Short & long Answer Question(Part-1) - Coordination Compounds Class 12 Notes | EduRev

Chemistry Class 12

Class 12 : Short & long Answer Question(Part-1) - Coordination Compounds Class 12 Notes | EduRev

The document Short & long Answer Question(Part-1) - Coordination Compounds Class 12 Notes | EduRev is a part of the Class 12 Course Chemistry Class 12.
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Q1. Ammonia has higher boiling point than phosphine. Why?

Ans. -Ammonia forms intermolecular H-bond.

Q2. Why does PCl3 fume in moisture?

Ans. In the presence of (H2O), PCl3 undergoes hydrolysis giving fumes of HCl.

PCl3 + 3H2O → H3PO3 + 3HCl

Q3. What Happens when H3PO3 is Heated?

Ans. It disproportionate to give orthophosphoric acid and Phosphine.

4H3PO3 → 3H3PO4 +PH3

Q4. Why H2S is acidic and H2S is neutral?

Ans. The S---H bond is weaker than O---H bond because the size of S atom is bigger than

that of O atom. Hence H2S can dissociate to give H+ Ions in aqueous solution.

Q5. Name two poisonous gases which can be prepared from chlorine gas?

Ans. Phosgene (COCl2), tear gas (CCl3NO2)

Q6. Name the halogen which does not exhibit positive oxidation state.

Ans. Flourine being the most electronegative element does not show positive oxidation

state.

Q7. Iodine forms I3-  but F2 does not form F-3 ions. why?

Ans. Due to the presence of vacant D-orbitals , I2 accepts electrons from I-ions to form I3- ions, but because of d-orbitals F2 does not accept electrons from F-ions to form Fions.

Q8. Phosphorous forms PCl5 but nitrogen cannot form NCl5 . Why?

Ans. Due to the availability of vacant d-orbital in p.

Q9. Why is HF acid stored in wax coated glass bottles?

Ans. This is because HF does not attack wax but reacts with glass. It dissolves SiO2

present in glass forming hydrofluorosilicic acid.

SiO2 +6HF → H2SiF6+2H2O

Q10. What is laughing gas? Why is it so called? How is it prepared?

Ans. Nitrous oxide (N2O) is called laughing gas, because when inhaled it produced

hysterical laughter. It is prepared by gently heating ammonium nitrate. 

NH4NO3 →N2O+2H2O

Q11. Give reasons for the following:

i. Conc.HNOturns yellow on exposure to sunlight.

ii. PCl5 behaves as an ionic species in solid state.

Ans. i. Conc HNO3 decompose to NO2 which is brown in colour & NO2 dissolves in HNO3

to it yellow.
ii. It exists as [PCl4 ] + [PCl6] - in solid state.

Q12. What happens when white P is heated with conc. NaOH solution in an atmosphere of CO2? Give equation.

Ans. Phosphorus gas will be formed.

P4+3NaOH + 3H2O → PH3 + 3NaH2PO2

Q13. How is ozone estimated quantitatively?

Ans. When ozone reacts with an excess of potassium iodide solution Buffered with a

borate buffer (pH9.2), Iodide is liberated which can be titrated against a standard

solution of sodium thiosulphate. This is a quantitative method for estimating Ogas.

Q14. Are all the five bonds in PCl5 molecule equivalent? Justify your answer.

Ans. PCl5 has a trigonal bipyramidal structure and the three equatorial P-Cl bonds are

equivalent, while the two axial bonds are different and longer than equatorial bonds.

Q15. NO2 is coloured and readily dimerises. Why?

Ans. NO2 contains odd number of valence electrons.It behaves as a typical odd

molecules. On dimerization, it is converted to stable N2O4 molecule with even number of

electrons.
Q16. Write the balanced chemical equation for the reaction of Cl2 with hot and concentrated NaOH. Is this reaction a dispropotionation reaction? Justify: 
Ans. 3Cl2 + 6NaOH → 5NaCl+NaClO3+3H2O
Yes, chlorine from zero oxidation state is changed to -1 and +5 oxidation states.
Q17. Account for the following. 
i. SF6 is less reactive than. 
ii. Of the noble gases only xenon chemical compounds.
Ans.
i. In SF6 there is less repulsion between F atoms than In SF4 .
 ii. Xe has low ionisation enthalpy & high polarising power due to larger atomic size.
Q18. With what neutral molecule is ClO-Isoelectronic? Is that molecule a Lewis base?
Ans.
CiF. Yes, it is Lewis base due to presence of lone pair of electron.

Q19. i. why is He used in diving apparatus?

ii. Noble gases have very low boiling points. Why?

iii. Why is ICl moe reactive than I2?

Ans. i. It is not soluble in blood even under high pressure.

ii. Being monoatomic they have weak dispersion forces.

iii. I-Cl bond is weaker than l-l bond

Q20. Complete the following equations.

i. XeF4 + H2O→

ii. Ca3P2 + H2O→

iii. AgCl(s) +NH3 (aq)

Ans i. 6XeF4+12H2O → 4Xe+2XeO3+24HF+3O2

ii. Ca2P2 + 6H2O → 3Ca (OH)2 +2PH3

iii. AgCl(s) +2NH3(aq→ [Ag(NH3)2]Cl(aq)
Q.21 i. How is XeOF4 prepared? Draw its structure.

ii. When HCL reacts with finely powdered iron, it forms ferrous chloride and not ferric chloride. Why?

Ans. i. Partial hydrolysis of XeOF4

XeF6 + H2O → XeOF4 + 2HF

Structure-square pyramidal.

ii. Its reaction with iron produces h2

Fe+2HCl → FeCl2+H2

Liberation of hydrogen prevents the formation of ferric chloride.
LONG ANSWER TYPE QUESTIONS 
Q.1 Account for the following.

i. Noble gas form compounds with F2 & O2 only.

ii. Sulphur shows paramagnetic behavior.

iii. HF is much less volatile than HCl.

iv. White phosphorous is kept under water.

v. Ammonia is a stronger base than phosphine.

Ans. i. F& O2 are best oxidizing agents.

ii. In vapour state sulphur partly exists as S2 molecule which has two unpaired electrons

in the antibonding pi *orbitals like Oand, hence, exhibit paramagnetism.

iii. HF is associated with intermolecular H bonding.

iv. Ignition temperature of white phosphorous is very low (303 K). Therefore on

explosure to air, it spontaneously catches fire forming P4O10 . Therefore to protect it

from air, it is kept under water.

v. Due to the smaller size of N, lone pair of electrons is readily available.
Q.2  When Conc. H2SO4 was added to an unknown salt present in a test tube, a brown gas

(A) was evolved. This gas intensified when copper turnings were added in to test

tube. On cooling gas (A) changed in to a colourless gas (B).

a. Identify the gases ‘A’ and ‘B’

b. Write the equations for the reactions involved

Ans. The gas ‘A’ is NO2 whereas ‘B’ is N2O4

XNO3 + H2SO4 → XHSO4 + HNO3

Salt (conc.)

Cu + 4HNO3 (Conc.) Cu (NO3)2 + 2NO2 + 2H2O

Blue Brown (A)

2NO2 (on cooling) → N2O4

Colourless (B)
Q.3  Arrange the following in the increasing order of the property mentioned.

i. HOCl, HClO2, HClO3, HClO4(Acidic strength)

ii. As2O3, ClO2, GeO3, Ga2O3(Acidity)

iii. NH3, PH3, AsH3, SbH3(HEH bond angle)

iv. HF, HCl, HBr, HI (Acidic strength)

v. MF, MCl, MBr, MI (ionic character)

Ans. i. Acidic strength: HOCl<HClO2<HCIO3<HCIO4

ii. Acidity: Ga2O3<GeO2<AsO3<CIO2

iii. Bond angle: SbH3<AsH3<PH3<NH3

iv. Acidic strength: HF<HCl<HBr<HI

v. Ionic character: MI<MBr<MCl<MF.

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