The P Block Elements
unit 7
The P-Block Elements
I. Multiple Choice Questions (Type-I)
- On addition of conc. $H_2 SO_4$ to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because
(i) $H_2 SO_4$ reduces $HI$ to $I_2$
(ii) $HI$ is of violet colour
(iii) HI gets oxidised to $I_2$
(iv) $HI$ changes to $HIO_3$
- In qualitative analysis when $H_2 S$ is passed through an aqueous solution of salt acidified with dil. $HCl$, a black precipitate is obtained. On boiling the precipitate with dil. $HNO_3$, it forms a solution of blue colour. Addition of excess of aqueous solution of ammonia to this solution gives
(i) deep blue precipitate of $Cu(OH)_2$
(ii) deep blue solution of $[Cu(NH_3)_4]^{2+}$
(iii) deep blue solution of $Cu(NO_3)_2$
(iv) deep blue solution of $Cu(OH)_2 \cdot Cu(NO_3)_2$
- In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present?
(i) 3 double bonds; 9 single bonds
(ii) 6 double bonds; 6 single bonds
(iii) 3 double bonds; 12 single bonds
(iv) Zero double bonds; 12 single bonds
- Which of the following elements can be involved in $p \pi-d \pi$ bonding?
(i) Carbon
(ii) Nitrogen
(iii) Phosphorus
(iv) Boron
- Which of the following pairs of ions are isoelectronic and isostructural?
(i) $CO_3^{2-}, NO_3^{-}$
(ii) $ClO_3^{-}, CO_3^{2-}$
(iii) $SO_3^{2-}, NO_3^{-}$
(iv) $ClO_3^{-}, SO_3^{2-}$
- Affinity for hydrogen decreases in the group from fluorine to iodine. Which of the halogen acids should have highest bond dissociation enthalpy?
(i) $HF$
(ii) $HCl$
(iii) $HBr$
(iv) $HI$
- Bond dissociation enthalpy of $E-H(E=$ element $)$ bonds is given below. Which of the compounds will act as strongest reducing agent?
Compound | $\mathbf{N H}_3$ | $\mathbf{P H}_3$ | $\mathbf{A s H}_3$ | $\mathbf{S b H}_3$ |
---|---|---|---|---|
$\Delta_{\text {diss }}(E-H) / kJ mol^{-1}$ | 389 | 322 | 297 | 255 |
(i) $NH_3$
(ii) $PH_3$
(iii) $AsH_3$
(iv) $SbH_3$
- On heating with concentrated $NaOH$ solution in an inert atmosphere of $CO_2$, white phosphorus gives a gas. Which of the following statement is incorrect about the gas?
(i) It is highly poisonous and has smell like rotten fish.
(ii) It’s solution in water decomposes in the presence of light.
(iii) It is more basic than $NH_3$.
(iv) It is less basic than $NH_3$.
- Which of the following acids forms three series of salts?
(i) $H_3 PO_2$
(ii) $H_3 BO_3$
(iii) $H_3 PO_4$
(iv) $H_3 PO_3$
- Strong reducing behaviour of $H_3 PO_2$ is due to
(i) Low oxidation state of phosphorus
(ii) Presence of two $-OH$ groups and one $P-H$ bond
(iii) Presence of one - $OH$ group and two $P-H$ bonds
(iv) High electron gain enthalpy of phosphorus
- On heating lead nitrate forms oxides of nitrogen and lead. The oxides formed are
(i) $N_2 O, PbO$
(ii) $NO_2, PbO$
(iii) $NO, PbO$
(iv) $NO, PbO_2$
- Which of the following elements does not show allotropy?
(i) Nitrogen
(ii) Bismuth
(iii) Antimony
(iv) Arsenic
- Maximum covalency of nitrogen is
(i) 3
(ii) 5
(iii) 4
(iv) 6
- Which of the following statements is wrong?
(i) Single $N-N$ bond is stronger than the single $P-P$ bond.
(ii) $PH_3$ can act as a ligand in the formation of coordination compound with transition elements.
(iii) $NO_2$ is paramagnetic in nature.
(iv) Covalency of nitrogen in $N_2 O_5$ is four.
- A brown ring is formed in the ring test for $NO_3^{-}$ion. It is due to the formation of
(i) $[Fe(H_2 O)_5(NO)]^{2+}$
(ii) $FeSO_4 \cdot NO_2$
(iii) $[Fe(H_2 O)_4(NO)_2]^{2+}$
(iv) $FeSO_4 \cdot HNO_3$
- Elements of group- 15 form compounds in +5 oxidation state. However, bismuth forms only one well characterised compound in +5 oxidation state. The compound is
(i) $Bi_2 O_5$
(ii) $BiF_5$
(iii) $BiCl_5$
(iv) $Bi_2 S_5$
- On heating ammonium dichromate and barium azide separately we get
(i) $N_2$ in both cases
(ii) $N_2$ with ammonium dichromate and $NO$ with barium azide
(iii) $N_2 O$ with ammonium dichromate and $N_2$ with barium azide
(iv) $N_2 O$ with ammonium dichromate and $NO_2$ with barium azide
- In the preparation of $HNO_3$, we get $NO$ gas by catalytic oxidation of ammonia. The moles of $NO$ produced by the oxidation of two moles of $NH_3$ will be
(i) 2
(ii) 3
(iii) 4
(iv) 6
- The oxidation state of central atom in the anion of compound $NaH_2 PO_2$ will be_______________-
(i) +3
(ii) +5
(iii) +1
(iv) -3
- Which of the following is not tetrahedral in shape?
(i) $NH_4^{+}$
(ii) $SiCl_4$
(iii) $SF_4$
(iv) $SO_4^{2-}$
- Which of the following are peroxoacids of sulphur?
(i) $H_2 SO_5$ and $H_2 S_2 O_8$
(ii) $H_2 SO_5$ and $H_2 S_2 O_7$
(iii) $H_2 S_2 O_7$ and $H_2 S_2 O_8$
(iv) $H_2 S_2 O_6$ and $H_2 S_2 O_7$
- Hot conc. $H_2 SO_4$ acts as moderately strong oxidising agent. It oxidises both metals and nonmetals. Which of the following element is oxidised by conc. $H_2 SO_4$ into two gaseous products?
(i) $Cu$
(ii) $S$
(iii) $C$
(iv) $Zn$
- A black compound of manganese reacts with a halogen acid to give greenish yellow gas. When excess of this gas reacts with $NH_3$ an unstable trihalide is formed. In this process the oxidation state of nitrogen changes from
(i) -3 to +3
(ii) -3 to 0
(iii) -3 to +5
(iv) 0 to -3
- In the preparation of compounds of $Xe$, Bartlett had taken $O_2^{+} Pt_6^{-}$as a base compound. This is because
(i) both $O_2$ and $Xe$ have same size.
(ii) both $O_2$ and $Xe$ have same electron gain enthalpy.
(iii) both $O_2$ and $Xe$ have almost same ionisation enthalpy.
(iv) both $Xe$ and $O_2$ are gases.
- In solid state $PCl_5$ is a
(i) covalent solid
(ii) octahedral structure
(iii) ionic solid with $[PCl_6]^{+}$octahedral and $[PCl_4]^{-}$tetrahedra
(iv) ionic solid with $[PCl_4]^{+}$tetrahedral and $[PCl_6]^{-}$octahedra
- Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power.
Ion
Reduction potential $E^{\ominus} / V$
(i) $ClO_4^{-}>IO_4^{-}>BrO_4^{-}$
(ii) $IO_4^{-}>BrO_4^{-}>ClO_4^{-}$
(iii) $BrO_4^{-}>IO_4^{-}>ClO_4^{-}$
(iv) $BrO_4^{-}>ClO_4^{-}>IO_4^{-}$
- Which of the following is isoelectronic pair?
(i) $ICl_2, ClO_2$
(ii) $BrO_2^{-}, BrF_2^{+}$
(iii) $ClO_2, BrF$
(iv) $CN^{-}, O_3$
II. Multiple Choice Questions (Type-II)
Note : In the following questions two or more options may be correct.
- If chlorine gas is passed through hot $NaOH$ solution, two changes are observed in the oxidation number of chlorine during the reaction. These are and_________
(i) 0 to +5
(ii) 0 to +3
(iii) 0 to -1
(iv) 0 to +1
- Which of the following options are not in accordance with the property mentioned against them?
(i) $F_2>Cl_2>Br_2>I_2$ $\quad$ Oxidising power.
(ii) $MI>MBr>MCl>MF$ $\quad$ Ionic character of metal halide.
(iii) $F_2>Cl_2>Br_2>I_2$ $\quad$ Bond dissociation enthalpy.
(iv) $HI<HBr<HCl<HF$ $\quad$ Hydrogen-halogen bond strength.
- Which of the following is correct for $P_4$ molecule of white phosphorus?
(i) It has 6 lone pairs of electrons.
(ii) It has six P-P single bonds.
(iii) It has three P-P single bonds.
(iv) It has four lone pairs of electrons.
- Which of the following statements are correct?
(i) Among halogens, radius ratio between iodine and fluorine is maximum.
(ii) Leaving $F-F$ bond, all halogens have weaker $X-X$ bond than $X-X^{\prime}$ bond in interhalogens.
(iii) Among interhalogen compounds maximum number of atoms are present in iodine fluoride.
(iv) Interhalogen compounds are more reactive than halogen compounds.
- Which of the following statements are correct for $SO_2$ gas?
(i) It acts as bleaching agent in moist conditions.
(ii) It’s molecule has linear geometry.
(iii) It’s dilute solution is used as disinfectant.
(iv) It can be prepared by the reaction of dilute $H_2 SO_4$ with metal sulphide.
- Which of the following statements are correct?
(i) All the three $N-O$ bond lengths in $HNO_3$ are equal.
(ii) All $P-Cl$ bond lengths in $PCl_5$ molecule in gaseous state are equal.
(iii) $P_4$ molecule in white phohsphorus have angular strain therefore white phosphorus is very reactive.
(iv) $PCl$ is ionic in solid state in which cation is tetrahedral and anion is octahedral.
- Which of the following orders are correct as per the properties mentioned against each?
(i) $As_2 O_3<SiO_2<P_2 O_3<SO_2 \quad$ Acid strength.
(ii) $AsH_3<PH_3<NH_3 \quad$ Enthalpy of vapourisation.
(iii) $S<O<Cl<F \quad$ More negative electron gain enthalpy.
(iv) $H_2 O>H_2 S>H_2 Se>H_2 Te \quad$ Thermal stability.
- Which of the following statements are correct?
(i) S-S bond is present in $H_2 S_2 O_6$.
(ii) In peroxosulphuric acid $(H_2 SO_5)$ sulphur is in +6 oxidation state.
(iii) Iron powder along with $Al_2 O_3$ and $K_2 O$ is used as a catalyst in the preparation of $NH_3$ by Haber’s process.
(iv) Change in enthalpy is positive for the preparation of $SO_3$ by catalytic oxidation of $SO_2$.
- In which of the following reactions conc. $H_2 SO_4$ is used as an oxidising reagent?
(i) $CaF_2+H_2 SO_4 \longrightarrow CaSO_4+2 HF$
(ii) $2 HI+H_2 SO_4 \longrightarrow I_2+SO_2+2 H_2 O$
(iii) $Cu+2 H_2 SO_4 \longrightarrow CuSO_4+SO_2+2 H_2 O$
(iv) $NaCl+H_2 SO_4 \longrightarrow NaHSO_4+HCl$
- Which of the following statements are true?
(i) Only type of interactions between particles of noble gases are due to weak dispersion forces.
(ii) Ionisation enthalpy of molecular oxygen is very close to that of xenon.
(iii) Hydrolysis of $XeF_6$ is a redox reaction.
(iv) Xenon fluorides are not reactive.
III. Short Answer Type
-
In the preparation of $H_2 SO_4$ by Contact Process, why is $SO_3$ not absorbed directly in water to form $H_2 SO_4$ ?
-
Write a balanced chemical equation for the reaction showing catalytic oxidation of $NH_3$ by atmospheric oxygen.
-
Write the structure of pyrophosphoric acid.
-
$PH_3$ forms bubbles when passed slowly in water but $NH_3$ dissolves. Explain why?
-
In $PCl_5$, phosphorus is in $s p^{3} d$ hybridised state but all its five bonds are not equivalent. Justify your answer with reason.
-
Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling it is diamagnetic?
-
Give reason to explain why $ClF_3$ exists but $FCl_3$ does not exist.
-
Out of $H_2 O$ and $H_2 S$, which one has higher bond angle and why?
-
$SF_6$ is known but $SCl_6$ is not. Why?
-
On reaction with $Cl_2$, phosphorus forms two types of halides ’ $A$ ’ and ’ $B$ ‘. Halide $A$ is yellowish-white powder but halide ’ $B$ ’ is colourless oily liquid. Identify $A$ and $B$ and write the formulas of their hydrolysis products.
-
In the ring test of $NO_3^{-}$ion, $Fe^{2+}$ ion reduces nitrate ion to nitric oxide, which combines with $Fe^{2+}$ (aq) ion to form brown complex. Write the reactions involved in the formation of brown ring.
-
Explain why the stability of oxoacids of chlorine increases in the order given below:
$HClO<HClO_2<HClO_3<HClO_4$
-
Explain why ozone is thermodynamically less stable than oxygen.
-
$P_4 O_6$ reacts with water according to equation $P_4 O_6+6 H_2 O \longrightarrow 4 H_3 PO_3$. Calculate the volume of $0.1 M NaOH$ solution required to neutralise the acid formed by dissolving $1.1 g$ of $P_4 O_6$ in $H_2 O$.
-
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 $62 g$ of white phosphorus with chlorine in the presence of water.
-
Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state.
-
Nitric acid forms an oxide of nitrogen on reaction with $P_4 O_{10}$. Write the reaction involved. Also write the resonating structures of the oxide of nitrogen formed.
-
Phosphorus has three allotropic forms - (i) white phosphorus (ii) red phosphorus and (iii) black phosphorus. Write the difference between white and red phosphorus on the basis of their structure and reactivity.
-
Give an example to show the effect of concentration of nitric acid on the formation of oxidation product.
-
$PCl_5$ reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous $NH_3$ solution. Write the reactions involved to explain what happens.
-
Phosphorus forms a number of oxoacids. Out of these oxoacids phosphinic acid has strong reducing property. Write its structure and also write a reaction showing its reducing behaviour.
IV. Matching Type
Note: Match the items of Column I and Column II in the following questions.
- Match the compounds given in Column I with the hybridisation and shape given in Column II and mark the correct option.
Column I
(A) $Xe F_6$
(B) $XeO_3$
(C) $Xe OF_4$
(D) $Xe F_4$
Column II
(1) $s p^{3} d^{3}$ - distorted octahedral
(2) $d sp^{3} d^{2}$ - square planar
(3) $sp^{3}$ - pyramidal
(4) $sp^{3} d^{2}$ - square pyramidal
Code :
(i) | A (1) | B (3) | C (4) | D (2) |
---|---|---|---|---|
(ii) | A (1) | B (2) | C (4) | D (3) |
(iii) | A (4) | B (3) | C (1) | D (2) |
(iv) | A (4) | B (1) | C (2) | D (3) |
- Match the formulas of oxides given in Column I with the type of oxide given in Column II and mark the correct option.
Column I
(A) $Pb_3 O_4$
(B) $N_2 O$
(C) $Mn_2 O_7$
(D) $Bi_2 O_3$
Column II
(1) Neutral oxide
(2) Acidic oxide
(3) Basic oxide
(4) Mixed oxide
Code :
(i) | A (1) | B (2) | C (3) | D (4) |
---|---|---|---|---|
(ii) | A (4) | B (1) | C (2) | D (3) |
(iii) | A (3) | B (2) | C (4) | D (1) |
(iv) | A (4) | B (3) | C (1) | D (2) |
- Match the items of Columns I and II and mark the correct option.
Column I
(A) $H_2 SO_4$
(B) $CCl_3 NO_2$
(C) $Cl_2$
(D) Sulphur
Column II
(1) Highest electron gain enthalpy
(2) Chalcogen
(3) Tear gas
(4) Storage batteries
Code :
(i) | A (4) | B (3) | C (1) | D (2) |
---|---|---|---|---|
(ii) | A (3) | B (4) | C (1) | D (2) |
(iii) | A (4) | B (1) | C (2) | D (3) |
(iv) | A (2) | B (1) | C (3) | D (4) |
- Match the species given in Column I with the shape given in Column II and mark the correct option.
Column I
(A) $SF_4$
(B) $BrF_3$
(C) $BrO_3^{-}$
(D) $NH_4^{+}$
Column II
(1) Tetrahedral
(2) Pyramidal
(3) Sea-saw shaped
(4) Bent T-shaped
Code :
(i) | A (3) | B (2) | C (1) | D (4) |
---|---|---|---|---|
(ii) | A (3) | B (4) | C (2) | D (1) |
(iii) | A (1) | B (2) | C (3) | D (4) |
(iv) | A (1) | B (4) | C (3) | D (2) |
- Match the items of Columns I and II and mark the correct option.
Column I
(A) Its partial hydrolysis does not change oxidation state of central atom
(B) It is used in modern diving apparatus
(C) It is used to provide inert atmosphere for filling electrical bulbs
(D) Its central atom is in $s p^{3} d^{2}$ hybridisation
Column II
(1) $He$
(2) $XeF_6$
(3) $XeF_4$
(4) $Ar$
Code :
(i) | A (1) | B (4) | C (2) | D (3) |
---|---|---|---|---|
(ii) | A (1) | B (2) | C (3) | D (4) |
(iii) | A (2) | B (1) | C (4) | D (3) |
(iv) | A (1) | B (3) | C (2) | D (4) |
V. Assertion and Reason Type
** Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.**
(i) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
(iii) Assertion is correct, but reason is wrong statement.
(iv) Assertion is wrong but reason is correct statement.
(v) Both assertion and reason are wrong statements.
- Assertion : $N_2$ is less reactive than $P_4$.
Reason : Nitrogen has more electron gain enthalpy than phosphorus.
- Assertion : $HNO_3$ makes iron passive.
Reason : $HNO_3$ forms a protective layer of ferric nitrate on the surface of iron.
- Assertion : $HI$ cannot be prepared by the reaction of KI with concentrated $H_2 SO_4$
Reason : $HI$ has lowest $H-X$ bond strength among halogen acids.
- Assertion : Both rhombic and monoclinic sulphur exist as $S_8$ but oxygen exists as $O_2$.
Reason : Oxygen forms $p \pi-p \pi$ multiple bond due to small size and small bond length but $p \pi-p \pi$ bonding is not possible in sulphur.
- Assertion : $NaCl$ reacts with concentrated $H_2 SO_4$ to give colourless fumes with pungent smell. But on adding $MnO_2$ the fumes become greenish yellow.
Reason : $MnO_2$ oxidises $HCl$ to chlorine gas which is greenish yellow.
- Assertion : $SF_6$ cannot be hydrolysed but $SF_4$ can be.
Reason : Six $F$ atoms in $SF_6$ prevent the attack of $H_2 O$ on sulphur atom of $SF_6$.
VI. Long Answer Type
-
An amorphous solid “A” burns in air to form a gas “B” which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aqueous $KMnO_4$ solution and reduces $Fe^{3+}$ to $Fe^{2+}$. Identify the solid “A” and the gas “B” and write the reactions involved.
-
On heating lead (II) nitrate gives a brown gas “A”. The gas “A” on cooling changes to colourless solid “B”. Solid “B” on heating with NO changes to a blue solid ’ $C$ ‘. Identify ’ $A$ ‘, ’ $B$ ’ and ’ $C$ ’ and also write reactions involved and draw the structures of ’ $B$ ’ and ’ $C$ ‘.
-
On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 mol of hydrogen $(H_2)$ in the presence of a catalyst gives another gas (C) which is basic in nature. Gas $C$ on further oxidation in moist condition gives a compound (D) which is a part of acid rain. Identify compounds (A) to (D) and also give necessary equations of all the steps involved.
ANSWERS
I. Multiple Choice Guestions (Type-I)
1. (iii) | 2. (ii) | 3. (i) | 4. (iii) | 5. (i) | 6. (i) |
---|---|---|---|---|---|
7. (iv) | 8. (iii) | 9. (iii) | 10. (iii) | 11. (ii) | 12. (i) |
13. (iii) | 14. (i) | 15. (i) | 16. (ii) | 17. (i) | 18. (i) |
19. (iii) | 20. (iii) | 21. (i) | 22. (iii) | 23. (i) | 24. (iii) |
25. (iv) | 26. (iii) | 27. (ii) |
II. Multiple Choice Questions (Type-II)
28. (i), (iii) | 29. (ii), (iii) | 30. (ii), (iv) | 31. (i), (iii), (iv) |
---|---|---|---|
32. (i), (iii) | 33. (iii), (iv) | 34. (i), (iv) | 35. (i), (ii) |
36. (ii), (iii) | 37. (i), (ii) |
III. Short Answer Type
-
Acid fog is formed, which is difficult to condense.
-
$4 NH_3+5 O_2 \xrightarrow[500 K, 9 \text { bar }]{Pt / Rh \text { gauge catalst }} 4 NO+6 H_2 O$
$$ \text {From air}$$
- v
Pyrophosphoric acid
-
$NH_3$ forms hydrogen bonds with water therefore it is soluble in it but $PH_3$ cannot form hydrogen bond with water so it escapes as gas.
-
[Hint : It has trigonal bipyramidal geometry]
-
In gaseous state $NO_2$ exists as monomer which has one unpaired electron but in solid state it dimerises to $N_2 O_4$ so no unpaired electron is left hence solid form is diamagnetic.
-
Because fluorine is more electronegative as compared to chlorine.
-
Bond angle of $H_2 O$ is larger, because oxygen is more electronegative than sulphur therefore bond pair electron of $O-H$ bond will be closer to oxygen and there will be more bond-pair bond-pair repulsion between bond pairs of two O-H bonds.
-
Due to small size of fluorine $six F^{-}$ion can be accomodated around sulphur whereas chloride ion is comparatively larger in size, therefore, there will be interionic repulsion.
-
A is $PCl_5$ (It is yellowish white powder)
$ P_4+10 Cl_2 \longrightarrow 4 PCl_5 $
$B$ is $PCl_3$ (It is a colourless oily liquid)
$ P_4+6 Cl_2 \longrightarrow 4 PCl_3 $
Hydrolysis products are formed as follows :
$ \begin{aligned} & PCl_3+3 H_2 O \longrightarrow H_3 PO_3+3 HCl \\ & PCl_5+4 H_2 O \longrightarrow H_3 PO_4+5 HCl \end{aligned} $
- $NO_3^{-}+3 Fe^{2+}+4 H^{+} \longrightarrow NO+3 Fe^{3+}+2 H_2 O$
$[Fe(H_2 O)_6]^{2+}+NO \longrightarrow[Fe(H_2 O)_5(NO)]^{2+}+H_2 O$ (brown complex)
- Oxygen is more electronegative than chlorine, therefore dispersal of negative charge present on chlorine increases from $ClO^{-}$to $ClO_4^{-}$ion because number of oxygen atoms attached to chlorine is increasing. Therefore, stability of ions will increase in the order given below :
$ ClO^{-}<ClO_2^{-}<ClO_3^{-}<ClO_4^{-} $
Thus due to increase in stability of conjugate base, acidic strength of corresponding acid increases in the following order
$ HClO<HClO_2<HClO_3<HClO_4 $
-
See the NCERT textbook for Class XII, page 186.
-
$P_4 O_6+6 H_2 O \longrightarrow 4 H_3 PO_3$
$H_3 PO_3+2 NaOH \longrightarrow Na_2 HPO_3+2 H_2 O O \times 4$ (Neutralisation reaction)
$P_4 O_6+8 NaOH \longrightarrow 4 Na_2 HPO_4+2 H_2 O$
$1 mol \quad 8 mol$
Product formed by $1 mol$ of $P_4 O_6$ is neutralised by 8 mols of $NaOH$
$\therefore$ Product formed by $\frac{1.1}{220} mol$ of $P_4 O_6$ will be neutralised by $\frac{1.1}{220} \times 8 mol$ of $NaOH$
Molarity of $NaOH$ solution is $0.1 M$
$\Rightarrow 0.1 mol NaOH$ is present in $1 L$ solution
$\therefore \frac{1.1}{220} \times 8 mol NaOH$ is present in $\frac{1.1 \times 8}{220 \times 0.1} L=\frac{88}{220} L=\frac{4}{10} L=0.4 L=$
$400 mL$ of $NaOH$ solution.
- $P_4+6 Cl_2 \longrightarrow 4 PCl_3$
$.PCl_3+3 H_2 O \longrightarrow H_3 PO_3+3 HCl] \times 4$
$P_4+6 Cl_2+12 H_2 O \longrightarrow 4 H_3 PO_3+12 HCl$
$1 mol$ of white phosphorus produces $12 mol$ of $HCl$ $62 g$ of white phosphorus has been taken which is equivalent to $\frac{62}{124}=\frac{1}{2} mol$.
Therefore $6 mol HCl$ will be formed.
Mass of $6 mol HCl=6 \times 36.5=219.0 g HCl$
- Three oxoacids of nitrogen are
(i) $HNO_2$, Nitrous acid
(ii) $HNO_3$, Nitric acid
(iii) Hyponitrous acid, $H_2 N_2 O_2$
$3 HNO_2 \xrightarrow{\text { Disproportionation }} HNO_3+H_2 O+2 NO$
-
$4 HNO_3+P_4 O_{10} \longrightarrow 4 HPO_3+2 N_2 O_5$
-
(a) - Structures (See NCERT textbook for Class XII)
- White phosphorus is discrete tetrahedral molecule. Thus it has tetrahedral structure with six P-P bonds.
- Red phosphorus has polymeric structure in which $P_4$ tetrahedra are linked together through $P-P$ bonds to form chain.
(b) Reactivity
White phosphorus is much more reactive than red phosphorus. This is because in white phosphorus there is angular strain in $P_4$ molecules because the bond angles are only of $60^{\circ}$.
- Dilute and concentrated nitric acid give different oxidation products on reaction with copper metal.
$ \begin{aligned} & .3 Cu+8 HNO_3 \text { (dil. }) \longrightarrow 3 Cu(NO_3)_2+2 NO+4 H_2 O \\ & Cu+4 HNO_3 \text { (Conc.) } \longrightarrow 3 Cu(NO_3)_2+2 NO+2 H_2 O \end{aligned} $
- $PCl_5+2 Ag \longrightarrow 2 AgCl+PCl_3$
$ AgCl+2 NH_3(aq) \longrightarrow[Ag(NH_3)_2]^{+} Cl^{-} $
$$ \text {soluble complex}$$
- Structure of phosphinic acid (Hypophosphorous acid) is as follows :
Reducing behaviour of phosphinic acid is observable in the reaction with silver nitrate given below :
$4 AgNO_3+2 H_2 O+H_3 PO_2 \longrightarrow 4 Ag+4 HNO_3+H_3 PO_4$
IV. Matching Type
-
(i)
-
(ii)
-
(i)
-
(ii)
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(iii)
V. Assertion and Reason Type
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(iii)
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(iii)
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(ii)
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(i)
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(i)
-
(i)
VI. Long Answer Type
- ‘A’ is $S_8 \quad$ ’ $B$ ’ is $SO_2$ gas
$ \begin{aligned} & S_8+8 O_2 \xrightarrow{\Delta} 8 SO_2 \\ & 2 MnO_4^{-}+5 SO_2+2 H_2 O \longrightarrow 5 SO_4^{2-}+4 H^{+}+2 Mn^{2+} \\ & \text { (violet) } & \quad \text { (colourless) } \\ & 2 Fe^{3+}+SO_2+2 H_2 O \longrightarrow 2 Fe^{2+}+SO_4^{2-}+4 H^{+} \end{aligned} $
- $Pb(NO_3)_2 \frac{\Delta}{673 K} 2 PbO+4 NO_2$
$ \quad $ (Brown colour)
$ 2 NO_2 \stackrel{\text { On cooling }}{\rightarrow} N_2 O_4 $
(B)
(Colourless solid)
$2 NO+N_2 O_4 \xrightarrow{\Delta 250 K} 2 N_2 O_3$
(Blue solid)
(Structure of $N_2 O_4$ )
(Structure of $N_2 O_3$ )
- $A=NH_4 NO_2 \quad B=N$
(i) $NH_4 NO_2 arrow N_2+2 H_2 O$
(ii) $N_2+3 H_2 arrow 2 NH_3$
(iii) $4 NH_3+5 O_2 arrow 4 NO+6 H_2 O$
$4 NO+O_2 arrow 2 NO_2$
$3 NO_2+H_2 O arrow 2 HNO_3+NO$