Coordination Compounds Test

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Coordination Compounds Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

Primary and secondary valency of $$Pt\ in\ \left[ Pt{ \left( en \right) }_{ 2 }{ Cl }_{ 2 } \right] { Cl}_{ 2 }$$ are:

A
4, 4

B
4, 6

C
6, 6

D
4, 2

Solution

Primary valency is the number of ions required to satisfy the charge on the metal ion.
Secondary valency is the number of ions of molecules that are coordinated to the metal ion.
So, in $$[Pt(en)_2Cl_2]Cl_2$$
Charge on $$Pt\Rightarrow x+2(0)+2(-1)=+2\Rightarrow x=+4$$
So, Primary valency is $$4$$
Secondary valency is $$2(2)+2=6$$
$$\therefore$$ Answer is option B.
Question 2
1 / -0

Which of the following represents the given mode of hybridization $$sp^2 - sp^2 - sp - sp$$ from left to right?

A
$$CH_2 = CH - C \equiv N$$

B
$$CH_2 = C = C = CH_2$$

C
$$CH_2 = CH - CH = CH_2$$

D
$$CH_2 = C = CH - CH_3$$

Solution

The type of hybridization of carbon atom is :
(A) $$\overset{{sp^2}}{CH_2}$$=$$\overset{sp^2}{CH}$$-$$\overset{sp}{CH}$$ = $$\overset{sp}{N}$$
(B) $$\overset{sp^2}{CH_2}$$ =$$\overset{sp}{CH}$$=$$\overset{sp}{CH}$$ =$$\overset{sp^2}{CH_2}$$
(C)$$\overset{sp^2}{CH_2}$$=$$\overset{sp^2}{CH}$$-$$\overset{sp^2}{CH}$$=$$\overset{sp^2}{CH_2}$$
(D)$$\overset{sp^2}{CH_2}$$=$$\overset{sp}{CH}$$=$$\overset{sp^2}{CH}$$-$$\overset{sp^3}{CH_3}$$
Therefore, (A) has the given mode of hybridization from left to right.
Question 3
1 / -0

Predict the order of $$\Delta_0$$ for the following compounds.
I. $$[Mn(H_2O)_6]^{2+}$$
II. $$[Mn(CN)_2(H_2O)_4]$$
III. $$[Mn(CN)_4(H_2O)_2]^{2-}$$

A
$$\Delta_0 (I) < \Delta_0 (II) < \Delta_0 (III)$$

B
$$\Delta_0 (II) < \Delta_0 (I) < \Delta_0 (III)$$

C
$$\Delta_0 (III) < \Delta_0 (I) < \Delta_0 (I)$$

D
$$\Delta_0 (I) < \Delta_0 (III) < \Delta_0 (II)$$

Solution

More is the number of strong field ligand higher is the crystal field stabilisation energy.
So, $$[Mn(Cn)_4(H_2O)_2]^{2-}$$ having $$4$$ $$CN^-$$ ligands will have the highest value of $$\triangle_0$$ then followed by $$[Mn(CN)_2(H_2O)_4]$$ then $$[Mn(H_2O)_6]^{2+}$$.
Question 4
1 / -0

1 mole of co-ordinate compound $$CoCl_3$$ on reaction with excess AgNCl $$143.5$$gms of AgCl ppt . Then number of chloride ions satisfying both primary valency and secondary valency are?

A
$$2$$

B
$$3$$

C
$$1$$

D
$$0$$

Solution

Moles of $$AgCl$$ formed $$=\cfrac{143.5}{143.5}=1$$
$$1$$ mole of $$CoCl_3$$ gives $$1$$ mol of $$AgCl$$ ppt.
Secondary valence is the number of ions that are coordinated to metal ion. Here, $$1Cl^-$$ ion is not coordinated to the metal ion, due to which it gives $$AgCl$$ ppt. Hence, no. of chloride ions satisfying the given condition $$=2.$$
Question 5
1 / -0

The CFSE of $$[Mn(H_2O)_6]^{2+}$$ is:

A
-1.2 $$\Delta_{\Theta}$$

B
-1.9 $$\Delta_{\Theta}$$

C
0

D
-2.4 $$\Delta_{\Theta}$$

Solution

The CFSE of a $$d^5$$octahedral complex is = -0.4$$\Delta_o$$x number of electrons in $$t_2g$$ orbitals + 0.6$$\Delta_o$$x number of electrons in $$eg$$ orbitals
=-0.4$$\Delta_o$$x3 + 0.6$$\Delta_o$$x 2 = 0. hence the correct option is C.
Question 6
1 / -0

The formula of the complex, tris- (ethylenediamine) cobalt (iii) sulphate is:

A
$$ [Co(en)_{2}SO_{4}] $$

B
$$ [Co(en)_{3}SO_{4}] $$

C
$$ [Co(en)_{3}]_{2}(SO_{4})_{3} $$

D
$$ [Co(en)_{3}]_{2}SO_{4} $$

Solution

$${ \left[ CO{ \left( en \right) }_{ 3 } \right] }_{ 2 }{ \left( { SO }_{ 4 } \right) }_{ 3 }$$
$$\rightarrow$$ Oxidation of $$en=0$$ $${ { SO } }_{ 4 }=-2$$ then we get $$CO=+3$$ as given
$$\rightarrow$$ And tris is used as two numbers are present at that same spot.
$$\rightarrow$$ where $$2$$ is called $$"di"$$ and $$3$$ is called $$"tris"$$.
Ans :- Option C.
Question 7
1 / -0

The number of isomers possible for square planar complex $$ K_{2}[PdClBr_{2}SCN] $$ is:

A
2

B
3

C
4

D
6

Solution

The given planar complex can have upto two optical isomes 1. Cis-$$[PdClBr_2SCN]^{2-}$$ and 2.trans-$$[PdClBr_2SCN]^{2-}$$.In first complex the Br-Br will be cis to each other and in second isomer they are trans to one another. Hence the correct option is A.
Question 8
1 / -0

The strongest -CO bond is present in:

A
$$ [Cr(CO)_{6}]^{+}$$

B
$$ [Fe(CO)_{5}]$$

C
$$ [V(CO)_{6}]^{-}$$

D
all have equal strength

Solution

The strongest CO bond is present in the complex$$ [Cr(CO)]^+$$. This is because as the oxidation state of metal increases the pi back bonding tendency becomes low and hence the strength of CO bond increases. In the given complexes the oxidation state increase in of the order $$V(-)<Fe(0)<Cr(+)$$. hence the correct option is A.
Question 9
1 / -0

Which of the following complexes has a geometry different from others?

A
$$\left [ NiCl_{4} \right ]^{2-}$$

B
$$Ni\left ( CO \right )_{4}$$

C
$$\left [ Ni\left ( CN \right )_{4} \right ]^{^{-2}}$$

D
$$\left [ Zn\left ( NH_{3} \right )_{4} \right ]^{2+}$$

Solution
Question 10
1 / -0

Type of isomerism shown by $$\left [ Cr(NH_{3})_{5}NO_{2} \right ]Cl_{2}$$ is

A
Optical

B
ionisation

C
geometrical

D
linkage

Solution

This complex has $$NO_2$$ in it,which is an ambidentate ligand,which means it has $$2$$ sites of attack to the central metal atom.
Therefore,the complex which has such type of ligand can show linkage isomerism and they are
(1)$$[Cr(NH_3)_5NO_2]Cl_2$$
(2)$$[Cr(NH_3)_5ONO]Cl_3$$
oxidation state of chromium

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Coordination Compounds Test - 33

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Coordination Compounds Test - 33

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Question 1

4, 4

4, 6

6, 6

4, 2

Solution

Question 2

$$CH_2 = CH - C \equiv N$$

$$CH_2 = C = C = CH_2$$

$$CH_2 = CH - CH = CH_2$$

$$CH_2 = C = CH - CH_3$$

Solution

Question 3

$$\Delta_0 (I) < \Delta_0 (II) < \Delta_0 (III)$$

$$\Delta_0 (II) < \Delta_0 (I) < \Delta_0 (III)$$

$$\Delta_0 (III) < \Delta_0 (I) < \Delta_0 (I)$$

$$\Delta_0 (I) < \Delta_0 (III) < \Delta_0 (II)$$

Solution

Question 4

$$2$$

$$3$$

$$1$$

$$0$$

Solution

Question 5

-1.2 $$\Delta_{\Theta}$$

-1.9 $$\Delta_{\Theta}$$

0

-2.4 $$\Delta_{\Theta}$$

Solution

Question 6

$$ [Co(en)_{2}SO_{4}] $$

$$ [Co(en)_{3}SO_{4}] $$

$$ [Co(en)_{3}]_{2}(SO_{4})_{3} $$

$$ [Co(en)_{3}]_{2}SO_{4} $$

Solution

Question 7

2

3

4

6

Solution

Question 8

$$ [Cr(CO)_{6}]^{+}$$

$$ [Fe(CO)_{5}]$$

$$ [V(CO)_{6}]^{-}$$

all have equal strength

Solution

Question 9

$$\left [ NiCl_{4} \right ]^{2-}$$

$$Ni\left ( CO \right )_{4}$$

$$\left [ Ni\left ( CN \right )_{4} \right ]^{^{-2}}$$

$$\left [ Zn\left ( NH_{3} \right )_{4} \right ]^{2+}$$

Solution

Question 10

Optical

ionisation

geometrical

linkage

Solution

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