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

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

313. Amongst $$Ni{\left( {CO} \right)_4},{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}{\text{and}}\,NiCl_4^{2 - }$$

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 $${K_3}\left[ {Cu{{\left( {CN} \right)}_4}} \right]$$
Answer :   $$\left[ {Co\left( {S{O_4}} \right)} \right]$$

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 $$\left[ {Co{{\left( {N{H_3}} \right)}_4}} \right]S{O_4}$$
Answer :   $$\left[ {Co{{\left( {N{H_3}} \right)}_4}} \right]S{O_4}$$

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 :   $$3d_{xy}^1,3d_{yz}^1,3d_{zx}^1$$

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

A $$18,000\,c{m^{ - 1}}$$
B $$16,000\,c{m^{ - 1}}$$
C $$8,000\,c{m^{ - 1}}$$
D $$20,000\,c{m^{ - 1}}$$
Answer :   $$8,000\,c{m^{ - 1}}$$

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

A 5
B 6
C 3
D 4
Answer :   4

319. The ionisation isomer of $$\left[ {Cr{{\left( {{H_2}O} \right)}_4}} \right]Cl\left( {N{O_2}} \right)Cl\,\,{\text{is}}$$

A $$\left[ {Cr{{\left( {{H_2}O} \right)}_4}\left( {{O_2}N} \right)} \right]C{l_2}$$
B $$\left[ {Cr{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]\left( {N{O_2}} \right)$$
C $$\left[ {Cr{{\left( {{H_2}O} \right)}_4}Cl\left( {ONO} \right)} \right]Cl$$
D $$\left[ {Cr{{\left( {{H_2}O} \right)}_4}C{l_2}\left( {N{O_2}} \right)} \right]{H_2}O$$
Answer :   $$\left[ {Cr{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]\left( {N{O_2}} \right)$$

320. Which of the following complexes will have tetrahedral shape?

A $${\left[ {PdC{l_4}} \right]^{2 - }}$$
B $${\left[ {Pd{{\left( {CN} \right)}_4}} \right]^{2 - }}$$
C $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$
D $${\left[ {NiC{l_4}} \right]^{2 - }}$$
Answer :   $${\left[ {NiC{l_4}} \right]^{2 - }}$$