Co - ordination Compounds MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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251.
The number of unpaired electrons in the complex ion $${\left[ {Co{F_6}} \right]^{3 - }}$$ is ( At. no. of $$Co = 27$$ )
A
3
B
2
C
4
D
0
Answer :
4
In complex ion $${\left[ {Co{F_6}} \right]^{3 - }},Co$$ is present in + 3 oxidation state
$$\eqalign{
& _{27}Co = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^7},4{s^2} \cr
& C{o^{3 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^6} \cr} $$
Therefore, the number of unpaired electrons in $$3d$$ subshell of $${\left[ {Co{F_6}} \right]^{3 - }}$$ is $$4.$$
252.
When one mole of each of the following complexes is treated with excess of $$AgN{O_3},$$ which will give maximum amount of $$AgCl?$$
A
$$\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]C{l_3}$$
B
$$\left[ {Co{{\left( {N{H_3}} \right)}_5}Cl} \right]C{l_2}$$
C
$$\left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]Cl$$
D
$$\left[ {Co{{\left( {N{H_3}} \right)}_3}C{l_3}} \right]$$
No explanation is given for this question. Let's discuss the answer together.
253.
In the separation of $$C{u^{2 + }}$$ and $$C{d^{2 + }}$$ of IInd group in qualitative analysis of cations, tetrammine copper (II) sulphate and tetrammine cadmium (II) sulphate react with $$KCN$$ to form the corresponding cyano complexes, which one of the following pairs of the complexes and their relative stability enables the separation of $$C{u^{2 + }}$$ and $$C{d^{2 + }}?$$
A
$${K_3}\left[ {Cu{{\left( {CN} \right)}_4}} \right]:$$ less stable and $${K_2}\left[ {Cd{{\left( {CN} \right)}_4}} \right]:$$ more stable
B
$${K_3}\left[ {Cu{{\left( {CN} \right)}_4}} \right]:$$ more stable and $${K_2}\left[ {Cd{{\left( {CN} \right)}_4}} \right]:$$ less stable
C
$${K_2}\left[ {Cu{{\left( {CN} \right)}_4}} \right]:$$ less stable and $${K_2}\left[ {Cd{{\left( {CN} \right)}_4}} \right]:$$ more stable
D
$${K_2}\left[ {Cu{{\left( {CN} \right)}_4}} \right]:$$ more stable and $${K_2}\left[ {Cd{{\left( {CN} \right)}_4}} \right]:$$ less stable
Answer :
$${K_3}\left[ {Cu{{\left( {CN} \right)}_4}} \right]:$$ more stable and $${K_2}\left[ {Cd{{\left( {CN} \right)}_4}} \right]:$$ less stable
$${K_3}\left[ {Cu{{\left( {CN} \right)}_4}} \right]$$ is more stable while in $${K_2}\left[ {Cd{{\left( {CN} \right)}_4}} \right]$$ is less stable.
Here, $$Cu$$ in + 1 oxidation state.
254.
The value of the 'spin only' magnetic moment for one of the following configurations is 2.84 $$BM.$$ The correct one is
A
$${d^4}$$ ( in strong ligand field )
B
$${d^4}$$ ( in weak ligand field )
C
$${d^3}$$ ( in weak as well as in strong fields )
D
$${d^5}$$ ( in strong ligand field )
Answer :
$${d^4}$$ ( in strong ligand field )
Spin only magnetic moment $$ = \sqrt {n\left( {n + 2} \right)} \,B.M.$$ Where, $$n=$$ no. of unpaired electrons.
Given, $$\sqrt {n\left( {n + 2} \right)} = 2.84$$
$$ \Rightarrow n = 2$$
In an octahedral complex, for a $${d^4}$$ configuration in a strong field ligand, number of unpaired electrons = 2
255.
Give reason for the statement, $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ is diamagnetic while $${\left[ {NiC{l_4}} \right]^{2 - }}$$ is paramagnetic in nature.
A
In $${\left[ {NiC{l_4}} \right]^{2 - }},$$ no unpaired electrons are present while in $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ two unpaired electrons are present.
B
In $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }},$$ no unpaired electrons are present while in $${\left[ {NiC{l_4}} \right]^{2 - }}$$ two unpaired electrons are present.
C
$${\left[ {NiC{l_4}} \right]^{2 - }}$$ shows $$ds{p^2}$$ hybridisation hence it is paramagnetic.
D
$${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ shows $$s{p^3}$$ hybridisation hence it is diamagnetic.
Answer :
In $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }},$$ no unpaired electrons are present while in $${\left[ {NiC{l_4}} \right]^{2 - }}$$ two unpaired electrons are present.
In $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ there is no unpaired electrons because $$C{N^ - }$$ is a strong field ligand thus it pair up the electrons.
In $${\left[ {NiC{l_4}} \right]^{2 - }},$$ there are two unpaired electrons because $$C{l^ - }$$ is a weak field ligand. Therefore, it does not pair up the electrons.
256.
$$CrC{l_3} \cdot 6{H_2}O$$ exists in different isomeric forms which show different colours like violet and green. This is due to
As $$3\,moles$$ of $$AgCl$$ are obtained when $$1\,mol$$ of $$CrC{l_3} \cdot 6{H_2}O$$ is treated with excess of $$AgN{O_3}$$ which shows that one molecule of the complex gives three chloride ions in solution. Hence, formula of the complex is $$\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]C{l_3}.$$
260.
The magnetic moment of the complex anion $${\left[ {Cr\left( {NO} \right)\left( {N{H_3}} \right){{\left( {CN} \right)}_4}} \right]^{2 - }}$$ is :