Coordination Compounds Test

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Coordination Compounds Test - 66

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Weekly Quiz Competition

Question 1
1 / -0

What is the valency of Fe which does not ionize in $${ K }_{ 3 }[Fe({ OX }_{ 3 })]$$?

A
Three

B
Four

C
Six

D
One
Question 2
1 / -0

$$NCl_3$$ and $$BCl_3$$ are mixed in a container. Which the following is true for hybridisation of $$N$$ and $$B$$ the product formed by bonding of $$NCl_3$$ and $$BCl_3$$?

A
Hybridisation of $$N$$ changes but for $$B$$ it remains same

B
Hybridisation of $$B$$ changes but for $$N$$ it remains same

C
Hybridisation of both $$N$$ and $$B$$ change

D
Hybridisation of neither $$N$$ nor $$B$$ changes
Question 3
1 / -0

Which of the following represent $$sp - sp - sp^2 - sp - sp^2 - sp^3$$ hybridization from left to right order?

A
$$CH_2= CH - C = \overset{H}{\overset{|}{C}} - C \equiv N$$

B
$$CH \equiv C - CH = C = CH-CH_3$$

C
$$N \equiv C - \overset{H}{\overset{|}{C}} - \underset{H}{\underset{|}{C}} - CH - CH_3$$

D

Solution

Carbon having single bond is $$sp^3$$ hybridised, carbon with double bond is $$sp^2$$ hybridised and the carbon having triple bond is $$sp$$ hybridised.
In option B $$CH$$ is $$sp$$ hybridised, $$C$$ is $$sp$$ hybridised, $$CH$$ is $$sp^2$$ hybridised, $$C$$ is $$sp$$ hybridised, $$CH$$ is $$sp2$$ hybridised and $$CH_3$$ is $$sp^3$$ hybridised.
Question 4
1 / -0

In $$CuSO_4.5H_2O$$ how many molecules of water are indirectly connected to Cu?

A
$$5$$

B
$$4$$

C
$$2$$

D
$$1$$

Solution

Solution:- (D) $$1$$
In $$CuSO_4.5H_2O$$, Four water molecules form coordinate bond with $$Cu_{2+}$$ ion while one water molecule is associated with $$H$$- bond
Question 5
1 / -0

Which compound to use for treatment of tumor?

A
$$Cis\left[ Pd{ Cl }_{ 2 }{ \left( { NH }_{ 3 } \right) }_{ 2 } \right] $$

B
$$Trans\left[ Pd{ Cl }_{ 2 }{ \left( { NH }_{ 3 } \right) }_{ 2 } \right] $$

C
$$Cis\left[ Pt{ Cl }_{ 2 }{ \left( { NH }_{ 3 } \right) }_{ 2 } \right] $$

D
$$Trans\left[ Pt{ Cl }_{ 2 }{ \left( { NH }_{ 3 } \right) }_{ 2 } \right] $$

Solution

The given compound is cis plating which is used in the treatment of cancer or tumor.
Question 6
1 / -0

When two or more complexes are combined, it is

A
Pattern

B
Theme

C
Complex

D
All of these
Question 7
1 / -0

The compound that inhibits the growth of tumors is:

A
$$cis-[Pd(Cl)_2 (NH_3)_2]$$

B
$$cis-[Pt(Cl)_2 (NH_3)_2]$$

C
$$trans-[Pt(Cl)_2 (NH_3)_2]$$

D
$$trans-[Pd(Cl)_2 (NH_3)_2]$$

Solution

Solution:- (B) cis-$$\left[ Pt({Cl})_{2} {\left( N{H}_{3} \right)}_{2} \right]$$

$$cis-[Pt(Cl)_2 (NH_3)_2]$$ is used in chemotherapy to inhibits the growth of tumors.
Question 8
1 / -0

The complex ion that will lose its crystal field stabilization energy upon oxidation of its metal to $$+3$$ state is:

A
$$[Fe(phen)_3]^{2+}$$

B
$$[Zn(phen)_3]^{2+}$$

C
$$[Ni (phen)_3]^{2+}$$

D
$$[Co(phen)_3]^{2+}$$

Solution

By oxidation of $$Fe^{2+}$$ into $$Fe^{3+}$$, the CFSE value decrease.

(2) $$[Zn(phen)_3]_2\xrightarrow {-e^−}[Zn(phen)_3]^{3+}$$

$$Zn^{2+}$$: $$3d^{10}$$ $$Zn^{3+}$$: $$3d^9$$

C.F.S.E $$=0$$ C.F.S.E=$$−0.6\Delta _o$$

By oxidation of $$Zn^{2+}$$ into $$Zn^{3+}$$ , the CFSE value increase.

(3) $$[Ni(phen)_3]^{2+} \xrightarrow {-e^−}[Ni(phen)_3]^{3+}$$

$$Ni^{2+}$$: $$3d^8$$ $$Ni^{3+}$$: $$3d^7$$

C.F.S.E $$=−1.2{\Delta}_0$$ C.F.S.E $$=−1.8{\Delta}_0$$

By oxidation of $$Ni^{2+}$$ into $$Ni^{3+}$$ , the CFSE value increase.

(4) $$[Co(phen)_3]^{2+} \xrightarrow {-e^−}[Co(phen)_3]^{3+}$$

$$Co^{2+}$$: $$3d^7$$ $$Co^{3+}$$: $$3d^6$$

$$C.F.S.E=−1.8{\Delta}_0$$ $$C.F.S.E=−2.4{\Delta}_0$$

By oxidation of $$Co^{2+}$$ into $$Co^{3+}$$ , the CFSE value increase.
Question 9
1 / -0

Complete removal of both the axial ligands (along the z-axis) from an octahedral complex leads to which of the following splitting patterns? (relative orbital energies not on scale).

A

B

C

D

Solution

If both ligands present along z-axis removed from octahedral field and converted into a square planar field, then the energy level of orbitals is shown.
Question 10
1 / -0

The degenerate orbitals of $${ \left[ Cr{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ 3+ }$$ are:

A
$${d}_{yz}$$ and $${d}_{{z}^{2}}$$

B
$${d}_{{z}^{2}}$$ and $${x}_{xz}$$

C
$${d}_{xz}$$ and $${d}_{yz}$$

D
$${d}_{{x}^{2}-{y}^{2}}$$ and $${d}_{xy}$$

Solution

Degenerate orbitals of $$[Cr(H_2O)_6]^{3+}$$ is shown in the diagram.

Hence according to the options given, degenerate orbitals are $$d_{xz}$$ and $$d_{yz}$$.

So, C option is correct.