrogen are : (i) \(HN{ O }_{ 2 },\) nitrous acid (ii) \(H{ N{ O } }_{ 3 },\) nitric acid and \({ H }_{ 2 }{ N }_{ 2 }{ O }_{ 2 },\) hyponitrous acid. O.S. of N in \(H{ NO }_{ 2 }\) = (+ 1) + x + 2 (- 2) = 0 or X= + 3
\(3H\overset { +3 }{ { NO }_{ 2 } } \xrightarrow { Disproportionation } H\overset { +3 }{ { NO }_{ 3 } } +2\overset { +2 }{ NO } +{ H }_{ 2 }O\)
