Coordination Compounds Test

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Coordination Compounds Test - 67

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Weekly Quiz Competition

Question 1
1 / -0

Which among the following is used in the treatment of cancer?

A
cis-$$[Pt(en)_2Cl_2]$$

B
cis-$$[PtCl_2(NH_3)_2]$$

C
trans-$$[Pt(en)_2Cl_2]$$

D
trans-$$[Pt(NH_3)_2Cl_2]$$

Solution

$$cis-[PtCl_2(NH_3)_2]$$ also known as cisplatin is used as an anticancer drug in the chemotherapy process of cancer treatment.

Hence, option B is correct.
Question 2
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A molecule $$(X)$$ has (i) four sigma bonds formed by the overlap of $$sp^{2}$$ and $$s$$ orbitals; (ii) one sigma bond formed by $$sp^{2}$$ and $$sp^{2}$$ - orbitals and (iii) one $$\pi-bond$$ formed by $$p_{z}$$ and $$p_{z}$$ orbitals. Which of the following is $$X$$?

A
$$C_{2}H_{6}$$

B
$$C_{2}H_{5}Cl$$

C
$$C_{2}H_{2}Cl_{2}$$

D
$$C_{2}H_{4}$$

Solution
Question 3
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The total number of possible coordination isomer for the given compound $$[Pt(NH_3)_4Cl_2][PtCl_4]$$ is?

A
$$2$$

B
$$4$$

C
$$5$$

D
$$3$$

Solution
Question 4
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The octahedral complex $$CoSO_4Cl_5\,NH_3$$ exists in two isomeric forms $$X$$ and $$Y$$. Isomers $$X$$ reacts with $$AgNO_3$$ to give a white precipitate, but does not react with $$BaCl_2$$. Isomer $$Y$$ gives white precipitate with $$BaCl_2$$ but does not react with $$AgNO_3$$.
Isomers $$X$$ and $$Y$$ are:

A
ionization isomers

B
linkage isomers

C
coordinating isomers

D
solvent isomers

Solution

$$\underset{\text{(X)}}{[Co(NH_3)_5SO_4]Cl}\xrightarrow{AgNO_3} \underset{\text{white ppt.}}{AgCl}\downarrow$$

$$\underset{\text{(Y)}}{[Co(NH_3)_5Cl]SO_4}\xrightarrow{BaCl_2}\underset{\text{white ppt.}}{BaSO_4}\downarrow$$

$$[X]$$ and $$[Y]$$ are ionization isomers.
Question 5
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Brown ring complex known as:

A
$$K_4[Fe(CN)_6]$$

B
$$[Fe(NO)(H_2O)_5]SO_4$$

C
$$[Fe(NO)_2)(H_2O)_4]Br_2$$

D
$$[FeF_6]^{3-}$$

Solution

$$6FeSO_4+3H_2SO_4+2HNO_3\rightarrow 3Fe_2(SO_4)_3+4H_2O+2NO$$

$$FeSO_4+NO+5H_2O\rightarrow [Fe(NO)(H_2O)_5]SO_4$$
Brown ring complex

Option B is correct.
Question 6
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Which complex among the following gives a white precipitate on treatment with an aqueous solution of barium chloride?

A
$$[Pt(NH_3)_4Br_2]Cl_2$$

B
$$[Co(NH_3)_5SO_4]NO_2$$

C
$$[Co(NH_3)_5NO_2]SO_4$$

D
$$[Pt(NH_3)_4Cl_2]Br_2$$

Solution

White precipitate on treatment with barium chloride is due to the formation of $$BaSO_4$$ and for this sulphate ion should be present in ionisable part.

$$[Co(NH_3)_5NO_2]SO_4 + BaCl_2(aq)\rightarrow [Co(NH_3)_5NO_2]Cl_2 + BaSO_4(s)\downarrow$$

$$BaSO_4$$ is a white precipitate.

The other compounds do not give white ppt. due to the absence of $$SO_4^{2-}$$ ions.

Hence, option $$C$$ is correct.
Question 7
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The complexes $$[Cr(NH_3)_6][Co(CN)_6]$$ and $$[Co(NH_3)_6][Cr(CN)_6]$$ are not the example of?

A
Geometrical isomers

B
Optical isomers

C
Coordination isomers

D
Linkage isomers
Question 8
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Directions For Questions

A metal complex having composition $$Cr(NH_3)_4Cl_2Br$$ has been isolated in two forms, (X) and (Y). The form (X) reacts with $$AgNO_3$$ to give a white precipitate readily soluble in dilute aqueous ammonia, whereas (Y) gives a pale yellow precipitate soluble in concentrate ammonia.

...view full instructions

Choose the correct statement regarding the complexes X and Y:
(I) Both complexes have the same six geometrical isomer.
(II) Both complexes have nearly the same conductivity in $$H_2O$$.
(III) In both complexes, $$NH_3$$ is involved only to secondary valency.
(IV) In both complexes, electronic configuration of $$t_{2g}$$ is same.

A
I, II and III

B
I, II, III and IV

C
II, III and IV

D
I, III and IV
Question 9
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Directions For Questions

Hybridization is the measure to express the molecular shapes reasonably. Hybridization, although theoretical, but is highly correlated with molecular shapes or inter orbital angles. If there is any departure in geometry in apparent departure in hybridization can be observed and the following characteristic relationship is the easiest way to interpret that departure. Any hybrid orbital made by s and p-orbitals is characterized by the following relationship.

$$ f_s = \dfrac {1}{i + 1}$$ or $$f_p = \dfrac {i}{i+1}$$ and $$i =\dfrac {f_p}{f_s} $$

For $$ sp^3, i= 3$$ and $$sp^2, i=2 $$ for an orbital, $$ f_s + f_p = 1 $$ while all the s-p hybrids of gein atom must satisfy the condition.

$$ \sum { f_x} = 1.00 $$

d and the inter orbital angle between two equivalent ( same $$f_s$$ and same $$f_p$$ ) hybrid orbitals is given by

$$ cos \theta = \dfrac {-1}{i} $$

...view full instructions

VSEPR theory suggest a small departure from $$ 109^0 28'$$ for $$ NH_3 $$ molecule. The H-N-H bond angle has been found to be $$ 107^0 $$ what kind of hybrid orbital will be occupied by the non -bonding pair of electron?

A
$$ sp^3 $$

B
$$ sp^{3.4} $$

C
$$ sp^{2.13} $$

D
$$ sp^2 $$

Solution

Question 10

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Match the complex species given in Column-I with the possible isomerism given in Column-II and assign the correct code.

Column-I(Complex species)	Column-II(Isomerism)
(a) $$[Co(NH_3)_4Cl_2]^+$$	(p) Optical
(b) cis-$$[Co(en)_2Cl_2]^+$$	(q) Ionization
(c) $$[Co(NH_3)_5(NO_2)]Cl_2$$	(r) Coordination
(d) $$[Co(NH_3)_6][Cr(CN)_6]$$	(s) Geometrical
	(t) Linkage

a(p) b(q) c(s) d(t)

a(s) b(r) c(q) d(p)

a(s) b(p) c(q,t) d(r)

a(s) b(p) c(q) d(r)

Solution

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Coordination Compounds Test - 67

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Coordination Compounds Test - 67

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Question 1

cis-$$[Pt(en)_2Cl_2]$$

cis-$$[PtCl_2(NH_3)_2]$$

trans-$$[Pt(en)_2Cl_2]$$

trans-$$[Pt(NH_3)_2Cl_2]$$

Solution

Question 2

$$C_{2}H_{6}$$

$$C_{2}H_{5}Cl$$

$$C_{2}H_{2}Cl_{2}$$

$$C_{2}H_{4}$$

Solution

Question 3

$$2$$

$$4$$

$$5$$

$$3$$

Solution

Question 4

ionization isomers

linkage isomers

coordinating isomers

solvent isomers

Solution

Question 5

$$K_4[Fe(CN)_6]$$

$$[Fe(NO)(H_2O)_5]SO_4$$

$$[Fe(NO)_2)(H_2O)_4]Br_2$$

$$[FeF_6]^{3-}$$

Solution

Question 6

$$[Pt(NH_3)_4Br_2]Cl_2$$

$$[Co(NH_3)_5SO_4]NO_2$$

$$[Co(NH_3)_5NO_2]SO_4$$

$$[Pt(NH_3)_4Cl_2]Br_2$$

Solution

Question 7

Geometrical isomers

Optical isomers

Coordination isomers

Linkage isomers

Question 8

I, II and III

I, II, III and IV

II, III and IV

I, III and IV

Question 9

$$ sp^3 $$

$$ sp^{3.4} $$

$$ sp^{2.13} $$

$$ sp^2 $$

Solution

Question 10

a(p) b(q) c(s) d(t)

a(s) b(r) c(q) d(p)

a(s) b(p) c(q,t) d(r)

a(s) b(p) c(q) d(r)

Solution

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