Coordination Compounds Test

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Coordination Compounds Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

The colour of a complex compound is due to :

A
promotion of $$3d$$-electrons of the central atom/ion to $$4s$$-orbital.

B
promotion of $$3d$$-electrons of the central atom/ion to $$4p$$-orbitals.

C
promotion of $$3d$$-electrons of the central atom/ion to $$d$$-orbitals.

D
promotion of $$4s$$-electrons of the central atom/ion to $$4p$$-orbitals.

Solution

Option (C) is correct.
The colour of a complex compound is due to promotion of $$3d$$-electrons of the central atom/ion to $$d$$-orbitals.
The colour of the complex compound is due to $$d-d$$ electronic transition.The $$d$$-orbitals of a free transition metal atom or ion are degenerate (all have the same energy.) When transition metals form coordination complexes, the d-orbitals of the metal interact with the electron cloud of the ligands in such a manner that the d-orbitals become non-degenerate (not all having the same energy).
Question 2
1 / -0

The complex used as an anticancer agent is :

A
$$trans$$-$$[Co(NH_3)_3Cl_3]$$

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

C
$$cis$$-$$K_2[PtCl_2Br_2]$$

D
$$Na_2CO_3$$

Solution

Cisplatin, or cis-diamminedichloroplatinum(II)[2] (CDDP) is a chemotherapy drug. It was the first member of a class of platinum-containing anti-cancer drugs, which now also includes carboplatin and oxaliplatin. These platinum complexes react in vivo, binding to and causing crosslinking of DNA, which ultimately triggers apoptosis (programmed cell death).
Question 3
1 / -0

If a transition-metal compound absorbs violet-indigo radiation in the visible region. Its color would be :

A
green.

B
yellow.

C
orange.

D
blue.

Solution

From color wheel, it can be seen that
if a transition-metal compound absorbs violet-indigo radiation in the visible region. Its color would be yellow
Question 4
1 / -0

Linear combination of two hybridzed orbitals, belonging to two atoms and each having one electron leads to

A
Sigma - bond

B
Double bond

C
Co-ordinate covalent bond

D
Pi - bond

Solution

Linear combination of two hybridized orbitals, belonging to two atoms and each having one electron leads to formation of sigma bond.
Question 5
1 / -0

Transition metal compounds are usually colored. This is due to the electronic transition :

A
from $$p$$-orbital to $$s$$-orbital.

B
from $$d$$-orbital to $$s$$-orbital.

C
from $$p$$-orbital to $$p$$-orbital.

D
within the $$d$$-orbitals.

Solution

coloured compound of transition elements is assosiated with partially filled (n-1)d orbitals. the transition metal ions containing unpaired d-electrons undergoes electronic transition from one d-orbital to another.

During this d-d transition process the electrons absorb certain energy from the radiation and emit the remainder of energy as colored light. the color of ion is complementary of the color absorbed by it.

hence, colored ion is formed due to d-d transition which falls in visible region for all transition elements.
Question 6
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Directions For Questions

The first ionization energy of Na, NO, Xe, and $$O_2$$ follows the order $$Na\, <\, NO\, <\, Xe\, =\, O_2.\, O_2$$ reacts with the powerful oxidizing agent $$PtF_6$$ to yield a compound $$O_2\, [PtF_6].$$

...view full instructions

Which of the following is first formed inert gas compound ?

A
$$Xe[Pt F_6]$$

B
$$Kr[Pt F_6]$$

C
$$XeF_2$$

D
$$XeO_3$$

Solution

The ionisation energy of Xe is lower than that of Kr. Hence upon reaction with $$\\ { [PtF }_{ 6 }]\\ $$, $$\\ { Xe[PtF }_{ 6 }]\\ $$ should get formed first.
Question 7
1 / -0

Find the ratio of geometrical isomers in [M(AA)$$_2$$b$$_2$$] and optical isomers of [M(AA)$$_3$$].

A
1

B
2

C
1.5

D
2.5

Solution

The complex [M(AA)$$_2$$b$$_2$$] has 2 geometrical isomers cis and trans isomers.
The complex [M(AA)$$_3$$] has two optical isomers d and l forms.
$$\displaystyle{\frac{2}{2}}$$ = 1
Thus, the ratio of geometrical isomers in [M(AA)$$_2$$b$$_2$$] and optical isomers of [M(AA)$$_3$$] is 1.
Question 8
1 / -0

Silver halides are used in photography because they are -

A
Photosensitive

B
Soluble in hypo solution

C
Soluble in $$NH_4OH$$

D
Insoluble in acids

Solution

Option (A) is correct.
Silver halides are used in photography because they are photosensitive as they react with light to form the image, silver halides being reduced to silver.They are also soluble in hyposolution.
Question 9
1 / -0

Directions For Questions

The first ionization energy of Na, NO, Xe, and $$O_2$$ follows the order $$Na\, <\, NO\, <\, Xe\, =\, O_2.\, O_2$$ reacts with the powerful oxidizing agent $$PtF_6$$ to yield a compound $$O_2\, [PtF_6].$$

...view full instructions

Which of the following inert gas has the smallest ionization energy ?

A
He

B
Ne

C
Ar

D
Xe

Solution

ionization energy decreases from top to bottom of each column because the outer shell electrons become farther removed from the nucleus.
Question 10
1 / -0

Mg is an important component of which biomolecule occurring extensively in living world ?

A
Hemoglobin

B
Chlorophyll

C
Florigen

D
ATP

Solution

Chlorophyll

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Coordination Compounds Test - 20

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Coordination Compounds Test - 20

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Question 1

promotion of $$3d$$-electrons of the central atom/ion to $$4s$$-orbital.

promotion of $$3d$$-electrons of the central atom/ion to $$4p$$-orbitals.

promotion of $$3d$$-electrons of the central atom/ion to $$d$$-orbitals.

promotion of $$4s$$-electrons of the central atom/ion to $$4p$$-orbitals.

Solution

Question 2

$$trans$$-$$[Co(NH_3)_3Cl_3]$$

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

$$cis$$-$$K_2[PtCl_2Br_2]$$

$$Na_2CO_3$$

Solution

Question 3

green.

yellow.

orange.

blue.

Solution

Question 4

Sigma - bond

Double bond

Co-ordinate covalent bond

Pi - bond

Solution

Question 5

from $$p$$-orbital to $$s$$-orbital.

from $$d$$-orbital to $$s$$-orbital.

from $$p$$-orbital to $$p$$-orbital.

within the $$d$$-orbitals.

Solution

Question 6

$$Xe[Pt F_6]$$

$$Kr[Pt F_6]$$

$$XeF_2$$

$$XeO_3$$

Solution

Question 7

1

2

1.5

2.5

Solution

Question 8

Photosensitive

Soluble in hypo solution

Soluble in $$NH_4OH$$

Insoluble in acids

Solution

Question 9

He

Ne

Ar

Xe

Solution

Question 10

Hemoglobin

Chlorophyll

Florigen

ATP

Solution

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