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Text Book Back Questions and Answers - Chapter 5 - Coordination Chemistry - 12th Chemistry Guide Samacheer Kalvi Solutions

Updated On 11-02-2025 By Lithanya


Coordination Chemistry
TextBook Evalution
I. Choose the correct answer.
Questions 1.

The sum of primary valance and secondary valance of the metal $\mathrm{M}$ in the complex $\left[\mathrm{M}(\mathrm{en})_2(\mathrm{Ox})\right] \mathrm{Cl}$ is
(a) 3
(b) 6
(c) -3
(d) 9
Answer:
(d) 9
Question 2.
An excess of silver nitrate is added to $100 \mathrm{ml}$ of a $0.01 \mathrm{M}$ solution of penta aquachlorido chromium (III) chloride. The number of moles of $\mathrm{AgCl}$ precipitated would be
(a) 0.02
(b) 0.002
(c) 0.01
(d) 0.2
Answer:
(b) 0.002
Question 3.
A complex has a molecular formula $\mathrm{MSO}_4 \mathrm{Cl} .6 \mathrm{H}_2 \mathrm{O}$. The aqueous solution of it gives white precipitate with Barium chloride solution and no precipitate is obtained when it is treated with silver nitrate solution. If the secondary valence of the metal is six, which one of the following correctly represents the complex?
(a) $\left[\mathrm{M}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}\right] \mathrm{SO}_2 \cdot 2 \mathrm{H}_2 2 \mathrm{O}$
(b) $\left[\mathrm{M}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{SO}_4$
(C) $\left[\mathrm{M}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{SO}_4 \cdot \mathrm{H}_2 \mathrm{O}$
(d) $\left[\mathrm{M}\left(\mathrm{H}_2 \mathrm{O}\right)_3 \mathrm{Cl}\right] \mathrm{SO}_4 \cdot 3 \mathrm{H}_2 \mathrm{O}$
Answer:
(c) $\left[\mathrm{M}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{SO}_4 \cdot \mathrm{H}_2 \mathrm{O}$
Question 4.
Oxidation state of Iron and the charge on the ligand $\mathrm{NO}$ in $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4$ are
(a) +2 and 0 respectively
(b) +3 and 0 respectively
(c) +3 and -1 respectively
(d) +1 and +1 respectively

Answer:
(d) +1 and +1 respectively
Question 5.
As per IUPAC guidelines, the name of the complex $\left[\mathrm{CO}(\mathrm{en})_2(\mathrm{ONO}) \mathrm{Cl}\right] \mathrm{Cl}$ is
(a) chlorobisethylenediaminenitritocobalt (III) chloride
(b chloridobis (ethane-1, 2-diamine) nitro $\mathrm{k}$ - Ocobaltate (III) chloride
(c) chloridobis (ethane-1, 2-diammine) nitrito $\mathrm{k}$ - Ocobalt (II) chloride
(d) chloridobis (ethane-1, 2-diamine) nitro $\mathrm{k}$ - Ocobalt (III) chloride
Answer:
(d) chloridobis (ethane-1, 2-diamine) nitro $\mathrm{k}$ - Ocobalt (III) chloride
Question 6.
IUPAC name of the complex $\mathrm{K}_3\left[\mathrm{Al}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]$ is
(a) potassiumtrioxalatoaluminium (III)
(b) potassiumtrioxalatoaluminate (II)
(c) potassiumtrisoxalatoaluminate (III)
(d) potassiumtrioxalatoaluminate (III)
Answer:
(d) potassiumtrioxalatoaluminate (III)
Question 7.
A magnetic moment of $1.73 \mathrm{BM}$ will be shown by one among the following
(a) $\mathrm{TiCl}_4$
(b) $\left[\mathrm{COCl}_6\right]^{4-}$
(c) $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}$
(d) $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$
Answer:
(c) $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}$
Question 8.
Crystal field stabilization energy for high spin $d^5$ octahedral complex is
(a) $-0.6 \Delta_0$
(b) 0
(c) $2\left(\mathrm{P}-\Delta_0\right)$
(d) $2\left(\mathrm{P}+\Delta_0\right)$
Answer:
(b) 0
Question 9.
In which of the following coordination entities the magnitude of $\Delta_0$ will be maximum?
(a) $\left[\mathrm{CO}(\mathrm{CN})_6\right]^{3-}$
(b) $\left[\mathrm{CO}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$
(c) $\left[\mathrm{CO}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$
(d) $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
Answer:
(a) $\left[\mathrm{CO}(\mathrm{CN})_6\right]^{3-}$
Question 10.
Which one of the following will give a pair of enantiomorphs?
(a) $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]\left[\mathrm{CO}(\mathrm{CN})_6\right]$
(b) $\left[\mathrm{CO}(\mathrm{en})_2 \mathrm{Cl}_2\right] \mathrm{Cl}$
(c) $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4\right]\left[\mathrm{PtCl}_4\right]$
(d) $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{NO}_2$
Answer:
(b) $\left[\mathrm{CO}(\mathrm{en})_2 \mathrm{Cl}_2\right] \mathrm{Cl}$
Question 11.
Which type of isomerism is exhibited by $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$ ?
(a) Coordination isomerism
(b) Linkage isomerism
(c) Optical isomerism
(d) Geometrical isomerism
Answer:

(d) Geometrical isomerism
Question 12 .
How many geometrical isomers are possible for $\left[\mathrm{Pt}(\mathrm{Py})\left(\mathrm{NH}_3\right)(\mathrm{Br})(\mathrm{Cl})\right]$ ?
(a) 3
(6) 4
(c) 0
(d) 15
Answer:
(a) 3
Question 13.
Which one of the following pairs represents linkage isomers?
(a) $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]\left[\mathrm{PtCl}_4\right]$ and $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4\right]\left[\mathrm{CuCl}_4\right]$
(b) $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_3\right)\right] \mathrm{SO}_4$ and $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5(\mathrm{ONO})\right]$
(c) $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_4(\mathrm{NCS})_2\right] \mathrm{Cl}$ and $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_4(\mathrm{SCN})_2\right] \mathrm{Cl}$
(d) both (b) and (c)
Answer:
(c) $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_4(\mathrm{NCS})_2\right] \mathrm{Cl}$ and $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_4(\mathrm{SCN})_2\right] \mathrm{Cl}$
Question 14.
Which kind of isomerism is possible for a complex $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_4 \mathrm{Br}_2\right] \mathrm{Cl}$ ?
(a) geometrical and ionization
(b) geometrical and optical
(c) optical and ionization
(d) geometrical only
Answer:
(a) geometrical and ionization
Question 15.
Which one of the following complexes is not expected to exhibit isomerism?
(a) $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_4\left(\mathrm{H}_2 \mathrm{O}\right)_2\right]^{2+}$
(b) $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$
(C) $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5 \mathrm{SO}_4\right] \mathrm{Cl}$
(d) $\left[\mathrm{Fe}(\mathrm{en})_3\right]^{3+}$
Answer:
(d) $\left[\mathrm{Fe}(\mathrm{en})_3\right]^{3+}$

Question 16.
A complex in which the oxidation number of the metal is zero is
(a) $\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$
(b) $\left[\mathrm{Fe}(\mathrm{CN})_3\left(\mathrm{NH}_3\right)_3\right]$
(c) $\left[\mathrm{Fe}(\mathrm{CO})_5\right]$
(d) both (b) and (c)
Answer:
(c) $\left[\mathrm{Fe}(\mathrm{CO})_5\right]$
Question 17.
Formula of tris (ethane-1, 2-diamine) iron (II) phosphate
(a) $\left[\mathrm{Fe}\left(\mathrm{CH}_3-\mathrm{CH}\left(\mathrm{NH}_2\right)_2\right)_3\right]\left(\mathrm{PO}_4\right)_3$
(b) $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NH}_2\right)_3\right]\left(\mathrm{PO}_4\right)$
(c) $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NH}_2\right)_3\right]\left(\mathrm{PO}_4\right)_2$
(d) $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NH}_2\right)_3\right]\left(\mathrm{PO}_4\right)_2$
Answer:
(d) $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NH}_2\right)_3\right]\left(\mathrm{PO}_4\right)_2$
Question 18.
Which of the following is paramagnetic in nature?
(a) $\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_4\right]^{2+}$
(b) $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
(c) $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
(d) $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$
Answer:
(c) $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
Question 19.
Facmer isomerism is shown by
(a) $\left[\mathrm{CO}(\mathrm{en})_3\right]^{3+}$
(b) $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_4(\mathrm{Cl})_2\right]^{+}$
(c) $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_3(\mathrm{Cl})_3\right]$
(d) $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}^{\mathrm{SO}} \mathrm{SO}_4\right.$
Answer:
(c) $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_3(\mathrm{Cl})_3\right]$

Question 20.
Choose the correct statement.
(a) Square planar complexes are more stable than octahedral complexes
(b) The spin only magnetic moment of $\left[\mathrm{Cu}(\mathrm{Cl})_4\right]^{2-}$ is $1.732 \mathrm{BM}$ and it has square planar structure.
(c) Crystal field splitting energy $\left(\Delta_0\right)$ of $\left[\mathrm{FeF}_6\right]^{4-}$ is higher than the $\left(\Delta_0\right)$ of $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$
(d) crystal field stabilization energy of $\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ is higher than the crystal field stabilization of $\left[\mathrm{Ti}^2\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ Answer:
(d) crystal field stabilization energy of $\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ is is higher than the crystal field stabilization of $\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
II. Answer the following questions
Question 1.

Write the IUPAC names for the following complexes.
1. $\mathrm{Na}_2[\mathrm{Ni}(\mathrm{EDTA})]$
2. $\left[\mathrm{Ag}(\mathrm{CN})_2\right]^{-}$
3. $\left[\mathrm{CO}(\mathrm{en})_3\right]_2\left(\mathrm{SO}_4\right)_3$
4. $\left[\mathrm{CO}(\mathrm{ONO})\left(\mathrm{NH}_3\right)_5\right]^{2+}$
5. $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}\left(\mathrm{NO}_2\right)\right]$
Answer:
1. $\mathrm{Na}_2[\mathrm{Ni}(\mathrm{EDTA})]$
= Sodium EthyicncdiaminetctraacetatonickcEate (II)
(or)
Sodium 2, 2', 2", 2" - (ethane - 1,2 - diyldinitrilo)
tetraacetatonickelate (II)
2. $\left[\mathrm{Ag}(\mathrm{CN})_2\right]^1$
$=$ dicyanidoargentate (I) ion
3. $\left[\mathrm{CO}(\mathrm{en})_3\right]_2\left(\mathrm{SO}_4\right)_3$
$=$ tris (ethylenediamine) cobait (III) sulphate
4. $\left[\mathrm{CO}(\mathrm{ONO})\left(\mathrm{NH}_3\right)_5\right]^{2+}$
$=$ Pentaammincnitrito $-\mathrm{kOCobalt}$ (III) ion.
5. $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}\left(\mathrm{NO}_2\right)\right]$
$=$ diamminedichloridonitrito $-\mathrm{kN}$ platinum (II)
Question 2.
Write the formula for the following coordination compounds.

1. potassiumhexacyanidoferrate (II)
2. petacarbonvliron(O)
3. pentaammineriitrito $-\mathrm{k}-\mathrm{N}-$ cobalt(III)ion
4. hexaamminecobalt (III) sulphate
5. sodiumtetrafluoridodihydroxidoch romate (III)
Answer:
1. potassiurnhexacyanidoferrate (11) $=\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$
2. petacarbonyliron $(\mathrm{O})=\left[\mathrm{Fe}(\mathrm{CO})_5\right]$
3. pentaamminenitrito $-\mathrm{KN}$ - cobalt (III) ion $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5 \mathrm{NO}_2\right]^{2-}$
4. hexaamminecobalt (III) sulphate $\left[\mathrm{CO}(\mathrm{CN})_6\right]_2\left(\mathrm{SO}_4\right)^3$
5. sodiumtetrafluoridodihyclroxidochromate (III) $=\mathrm{Na}_3\left[\mathrm{CrF}_4(\mathrm{OH})_2\right]$
Question 3 .
Arrange the following in order of increasing molar conductivity
1. $\mathrm{Mg}\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)(\mathrm{Cl})_5\right]$
2. $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}_3\left[\mathrm{COF}_6\right]_2\right.$
3. $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]$
Answer:
These complexes can ionise in solution as:
1. $\mathrm{Mg}\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)(\mathrm{Cl})_5\right]=\mathrm{Mg}^{2+}\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)(\mathrm{Cl})_5\right]^{2-}$
2. $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}_3\left[\mathrm{COF}_6\right]_2=\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}^{2+}+\left[\mathrm{COF}_6\right]^{3-}\right.\right.$
3. $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]=$ does not ionize
As the number of ions in solution increases, their molar conductivity also increases. Therefore, conductivity follows the order:
$
\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]<\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}_3\left[\mathrm{COF}_6\right]_2<\mathrm{Mg}\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)(\mathrm{Cl})_5\right]\right.
$
Question 4.
$\mathrm{Ni}^{2+}$ is identified using alcoholic solution of dimethyl glyoxime. Write the structural formula for the rosy red precipitate of a complex formed in the reaction.
Answer:
1. $\mathrm{Ni}^{2+}$ ions present in Nickel chloride solution is estimated accurately for forming an insoluble complex called $\left[\mathrm{Ni}(\mathrm{DMG})_2\right]$.

2. Nickel ion reacts with alcoholic solution of DMG in the presence of ammonical medium, to give rosy red precipitate of $\left[\mathrm{Ni}(\mathrm{DMG})_2\right]$ complex.

Question 5.
$\left[\mathrm{CuCl}_4\right]^{2-}$ exists while $\left[\mathrm{CuI}_4\right]^{2-}$ does not exist why?
Answer:
1. In $\left[\mathrm{CuI}_4\right]^{2-}$ complex, the size of chloride ion is less hence exist. But in $\left[\mathrm{CuI}_4\right]^{2-}$ the bigger iodide ion makes the compound unstable.
2. When copper cation comes in contact with iodide anion, iodide get oxidised to iodine molecule hence the formation of the above complex ion does not take place. Hence $\left[\mathrm{CuI}_4\right]^{2-}$ exists while $\left[\mathrm{CuI}_4\right]^{2-}$ does not exist.
Question 6.
Calculate the ratio
$\begin{gathered}
\left\lceil\mathrm{Ag}^{+}\right\rceil \\
{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}}
\end{gathered}$
in $0.2 \mathrm{M}$ solution of $\mathrm{NH}_3$. If the stability constant for the complex $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]+$ is $1.7 \times 10^7$
Answer:
The stability constant for the complex $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$is $1.7 \times 10^7$, overall dissociation constant is the reciprocal of overall stability constant
$
\mathrm{K}=\frac{1}{\beta} \Rightarrow \mathrm{K}=\frac{1}{1.7 \times 107} \Rightarrow \mathrm{K}=0.588 \times 10^7 \Rightarrow \mathrm{K}=5.88 \times 10^7
$

Question 7.
Give an example of coordination compound used in medicine and two examples of biologically important coordination compounds.
Answer:
Medical uses of coordination compounds:
1. Ca-EDTA chelate, is used in the treatment of lead and radioactive poisoning. That is for removing lead and radiactive metal ions from the body.
2. Cis-platin is used as an antitumor drug in cancer treatment.
Biological important of coordination compounds:
1. A red blood corpuscles $(\mathrm{RBC})$ is composed of heme group, which is $\mathrm{Fe}^{2+}$ Porphyrin complex.it plays an important role in carrying oxygen from lungs to tissues and carbon dioxide from tissues to lungs.
2. Chlorophyll, a green pigment present in green plants and algae, is a coordination complex containing $\mathrm{Mg}^{2+}$ as central metal ion surrounded by a modified Porphyrin ligand called corrin ring. It plays an important role in photosynthesis, by which plants converts $\mathrm{CO}$, and water into carbohydrates and oxygen.
3. Vitamin $B_{12}$ (cyanocobalamine) is the only vitamin consist of metal ion. it is a coordination complex in which the central metal ion is $\mathrm{CO}^{+}$surrounded by Porphyrin like ligand.
4. Many enzymes are known to be metal complexes, they regulate biological processes. For example, Carboxypeptidase is a protease enzyme that hydrolytic enzyme important in digestion, contains a zinc ion coordinated to the protein.
Question 8.
Based on VB theory explain why $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ is paramagnetic, while $\left[\mathrm{Cr}(\mathrm{NH})_4\right]^{2-}$ is diamagnetic.
Answer:
1. $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
In this complex $\mathrm{Cr}$ is in the +3 oxidation state. Electronic configuration of $\mathrm{Cr}$ atom. Electronic configuration of $\mathrm{Cr}^{3+}$ ion 172 Chemistry 12

Hybridisation and formation of $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ Complex Due to the presence of three unpaired electrons in $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ it behaves as a paramagnetic substance. The spin magnetic moment, $\mu_{\mathrm{s}}=g \sqrt{3(3+2)}=g \sqrt{15}=3.87 \mathrm{BM}$ $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ is an inner orbital octahedral complex.

2. $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$
in this complex $\mathrm{Ni}$ is in the +2 oxidation state. Electronic configuration of $\mathrm{Ni}$ atom. Electronic configuration of $\mathrm{Ni}^{2+}$ ion. Hybridisation and formation of $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ Complex

Since $\mathrm{CN}^{-}$is strong field ligand, hence the electrons in $3 \mathrm{~d}$ orbitais are forced to pair up and there is no unpaired electron in $\left[\mathrm{Ni}(\mathrm{CN})_4\right]_2$, hence it should be diamagnetic substance.
Question 9.
Draw all possible geometrical isomers of the complex $\left[\mathrm{CO}(\mathrm{en})_2 \mathrm{CI}_2\right]^{+}$and identify the optically active isomer.
Answer:
1. $\mathrm{Cis}-\left[\mathrm{CO}(\mathrm{en})_2 \mathrm{CI}_2\right]^{+}$

2. Trans $\left[\mathrm{CO}(\mathrm{en})_2 \mathrm{CI}_2\right]^{+}$


The coordination complex $\left[\mathrm{CO}(\mathrm{en})_2 \mathrm{CI}_2\right]^{+}$has three isomers two optically active cis forms and the optically inactive trans form.

Question 10.
$\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ is coloured, while $\left[\mathrm{Sc}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ is colourless- explain.
Answer:
$\mathrm{Ti}$ in $\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ is in +3 oxidation state. $\mathrm{Sc}$ in $\left[\mathrm{Sc}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ is in +3 oxidation state. The outer electronic configuration of $\mathrm{Sc}$, $\mathrm{Ti}$ and their trivalent ions are,
$\mathrm{SC}: 3 \mathrm{~d}^1 4 \mathrm{~S}^2$
$\mathrm{SC}^{3+}: 3 \mathrm{~d}^0$
$\mathrm{Ti}: 3 \mathrm{~d}^2 4 \mathrm{~S}^2$
$\mathrm{Ti}^{3+}: 3 \mathrm{~d}^1$
$\mathrm{Ti}^{3+}{ }^{+}$has one unpaired electron in $3 \mathrm{~d}$ orbital and they undergoes $\mathrm{d}-\mathrm{d}$ transition. This electron can be promoted to a higher energy level by light absorption. Therefore $\left[\mathrm{Ti}_2\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ is coloured. In the case of $\left[\mathrm{Sc}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ there is no electron in $3 \mathrm{~d}$ orbital of $\mathrm{Sc}^{3+}$, hence there is no possibility of light absorbance. Therefore $\left[\mathrm{Sc}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}{ }^3$ is colourless.
Question 11.
Give an example for complex of the type $\left[\mathrm{Ma}_2 \mathrm{~b}_2 \mathrm{c}_2\right]$ where $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are monodentate ligands and give the possible isomers.
Answer:
The octahedral complexes of $\left[\mathrm{Ma}_2 \mathrm{~b}_2 \mathrm{c}_2\right]$ type can exist in five geometrical isomers. The five geometrical isomers for the complex ion $\left[\mathrm{PtCl}_2\left(\mathrm{NH}_3\right)_2(\mathrm{py})_2\right]^{2+}$ are shown below.

Question 12.
Give one test to differentiate $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{SO}_4$ and $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5 \mathrm{SO}_4\right] \mathrm{Cl}$.
Answer:
- $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}^{\mathrm{SO}} \mathrm{SO}_4 \rightarrow\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{+2}+\mathrm{SO}_4{ }^{2-}\right.$
- $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5 \mathrm{SO}_4\right] \mathrm{Cl} \rightarrow\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{SO}_4$
1. Aqueous solution of (a) gives sulphate ion. When an addition of $\mathrm{BaCL}$, solution (a) gives white precipitate of $\mathrm{BaS04}$. But (b) does not give any precipitate.
2. Aqueous solution of (b) gives chloride ion. When an addition of $\mathrm{AgNO}_3$ solution (b) gives curdy white precipitate of $\mathrm{AgCl}$. But (a) does not give any precipitate.
Question 13.
In an octahedral crystal field, draw the figure to show splitting of $\mathrm{d}$ orbitals.
Answer:
Step 1:
In an isolated gaseous state, all the five d orbitals of the central metal ion are degenerate. Initially, the ligands form a spherical field of negative charge around the metal. In this filed, the energies of all the five $d$ orbitals will increase due to the repulsion between the electrons of the metal and the ligand.


Step 2:
The ligands are approaching the metal atom in actual bond directions. To illustrate this let us consider an octahedral field, in which the central metal ion is located at the origin and the six ligands are coming from the $+x,-x,+y,-y,+z$ and $-z$ directions as shown below. As shown in the figure, the orbitals lying along the axes $\mathrm{dx}^2-$ $y^2$ and $\mathrm{dz}^2$ orbitals will experience strong repulsion and raise in energy to a greater extent than the orbitals with lobes directed between the axes $\left(d_{x y}, d_{y z}\right.$ and $\left.d_{z x}\right)$. Thus the degenerate $d$ orbitals now split into two sets and the process is called crystal field splitting.

Step 3:
$\mathrm{Up}$ to this point the complex formation would not be favoured. However, when the ligands approach further, there will be an attraction between the negatively charged electron and the positively charged metal ion, that results in a net decrease in energy. This decrease in energy is the driving force for the complex formation.

During crystal field splitting in octahedral field, in order to maintain the average energy of the orbitals (barycentre) constant, the energy of the orbitals $d_{x^2}-y^2$ and $d_z{ }^2$ (represented as $t_{2 g}$ orbitals) will increase by $3 / 5 \Delta_0$ while that of the other three orbitals $d_{x y} \cdot d_{y z}$ and $d_{z x}$ (represented as $t_{2 g}$ orbitals) decrease by $2 / 5 \Delta_0 \cdot H e r e, \Delta_0$ represents the crystal field splitting energy in the octahedral field.
Question 14.
What is linkage isomerism? Explain with an example.
Answer:
This type of isomers arises when an ambidentate ligand is bonded to the central metal atom/ ion through either of its two different donor atoms. For examples - $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5 \mathrm{ONO}\right] \mathrm{Cl}_2$ - (Pentaammine nitrito cobalt (III)
chloride) $-\mathrm{O}$ - attached. (Red in colour). $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5 \mathrm{NO}_2\right] \mathrm{Cl}_2$ - (Pentaammine nitro cobalt (III) chloride) $-\mathrm{N}-$ attached (Yellow-brown in colour).
Question 15 .
Write briefly about the applications of coordination compounds in volumetric analysis.
Answer:
Hardness of water is due to the presence of $\mathrm{Ca}^{2+}$ and $\mathrm{Mg}^{2+}$ ions in water. EDTA forms stable complexes with $\mathrm{Ca}^{2+}$ and $\mathrm{Mg}^{2+}$. So the total hardness of water can be estimated by simple volumetric titration of water with EDTA.
Question 16.
Classify the following ligand based on the number of donor atoms,
1. $\mathrm{NH}_3$
2. en
3. $o x^{2-}$
4. triaminotriethylamine
5. pyridine
Answer:
1. $\mathrm{NH}_3$ - Monodentate ligands $(\mathrm{N}$ - Donor atom)
2. en - Bidentate ligand ( $2 \mathrm{~N}$ - Donor atom)
3. $\mathrm{ox}^{2-}-\mathrm{Bidentate}$ ligand $(2 \mathrm{O}-$ Donor atom)
4. triaminotriethylamine - Tridentate ligand ( $3 \mathrm{~N}$ - Donor atom)
5. pyridine - Monodentate ligand $(\mathrm{N}-$ Donor atom)
Question 17.
Give the difference between double salts and coordination compounds.
Answer:
Double Salt:

1. A double salt is' a compound prepared by the combination of two different salt components.
2. Completely dissociate into its ions in water.
3. Give simple ions when added to water.
4. It can be easily analyzed by determining the ions present in the aqueous solution.
Example : Potash alum $\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3 \cdot 24 \mathrm{H}_2 \mathrm{O}$
Coordination compounds (Complex salt):
1. A complex salt is a compound composed of a central metal atom having coordination bonds with ligands around it.
2. Do not completely dissociate into its ions in water.
3. Do not give simple ions.
4. It cannot be easily analyzed by determining the ions in the aqueous solution.
Example: Potassium ferro cyanide $\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$
Question 18.
Write the postulates of Werner's theory.
Answer:
1. The central metal ion in any complex ion/compound exhibits two types of valencies, these are (a) Primary valency (b) Secondary valency.
2. The primary valency is ionisable and corresponds to the oxidation state of the metal joining the central ion.
3. The secondary valency is non - ionisable. Every central ion has a fixed number of secondary valencies. This number is called the coordination number of the central ion.
4. The primary valency of the metal ion is always satisfied by a negative ion. The attachment of the central metal ion to the negative ligand is shown by dotted lines.
5. The secondary valencies are satisfied by either negative ions or neutral molecules. The secondary valencies are shown by their lines. The molecules or ions that satisfy secondary valency are called ligands.
6. The ligands which satisfy the secondary valencies must point out in the definite directions in space. Whereas the primary valencies are non - directional in nature. The spatial arrangement of the secondary valencies around the central metal ion is called coordination polyhedron.
7. The secondary valencies are responsible for isomerism in the coordination compounds,

8. Werner's representation of $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$


Question 19.
$\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is diamagnetic, while $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is paramagnetic, explain using crystal field theory.
Answer:
1. $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$
$
\begin{aligned}
& \mathrm{Ni}=3 \mathrm{~d}^8 4 \mathrm{~s}^2 \\
& \mathrm{Ni}^{2+}=3 \mathrm{~d}^8
\end{aligned}
$

Nature of the complex - Low spin (Spin paired)
Ligand filled elelctronic configuration of central metla ion, $\mathrm{t}_{2 \mathrm{~g}}{ }^6 \mathrm{e}_{\mathrm{g}}{ }^6$. Magnetic property - No unpaired electron ( $\mathrm{CN}^{-}$is strong filled ligand), hence it is diamagnetic Magnetic moment $-\mu_{\mathrm{s}}=0$
$
\begin{aligned}
& \text { 2. }\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-} \\
& \mathrm{Ni}=3 \mathrm{~d}_8 4 \mathrm{~S}_2 \\
& \mathrm{Ni}^{2+}=3 \mathrm{~d}^8
\end{aligned}
$

Nature of the complex - high spin
Ligand filled electronic configuration of central metal ion, $\mathrm{t}_{2 \mathrm{~g}}{ }^6 \mathrm{e}_{\mathrm{g}}{ }^6$. Magnetic property - Two unpaired electron ( $\mathrm{CL}^{-}$is weak field ligand). Hence it is paramagnetic Magnetic moment - it is paramagnetic
Question 20.
Why tetrahedral complexes do not exhibit geometrical isomerism.
Answer:
In tetrahedral geometry
1. All the four ligands are adjacent or equidistant to one another.
2. The relative positions of donor atoms of ligands attached to the central metal atom are same with respect to each other.
3. It has plane of symmetry. Therefore, tetrahedral complexes do not exhibit geometrical isomerism.
Question 21.
Explain optical isomerism in coordination compounds with an example.
Answer:
1. Coordination compounds which possess chairality exhibit optical isomerism similar to organic compounds.
2. The pair of two optically active isomers which are mirror images of each other are called enantiomers.
3. Their solutions rotate the plane of the plane polarised light either clockwise or anticlockwise and the corresponding isomers are called $\mathrm{d}$ (dextrorotatory) and 1 (levorotatory) forms respectively.
4. The octahedral complexes of type $[M(x x) 3]^{\mathrm{n} \pm},\left[M(x x)_2 A B\right]^{\mathrm{n} \pm}$ and $\left[M(x x)_2 B_2\right]^{\mathrm{n} \pm}$ exhibit optical isomerism.
Eamples:
1. The optical isomers of $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$ are shown below.

Optical isomer
2. The coordination complex $\left[\mathrm{COCl}_2(\mathrm{en})_2\right]^{+}$has three isomers, two optically active cis forms and one optically
inactive trans form. These structures are shown below.

3. In a coordination compound of type $\left[\mathrm{Pt} \mathrm{Cl}_2(\mathrm{en})_2\right]^{2+}$, two geometrical isomers are possible. They are cis and trans. Among these two isomers, cis isomer shows optically active isomerism because the whole molecule is asymmetric.

Question 22.
What are hydrate isomers? Explain with an example.
Answer:
The exchange of free solvent molecules such as water, ammonia, alcohol etc., in the crystal lattice with a ligand in the coordination entity will give different isomers. These type of isomers are called solvate isomers. If the solvent molecule is water, then these isomers are called hydrate isomers. For example, the complex with
chemical formula $\mathrm{CrCl}_3 \cdot 6 \mathrm{H}_2 \mathrm{O}$ has three hydrate isomers as shown below.

Question 23.
What is crystal field splitting energy?
Answer:
1. In an octahedral complex, the $\mathrm{d}$ - orbitals of the central metal ion divide into two sets of different energies. The separation in energy is the crystal field splitting energy.
2. The $\mathrm{d}$ orbitals lying along the axes $\mathrm{dx}^2, \mathrm{dy}^2$ and $\mathrm{dz} z^2$ orbitals will experience strong repulsion and raise in energy to a greater extent than the orbitals with lobes directed between the axes $\left(\mathrm{d}_{\mathrm{xy}}, \mathrm{d}_{\mathrm{yz}}\right.$ and $\left.\mathrm{d}_{\mathrm{zx}}\right)$. Thus the degenerate $\mathrm{d}$ - orbitals now split into two sets and the process is called crystal filled splitting.
Question 24
What is crystal field stabilization energy (CFSE)?
Answer:
The crystal field stabilisation energy is defined as the energy difference of electronic configurations in the ligand field (ELF) and the isotropic field (Eiso).
$\operatorname{CFSE}\left(\Delta \mathrm{E}_0\right)=\left\{\mathrm{E}_{\mathrm{LF}}\right\}-\left\{\mathrm{E}_{\text {iso }}\right\}$
$=\left\{\mathrm{n}_{\mathrm{t}_{2 \mathrm{~g}}}(-0.4)+\mathrm{n}_{6 \mathrm{~g}}(0.6) \Delta_0-\mathrm{n}_{\mathrm{p}} \mathrm{P}\right\}-\left\{\mathrm{n}_{\mathrm{p}} \mathrm{P}\right\}$
Here $\mathrm{n}_{\mathrm{t}}$ is the number of electrons in $\mathrm{t}$, orbitals
$\mathrm{n}_{\mathrm{e}}$ is the number of electrons in e orbitals
$\mathrm{n}_{\mathrm{p}}$ is the number of electrons in the ligand field
$\mathrm{n}_{\mathrm{p}}$, is the number of electrons in the isotropic field
Question 25.
A solution of $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ is green, whereas a solution of $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is colorless - Explain.
Answer:
1. In $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}, \mathrm{Ni}$ is in +2 oxidation state with the configuration $3 \mathrm{~d}^8$, i.e., it has two unpaired electrons which do not pair up in the presence of weak $\mathrm{H}_2 \mathrm{O}$ ligand. Hence, it is coloured. The $\mathrm{d}-\mathrm{d}$ transtion absorbs red light and the complementary light emitted is green.
2. In the case of $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-} \mathrm{Ni}$ is again in +2 oxidation state with the configuration $3 \mathrm{~d}^8$, but in the presence of strong $\mathrm{CN}^{-}$ligand the two impaired electrons in the $3 \mathrm{~d}$ orbitals pair up. Thus there is no unpaired electron present. Hence it is colourless. Therefore, a solution of $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ is green, whereas a solution of $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is colourless.
Question 26.
Discuss briefly the nature of bonding in metal carbonyls.
Answer:
1. In metal carbonyls, the bond between metal atom and the carbonyl ligand consists of two components.

2. The first component is an electron pair donation from the carbon atom of carbonyl ligand into a vacant $\mathrm{d}-$ orbital of central metal atom. This electron pair donation forms  sigma bond.
3. This sigma bond formation increases the electron density in metal d-orbitals and makes the metal electron rich.
4. In order to compensate for this increased electron density, a filled metal d-orbital interacts with the empty $\pi^*$ orbital on the carbonyl ligand and transfers the added electron density back to the ligand. This second component is called $\pi$ - back bonding. Thus in metal carbonyls, electron density moves from ligand to metal through sigma bonding and from metal to ligand through pi bonding, this synergic effect accounts for strong $M$ $\leftarrow$ CO bond in metal carbonyls. This phenomenon is shown diagrammatically as follows.

Question 27.
What is the coordination entity formed when excess of liquid ammonia is added to an aqueous solution copper sulphate?
Answer:
When excess of liquid ammonia is added to an aqueous solution of copper sulphate to give tetraamminecopper (II) sulphate

Therefore, the coordination entity is $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}$
Question 28.
On the basis of VB theory explain the nature of bonding in $\left[\mathrm{CO}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$.
Answer:
In the complex entity $\left[\mathrm{CO}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$, the $\mathrm{Co}$ is in +3 oxidation state. The outer electronic configuration of $\mathrm{CO}^{3+}$ is $3 \mathrm{~d}^6$. The oxalato ligand is fairly strong field ligand. So it faces the $3 \mathrm{~d}$ electrons in $\mathrm{CO}^{3+}$ to pair up and make two of the $3 \mathrm{~d}$ orbitals available for bonding. As a result, $\mathrm{CO}^{3+}$ shows $\mathrm{d}^2 \mathrm{sp}^2$ hybridisation. Electronic configuration of Co atom Electronic configuration of $\mathrm{CO}^{3+}$ ion Hybridisation and formation of $\left[\mathrm{CO}_2\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$
- There is no unpaired electron $\operatorname{in}\left[\mathrm{CO}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$
$
\text { Thus }\left[\mathrm{CO}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}
$
- During the formtion of $\left[\mathrm{CO}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$, two of the $3 \mathrm{~d}$-orbitals are used in bonding. Therefore it is an inner orbital (low spin) complex.
- The $\left[\mathrm{CO}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$ has the octahedral geometry

Question 29.
What are the limitations of VB theory?
Answer:
Limitations of VB - Theory:
1. It does not explain the colour of the complex
2. It considers only the spin only magnetic moments and does not consider the other components of magnetic moments.
3. It does not provide a quantitative explanation as to why certain complexes are inner orbital complexes and the others are outer orbital complexes for the same metal. For example, $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ is diamagnetic (low spin) whereas $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ is paramagnetic (high spin).
Question 30.
Write the oxidation state, coordination number, nature of ligand, magnetic property and electronic configuration in octahedral crystal field for the complex $\mathrm{K}_4\left[\mathrm{Mn}(\mathrm{CN})_6\right]$.
Answer:

Question 1.
When a coordination compound $\mathrm{CrCl}_3 \cdot 4 \mathrm{H}_2 \mathrm{O}$ is mixed with silver nitrate solution, one mole of silver chloride is precipitated per mole of the compound. There are no free solvent molecules in that compound. Assign the secondary valence to the metal and write the structural formula of the compound.
Answer:
1. When a coordination compound $\mathrm{CrCl}_3 .4 \mathrm{H}_2 \mathrm{O}$ is mixed with silver nitrate solution, one mole of silver chloride is precipitated per mole of the compound. This shows $\mathrm{CrCl}_3 \cdot 4 \mathrm{H}_2 \mathrm{O}$ complex compound contains one $\mathrm{Cl}^{\sqrt{8}}$ counter ion.
2. There are no free solvent molecules in $\mathrm{CrCl}_3 .4 \mathrm{H}_2 \mathrm{O}$ compound, this shows water molecules are coordinated with central metal ion.
3. Therefore, coordination complex is $\left[\mathrm{CrCl}_3 \cdot 4 \mathrm{H}_2 \mathrm{O}\right] \mathrm{Cl}$ Secondary value of the central metal ion is $2 \mathrm{Cl}^{(8)}$ and 4 $\mathrm{H}_2 \mathrm{O}$. Hence coordination number is 6 .
4. Werner's structure of $\left[\mathrm{CrCl}_2 \cdot\left(\mathrm{H}_2 \mathrm{O}\right)_4\right] \mathrm{Cl}$

Question 2.
In the complex, $\left[\mathrm{Pt}\left(\mathrm{NO}_2\right)\left(\mathrm{H}_2 \mathrm{O}\right)\left(\mathrm{NH}_3\right)_2\right] \mathrm{Br}$, identify the following
1. Central metal atom/ion
2. Ligands(s) and their types
3. Coordination entity
4. Coordination number
5. Oxidation number of the central metal ion
Answer:
$
\left[\mathrm{Pt}\left(\mathrm{NO}_2\right)\left(\mathrm{H}_2 \mathrm{O}\right)\left(\mathrm{NH}_3\right)_2\right] \mathrm{Br}
$
1. Central metal ion $-\mathrm{Pt}^{2+}$
2. Ligands and their types $-\mathrm{NO}_2-\mathrm{Mono}$ dendata ligand $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3$ - neutral monodendate ligand
3. Coordination entity $-\left[\mathrm{pt}\left(\mathrm{NO}_2\left(\mathrm{H}_2 \mathrm{O}\left(\mathrm{NH}_3\right)_2\right]^{3+}\right.\right.$
4. Oxidation number of the central metal ion $-\mathrm{x}+1(-1)+1(0)+1(0)=+1 \Rightarrow \mathrm{x}-1=+1 \Rightarrow \mathrm{x}=+2$
5. Coordination number -4
Question 3.
Write the IUPAC name for the following compounds.
1. $\mathrm{K}_2\left[\mathrm{Fe}(\mathrm{CN})_3(\mathrm{Cl})_2\left(\mathrm{NH}_3\right)\right]$
2. $\left[\mathrm{Cr}(\mathrm{CN})_2\left(\mathrm{H}_2 \mathrm{O}\right)\right]\left[\mathrm{CO}(\mathrm{ox})_2(\mathrm{en})\right]$
3. $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$
4. $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_3(\mathrm{NC})_2\left(\mathrm{H}_2 \mathrm{O}\right)\right]^{+}$
5. $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$
Answer:
1. $\mathrm{K}_2\left[\mathrm{Fe}(\mathrm{CN})_3(\mathrm{Cl})_2\left(\mathrm{NH}_3\right)\right]$ - Potassium amminedichloridotricyanidoferrate (III)
2. $\left[\mathrm{Cr}(\mathrm{CN})_2\left(\mathrm{H}_2 \mathrm{O}\right)\right]\left[\mathrm{CO}(\mathrm{ox})_2\right.$ (en) $]$ - Tetraaquadicyanidochromium (II) ethenel,2diaminebis (oxalato) cobalate (II)
3. $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$ - diamminedichloro copper (II)
4. $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_3(\mathrm{NC})_2\left(\mathrm{H}_2 \mathrm{O}\right)\right]^{+}-$triammineaquodicyanido - $\mathrm{KN}$ Chromium (Ili)ion
5. $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}-$ Hexacyanidoferrate (II) ion

Question 4.
Give the structure for the following compounds.
1. diamminesilver (I) dicyanidoargentate(I)
2. Pentaammine nitrito-KNcobalt (III) ion
3. hexafluorido cobaltate (III) ion
4. dichloridobis(ethylenediamine) Cobalt (III) sulphate
5. Tetracarbonylnickel $(0)$
Answer:
1. diamminesilver(I) dicyanidoargentate(I) - $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_9\right]\left[\mathrm{Ag}\left(\mathrm{CN}_2\right]\right.$
2. Pentaammine nitrito-KNcobalt (III) ion $-\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5 \mathrm{NO}_2\right]^{+}$
3. hexafluorido cobaltate (III) ion $-\left[\mathrm{COF}_6\right]^{3-}$
4. dichloridobis(ethylenediamine) Cobalt (III) sulphate $-\left[\mathrm{CO}(\mathrm{en})_2 \mathrm{Cl}_2\right] \mathrm{SO}_4$
5. Tetracarbonylnickel $(0)-\left[\mathrm{Ni}(\mathrm{CO})_4\right]$
Question 5.
A solution of $\left[\mathrm{CO}\left(\mathrm{NH}_3\right) 4 \mathrm{I}_2\right] \mathrm{Cl}$ when treated with $\mathrm{AgNO}_3$ gives a white precipitate. What should be the formula of isomer of the dissolved complex that gives yellow precipitate with $\mathrm{AgNO}_3$. What are the above isomers called?
Answer:
1. A solution of $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_4 \mathrm{I}_2\right] \mathrm{Cl}$ when treated with $\mathrm{AgNO}_3$ gives a white precipitate, because $\mathrm{Cl}$-ion is counter ion.
2. Formula of isomer of the dissolved complex that gives yellow precipitate with $\mathrm{AgNO}_3$ is, $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}\right.$ I] I because $\mathrm{I}^{\Theta}$ is counter ion
3. $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_4 \mathrm{I}_2\right] \mathrm{Cl}$ and $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl} \mathrm{I}\right] \mathrm{I}$ both are ionisation isomers.
Question 6.
Three compounds $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ have empirical formula $\mathrm{CrCl}_3 \cdot 6 \mathrm{H}_2 \mathrm{O}$. they are kept in a container with a dehydrating agent and they lost water and attaining constant weight as shown below.

Answer:
1. $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_3 \mathrm{Cl}_3\right] .3 \mathrm{H}_2 \mathrm{O}$
2. $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl} \cdot 2 \mathrm{H}_2 \mathrm{O}$
3. $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3$
Question 7.
Indicate the possible type of isomerism for the following complexes and draw their isomers
1. $\left[\mathrm{CO}(\mathrm{en})_3\right]\left[\mathrm{Cr}(\mathrm{CN})_6\right]$
2. $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right)\right]^{2+}$
3. $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_3\left(\mathrm{NO}_2\right)\right] \mathrm{Cl}$
Answer:
1. $\left[\mathrm{CO}(\mathrm{en})_3\right]\left[\mathrm{Cr}(\mathrm{CN})_6\right]$ - Exhibits coordination isomerism
(a) $\left[\mathrm{CO}(\mathrm{en})_3\right]\left[\mathrm{Cr}(\mathrm{CN})_6\right]$

(b) $\left[\mathrm{Cr}(\mathrm{en})_3\right]\left[\mathrm{CO}(\mathrm{CN})_6\right]$


2. $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right)\right]^{2+}$ - Exhibits linkage isomerism
(a) $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right)\right]^{2+}-\mathrm{N}$ attached

(b) $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_5\left(\mathrm{ONO}_2\right)\right]^{2+}-\mathrm{O}$ attached


3. $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_3\left(\mathrm{NO}_2\right)\right] \mathrm{Cl}$ - Exhibits ionisation isomerism
(a) $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_3\left(\mathrm{NO}_2\right)\right] \mathrm{Cl}$


(b) $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}\right] \mathrm{NO}_2$


Question 8.
Draw all possible stereo isomers of a complex $\mathrm{Ca}\left[\mathrm{CO}\left(\mathrm{NH}_3\right) \mathrm{Cl}(\mathrm{Ox})_2\right]$ Answer:

$\text { possible stereo isomers of a complex } \mathrm{Ca}\left[\mathrm{CO}\left(\mathrm{NH}_3\right) \mathrm{Cl}(\mathrm{Ox})_2\right]$

Question 9.
The spin only magnetic moment of Tetrachloridomanganate(II)ion is $5.9 \mathrm{BM}$. On the basis of VBT, predict the
type of hybridisation and geometry of the compound. $\left[\mathrm{Mn} \mathrm{Cl}_4\right]^{2-}$
Answer:
Electronic configuration of $\mathrm{Mn}$ atom Electronic configuration of $\mathrm{Mn}^{2+}$ ion. Hybridisation and formation of [Mn $\left.\mathrm{Cl}_4\right]^{2-}$ complex

$\mathrm{Cl}^{-}$is weak field ligand no electrons pairing occurs. $\mathrm{sp}^3$ hybridisation, It has 5 unpaired electrons. Hence paramagnetic Magnetic moment,
$$
\begin{aligned}
& \mu_{\mathrm{s}}=g \sqrt{n(n+2)} \\
& =g \sqrt{5(5+2)} \\
& =g \sqrt{5(7)} \\
& =g \sqrt{(35)} \\
& =5.9 \mathrm{BM}
\end{aligned}
$$
It has tetrahedral geometry.
Question 10.
Predict the number of unpaired electrons in $\left[\mathrm{COCl}_4\right]^{2-}$ ion on the basis of $\mathrm{VBT} .\left[\mathrm{COCl}_4\right]^{2-}$
Answer:
Electronic configuration of $\mathrm{CO}$ atom
Electronic configuration of $\mathrm{CO}^{2+}$ ion
Hybridisation and formation of $\left[\mathrm{COCl}_4\right]^{2-}$ complex

$\mathrm{Cl}^{-}$is weak field ligand, therefore no electrons pairing occurs. $\mathrm{sp}^3$ hybridization. It has 3 unpaired electrons, hence it is paramagnetic. Magnetic moment,
$
\begin{aligned}
& \mu_{\mathrm{s}}=g \sqrt{n(n+2)} \\
& =g \sqrt{3(3+2)} \\
& =g \sqrt{15} \\
& =3.87 \mathrm{BM}
\end{aligned}
$
It has tetrahedral geometry.

Question 11.
A metal complex having composition $\mathrm{CO}(\mathrm{en}) 2 \mathrm{Cl}_2 \mathrm{Br}$ has been isolated in two forms $\mathrm{A}$ and $\mathrm{B}$. (B) reacted with silver nitrate to give a white precipitate readily soluble in ammonium hydroxide. Whereas A gives a pale yellow precipitate. Write the formula of $\mathrm{A}$ and $\mathrm{B}$. state the hybridization of $\mathrm{CO}$ in each and calculate their spin only magnetic moment.
Answer:
A metal complex having composition $\mathrm{CO}(\mathrm{en}) 2 \mathrm{Cl}_2 \mathrm{Br}$ has been isolated in two forms $\mathrm{A}$ and $\mathrm{B}$.
1.(B) reacts with silver nitrate to give a white precipitate readily soluble in ammonium hydroxide. This shows (B) has $\mathrm{Cl}^{\Theta}$ counter ion. Hence $\mathrm{B}$ is $\left[\mathrm{CO}(\mathrm{en})_2 \mathrm{Cl} \mathrm{Br}\right] \mathrm{Cl}$ form.
2. (A) reacts with silver nitrate to give a pale yellow precipitate. This shows (A) has $\mathrm{Br}^{\Theta}$ counter ion. Hence $\mathrm{A}$ is $\left[\mathrm{CO}(\mathrm{en}) 2 \mathrm{Cl}_2\right] \mathrm{Br}$
3. Formula of $A$ and $B$
1. $\left[\mathrm{CO}(\mathrm{en})_2 \mathrm{Cl}_2\right] \mathrm{Br}$
2. $\left[\mathrm{CO}(\mathrm{en})_2 \mathrm{Cl} \mathrm{Br}\right] \mathrm{Cl}$
1. $\mathrm{A}-\left[\mathrm{CO}(\mathrm{en})_2 \mathrm{Cl}_2\right] \mathrm{Br}$
Electronic configuration of $\mathrm{CO}$ atom Electronic configuration of $\mathrm{CO}^{3+}$ atom Hybridisation and formation of $\left[\mathrm{CO}(\mathrm{en})_2 \mathrm{Cl}_2\right] \mathrm{Br}$ complex

$\mathrm{d}^2 \mathrm{sp}^3$ hybridisation en is strong field ligand. No unpaired electrons, hence it is diamagnetic. Magnetic moment
$
\begin{aligned}
& \mu \mathrm{s}=g \sqrt{n(n+2)} \\
& \mathrm{n}=0 \\
& \mu_{\mathrm{s}}=0
\end{aligned}
$
2. $\mathrm{A}-\left[\mathrm{CO}(\mathrm{en})_2 \mathrm{Cl} \mathrm{Br}\right] \mathrm{Cl}$
Electronic configuration of $\mathrm{CO}$ atom
Electronic configuration of $\mathrm{CO}^{3+}$ atom

$\mathrm{d}^2 \mathrm{sp}^3$ hybridisation en is strong field ligand. No unpaired electrons, hence it is diamagnetic. Magnetic moment,
$
\begin{aligned}
& \mu_{\mathrm{s}}=0 \\
& \because \mathrm{n}=0
\end{aligned}
$
Question 12 .
The mean pairing energy and octahedral field splitting energy of $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$ are $28,800 \mathrm{~cm}^{-1}$ and $38500 \mathrm{~cm}^{-1}$ respectively. Whether this complex is stable in low spin or high spin?
Answer:
Mean pairing energy $=28,800 \mathrm{~cm}^{-1}$
Octahedral field splitting energy $=38,500 \mathrm{~cm}^{-1}$
$\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$
$
\mathrm{Mn}=3 \mathrm{~d}^5 4 \mathrm{~s}^2 \mathrm{M}_{\mathrm{n}}^{3+}=3 \mathrm{~d}^4
$

Question 13.
Draw energy level diagram and indicate the number of electrons in each level for the complex $\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$. Whether the complex is paramangnetic or diamagnetic ?
Answer:
$\mathrm{Cu}$ in $\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ has +2 oxidation state.
Electronic configuration of $\mathrm{Cu}$ atom $-3 \mathrm{~d}^{10} 4 \mathrm{~s}^1$
Electronic configuration of $\mathrm{Cu}^{2+}$ ion $-3 \mathrm{~d}^9$
Distortions in Octahedral Geometry Observed in $\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
If the ground electronic configuration of a non-linear complex is orbitally degenerate, the complex will distort so as to remove the degeneracy and achieve a lower energy. This is called the Jahn-Effect.

$\mathrm{Cu}^{2+}-$ Two ways of filling the eg orbitals; there is degeneracy and jahn - Teller Distortion is observed
$
\text { Jahn - Teller Distortion in Cu(III) Complexs }\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}
$

- $t_{2 g}{ }^6 \mathrm{eg}^3$
- It contains one unpaired electron in $\mathrm{e}_{\mathrm{g}}\left(\mathrm{dx}^2-\mathrm{y}^2\right.$ orbital)
- Hence it is paramagnetic
Question 14.
For the $\left[\mathrm{COF}_6\right]^{3-}$ ion the mean pairing energy is found to be $21000 \mathrm{~cm}^{-1}$. The magnitude of $\Delta_0$ is $13000 \mathrm{~cm}^{-1}$. Calculate the crystal field stabilization energy for this complex ion corresponding to low spin and high spin states.
Answer:
Mean pairing energy $=21,000 \mathrm{~cm}^{-1} \Delta_0=13000 \mathrm{~cm}^{-1}$.
$\left[\mathrm{COF}_6\right]^{3-}, \mathrm{CO}=3 \mathrm{~d}^7 4 \mathrm{~s}^2 ; \mathrm{CO}^{3+}=3 \mathrm{~d}^6$