Coordination Compounds Test

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Coordination Compounds Test - 52

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Weekly Quiz Competition

Question 1

1 / -0

The colour of the solution/precipitate obtained in the elemental analysis of an organic compound and the molecule/ion responsible for the colour are given below. Choose the incorrectly matched pair.

Prussian blue - $$Fe_{4}[Fe(CN)_{6}]_{3}]\cdot xH_{2}O$$

Black - $$PbS$$

Violet colour - $$[Fe(CN)_{5}NOS]^{4-}$$

Blood red colour - $$[Fe(SCN)^{2+}$$

Yellow - $$(NH_{4})_{2}MoO_{4}$$

Solution

In the elemental analysis of an organic compound, presence of element may be confirmed with following colour of the solution/ precipitate.

Element	Colour of solution/ precipitate	Molecule ion formed
$$N$$	Prussian blue	$$Fe_{4}[Fe(CN)_{6}]_{3}]\cdot xH_{2}O$$
$$S$$	black	$$PbS$$
$$S$$	violet colour	$$[Fe(CN)_{5}NOS]^{4-}$$
$$N$$ and $$S$$	blood red colour	$$[Fe(SCN)]^{2+}$$
$$P$$	yellow	$$(NH_{4})_{3}PO_{4}\cdot 12MoO_{3}$$

Question 2
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What is the oxidation number of Pt in $$K[PtNH_3Cl_5]$$?

A
0

B
+1

C
+2

D
+4

Solution

Let X be the oxidation number of $$\displaystyle Pt$$ in $$\displaystyle K[PtNH_3Cl_5]$$. The oxidation numbers of $$\displaystyle K, NH_3 $$ and $$\displaystyle Cl$$ are $$\displaystyle +1$$, $$\displaystyle 0$$ and $$\displaystyle -1$$ respectively.

$$\displaystyle 1 + X + 0+5(-1) =0$$

$$\displaystyle 1+X-5=0$$

$$\displaystyle X = +4$$

Hence, the oxidation number of $$\displaystyle Pt$$ in $$\displaystyle K[PtNH_3Cl_5]$$ is $$\displaystyle +4$$.

Option $$D$$ is the correct answer.
Question 3
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Excess of silver nitrate solution is added to 100 mL of 0.01 M pentaaquachlorochromium (III) chloride solution. The mass of silver chloride obtained in grams is: (atomic mass of silver is 108).

A
$$287\times10^{-3}$$

B
$$28.7 \times10^{-3}$$

C
$$2\times10^{-3}$$

D
$$2870\times10^{-3}$$
Question 4
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For square planar complex of platinum (II), $${ \left[ Pt\left( { NH }_{ 3 } \right) \left( Br \right) \left( Cl \right) py \right] }^{ 0 }$$, how many isomeric forms are possible?

A
Two

B
Three

C
Four

D
Six

Solution

Square planar complexes of the type $$[MABCD]$$ occur in three isomeric forms.
Considering one ligand as a reference, here Bromine ($$Br$$), a total of three isomeric forms of $$[Pt(NH_3)(Br)(Cl)py]^0$$ are possible as shown in the figure.

Hence option B is correct.
Question 5
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The energies of $$d_{xy}$$ and $$d^2_z$$ orbitals in octahedral and tetrahedral transition metal complexes are such that:

A
$$E(d_{xy})> E(d^2_z)^{xy}$$ in both tetrahedral and octahedral complexes

B
$$E(d_{xy})< E(d^2_z)$$ in both tetrahedral and octahedral complexes

C
$$E(d_{xy})>E(d^2_z)$$ in tetrahedral a but $$E(d_{xy})< E(d^2_z)$$ in octahedral complexes

D
$$E(d_{xy})< E(d^2_z)$$ in tetrahedral a but $$E(d_{xy})< E(d^2_z)$$ in octahedral complexes

Solution

Hence option $$C$$ is correct.
Question 6
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Which one of the following is not correct in respect of hybridization of orbitals?

A
The orbitals presentt in valence shell only are hybridized

B
The orbitals undergoing hybridization have almost equal energy

C
Promotion of electron is not essential condition for hybridization

D
It is not always that only partially filled orbitals participate in hybridization in some cases even filled orbitals in valence shell take part

E
Pure atomic orbitals are more effective in forming stable bonds than hybrid orbitals
Question 7
1 / -0

Hybridisation of carbon in $${ CH }_{ 2 }=CH-\overset { \oplus }{ { CH }_{ 2 } }$$ is:

A
$${ sp }^{ 2 }$$

B
$$sp$$

C
$${ sp }^{ 3 }$$

D
all of these

Solution

$$S{ N }_{ 1 }=\dfrac { 4+2 }{ 2 } =\dfrac { 6 }{ 2 } =3-s{ p }^{ 2 }\\ S{ N }_{ 2 }=\dfrac { 4+2 }{ 2 } =\dfrac { 6 }{ 2 } =3-s{ p }^{ 2 }\\ S{ N }_{ 3 }=\dfrac { 4+3-1 }{ 2 } =3-sp^{ 2 }$$
Question 8
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A 0.001 mol sample of $$Co(NH_3)_5(NO_3)(SO_4)$$, is passed through a cation exchanger and the acid coming out of it requires 20 mL of 0.1 M NaOH for neutralisation. Hence, the complex is:

A
$$[Co(NH_3)_5(SO_4)NO_3]$$

B
$$[Co(NH_3)_5(NO_3)SO_4]$$

C
$$[Co(NH_3)_5NO_3SO_4]$$

D
None of these
Question 9
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Which of the following order is not correct?

A
$$\displaystyle\left[Co(H_2O)_6\right]^{+2}\,<$$ $$\displaystyle\left[Fe\left(H_2O\right)_6\right]^{+3}$$ < $$\displaystyle\left[Fe\left(CN\right)_6\right]^{-3}$$: $$\Delta_{\circ} $$ order

B
$$\displaystyle\left[Fe(CN)_6\right]^{4-}\,>$$ $$\displaystyle\left[Fe\left(CN\right)_6\right]^{3-}$$: $$Fe-C$$ Bond length

C
$$CN^{-}$$ $$< CO$$: $$\pi$$ accepting tendency

D
$$C_2$$$$H_4$$ > $$K[PtCl_3$$$$(C_2$$$$H_4$$)]: $$C-C$$ bond length

Solution

$$(A)$$ $$i)$$  $${ \left[ { Fe\left( CN \right) }_{ 6 } \right] }^{ -3 }$$
$${ Fe }^{ +3 }\longrightarrow { 3d }^{ 5 }$$
$$CFSE:--0.4{ \Delta }_{ 0 }5+0.6{ \Delta }_{ 0 }\times 0$$
$$=-2.0{ \Delta }_{ 0 }$$
$$ii)$$  $${ \left[ { Fe\left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ +3 }$$
$${ Fe }^{ +3 }\longrightarrow { 3d }^{ 5 }$$
$$CFSE=-0.4{ \Delta }_{ 0 }\times 3+0.6\times 2$$
$$=-1.2+1.2$$
$$=0$$
$$iii)$$  $${ \left[ Co{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ +2 }$$
$${ Co }^{ +2 }\longrightarrow { 3d }^{ 7 }$$
$$CFSE=-0.4\times 5+0.6\times 2$$
$$=-2.0+1.2$$
$$=-0.8{ \Delta }_{ 0 }$$
Correct order $$\rightarrow$$  $${ \left[ Fe{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ 3+ }<{ \left[ { Co\left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ 2+ }<{ \left[ { Fe\left( { CN }_{ 6 } \right) }_{ 6 } \right] }^{ -3 }$$
Question 10
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Which of the following statements is incorrect?

A
$${ K }_{ 4 }\left[ Ni{ \left( CN \right) }_{ 4 } \right] $$ and $${ K }_{ 2 }\left[ Ni{ \left( CN \right) }_{ 4 } \right] $$ both have same magnetic moment

B
$${ K }_{ 4 }\left[ Ni{ \left( CN \right) }_{ 4 } \right] $$ is a diamagnetic while $${ K }_{ 4 }\left[ Ni{ \left( Cl \right) }_{ 4 } \right] $$ is paramagnetic

C
$${ K }_{ 4 }\left[ Ni{ \left( CN \right) }_{ 4 } \right] $$ is diamagnetic while $${ K }_{ 2 }\left[ Ni{ \left( CN \right) }_{ 4 } \right] $$ is tetrahedral

D
$${ K }_{ 4 }\left[ Ni{ \left( Cl \right) }_{ 4 } \right] $$ and $${ K }_{ 4 }\left[ Ni{ \left( CN \right) }_{ 4 } \right] $$ both have same geometrical shapes

Solution

$${ K }_{ 4 }\left[ Ni{ \left( CN \right) }_{ 4 } \right] \rightarrow Ni (0), 4s^23d^8$$, strong field ligand, 0 unpaired electrons, $$dsp^2$$ & square planer, diamagnetic

$${ K }_{ 2 }\left[ Ni{ \left( CN \right) }_{ 4 } \right] \rightarrow Ni (II),$$ $$4s^03d^8$$, strong field ligand, 0 unpaired electrons, $$dsp^2$$ & square planer, diamagnetic

$${ K }_{ 4 }\left[ Ni{ \left( Cl \right) }_{ 4 } \right] \rightarrow Ni (0),$$ $$4s^23d^8$$, weak field ligand, 2 unpaired electrons, $$sp^3$$ & tetrahedral, paramagnetic.

The statement given in option C is incorrect.

Ans- Option C