Coordination Compounds Test

Result

Coordination Compounds Test - 31

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The value of X in $$Fe{ \left( CO \right) }_{ 2 }\left( NO \right) x,$$ is:

A
3

B
4

C
2

D
1

Solution

The coordination sphere of a coordination compound or complex consists of the central metal atom/ion plus its attached ligands. The coordination sphere is usually enclosed in brackets when written in a formula. The coordination number is the number of donor atoms bonded to the central metal atom/ion.
Most metal complexes or compounds except for alloys. Specific examples include hemoglobin and $$Ru_3(CO)_{12}$$.
So the total electron should be 18
valence electron of iron is = 8
CO gives = $$2e^-$$ , NO gives $$3e^-$$
18 = $$8+(2\times2)+(3\times x)$$
18 - 8 - 4 = $$3x$$
$$x$$ = 2
Question 2
1 / -0

Order of the bond strength of $$C-H$$ bonds involving sp, $$sp^2$$ and $$sp^3$$ hybridized carbon atoms is:

A
$$sp \, > \, sp^2 \, > \, sp^3$$

B
$$sp^3 \, > \, sp^2 \, > \, sp$$

C
$$sp^2 \, > \, sp^3 \, > \, sp$$

D
$$sp^2 \, > \, sp \, > \, sp^3$$

Solution

$$\equiv \overset {sp}{C}-H$$ $$=\overset {sp^2}{CH}-H$$ $$-\overset {sp^3}{CH_2}-H$$
In $$sp, sp^2$$ and $$sp^3$$ hybridized central carbon atoms, the percentage of $$s$$ character in bonding orbitals (i.e. $$C-H$$ bonds) are $$50$$%, $$33.3$$% and $$25$$% respectively. We know that percentage of $$s$$ character increases the bond strength because the shape of $$s$$ orbitals is spherical and it is more closer to nucleus of an atom. Therefore order of bond strength of $$C-H$$ bonds will be:
$$sp >sp^2 > sp^3$$
Question 3
1 / -0

Which of the following compounds is most stable ?

A
$$LiCl$$

B
$$LiI$$

C
$$LiBr$$

D
$$LiF$$

Solution

$$\bf{Explanation-}$$

Lithium is an alkali metal and it is electro-positive in nature.
The order of electronegativity of halides present in these compounds are as follow:
$$F>Cl>Br>I$$
Hence, lithium fluoride $$(LiF)$$ is the most stable compound because of the small size of $$Li$$.
As we go down the group in the periodic table the size of the element increases so, the $$Li-X$$ bond becomes weak.
As $$Li$$ and $$F$$ are of comparable size that is why $$LiF$$ is the most stable.
Also, $$LiF$$ has the highest lattice enthalpy.

$$\bf{Conclusion-}$$ Hence, the correct answer is option D.
Question 4
1 / -0

Which of the following will exhibit more than ten isomers?

A
$$\left[ Co\left( en \right) Cl\ BrI\left( CN \right) \right]$$

B
$$\left[ Co\left( gly \right) Cl\ BrI\left( NC \right)\right]$$

C
$$\left[ Co\left( gly \right) Cl\ Br\left( CN \right)NO_2 \right]$$

D
All the above

Solution

There are five types of ligand present in which one is bidentate and one is ambidentate i.e. $$(CN)$$.
(Refer to Image)
These are $$6$$ basic isomers. As one ambidentate ligand is present i.e. $$CN$$ or $$NO_2$$ which can attach through two site. Thus show linkage isomerism.
Thus the total number of isomers will be greater than $$10$$. Thus all will give more than $$10$$ isomers.
Question 5
1 / -0

Which of the following can show coordination isomerism ?

A
$$[Cu(NH_3)_4][PtCl_4]$$

B
$$[Fe(NH_3)_6]^{-2}[Pt(CN)_6]^{-3}$$

C
Both A and B

D
$$[Pt[en]_3] [SO_4]_2$$

Solution
Question 6
1 / -0

If the thio-cyanide ion is added to potash-ferric alum then red colour appears. This colour is due to the formation of:

A
$$KSCN$$

B
$$Fe(SCN)_3$$

C
$$Fe(SCN)_2$$

D
$$Fe(SCN)$$

Solution

The $$Fe^{3+}$$ ions react with the thiocyanide ion present in the solution to form a dark red colored complex of iron thiocyanate, i.e;
$$Fe^{3+}+SCN^-+5H_2O\longrightarrow [Fe(SCN)$$3$$(H_2O)_5]^{2+}$$
Question 7
1 / -0

Number of stereoisomer's possible for coordination complex $$Na[Cr(en)Cl_2Br_2]$$ is:

A
2

B
3

C
4

D
5

Solution

There are four stereoisomers are possible for the given coordination compound.
It shows optical as well as geometrical isomerism.
Question 8
1 / -0

The co-ordination number of a metal in co-ordination compound is:

A
same as primary valency

B
sum of primary and secondary valences

C
same as secondary valency

D
none of the above

Solution

The secondary valency is equal to the coordination number the secondary valency are non ionizable valencies. These are satisfied by neutral molecules or negative ions. For example in $$[Ni(CO)_4]$$ the coordination number of Ni metal is four and its secondary valency is also four.
Question 9
1 / -0

An ion $${ M }^{ 2+ }$$, forms the complexes $${ \left[ M{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ 2+ },{ \left[ M{ \left( en \right) }_{ 3 } \right] }^{ 2+ }$$ and $${ \left[ { MBr }_{ 6 } \right] }^{ 4- }$$. The colour of the complexes will be _____________ respectively.

A
green, blue and red

B
blue, red and green

C
green, red and blue

D
red, blue and green

Solution

Crystal Field Stabilization Energy(CFSE) is proportional to the frequency of the absorbed light. The emitted colors are red, green and blue. The corresponding absorbed colors are green, red and orange. So the first complex will be blue, second is red and third is green.
Question 10
1 / -0

The correct representation of $$CuSO_4\cdot 5H_2O$$ is:

A
$$[Cu(H_2O)_5]SO_4$$

B
$$[Cu(H_2O)_3SO_4]\cdot 2H_2O$$

C
$$[CuSO_4\cdot H_2O]\cdot 4H_2O$$

D
$$[Cu(H_2O)_4]SO_4\cdot H_2O$$

Solution

For,
$$CuS{O_4}.5{H_2}O$$ in this $$4{H_2}O$$ are bonded by co-ordinate bond and $$1 {H_2}O$$ by covalent bonding.
Therefore,
Its formula of the compound is $$\left[ {Cu{{\left( {{H_2}O} \right)}_4}} \right]S{O_4}.{H_2}O$$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Coordination Compounds Test - 31

Result

Coordination Compounds Test - 31

Score

-

Rank

-

SHARING IS CARING

Question 1

3

4

2

1

Solution

Question 2

$$sp \, > \, sp^2 \, > \, sp^3$$

$$sp^3 \, > \, sp^2 \, > \, sp$$

$$sp^2 \, > \, sp^3 \, > \, sp$$

$$sp^2 \, > \, sp \, > \, sp^3$$

Solution

Question 3

$$LiCl$$

$$LiI$$

$$LiBr$$

$$LiF$$

Solution

Question 4

$$\left[ Co\left( en \right) Cl\ BrI\left( CN \right) \right]$$

$$\left[ Co\left( gly \right) Cl\ BrI\left( NC \right)\right]$$

$$\left[ Co\left( gly \right) Cl\ Br\left( CN \right)NO_2 \right]$$

All the above

Solution

Question 5

$$[Cu(NH_3)_4][PtCl_4]$$

$$[Fe(NH_3)_6]^{-2}[Pt(CN)_6]^{-3}$$

Both A and B

$$[Pt[en]_3] [SO_4]_2$$

Solution

Question 6

$$KSCN$$

$$Fe(SCN)_3$$

$$Fe(SCN)_2$$

$$Fe(SCN)$$

Solution

Question 7

2

3

4

5

Solution

Question 8

same as primary valency

sum of primary and secondary valences

same as secondary valency

none of the above

Solution

Question 9

green, blue and red

blue, red and green

green, red and blue

red, blue and green

Solution

Question 10

$$[Cu(H_2O)_5]SO_4$$

$$[Cu(H_2O)_3SO_4]\cdot 2H_2O$$

$$[CuSO_4\cdot H_2O]\cdot 4H_2O$$

$$[Cu(H_2O)_4]SO_4\cdot H_2O$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.