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Block D and F Elements

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Block D and F Elements Class 12th

The total number of Mn = O bonds in Mn₂O₇ is________.

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alok kumar singh

Contributor-Level 10

Number of Mn = O bonds = 6

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Block D and F Elements Class 12th

Given below are two statements.

Statement I: Iron (III) catalyst, acidified K₂Cr₂O₇ and neutral KMnO₄ have the ability to oxidize I^- to I₂ independently.

Statement II: Magnate ion is paramagnetic in nature and involves pp-pp bonding

In the light of the above statements, choose the most appropriate answer from the options given below.

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alok kumar singh

Contributor-Level 10

Neutral KMnO₄ oxidises I^- to and magnate ion has dpPp bonding

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Block D and F Elements Class 12th

Match List-I with List-II

Choose the correct answer from the options given below

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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Block D and F Elements Class 12th

Which of the following 3d- metal ion will give the lowest enthalpy of hydration $(Δ_{h y d} H)$ when dissolved in water?

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alok kumar singh

Contributor-Level 10

Following are the $Δ H °$ hydration of given ions

Cr²⁺ - 1926 KJ/mole

Mn²⁺ - 1860 KJ/mole

Fe²⁺ - 2000 KJ/mole

Co²⁺ - 2078 KJ/mole

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Block D and F Elements Class 12th

The oxidation state of manganese in the product obtained in a reaction of potassium permanganate and hydrogen peroxide in basic medium is___________.

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Vishal Baghel

Contributor-Level 10

$2 K M n O_{4} + 3 H_{2} O_{2} \to_{}^{B a s i c m e d i u m} 2 M n O_{2} + 3 O_{2} + 2 H_{2} O + 2 K O H$

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Block D and F Elements Class 12th

The spin-only magnetic moment value of M³+ ion (in gaseous state) from the pairs Cr³⁺/Cr²⁺, Mn³⁺/Mn²⁺, Fe³⁺/Fe²⁺ and Co³⁺/Co²⁺ that has negative standard electrode potential, is________ B.M [Nearest integer]

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Payal Gupta

Contributor-Level 10

In given electrodes, only $E_{C r^{3 +} / C r^{2 +}}^{0}$ is negative

Cr (24) – [Ar] 4s¹3d⁵4p⁰

Number of unpaired electrons = 3

$μ = \sqrt[]{n (n + 2)} B M = \sqrt[]{3 (3 + 2)} B M = 3.97 B M \approx θ 4 B M$

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Block D and F Elements Class 12th

Among Co³⁺, Ti²⁺, V²⁺ and Cr²⁺ ions, one if used as a reagent cannot liberate H₂ from dilute mineral acid solution, its spin-only magnetic moment in gaseous state is _________ B.M. (Nearest integer)

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Vishal Baghel

Contributor-Level 10

$C o^{3 +}$ $^{}$ will not get oxidized, so will not liberate H2 gas upon reaction with acidic solution.

No. of unpaired electron = 4

Spin only magnetic moment = $\sqrt[]{4 (4 + 2)} B M = \sqrt[]{24} B M \approx 5 B M$

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Block D and F Elements Class 12th

Consider the following metal complexes:

[Co(NH₃)₆]³⁺

[CoCl(NH₃)₅]²⁺

[Co(CN)₆]^3-

[Co(NH₃)₅(H₂O)]³⁺

The spin-only magnetic moment value of the complex that absorbes light with shortest wavelength is _____________. B.M (Nearest integer)

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Vishal Baghel

Contributor-Level 10

Complex absorbing minimum wavelength of light means the value of $Δ$ for it is maximum.

That complex will be ${[C o {(C N)}_{6}]}^{3 -}$ ${_{}}^{}$

$C o^{3 +} \Rightarrow 3 d^{6} 4 s^{0} \Rightarrow t_{2} g^{6} e g^{0}$

No. of unpaired electron = 0

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Block D and F Elements Class 12th

The disproportionation of $M n O_{4}^{2 -}$ in acidic medium resultant in the formation of two manganese compounds A and B. If the oxidation state of Mn is B is smaller than that of A, then the spin-only magnetic moment $(μ)$ value of B in BM is___________.

(Nearest integer)

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Vishal Baghel

Contributor-Level 10

$M n O_{4}^{- 2} \to A + B$

Oxidation state of Mn in B < A

$M n O_{4}^{- 2} + H^{+} \to M n O_{4}^{-} + M n O_{2}$

B is MnO₂

$∴$ Oxidation state of Mn = +4

$∴_{25} M n^{+ 4} = 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{0} 3 d^{3}$

$∴$ unpaired electron = 3

Spin only magnetic moment $(μ)$

$= \sqrt[]{n (n + 2)} = \sqrt[]{3 (3 + 2)} = \sqrt[]{15}$

$= 3.87 \approx 4$

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Block D and F Elements Class 12th

If ${[C u {(H_{2} O)}_{4}]}^{2 +}$ absorbs a light of wavelength 600nm d-d transition, then the value of octahedral crystal field splitting energy for ${[C u {(H_{2} O)}_{6}]}^{2 +}$ will be_________* 10^-^21J. (Nearest Integer)

(Given: h = 6.63 * 10^-³⁴ Js and c = 3.08 * 10⁸ ms^-¹)

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Payal Gupta

Contributor-Level 10

$Δ_{t} = \frac{h c}{λ} = \frac{6.63 * 1 0^{- 34} * 3.08 * 1 0^{8}}{600 * 1 0^{- 9}}$

$= \frac{6.63 * 3.08 * 1 0^{- 17}}{600}$ J

$= 765.765 * 1 0^{- 21} J$

$? 766 * 1 0^{- 21}$ J

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