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Ncert Solutions Chemistry Class 12th

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Ncert Solutions Chemistry Class 12th Class 12th

The separation of two coloured substances was done by paper chromatography. The distance travelled by solvent front, substance A and substance B from the base line are 3.25cm, 2.08cmand 1.05cm, respectively. The ratio of R_f values of A to B is ________.

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Payal Gupta

Contributor-Level 10

$R_{f}^{0} = distance travelled by the solute$ $/$ $/ distance travelled by the solvent$

${(R_{f})}_{A} = \frac{2.08}{3.25} {(R_{f})}_{B} = \frac{1.05}{3.25}$

$\frac{{(R_{f})}_{A}}{{(R_{f})}_{B}} = \frac{2}{1}$

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Ncert Solutions Chemistry Class 12th Class 12th

A sample of 4.5mg of an unknown monohydric alcohol, R-OH was added to methylmagnesium iodide. A gas is evolved and is collected and its volume measured to be 3.1 mL. The molecular weight of the unknown alcohol is ____________ g/mol.

[Nearest integer]

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Payal Gupta

Contributor-Level 10

(Calculation is done considering STP condition)

$R O H + C H_{3} M g l \to C H_{4} + R O M g l$

No. of moles ROH = no. of moles of CH4

$\frac{4.5 * 1 0^{- 3}}{M} = \frac{31}{22400} \Rightarrow 32.52 \approx 33 g m / m o l e$

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Ncert Solutions Chemistry Class 12th 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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Ncert Solutions Chemistry Class 12th Class 12th

The sum of number of lone pairs of electrons present on the central atoms of XeO₃, XeOF₄ and XeF₆, is___________.

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Payal Gupta

Contributor-Level 10

XeO₃ ⇒ Number of lone pair = 1

XeOF₄ ⇒ Number of lone pair = 1

XeF₆ ⇒ Number of lone pair = 1

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Ncert Solutions Chemistry Class 12th Class 12th

When the excited electron of a H atom from n = 5 drops to the ground state, the maximum number of emission lines observed are____________.

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Payal Gupta

Contributor-Level 10

For single H- atom, maximum number of spectral lines = (n – 1), n = orbit number of excited electron.

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Ncert Solutions Chemistry Class 12th Class 12th

In the given reaction

(Where Et is -C₂H₅)

The number of chiral carbon/s in product A is __________.

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Vishal Baghel

Contributor-Level 10

2 (only 2 carbon will chiral in product form)

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Ncert Solutions Chemistry Class 12th 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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Ncert Solutions Chemistry Class 12th 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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Ncert Solutions Chemistry Class 12th Class 12th

The number of chlorine atoms in bithionol is_____________.

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Payal Gupta

Contributor-Level 10

Structure of Bithinol is;

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Ncert Solutions Chemistry Class 12th Class 12th

A 1.84 mg sample of polyhydric alcoholic compound 'X' of molar mass 92.0 g/mol gave 1.344mL of H₂ gas at STP. The number of alcoholic hydrogens present in compound 'X' is_________.

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Payal Gupta

Contributor-Level 10

Mole of polyhydric alcohol = $\frac{1.84 * 1 0^{- 3}}{92} = 2.0 * 1 0^{- 5} m o l e .$

Mole of H₂ gas produced = $\frac{1.344}{22400} = 6.0 * 1 0^{- 5} m o l e .$

No of OH gp present = $\frac{6.0 * 1 0^{- 5}}{2.0 * 1 0^{- 5}} = 3$

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