Coordination Compounds Test

Result

Coordination Compounds Test - 32

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

$$NiCl^+_2\overset{KCN}{\underset{excess}{\rightarrow}}Complex(A)$$
$$NiCl^+_2\overset{KCl}{\underset{excess}{\rightarrow}}Complex (B)$$
The coordination number of nickel in both complexes (A) and (B) is $$4$$.
The IUPAC name of the complex (A) and (B) are:

A
Potassium tetracyanonicklate (II) and Potassium tetrachloronicklate (II)

B
Potassium tetracyanonickel (II) and Potassium tetrachloronickel (II)

C
Potassium cyanonicklate (II) and Potassium chloronicklate (II)

D
Potassium cyanonickel (II) and Potassium Chloronickel (II)

Solution
Question 2
1 / -0

The coordination number and oxidation number of the central metal ion in the complex $$[Pt(en)_2]^{+2}$$ is:

A
C.N. $$=2$$, O.N. $$=+2$$

B
C.N. $$=6$$, O.N. $$=+4$$

C
C.N. $$=4$$, O.N. $$=+4$$

D
C.N. $$=4$$, O.N. $$=+2$$

Solution

Ethylenediamine which is bidentate ligand is of two in number hence 2X2 =4
Hence co-ordination number of Platinum will be 4
Oxidation state = [x + 2 (0)] = 2+
x =+2
Thus, coordination number is 4 and oxidation number is +2
Question 3
1 / -0

How many of the following metals when heated in an atmosphere of $$N_2$$ gas form nitrides?

Li, Na, K, Rb, Cs, Mg, Ca, Sr, Ba

A
9

B
5

C
3

D
6
Question 4
1 / -0

$$FeCl_3. 4H_2O$$ is actually:

A
$$[Fe(H_2O)_4]Cl_3$$

B
$$[Fe(H_2O)_3Cl]Cl_2.H_2O$$

C
$$[Fe(H_2O)_4Cl_2]Cl$$

D
$$[Fe(H_2O)_3Cl_2]Cl.H_2O$$

Solution

$$Fe{Cl}_{3}$$.$$4{H}_{2}O$$ is actually $$[Fe({H}_{2}O)_{4}]{Cl}_{3}$$.

The water molecules are coordinated with the iron atom, are inside the coordination sphere whereas the chloride ions are outside the Ionization sphere.
Hence, the correct option is A
Question 5
1 / -0

Which of the following is $$correct\ IUPAC$$ name for coordination isomer of $$[Pt(NH_{3})_{4}][Pd(C_{2}O_{4})_{2}]$$?

A
Diammine (oxalato) platinum (II) diammine (oxalato) paladate (II)

B
Diammine (oxalato) platinum (II) diammine (oxalato) paladium (II)

C
Tetraammineplatinum (IV) bis(oxalato) paladate (II)

D
Tetraammineplatinum (IV) bis(oxalato) paladium (II)

Solution

Rules for writing IUPAC name in Coordination Chemistry:-
Rule 1- While naming a coordination compound cation is always named before the anion irrespective of the fact that whether complexion is cation or anion.
Rule 2- If there are more than one type of ligands present in any coordination compound, the name of the ligands is to be done in alphabetic order followed by the name of central metal atom/ion.
Name of anionic ligands ends with $$O$$.
For neutral ligands, the common name is used.
Rule 3- If the name of ligand already contains a numerical prefix, then the terms, bis, tris, tetrakis are used and the ligand to which they refer to being placed in parentheses.
Rule 4- After naming the ligand in alphabetical order name of the central metal atom/ion is written.
If the complex is an anion, the name of metal ends with the suffix -ate for Latin name.
Rule 5- The oxidation state of the metal in the complex is given as a Roman numeral in parentheses.
Rule 6- The neutral complex molecule is named similar to that of the complex cation.
Rule 7- These are some ligands that may be attached to the central metal atom/ion through different atoms. For eg- $$NO_2,CN,$$etc.
Like,
$$M-NO\longrightarrow$$ nitro
$$M-ONO^2\longrightarrow$$ nitrito
$$M-SCN\longrightarrow$$ thiocyanato
$$M-NCS\longrightarrow$$ isothiocyanato

IUPAC name of $$[Pt(NH_3)_4][Pd(C_2O_4)_2]$$
Tetraamine platinum (IV) bis-oxalatopaladate (II)
Question 6
1 / -0

Consider the reaction.
$$[Ni(H_2O)_6]^{2+}(aq)+3en(aq) \to [Ni(en)_3]^{2+}+2H_2O$$
Which of the following property of compound is changed in reaction?

A
Co-ordination number

B
Geometry of complex

C
Hybridization of central atom

D
Wavelength of absorbed light

Solution

In the given reaction the only property that is changing is the wavelength of absorbed light because at the reactant side the complex have water as a ligand which is a weak field ligand on the other side the the product formed has the ethylenediammine ligand attached to metal which is a strong field ligand. As the strength of the ligand changes the $$\Delta_o$$ value changes and hence the wavelegth which is indirectly proportional to it also changes. Hence D is the correct option
Question 7
1 / -0

Which of the following complex ion is expected to absorb light in the visible region?

A
$$[Ti(en)_{3}]^{4+}$$

B
$$[Sc(NH_{3})_{4}(H_{2}O)_{2}]^{3+}$$

C
$$[CrO_{4}]^{2-}$$

D
$$[Zn(CN)_{2}]^{2-}$$

Solution

Only those transition metal complexes are expected to absorb visible light, in which d- subshell is incomplete and excitation of an electron from a lower energy orbital to higher energy orbital is possible.

(a) In $$\left[ Sc({ H }_{ 2 }O)({ NH }_{ 3 })_{ 3 } \right] ^{ 3+ }$$ Sc is present in $$ { Sc }^{ 3+ }$$=$$[Ar]3{ d }^{ 0 },4{ s }^{ 0 }$$

Since in this complex excitation of the electron is not possible, it will not absorb visible light.

(b) $$\left[ Ti(en)_{ 2 }(N{ H }_{ 3 })_{ 2 } \right] ^{ 4+ },$$ Ti is present as $${ Ti }^{ 4+ }$$

$${ Ti }^{ 4+ }=[Ar]3{ d }^{ 0 },4{ s }^{ 0 }$$
Hence it will not absorb visible light

(c) In $$\left[ Cr({ NH }_{ 3 })_{ 6 } \right] ^{ 3+ }$$ Cr is present as $${ Cr }^{ 3+ }$$

$${ Cr }^{ 3+ }$$=$$[Ar]3{ d }^{ 3 },4{ s }^{ 0 }$$

Since this complex has three unpaired electrons excitation of electrons is possible and thus, it is expected that this complex will absorb visible light.

(d) In $$\left[ Zn(N{ H }_{ 3 })_{ 6 } \right] ^{ 2+ }$$ Zn is present as $${ Zn }^{ 2+ }$$

$${ Zn }^{ 2+ }$$=$$[Ar]3{ d }^{ 10 },4{ s }^{ 0 }$$

Hence , this complex will not absorb visible light.

The correct option is C.
Question 8
1 / -0

In the compound $$CoCl_{3}.5NH_{3}$$:

A
all the $$Cl$$ show primary valency only

B
Two $$Cl$$ show primary valency and one $$Cl$$ shows secondary valency

C
Two $$Cl$$ show primary valency and one $$Cl$$ shows primary valency as well as secondary valency

D
All the $$Cl$$ show secondary valency

Solution

In the given coordination complex 5 $$NH_3$$ molecule are directly attached to metal whereas only one one Cl is directly attached to metal and rest 2 are satisfying the oxidation state of metal and called as primary valency of the metal.Therefore it can be seen from the image and the formula of the complex which is $$[Co(NH_3)_5Cl]Cl_2$$ that the one Cl is acting as a secondary valency and 2 Cl act as primary valency.Hence option B is correct.
Question 9
1 / -0

Which of the following will given maximum number of isomers?

A
$$[Co(NH_3)_4 Cl_2]^+$$

B
$$[Ni(en)(NH_3)_4]^{2+}$$

C
$$[Cr(SCN)_2(NH_3)^+_4]$$

D
$$[Ni(C_2O_4)(en)_2]^{2+}$$

Solution

It can cis-trans isomers and also has stereo isomers in the trans state while others have only two isomeric structures.
Question 10
1 / -0

If a complex $$[M(H_2O)_6]^{2+}$$ absorbs energy corresponding to $$\lambda$$ nm, then the wavelength (in nm) absorbed by tetrahedral, complex $$[m(H_2O)_4]^{2+}$$ will be:

A
$$\frac{4}{9}\lambda$$

B
$$\frac{5}{9}\lambda$$

C
$$\frac{9}{4}\lambda$$

D
$$\frac{1}{\lambda}$$

Solution

In coordination complex Energy is directly related to $$\Delta$$ and inversely to $$ \lambda $$. and it is known that $$\Delta_t$$ =(4/9)$$\Delta_o$$ then,$$1/\lambda_t$$ =(4/9)$$\frac{1}{\lambda _o}$$ Hence,$$\lambda _t = \frac{9}{4}\lambda _o$$. this means The wavelegth absorbed by tetrahedral comlex will be 9/4$$ \lambda $$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Coordination Compounds Test - 32

Result

Coordination Compounds Test - 32

Score

-

Rank

-

SHARING IS CARING

Question 1

Potassium tetracyanonicklate (II) and Potassium tetrachloronicklate (II)

Potassium tetracyanonickel (II) and Potassium tetrachloronickel (II)

Potassium cyanonicklate (II) and Potassium chloronicklate (II)

Potassium cyanonickel (II) and Potassium Chloronickel (II)

Solution

Question 2

C.N. $$=2$$, O.N. $$=+2$$

C.N. $$=6$$, O.N. $$=+4$$

C.N. $$=4$$, O.N. $$=+4$$

C.N. $$=4$$, O.N. $$=+2$$

Solution

Question 3

9

5

3

6

Question 4

$$[Fe(H_2O)_4]Cl_3$$

$$[Fe(H_2O)_3Cl]Cl_2.H_2O$$

$$[Fe(H_2O)_4Cl_2]Cl$$

$$[Fe(H_2O)_3Cl_2]Cl.H_2O$$

Solution

Question 5

Diammine (oxalato) platinum (II) diammine (oxalato) paladate (II)

Diammine (oxalato) platinum (II) diammine (oxalato) paladium (II)

Tetraammineplatinum (IV) bis(oxalato) paladate (II)

Tetraammineplatinum (IV) bis(oxalato) paladium (II)

Solution

Question 6

Co-ordination number

Geometry of complex

Hybridization of central atom

Wavelength of absorbed light

Solution

Question 7

$$[Ti(en)_{3}]^{4+}$$

$$[Sc(NH_{3})_{4}(H_{2}O)_{2}]^{3+}$$

$$[CrO_{4}]^{2-}$$

$$[Zn(CN)_{2}]^{2-}$$

Solution

Question 8

all the $$Cl$$ show primary valency only

Two $$Cl$$ show primary valency and one $$Cl$$ shows secondary valency

Two $$Cl$$ show primary valency and one $$Cl$$ shows primary valency as well as secondary valency

All the $$Cl$$ show secondary valency

Solution

Question 9

$$[Co(NH_3)_4 Cl_2]^+$$

$$[Ni(en)(NH_3)_4]^{2+}$$

$$[Cr(SCN)_2(NH_3)^+_4]$$

$$[Ni(C_2O_4)(en)_2]^{2+}$$

Solution

Question 10

$$\frac{4}{9}\lambda$$

$$\frac{5}{9}\lambda$$

$$\frac{9}{4}\lambda$$

$$\frac{1}{\lambda}$$

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.