d and f Block Elements — Page 7 | JEE Chemistry

Q61

The correct electronic configuration and spin-only magnetic moment (BM) of Gd3+ (Z = 64), respectively, are :

A [Xe]5f7 and 8.9

B [Xe]4f7 and 7.9

C [Xe]5f7 and 7.9

D [Xe]4f7 and 8.9

Correct Answer

Option B

Solution

Gd3+ (Z = 64) = [Xe] 4f7 Magnetic moment ( $\mu$ ) =

\sqrt {n\left( {n + 2} \right)}

B.M =

\sqrt {7\left( {7 + 2} \right)}

= 7.9 B.M

Q62

The incorrect statement(s) among (a) - (c) is (are) (a) W(VI) is more stable than Cr(VI). (b) In the presence of HCl, permanganate titrations provide satisfactory results. (c) Some lanthanoid oxides can be used as phosphors.

A (a) and (b) only

B (a) only

C (b) only

D (b) and (c) only

Correct Answer

Option C

Solution

W(VI) is more stable than Cr(VI) Permanganate titrations in presence of HCl are unsatisfactory as HCl is oxidised to Cl2 Lanthanoid oxides are used as phosphors.

Q63

The common positive oxidation states for an element with atomic number 24, are :

A +1 and +3

B +2 to +6

C +1 to +6

D +1 and +3 to +6

Correct Answer

Option B

Solution

24Cr = [Ar]3d54s1 The element with atomic number 24 is Chromium (Cr).

Chromium commonly exhibits several positive oxidation states, which are due to the loss of electrons from both the 4s and 3d orbitals.

Here are the common oxidation states for Chromium : +2 (as in ${Cr}^{2+}$ ) : This is observed where the electron configuration is $[{Ar}] 3d^4$ +3 (as in ${Cr}^{3+} )$ : This is the most stable state, with the electron configuration $[ {Ar} ] 3d^3$ +6 (as in ${CrO}_4^{2-}$ and ${Cr}_2\text{O}_7^{2-} )$ : In these cases, chromium exhibits an oxidation state of +6.

Hence, the common positive oxidation states for Chromium are +2, +3, and +6.

Therefore, the correct option is : Option B : +2 to +6.

Q64

Match with

List - I		List - II
(a)	Chlorophyll	(i)	Ruthenium
(b)	Vitamin- ${B_{12}}$	(ii)	Platinum
(c)	Anticancer drug	(iii)	Cobalt
(d)	Grubbs catalyst	(iv)	Magnesium Choose the most appropriate answer from the options given below :

A (a) - (iii), (b) - (ii), (c) - (iv), (d) - (i)

B (a) - (iv), (b) - (iii), (c) - (ii), (d) - (i)

C (a) - (iv), (b) - (iii), (c) - (i), (d) - (ii)

D (a) - (iv), (b) - (ii), (c) - (iii), (d) - (i)

Correct Answer

Option B

Solution

Chlorophyll is a coordination compound of magnesium.

Vitamin B-12, cyanocobalamine is a coordination compound of cobalt.

Cisplatin is used as an anti-cancer drug and is a coordination compound of platinum.

Grubbs catalyst is a compound of Ruthenium.

Q65

The correct order of following 3d metal oxides, according to their oxidation numbers is : (a) CrO3, (b) Fe2O3, (c) MnO2, (d) V2O5, (e) Cu2O

A (d) > (a) > (b) > (c) > (e)

B (a) > (c) > (d) > (b) > (e)

C (a) > (d) > (c) > (b) > (e)

D (c) > (a) > (d) > (e) > (b)

Correct Answer

Option C

Solution

(a)

\mathop C\limits^{ + 6} r{O_3}

: Since there are three oxygen atoms each with an oxidation number of -2, the total oxidation number from the oxygen atoms is -6.

To make the compound neutral, Chromium (Cr) must therefore have an oxidation number of +6. (b)

\mathop {F{e_2}}\limits^{ + 3} {O_3}

: Here, there are three oxygen atoms, contributing a total oxidation number of -6.

To balance this, the total oxidation number for the two iron (Fe) atoms must be +6, meaning each Fe atom has an oxidation number of +3. (c)

\mathop {Mn}\limits^{ + 4} {O_2}

: With two oxygen atoms, the total oxidation number is -4.

Thus, the manganese (Mn) atom must have an oxidation number of +4 to balance this. (d)

\mathop {{V_2}}\limits^{ + 5} {O_5}

: In this case, there are five oxygen atoms for a total oxidation number of -10.

To balance this, the total oxidation number for the two vanadium (V) atoms must be +10, meaning each V atom has an oxidation number of +5. (e)

\mathop {C{u_2}}\limits^{ + 1} O

: Here, there's one oxygen atom with an oxidation number of -2.

To balance this, the total oxidation number for the two copper (Cu) atoms must be +2, meaning each Cu atom has an oxidation number of +1.

So, the order of the oxidation numbers from highest to lowest is : (a) CrO $_3$ > (d) V $_2$ O $_5$ > (c) MnO $_2$ > (b) Fe $_2$ O $_3$ > (e) Cu $_2$ O.

Therefore, the answer is Option C : (a) > (d) > (c) > (b) > (e).

Q66

The spin only magnetic moments (in BM) for free Ti3+, V2+ and Sc3+ ions respectively are : (At.No. Sc : 21, Ti : 22, V : 23)

A 3.87, 1.73, 0

B 1.73, 3.87, 0

C 1.73, 0, 3.87

D 0, 3.87, 1.73

Correct Answer

Option B

Solution

\mu = \sqrt {n(n + 2)} BM

\begin{array}{lll}{T{i^{ + 3}} = [Ar]3{d^1}} & {n = 1} & {\mu = 1.73\,BM} \\ {{V^{ + 2}} = [Ar]3{d^3}} & {n = 3} & {\mu = 3.87\,BM} \\ {S{c^{ + 3}} = [Ar]3{d^0}4{s^0}} & {n = 0} & {\mu = 0} \end{array}

Q67

Which one of the following metals forms interstitial hydride easily?

A Cr

B Fe

C Mn

D Co

Correct Answer

Option A

Solution

Elements of group 7, 8, 9 do not form hydrides thus Cr will only form hydride among the given elements (Fe, Mn, Co).

Q68

Which one of the lanthanoids given below is the most stable in divalent form?

A Ce (Atomic Number 58)

B Sm (Atomic number 62)

C Eu (Atomic Number 63)

D Yb (Atomic Number 70)

Correct Answer

Option C

Solution

The stability of a divalent state can often be attributed to the stability of the electron configuration of that state.

Among the options given, we can evaluate the electron configurations of each element in their divalent state to find out which is the most stable.

The electronic configurations of the neutral atoms are : Ce (Atomic Number 58) : $[Xe] 4f^1 5d^1 6s^2$ Sm (Atomic Number 62) : $[Xe] 4f^6 6s^2$ Eu (Atomic Number 63) : $[Xe] 4f^7 6s^2$ Yb (Atomic Number 70) : $[Xe] 4f^{14} 6s^2$ In their divalent state (+2 oxidation state), the electron configurations would be : Ce : $[Xe] 4f^1 5d^1$ (removal of the 2 electrons from the 6s orbital) Sm : $[Xe] 4f^6$ (removal of the 2 electrons from the 6s orbital) Eu : $[Xe] 4f^7$ (removal of the 2 electrons from the 6s orbital) Yb : $[Xe] 4f^{14}$ (removal of the 2 electrons from the 6s orbital) Among these, Eu in its divalent state has a half-filled (4f) shell, which offers extra stability due to symmetrical electron distribution.

Therefore, Eu (Atomic Number 63) is the most stable in the divalent form.

So the correct answer is Option C : Eu (Atomic Number 63).

Q69

Cerium (IV) has a noble gas configuration. Which of the following is correct statement about it?

A It will not prefer to undergo redox reactions.

B It will prefer to gain electron and act as an oxidizing agent.

C It will prefer to give away an electron and behave as reducing agent.

D It acts as both, oxidizing and reducing agent.

Correct Answer

Option B

Solution

Cerium exists in two different oxidation state

+3

,

+4

\begin{array}{ll} \mathrm{Ce}^{+4}+\mathrm{e}^{-} \rightarrow \mathrm{Ce}^{3+} & \mathrm{E}^{0}=+1.61 \mathrm{~V} \\ \mathrm{Ce}^{+3}+3 \mathrm{e}^{-} \rightarrow \mathrm{Ce} & \mathrm{E}^{0}=-2.336 \mathrm{~V} \end{array}

It shows

\mathrm{Ce}^{+4}

acts as a strong oxidising agent & accepts electron.

Q70

Given below are two statements. Statement I : Iron (III) catalyst, acidified

\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}

and neutral

\mathrm{KMnO}_{4}

have the ability to oxidise

\mathrm{I}^{-}

to

\mathrm{I}_{2}

independently. Statement II : Manganate ion is paramagnetic in nature and involves

\mathrm{p} \pi-\mathrm{p} \pi

bonding. In the light of the above statements, choose the correct answer from the options given below.

A Both Statement I and Statement II are true.

B Both Statement I and Statement II are false.

C Statement I is true but Statement II is false.

D Statement I is false but Statement II is true.

Correct Answer

Option B

Solution

Manganate ion $\mathrm{MnO}_{4}^{2-}$ has tetrahedral structure has only $\mathrm{d} \pi-\mathrm{p} \pi\,\, \pi$ -bonds. $\mathrm{Fe}^{3+}$ is not used as a catalyst in the conversion of I- to $\mathrm{I}_{2}$ by $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \cdot \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ oxidise $\mathrm{I}^{-}$ in acidic medium easily