3 Marks Question

Question 13 Marks

Define isomerism and tell its types.

Answer

Isomerism : Two or more such compounds which have the same chemical formula (molecules) but have different arrangement of atoms are called isomers of each other and this property is called isomerism. Due to different arrangements of atoms, there is variation in one or more of their physical or chemical properties. There are two main types of isomerism in coordination compounds, which can again be classified into several parts :
(a) Stereoisomerism :
(1) Geometric isomerism
(2) Polarization isomerism
(b) Structural isomerism :
(i) Bond isomerism
(ii) Subcovalent isomerization or coordination isomerism
(iii) Ionization isomerism
(iv) Solvation isomerism or hydrate isomerism
(v) Ligand isomerism
(vi) Polymerization isomerism
(vii) Coordination position isomerism.

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Question 23 Marks

$Pt ( IV ), NH _3, Cl ^{-}$and $Na ^{+}$combine with each other to form seven types of complex compounds.
The following is :
$
\left[Pt\left(NH_3\right)_6\right] Cl_4
$
(i) Write the formulas of other six complex compounds.
(ii) Write the IUPAC names of these complex compounds.
(iii) Which of these complex aqueous solutions will have the highest conductivity?
(iv) Which of these complexes is non-ionic?
(v) Oxidation number of Pt in these complexes and also tell the coordination number.

Answer

(i) (a) $\left[ Pt \left( NH _3\right)_5 Cl ^2\right] Cl _3$
(b) $\left[ Pt \left( NH _3\right)_4 Cl _2\right] Cl _2$
(c) $\left[ Pt \left( NH _3\right)_3 Cl _3\right] Cl$
(d) $\left[ Pt \left( NH _3\right)_2 Cl _4\right]$
(e) $Na \left[ Pt \left( NH _3\right) Cl _5\right]$
(f) $Na _2\left[ PtCl _6\right]$
(ii) (a) Pentaammine platinum (IV) chloride
(b) Tetraammine dichlorido platinum (IV) chloride
(c) Triammine trichlorido platinum (IV) chloride
(d) Diammine tetrachlorido platinum (IV)
(e) Sodium ammine pentachlorido platinate (IV)
(f) Sodium hexachlorido platinate (IV)
(iii) The conductivity of complex $\left[ Pt \left( NH _3\right)_6\right] Cl _4$ will be highest because it gives maximum (five ion) in solution.
(iv) $\left[ Pt \left( NH _3\right)_6\right] Cl _4$ is non-polar.
(v) In all these complexes the oxidation number of Pt is 4 and coordination number is 6 .

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Question 33 Marks

How many geometrical isomers are possible in the following coordination entities?
(i) $\left[ Cr \left( C _2 O _4\right)_3\right]^{3-}$
(ii) $\left[ Co \left( NH _3\right)_3 Cl _3\right]$

Answer

(i) $\left[ Cr \left( C _2 O _4\right)_3\right]^{3-}$ : Geometric isomerism is not possible in this complex because only one type of arrangement is possible.
(ii) Two special types of geometrical isomers of [Co(NH₃)₃Cl₃] are possible : (i) Facial (Fac) and (ii) Meridional (Mer).

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Question 43 Marks

What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.

Answer

Unidentate Ligand : The ligands which is attached to the metal ion by a donor atom is called unindentate ligand. Like- Cl^-, H₂O and NH₃.
Didentate Ligand : The ligand which is attached to the metal ion by two donor atoms is called a didentate ligand. Like- C₂O₂ (oxalate) and H₂N-CH₂-CH₂-NH₂ (Ethane 1, 2-diamine).
Ambidentate Ligand : The ligand with can bound to a metal through two different atoms is called ambidentate ligands.
Example : NO₂ and SCN^- ions. NO₂^- ion can bind to the central atom through nitrogen and oxygen. Similarly, SCN^- can bind to sulphur and nitrogen atom.

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Question 53 Marks

What is meant by the chelate effect? Give an example.

Answer

Chelate effect : In a complex, when a didentate and polydentate ligand forms a bond with the same metal ion through its two or more donor atoms, then it is called a chelate ligand and the number of atoms forming the bond is called the denticity of the ligard. This process of forming bonds is called chelation and such complexes are called chelate complexes. The stability of such complexes is relatively higher. Stabilization of a complex (coordination compound) by a chelation is called chelate effect, while forming a chelate complex, a ring is formed, it is called a chelate ring and due to the formation of this ring, the stability of the complex increases.

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Question 63 Marks

Explain the violet colour of the complex $\left[ Ti \left( H _2 O \right)_6\right]^{3+}$ on the basis of crystal field theory.

Answer

$\left[ Ti \left( H _2 O \right)_6\right]^{3+}$ is an octahedral complex in which the electron of d-orbital of the metal is in the t_2g orbitals with the lowest energy state of the complex. The next higher state $e _{ g }$ orbital available for the electron is vacant. This comlex absorbs light corresponding to the energy of the yellow-green region due to which the electron gets excited from the t_2g level to the $e _{ g }$ level $\left(t_{2 g}{ }^1 e _{ g }^0 \rightarrow t_{2 g}^0 e _{ g }^1\right)$ due to which the complex occurs purple in colour. According to crystal field theory, this is the d-d transition of the electron. Crystal field splitting does not occurs in the absence of a ligand. Therefore when $\left[ Ti \left( H _2 O \right)_6\right]^{3+}$ is heated it becomes colourless due to the loss of water from it.

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Question 73 Marks

Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes :
(i) $K _3\left[ Co \left( C _2 O _4\right)_3\right]$
(ii) cis - $\left[ Cr ( en )_2 Cl _2\right] Cl$
(iii) $\left( NH _4\right)_2\left[ CoF _4\right]$
(iv) $\left[ Mn \left( H _2 O \right)_6\right] SO _4$

Answer

(i) In K₃[Co(C₂O₄)], the oxidation state of metal is +3 and coordination number is 6 because $C _2 O _4^{2-}$ (oxalate) is didentate stronge ligand and hence the electronic configuration of $Co ^{3+}$ is $t _{2 g}^6 e _{ g }^0\left[ Co ^{+3}=3 d^6\right]$.
(ii) In cis- $\left[ Cr ( en )_2 Cl _2\right] Cl$, the oxidation state of the metal is +3 and coordination number is 6. The electronic configuration of metal ion $t_{2 g}^3 \cdot\left[ Cr ^{3+}=3 d^3\right]$
(iii) In $\left( NH _4\right)_2\left[ CoF _4\right]$, the oxidation state of the metal is +2 and the coordination number is 4. The electronic configuration of the metal ion is $t_{2 g}^5 e_g^2 \cdot\left[ Co ^{+2}=3 d^7\right]$
(iv) In $\left[ Mn \left( H _2 O \right)_6\right] SO _4$, the oxidation state of metal is +2 and the coordination number is 6. The electronic configuration of mental ion is $t_{2 g}^3 e_g^2 \cdot\left[Mn^{3+}=3 d^5\right]$

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Question 83 Marks

[Fe(CN)₆]^4- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions. Why?

Answer

In [Fe(CN)₆]^4-, CN^- is a strong ligand and all the electrons in it are paired (d²sp³ hybridization), hence there is no d-d transition and the complex is colourless.
Whereas in [Fe(H₂O)₆]²⁺, H₂O is a weak ligand and hence it has four unpaired electrons even after d²sp³hybridization, due to which d-d transition occurs easily. For this reason, the complex is colourful.
Due to different ligands, the crystal field splitting energy also varies due to which there is a difference in colours despite being the same metal ion.

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Question 93 Marks

What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.

Answer

According to the crystal field theory, the crystal field splitting, $\Delta_0$ , in octahedral complexes depends on the field generated by the charge located on the ligand and the metal ion. Some ligands produce strong field and in this case, the splitting is more, these are called strong field ligands. Other ligands produce weak fields due to which the splitting of d-orbitals is less, these are called weak field ligands.
Ligands are arranged in a series in order of their increasing field strength as follows:
$\begin{array}{l}\quad I ^{-}< Br ^{-}< SCN ^{-}< Cl ^{-}< S ^{2-}< F ^{-}< OH ^{-}< C2O_4^{2-} \\ < H _2 O < NCS ^{-}<\text {edta }^{4-}< NH _3<\text { en }< CN -< CO \end{array}$
This series is known as spectrochemical series and it is based on complexes formed with different ligands. On the basis of facts obtained from experimental measurements of absorption of light by complexes formed with different ligands.

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Question 103 Marks

Aqueous copper sulphate solution (blue in colour) gives :
(i) a green precipitate with aqueous potassium fluoride and
(ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.

Answer

In aqueous solution CuSO₄, is found in form of [Cu(H₂O)₄]SO₄ whose blue colour is due to [Cu(H₂O)₄]²⁺ ions.
(i) On adding aqueous KF to it, the weak ligand H₂O is replaced by F^- and [CuF₄]^2- ion is formed which is green precipitate.
$\left[Cu\left(H_2 O\right)_4\right]^{2+}+4 F^{-} \rightarrow\left[CuF_4\right]^{2-}+4 H_2 O$
$\qquad\qquad\qquad\qquad\qquad$Tetrafluorocuprate (II) ion
$\qquad\qquad\qquad\qquad\qquad\quad$(green precipitate)
(ii) When aqueous KCl is added to [Cu(H₂O₄)]²⁺, the Cl^- replaces the weak ligand (H₂O) to form (CuCl₄)²⁺ which is bright green in colour.
$
\left[Cu\left(H_2 O\right)_4\right]^{2+}+4 Cl^{-} \rightarrow\left[CuCl_4\right]^{2-}+4 H_2 O
$
$\qquad\qquad\qquad\qquad\qquad$Tetrachloro cuprate (II) ion
$\qquad\qquad\qquad\qquad\qquad\qquad$(bright green)

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