Co - ordination Compounds MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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311.
Nickel $$\left( {Z = 28} \right)$$ combines with a uninegative monodentate ligand $${X^ - }$$ to form a paramagnetic complex $${\left[ {Ni{X_4}} \right]^{2 - }}.$$ The number of unpaired electron(s) in the nickel and geometry of this complex ion are, respectively :
A
one, square planar
B
two, square planar
C
one, tetrahedral
D
two, tetrahedral
Answer :
two, tetrahedral
$${\left[ {Ni{X_4}} \right]^{2 - }},$$ the electronic configuration of $$N{i^{2 + }}$$ is
It contains two unpaired electrons and the hybridisation is $$s{p^3}$$ (tetrahedral).
312.
According to Werner's theory of coordination compounds
A
primary valency is ionisable
B
secondary valency is ionisable
C
primary and secondary valencies are ionisable
D
neither primary nor secondary valency is ionisable
Answer :
primary valency is ionisable
No explanation is given for this question. Let's discuss the answer together.
A
$$Ni{\left( {CO} \right)_4}\,{\text{and}}\,NiCl_4^{2 - }$$ are diamagnetic and $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ is paramagnetic
B
$$NiCl_4^{2 - }\,{\text{and}}\,{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ are diamagnetic and $$Ni{\left( {CO} \right)_4}$$ is paramagnetic
C
$$Ni{\left( {CO} \right)_4}\,{\text{and}}\,{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ are diamagnetic and $$NiCl_4^{2 - }$$ is paramagnetic
D
$$Ni{\left( {CO} \right)_4}$$ is diamagnetic and $$NiCl_4^{2 - }\,{\text{and}}\,{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ are paramagnetic
Answer :
$$Ni{\left( {CO} \right)_4}\,{\text{and}}\,{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ are diamagnetic and $$NiCl_4^{2 - }$$ is paramagnetic
314.
Among the following compounds which one is both paramagnetic and coloured?
A
$${K_2}C{r_2}{O_7}$$
B
$$\left[ {Co\left( {S{O_4}} \right)} \right]$$
C
$${\left( {N{H_4}} \right)_2}\left[ {TiC{l_6}} \right]$$
D
$${K_3}\left[ {Cu{{\left( {CN} \right)}_4}} \right]$$
In $$\left[ {Co\left( {S{O_4}} \right)} \right],$$ the oxidation state of $$Co$$ is +2. Configuration of $$C{o^{2 + }} = 3{d^7},$$ it has unpaired electrons in $$3d$$ - orbitals so it is paramagnetic. Because of incompletely filled $$d$$ - orbitals it is coloured.
315.
Pick a poor electrolytic conductor complex in solution
A
$${K_2}\left[ {PtC{l_6}} \right]$$
B
$$\left[ {Co{{\left( {N{H_3}} \right)}_3}} \right]{\left( {N{O_2}} \right)_3}$$
C
$${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$$
D
$$\left[ {Co{{\left( {N{H_3}} \right)}_4}} \right]S{O_4}$$
The complex furnishing least number of ions in solution will be poor electrolytic
conductor.
316.
$$\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]C{l_3}$$ ( at. no. of $$Cr=24$$ ) has a magnetic moment of $$3.83\,BM,$$ the correct distribution of $$3d$$ electrons in the chromium of the complex is
A
$$3d_{xy}^1,3d_{yz}^1,3d_{{z^2}}^1$$
B
$$3{d_{\left( {{x^2} - {y^2}} \right)}},3d_{{z^2}}^1,3d_{xz}^1$$
C
$$3{d_{xy}},3{d_{\left( {{x^2} - {y^2}} \right)}},3d_{yz}^1$$
D
$$3d_{xy}^1,3d_{yz}^1,3d_{zx}^1$$
Answer :
$$3d_{xy}^1,3d_{yz}^1,3d_{zx}^1$$
Magnetic moment $$\left( \mu \right) = \sqrt {n\left( {n + 2} \right)} \,\,BM$$
$$\eqalign{
& {\text{or}}\,\,\,3.83 = \sqrt {n\left( {n + 2} \right)} \cr
& {\text{or}}\,\,\,3.83 \times 3.83 = {n^2} + 2n \cr
& 14.6689 = {n^2} + 2n \cr
& n \simeq 3 \cr} $$
Hence, number of unpaired electrons in $$d$$ - subshell of chromium $$(Cr= 24) =3.$$
So, the configuration of chromium ion is $$C{r^{3 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^3}$$
In $$\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]C{l_2},$$ oxidation state of $$Cr$$ is + 3.
Hence, in $$3{d^3}$$ the distribution of electrons $$3d_{xy}^1,3d_{yz}^1,3d_{zx}^1,3d_{{x^2} - {y^2}}^0,3d_{{z^2}}^0$$
317.
The $$CFSE$$ for octahedral $${\left[ {CoC{l_6}} \right]^{4 - }}$$ is $$18,000\,c{m^{ - 1}}.$$ The $$CFSE$$ for tetrahedral $${\left[ {CoC{l_4}} \right]^{2 - }}$$ will be
318.
The total number of possible isomers for the complex compound $$\left[ {C{u^{{\text{II}}}}{{\left( {N{H_3}} \right)}_4}} \right]\left[ {Pt{\,^{{\text{II}}}}C{l_4}} \right]$$ are