Short & long Answer Questions: Coordination Compounds | Inorganic Chemistry for NEET PDF Download

Q1. Ammonia has a higher boiling point than phosphine. Why?
Ans. Ammonia forms an 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.

H₂S Structure

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 F₃ ions.

Q8. Phosphorous forms PCl₅ but nitrogen cannot form NCl₅ . Why?
Ans. Due to the availability of vacant d-orbital in p.

Q9. Using IUPAC norms write the formula for the following: Tetrahydroxozincate(II)
Ans. [Zn(OH₄)^2-

Q10. Using IUPAC norms write the formula for the following: Hexaamminecobalt(III) sulphate
Ans. [Co(NH₃)₆]₂(SO₄)₃

Q11. Using IUPAC norms write the formula for the following: Pentaamminenitrito-cobalt(III)
Ans. [Co(ONO) (NH₃)₅]²⁺

Q12. Using IUPAC norms write the systematic name of the following: [Co(NH₃)₆]Cl₃
Ans. Hexaamminecobalt(III) chloride

Q13. Using IUPAC norms write the systematic name of the following: [Pt(NH₃)₂Cl(NH₂CH₃)]Cl
Ans. Diamminechlorido(methylamine) platinum(II) chloride

Q14. Using IUPAC norms write the systematic name of the following: c[Cr(C₂O₄)₃]³⁺
Ans. Tris(ethane-1, 2-diammine) cobalt(III) ion 

Q15. What is Spectro chemical series? Explain the difference between a weak field ligand and a strong field ligand.
Ans. A Spectro chemical series is the arrangement of common ligands in the increasing order of their crystal-field splitting energy (CFSE) values.
I- < BR - < S₂- < SCN- < Cl-<N₃,<F-<OH-<C₂O₄₂-~H₂O-~NCS-~H-<CN-<NH₃<en~ SO₃₂-<NO₂-<phen<CO

Q16. [Cr(NH₃)₆]³⁺is paramagnetic while [Ni(CN)₄]²⁻is diamagnetic. Explain why?
Ans. Cr is in the +3 oxidation state i.e., d3 configuration. Also, NH₃ is a weak field ligand

that does not cause the pairing of the electrons in the 3d orbital. Cr³⁺:

Therefore, it undergoes d²sp³ hybridization and the electrons in the 3d orbitals remain
unpaired. Hence, it is paramagnetic in nature.
In [Ni(CN)₄]²⁻, Ni exists in the +2 oxidation state i.e., d⁸configuration. Ni²⁺:

CN⁻ is a strong field ligand. It causes the pairing of the 3d orbital electrons. Then, Ni²⁺undergoes dsp² hybridization.

Q17. A solution of [Ni(H₂O)₆ ] ²⁺ is GREEN BUT A SOLUTION OF [Ni(CN)₄ ] ^2- colourless. Explain.
Ans. In [Ni(H₂O)₆]²⁺is a weak field ligand. Therefore, there are unpaired electrons in Ni²⁺. In this complex, the d electrons from the lower energy level can be excited to the higher energy level i.e., the possibility of d−d transition is present. Hence, Ni(H₂O)₆]²⁺is n coloured.
In [Ni(CN)₄]^2-, the electrons are all paired as CN^- is a strong field ligand. Therefore, d-d transition is not possible in [Ni(CN)₄]²⁻. Hence, it is colourless. As there are no unpaired electrons, it is diamagnetic.

Q18.  What is meant by stability of a coordination compound in solution? State the factors which govern stability of complexes.
Ans. The stability of a complex in a solution refers to the degree of association between the two species involved in a state of equilibrium. Stability can be expressed quantitatively in terms of stability constant or formation constant.
M + 3L ↔ ML₃ 
Stability constant
For this reaction, the greater the value of the stability constant, the greater is the proportion of ML₃ in the solution.

Q19. 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

Q20. 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

Q21. Give reasons for the following:
i. Conc.HNO₃ turns 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.

Q22. 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

Q23. 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 O₃ gas.

Q24. 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.

Q25. 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.

Q26. 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.

Q27. 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.

Q28. 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.

Q29. 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

Q30. 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) +2NH_3(aq) → [Ag(NH₃)₂]Cl_(aq)

Q.31 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. F₂ & 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 O₂ and, 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.

Q4. Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:

i. [Fe(CN)₆]⁴⁻

ii. [FeF₆]³⁻

iii. [Co(C₂O₄)₃]³⁻

iv. [CoF₆]³⁻

Ans. i. [Fe(CN)₆]⁴⁻ In the above coordination complex, iron exists in the +II oxidation

state. Fe²⁺: Electronic configuration is 3d₆ Orbitals of Fe²⁺ion:

As CN⁻is a strong field ligand, it causes the pairing of the unpaired 3d electrons. Since

there are six ligands around the central metal ion, the most feasible hybridization is d²sp³. d²sp³ hybridized orbitals of Fe²⁺ are:

 
6 electron pairs from CN⁻ions occupy the six hybrid d₂sp³ orbitals.Then,

Hence, the geometry of the complex is octahedral and the complex is diamagnetic (as there are no unpaired electrons).

ii. [FeF₆]³⁻ In this complex, the oxidation state of Fe is +3.

Orbitals of Fe⁺³ ion:

There are 6 F− ions. Thus, it will undergo d²sp³ or sp³d² hybridization. As F⁻is a weak

field ligand, it does not cause the pairing of the electrons in the 3d orbital. Hence, the

most feasible hybridization is sp³d².sp³d² hybridized orbitals of Fe are:

Hence, the geometry of the complex is found to be octahedral.

iii. [Co(C₂O₄)₃]³⁻ Cobalt exists in the +3 oxidation state in the given complex. Orbitals of

Co³⁺ion: Oxalate is a weak field ligand. Therefore, it cannot cause the pairing of the

3d orbital electrons. As there are 6 ligands, hybridization has to be either sp³d² or d²sp³ hybridization. sp³d² hybridization of Co³⁺ :

The 6 electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand)

occupy these sp³d² orbitals.

Hence, the geometry of the complex is found to be octahedral.

iv. [CoF₆]³⁻ Cobalt exists in the +3 oxidation state.
Orbitals of Co³⁺ion:

Again, fluoride ion is a weak field ligand. It cannot cause the pairing of the 3d

electrons. As a result, the Co³⁺ion will undergo sp³d² hybridization.sp³d² hybridized

orbitals of Co³⁺ion are:

Hence, the geometry of the complex is octahedral and paramagnetic.