The d- and f- Block Elements Test

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The d- and f- Block Elements Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

Chrome yellow is chemically known as

A
lead chromate

B
lead sulphate

C
lead iodide

D
basic lead acetate

Solution

Chrome yellow is lead(II) chromate (PbCrO4). It occurs naturally as the mineral crocoite but the mineral ore was never used as a pigment for paint.
lead cromate is correct answer
Question 2
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A blood red colour is obtained when ferric chloride solution reacts eith:

A
$$KCN$$

B
$$KSCN$$

C
q$$K_4[Fe(CN)_6]$$

D
$$K_3[Fe(CN)_6]$$

Solution

$$FeCl_3$$+3KSCN → blood red colour $$Fe(SCN)_3+3KCl$$
Question 3
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In the dichromate anion $$(Cr_{2}O_{7}^{2-}),$$

A
all $$Cr - O$$ bonds are equivalent

B
$$6 \,Cr - O$$ bonds are equivalent

C
$$3 \,Cr - O$$ bonds are equivalent

D
no bonds in $$Cr_{2}O_{7}^{2-}$$ are equivalent.

Solution
Question 4
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Which of the following statements is wrong?

A
$$Ti^{4+}$$ and $$Ag^{+}$$ are repelled by magnetic field

B
$$Mn^{2+}$$ shows maximum magnetic character among the first transition series.

C
$$Fe^{2+}$$ is more stable than $$Mn^{2+}$$ towards oxidation to $$+3$$ state.

D
$$Cr$$ in $$Cr_{2}O_{7}^{2-}$$ ion involves $$sp^{3}d^{2}$$ hybridisation

Solution
Question 5
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$$CuSO_{4}$$ is paramagnetic while $$ZnSO_{4}$$ is diamagnetic because

A
$$Cu^{2+}$$ ion has $$3d^{9}$$ configuration while $$Zn^{2+}$$ ion has $$3d^{10}$$ configuration

B
$$Cu^{2+}$$ ion has $$3d^{5}$$ configuration while $$Zn^{2+}$$ ion has $$3d^{6}$$ configuration

C
$$Cu^{2+}$$ has half filled orbitals while $$Zn^{2+}$$ has fully filled orbitals

D
$$CuSO_{4}$$ is blue in colour while $$ZnSO_{4}$$ is white.

Solution
Question 6
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Which of the following complex ion has a magnetic moment same as $$[Cr(H_2O)_6]^{3+}$$?

A
$$[Mn(H_2O)_6]^{4+}$$

B
$$[Mn(H_2O)_6]^{3+}$$

C
$$[Fe(H_2O)_6]^{3+}$$

D
$$[Cu(H_3H)_4]^{2+}$$

Solution

compound $$[Mn(H_2O)_6]^{4+}$$ have same properties to like magnetic moments like $$[Cr(H_2O)_6]^3+$$
Question 7
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These questions consists of two statements each, printed as assertion and reason, while answering these questions you are required to choose any one of the following responses.

Assertion : $$CrO_3$$ reacts with $$HCl$$ to form thornyl chloride gas.
Reason : Chromyl chloride $$(CrO_2Cl_2)$$ has tetrahedral shape.

A
If both assertion and reason are true and the reason is a correct explanation of assertion

B
If both assertion and reason are true but reason is not a correct explanation of assertion

C
If assertion is true but the reason is false

D
If assertion is false but the reason is true

Solution

Assertion : $$CrO_3$$ reacts with $$HCl$$ to form thornyl chloride gas.
Reason : Chromyl chloride $$(CrO_2Cl_2)$$ has tetrahedral shape.
both are true but not show correct explanation
Question 8
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Transition metal and their compounds are used as catalysts in industry and in biological system. For example, in the contact process, vanadium compounds in the $$+5$$ state $$(V_2O_5$$ or $$VO_3^$$) are used to oxidise $$SO_2$$ to $$SO_3$$:
$$SO_2 + \frac{1}{2}O_2 \overset{V_2O_5}{\rightarrow}SO_3$$
It is thought that the actual oxidation process takes place in two stages. In the first step, $$V^{5+}$$ in the presence of oxide ions converts $$SO_2$$ to $$SO_3$$. At the same time, $$V^{5+}$$ is reduced to $$V^{4+}$$. $$2V^{5+}+O^{2-} + SO_2 2V^{4+}+ SO_3$$. In the second step, $$V^{5+}$$ is regenerated from $$V^{4+}$$ by oxygen :
$$2V^{4+}+\frac{1}{2}O_2 \rightarrow 2V^{5+} + O^{2-}.$$
The overall process is, of course, the sum of these two steps:
$$SO_2 + \frac{1}{2}O_2 \rightarrow SO_3$$

During the course of the reaction

A
catalyst undergoes changes in oxidation state

B
catalyst increases the rate constant

C
catalyst is regenerated in its original form when the reactants form the products

D
all are correct

Solution

during this reaction
1, catalyst undergoes changes in oxidation state
2, catalyst increases the rate constant
3, catalyst is regenerated in its original form when the reactants form the products
all are A,B , C is correct
Question 9
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Amongst $$Tif_{6}^{2-}, CoF_{6}^{3-}, Cu_2Cl_2 and NiCl_{4}^{2-}$$ the colourless species are:

A
$$CoF_{6}^{3-}$$ and $$NiCl_{4}^{2-}$$

B
$$TiF_{6}^{2-}$$ and $$CoF_{6}^{3-}$$

C
$$Cu_2Cl_2$$ and $$NiCl_{4}^{2-}$$

D
$$TiF_{6}^{2-}$$ and $$Cu_2Cl_2$$

Solution

$$TiF_{6}^{2-}$$ and $$Cu_2Cl_2$$ these are two molecule show coloueless species
Question 10
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A larger number of oxidation states are exhibited by the actinides than those by the lanthanides. The main reason being:

A
more reactive nature of the actinides than the lanthanides

B
$$4f$$ orbitals are more diffused than the $$5f$$ orbitals

C
lesser energy difference between $$5f$$ and $$6d$$ orbitals than between $$4f$$ and $$5d$$ orbitals

D
more energy difference between $$5f$$ and $$6d$$ orbitals than between $$4f$$ and $$5d$$ orbitals

Solution

A larger number of oxidation states are exhibited by the actinides than those by the lanthanides because energy difference between $$5f$$ and $$6d$$ orbitals is less than that between $$4f$$ and $$5d$$ orbitals.

Hence, excitation of electrons can take place easily in actinides and thus, they exhibt a large number of oxidation states.

So, option C is correct.

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Re-Attempt Weekly Quiz Competition

The d- and f- Block Elements Test - 56

Result

The d- and f- Block Elements Test - 56

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SHARING IS CARING

Question 1

lead chromate

lead sulphate

lead iodide

basic lead acetate

Solution

Question 2

$$KCN$$

$$KSCN$$

q$$K_4[Fe(CN)_6]$$

$$K_3[Fe(CN)_6]$$

Solution

Question 3

all $$Cr - O$$ bonds are equivalent

$$6 \,Cr - O$$ bonds are equivalent

$$3 \,Cr - O$$ bonds are equivalent

no bonds in $$Cr_{2}O_{7}^{2-}$$ are equivalent.

Solution

Question 4

$$Ti^{4+}$$ and $$Ag^{+}$$ are repelled by magnetic field

$$Mn^{2+}$$ shows maximum magnetic character among the first transition series.

$$Fe^{2+}$$ is more stable than $$Mn^{2+}$$ towards oxidation to $$+3$$ state.

$$Cr$$ in $$Cr_{2}O_{7}^{2-}$$ ion involves $$sp^{3}d^{2}$$ hybridisation

Solution

Question 5

$$Cu^{2+}$$ ion has $$3d^{9}$$ configuration while $$Zn^{2+}$$ ion has $$3d^{10}$$ configuration

$$Cu^{2+}$$ ion has $$3d^{5}$$ configuration while $$Zn^{2+}$$ ion has $$3d^{6}$$ configuration

$$Cu^{2+}$$ has half filled orbitals while $$Zn^{2+}$$ has fully filled orbitals

$$CuSO_{4}$$ is blue in colour while $$ZnSO_{4}$$ is white.

Solution

Question 6

$$[Mn(H_2O)_6]^{4+}$$

$$[Mn(H_2O)_6]^{3+}$$

$$[Fe(H_2O)_6]^{3+}$$

$$[Cu(H_3H)_4]^{2+}$$

Solution

Question 7

If both assertion and reason are true and the reason is a correct explanation of assertion

If both assertion and reason are true but reason is not a correct explanation of assertion

If assertion is true but the reason is false

If assertion is false but the reason is true

Solution

Question 8

catalyst undergoes changes in oxidation state

catalyst increases the rate constant

catalyst is regenerated in its original form when the reactants form the products

all are correct

Solution

Question 9

$$CoF_{6}^{3-}$$ and $$NiCl_{4}^{2-}$$

$$TiF_{6}^{2-}$$ and $$CoF_{6}^{3-}$$

$$Cu_2Cl_2$$ and $$NiCl_{4}^{2-}$$

$$TiF_{6}^{2-}$$ and $$Cu_2Cl_2$$

Solution

Question 10

more reactive nature of the actinides than the lanthanides

$$4f$$ orbitals are more diffused than the $$5f$$ orbitals

lesser energy difference between $$5f$$ and $$6d$$ orbitals than between $$4f$$ and $$5d$$ orbitals

more energy difference between $$5f$$ and $$6d$$ orbitals than between $$4f$$ and $$5d$$ orbitals

Solution

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