d and f Block Elements

JEE Chemistry · 278 questions · Page 18 of 28 · Click an option or "Show Solution" to reveal answer

Q171

Wilkinson catalyst is : (Et = C2H5)

A [(Ph3P)3 RhCl]

B [(Et3P)3RhCl]

C [(Et3P)3IrCl]

D [(Ph3P)3IrCl]

Correct Answer

Option A

Solution

Wilkinsion catalyst is [(Ph3P)3RhCl]

Q172

The incorrect statement is :

A In manganate and permanganate ions, the p-bonding takes place by overlap of p-orbitals of oxygen and d-orbitals of manganese

B Manganate ion is green in colour and permanganate ion in purple in colour

C Manganate and permanganate ions are paramagnetic

D Manganate and permanganate ions are tetrahedral

Correct Answer

Option C

Solution

To identify the incorrect statement, let's analyze each option : Option A : "In manganate and permanganate ions, the π-bonding takes place by overlap of p-orbitals of oxygen and d-orbitals of manganese." This statement is true.

In both manganate ${MnO}_4^{2-}$ and permanganate ${MnO}_4^-$ ions, the manganese atom is in a high oxidation state (+6 and +7, respectively) and forms bonds with oxygen atoms involving the overlap of p-orbitals of oxygen and d-orbitals of manganese.

Option B : "Manganate ion is green in colour and permanganate ion in purple in colour." This statement is true.

Manganate ions are indeed green and permanganate ions are purple in aqueous solution.

Option C : "Manganate and permanganate ions are paramagnetic." This statement is incorrect.

Both manganate and permanganate ions are diamagnetic, not paramagnetic.

In these ions, manganese is in +6 and +7 oxidation states, respectively, and has no unpaired electrons.

Manganate(

\mathop {Mn}\limits^{ + 6}

O4 2–) , Mn(25) = [Ar]3d64s2 Mn+6(25) = [Ar]3d1 $\to$ Paramagnetic Permanganate (

\mathop {Mn}\limits^{ + 7}

O4-) Mn(25) = [Ar]3d64s2 Mn+7(25) = [Ar]3d0 $\to$ Diamagnetic Option D : "Manganate and permanganate ions are tetrahedral." This statement is true.

Both manganate and permanganate ions have a tetrahedral geometry, with the manganese atom at the center and the four oxygen atoms surrounding it.

Therefore, the incorrect statement is option C.

Q173

Given below are two statements : Statement I : Colourless cupric metaborate is reduced to cuprous metaborate in a luminous flame. Statement II : Cuprous metaborate is obtained by heating boric anhydride and copper sulphate in a non-luminous flame. In the light of the above statements, choose the most appropriate answer from the options given below.

A Statement I is true but Statement II is false

B Statement I is false but Statement II is true

C Both Statement I and Statement II are false

D Both Statement I and Statement II are true

Correct Answer

Option C

Solution

In presence of luminous flame, blue cupric metaborate is reduced to colourless cuprous metaborate.

If non-luminous flame is present in reaction, cupric metaborate is obtained by heating boric anhydride with copper sulphate.

So, both statements are false.

Q174

The incorrect statement among the following is :

A RuO4 is an oxidizing agent

B Cr2O3 is an amphoteric oxide.

C VOSO4 is a reducing agent

D Red colour of ruby is due to the presence of Co3+

Correct Answer

Option D

Solution

Red colour of ruby is due to presence of ${Cr}^{3+}$ (chromium ions), not ${Co}^{3+}$ (cobalt ions). in Al2O3.

Chromium is the trace element that causes ruby’s red colour, which ranges from an orange red to a publish red.

The strength of ruby’s red depends on how much chromium is present.

Q175

Given below are two statements : Statement I : CeO2 can be used for oxidation of aldehydes and ketones. Statement II : Aqueous solution of EuSO4 is a strong reducing agent. In the light of the above statements, choose the correct answer from the options given below :

A Statement I is true but Statement II is false.

B Statement I is false but Statement II is true

C Both Statement I and Statement II are false

D Both Statement I and Statement II are true

Correct Answer

Option D

Solution

The +3 oxidation state of lanthanide is most stable and therefore lanthanide in +4 oxidation state has strong tendence to gain e– and converted into +3 and therefore act as strong oxidizing agent. eg Ce+4 And therefore CeO2 is used to oxidized alcohol aldehyde and ketones.

Lanthanide in +2 oxidation state has strong tendency to loss e– and converted into +3 oxidation state therefore act as strong reducing agent.

$\therefore$ EuSO4 act as a strong reducing agent.

Q176

In which of the following pairs, the outer most electronic configuration will be the same?

A Cr+ and Mn2+

B Ni2+ and Cu+

C V2+ and Cr+

D Fe2+ and Co+

Correct Answer

Option A

Solution

Cr+ $\to$ [Ar]3d5 Mn2+ $\to$ [Ar]3d5

Q177

Which one of the following lanthanoids does not form MO2? [M is lanthanoid metal]

A Nd

B Dy

C Yb

D Pr

Correct Answer

Option C

Solution

Nd (60) = 4f4 6s2 Pr (59) = 4f3 6s2 Dy (66) = 4f10 6s2 Yb (70) = 4f14 6s2 Yb+2 has fully-filled 4f orbital, it will require very large amount of energy to reach +4 oxidation state.

Q178

Given below are two statement : one is labelled as Assertion A and the other is labelled as Reason R : Assertion A : Size of Bk3+ ion is less than Np3+ ion. Reason R : The above is a consequence of the lanthanoid contraction. In the light of the above statements, choose the correct answer from the options given below :

A Both A and B are true and R is the correct explanation of A

B A is false but R is true

C A is true but R is false

D Both A and B are true but R is not the correct explanation of A

Correct Answer

Option C

Solution

Size of Bk3+ is 98 pm Size of Np3+ is 101 pm So size of Np3+ is more than Bk3+ ion. there is a gradual decrease in the size of M3+ ions across the series.

This may be referred to as the actinoid contraction.

Q179

Fex2 and Fey3 are known when x and y are :

A x = F, Cl, Br, I and y = F, Cl, Br, I

B x = F, Cl, Br and y = F, Cl, Br, I

C x = F, Cl, Br, I and y = F, Cl, Br

D x = Cl, Br, I and y = F, Cl, Br, I

Correct Answer

Option C

Solution

FeI3 does not exist as I– reduces Fe3+ to Fe2+. 2Fe3+ + 2I – $\to$ 2Fe2+ + I2.

Q180

Arrange the following metal complex/compounds in the increasing order of spin only magnetic moment. Presume all the three, high spin system. (Atomic numbers Ce = 58, Gd = 64 and Eu = 63) (a) (NH4)2[Ce(NO3)6] (b) Gd(NO3)3 and (c) Eu(NO3)3

A (a) < (b) < (c)

B (b) < (a) < (c)

C (a) < (c) < (b)

D (c) < (a) < (b)

Correct Answer

Option C

Solution

(A)

{}_{58}Ce \to [Xe]4{f^2}5{d^0}6{s^2}

In complex

C{e^{ + 4}} \to [Xe]4{f^0}5{d^0}6{s^0}

There is no unpaired electron, so, $\mu$ m = 0 BM (B)

{}_{64}Gd \to [Xe]4{f^7}5{d^1}6{s^2}

In complex,

{}_{64}G{d^{2 + }} \to [Xe]4{f^7}5{d^1}

There are 8 unpaired electrons. $\therefore$

{\mu _m} = \sqrt {8(8 + 2)} = \sqrt {80} = 8.94BM

(C)

{}_{63}Eu \to [Xe]4{f^9}5{d^0}6{s^0}

In complex

{}_{63}E{u^{ + 3}} \to [Xe]4{f^6}5{d^0}6{s^0}

contain six unpaired electrons. So,

{\mu _m} = \sqrt {6(6 + 2)} = \sqrt {48} BM = 6.93BM

Hence, order of spin only magnetic moment

B > C > A

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