Coordination Compounds Test

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Coordination Compounds Test - 16

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Question 1
1 / -0

The slight irregularities in size have been ascribed to:

A
crystal field effects

B
shapes of d orbitals

C
charge transfer

D
splitting of d orbitals

Solution

Only crystal field effects has been effected by size so irregularities in size have been ascribed by this.
Question 2
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$$[Co(NH_{3})_{4}(NO_{2})_{2}]Cl$$ exhibits:

A
linkage, ionization and geometrical isomerism

B
ionisation, geometrical and optical isomerism

C
linkage, geometrical and optical isomerism

D
linkage, ionization and optical isomerism

Solution

By changing $$Cl$$ and $$NO_2$$, it will show ionization isomerism
Due to $$-NO_2$$ and $$-ONO$$, it will show linkage isomerism.
Also, it show cis/trans isomerism.
But due to symmetry it will not show optical isomerism.
Question 3
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The correct order of ligands for writing the formula of complex compounds is _________.

A
neutral, anionic, cationic

B
anionic, neutral, cationic

C
anionic, cationic, neutral

D
cationic, neutral, anionic

Solution

The formula of a coordination complex is written in a different order than its name. The chemical symbol of the metal center is written first. The ligands are written next, with neutral ligands coming before anionic ligands. If there is more than one anion or neutral ligand, they are written in alphabetical order according to the first letter in their chemical formula.
Question 4
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Which of the following is considered to be ananticancer species :

A

B

C

D

Solution

Cisplatin i.e ($$cis-diamminedichloroplatinum$$) is considered as anticancer species. Hence the option B is correct
Question 5
1 / -0

Coordination compounds have great importance in biological systems. In this context which of the following statements is incorrect?

A
Chlorophylls are green pigments in plants and contain calcium

B
Haemoglobin is the red pigment of blood and contains iron

C
Cyanocobalamin is $$B_{12}$$ and contains cobalt

D
Carboxypeptidase-A is an enzyme and contains zinc

Solution
Question 6
1 / -0

The correct statement in respect of protein haemoglobin is that it :-

A
Maintains blood sugar level

B
Acts as an oxygen carrier in the blood

C
Forms antibodies and offers resistance to diseases

D
Functions as a catalyst for biological reactions

Solution

Haemoglobin is transports protein. It transports oxygen to the cells.
Enzymes are the proteins which act as a biological catalyst. The maintenance of blood sugar is done by the insulin and the various proteins in the form of the antibodies offer resistance to the disease.
Question 7
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In which of the compound given below contains more than one kind of hybridization ($$sp,sp^{2}$$, $$sp^{3}$$ ) for carbon:

A
$$CH_{3}CH_{2}CH_{2}CH_{3}$$

B
$$CH_{3}-CH=CH-CH_{3}$$

C
$$CH_{2}=CH-CH=CH_{2}$$

D
$$H-C\equiv C-H$$

Solution

(i) $$\overset{sp^3}{CH_3}-\overset{sp^3}{CH_2}-\overset{sp^3}{CH_2}-\overset{sp^3}{CH_3}$$

Only $$sp^3$$ hybridized carbon.

(ii) $$\overset{sp^3}{CH_3}-\overset{sp^2}{CH}=\overset{sp^2}{CH}-\overset{sp^3}{CH_3}$$

Both $$sp^2$$ and $$sp^3$$ hybridized carbon.

(iii) $$\overset{sp^2}{CH_2}=\overset{sp^2}{CH}-\overset{sp^2}{CH}=\overset{sp^2}{CH_2}$$

Only $$sp^2$$ hybridized carbon.

(iv) $$H- \underset{sp}{C}\equiv \underset{sp}{C}-H$$

Only $$sp$$ hybridized carbon.

Option B is correct.
Question 8
1 / -0

The neutral complex, diamine dibromo dichloro platinum (IV) is best represented as:

A
$$[Pt(NH_{3})_{2}Br_{2}Cl_{2}]$$

B
$$[PtCl_{2}Br_{2}(NH_{3})_{2}]$$

C
$$[PtBr_{2}Cl_{2}(NH_{3})_{2}]$$

D
$$[Pt(NH_{3})_{2}Cl_{2}Br_{2}]$$

Solution

In a complex compound, the neutral ligand is written first followed by an anionic ligand. But if there are more than one anionic ligand then their name is written in alphabetical order thus bromo is written first then chloro.
Question 9
1 / -0

Total numbers of hybrid orbitals present in $$CH_{3}-C\equiv CH$$ are:

A
$$8$$

B
$$6$$

C
$$4$$

D
$$2$$

Solution

Propyne has $$8$$ hybrid orbitals.
Question 10
1 / -0

$$CuSO_{4}$$ when reacts with $$KCN$$ forms $$CuCN$$ which is insoluble in water. It is soluble in excess of $$KCN$$, due to formation of the following complex:

A
$$K_{2}[Cu(CN)_{4}]$$

B
$$K_{3}[Cu(CN)_{4}]$$

C
$$CuCN_{2}$$

D
Cu$$[KCu(CN)_{4}]$$

Solution

Here, $$CuCN$$ is insoluble in water. This means $$Cu$$ in $$CuCN$$ is in $$+1$$ state. Because $$Cu$$ is unstable in water only when it is in $$+1$$ state.
Co-ordination sphere exactly contains $$4 \ CN^-$$ ions. So, its structure would be $$[Cu(CN)_4]^{x}$$
Calculating $$x$$, $$Cu$$ is $$+1$$ state
$$1 + 4(-1)=x$$
$$1-4=x$$
$$x=-3$$
So, $$[Cu(CN)_4]^{-3}$$
It combines with $$3K^+$$ ions to form $$K_3[Cu(CN)_4]$$