Chemistry

Explanation:

Partial Pressure Ratio of Nitrogen to Oxygen :

The partial pressure of a gas is calculated using its mole fraction. The mole fraction is the ratio of the number of moles of the gas to the total number of moles of all gases.

For nitrogen ($ \text{N}_2 $), the mole fraction $\frac{n_{\text{N}_2}}{n_{\text{N}_2} + n_{\text{O}_2} + n_{\text{Ar}}}$ can be simplified as $\frac{70/28}{70/28 + 27/32 + 3/40}$.

For oxygen ($ \text{O}_2 $), the mole fraction is $\frac{27/32}{70/28 + 27/32 + 3/40}$.

Therefore, the ratio of the partial pressure of nitrogen to oxygen is :

$ \frac{P_{\text{N}_2}}{P_{\text{O}_2}} = \frac{\text{mole fraction of } \text{N}_2}{\text{mole fraction of } \text{O}_2} = \frac{70/28}{27/32} = 2.96 $

Partial Pressure Ratio of Oxygen to Argon :

For argon ($ \text{Ar} $), the mole fraction is $\frac{3/40}{70/28 + 27/32 + 3/40}$.

The ratio of the partial pressure of oxygen to argon is :

$ \frac{P_{\text{O}_2}}{P_{\text{Ar}}} = \frac{\text{mole fraction of } \text{O}_2}{\text{mole fraction of } \text{Ar}} = \frac{27/32}{3/40} = 11.25 $

Explanation:

The compound given is hex - 1 - en - 4 - yne

Structure is

a double bond has one $\pi$ bond and one sigma bond.

A triple bond has one $\sigma$ bond and two $\pi$ bonds.

So, the number of $\pi$ bonds $=1+2=3$

For sigma bonds, count all carbon $-$ carbon bonds and carbon $-$ hydrogen bonds except the $\pi$ bonds in double bond and triple bond.

The number of sigma bonds = 13

So, total number of bonds $\Rightarrow \sigma=13\quad \pi=3$

Answer : option (C) 13 and 3

Explanation:

Molecular formula of isomeric hydrocarbon $\to C_9H_{12}$ - Negative Baeyer's test

Baeyers test is given by compounds that contain readily active carbon-carbon double bond.

So, in a negative Baeyer's test, there is no readily active c = c.

Compounds that give negative Baeyer's test are alkanes and aromatic compounds.

C$_9$H$_{12}$ compound is not an alkane

(Alkane general formula C$_n$H$_{2n+2}$)

Hydrocarbons with the formula C$_9$H$_{12}$ are considered as aromatic and isomers of substituted benzene rings.

The possible isomers of C$_9$H$_{12}$ are

From these, total number of the isomers with four different non-aliphatic substitution sites are 2

Four positions (1,2,3,4) are different. So, it contain four different non-aliphatic substitution sites.	Four position (1,2,3,4) are different. So, it contain four different non-aliphatic substitution sites.	Four position (1,2,3,4) are not different. So, it contain four different non-aliphatic substitution sites.

Explanation:

The compound given is hex-1,3-dien-5-yne structure is

A doble bond has one $\pi$ bond and one sigma bond.

A triple bond has one sigma bond and two $\pi$ bonds.

For the one triple bond, number of $\pi$ bonds = 2

For one double bond, number of $\pi$ bond = 1

For two double bonds, number of $\pi$ bonds = 2

So, total number of $\pi$ bonds = 2 + 2 = 4

For sigma bonds, count all carbon - carbon bonds and carbon - hydrogen bonds except the $\pi$ bonds in double bonds and triple bond.

The total number of sigma bonds = 11

So, total number of bonds:

$\sigma=11$

$\pi=4$

Sum of $\sigma$ and $\pi$ bonds

$=11+4=15$

Explanation:

Statement I

“One mole of propyne reacts with excess of sodium to liberate half a mole of $\mathrm{H}_2$ gas.”

Reaction and Stoichiometry

Propyne (terminal alkyne): $\mathrm{CH_3C \equiv CH}$.

Reaction with sodium:

$ 2\,\mathrm{CH_3C \equiv CH} \;+\; 2\,\mathrm{Na} \;\longrightarrow\; 2\,(\mathrm{CH_3C \equiv C^-Na^+}) \;+\; \mathrm{H_2}. $

From this balanced equation, 2 moles of propyne produce 1 mole of $\mathrm{H_2}$.

Hence, 1 mole of propyne will produce $\tfrac{1}{2}$ mole of $\mathrm{H_2}$.

$ \boxed{\text{Statement I is correct.}} $

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Statement II

“Four grams of propyne reacts with $\mathrm{NaNH_2}$ to liberate $\mathrm{NH_3}$ gas which occupies 224 mL at STP.”

Analysis

Moles of propyne

Molecular mass of propyne ($\mathrm{C_3H_4}$):

$ 3 \times 12 + 4 \times 1 = 36 + 4 = 40\,\mathrm{g/mol}. $

Four grams of propyne is:

$ \frac{4\,\mathrm{g}}{40\,\mathrm{g/mol}} = 0.1\,\mathrm{mol}. $

Reaction with $\mathrm{NaNH_2}$

For a terminal alkyne:

$ \mathrm{CH_3C \equiv CH} \;+\; \mathrm{NaNH_2} \;\longrightarrow\; \mathrm{CH_3C \equiv C^-Na^+} \;+\; \mathrm{NH_3}. $

1 mole of propyne produces 1 mole of $\mathrm{NH_3}$.

Moles of $\mathrm{NH_3}$ produced

With $0.1\,\mathrm{mol}$ of propyne, we get $0.1\,\mathrm{mol}$ of $\mathrm{NH_3}$.

Volume of $\mathrm{NH_3}$ at STP

1 mole of any ideal gas at STP $\approx 22.4\,\mathrm{L} = 22400\,\mathrm{mL}.$

$0.1\,\mathrm{mol}$ of $\mathrm{NH_3}$ occupies $0.1 \times 22.4\,\mathrm{L} = 2.24\,\mathrm{L} = 2240\,\mathrm{mL}.$

However, Statement II says the liberated $\mathrm{NH_3}$ occupies only 224 mL at STP, which corresponds to $0.01\,\mathrm{mol}$ of $\mathrm{NH_3}$, not $0.1\,\mathrm{mol}$. Therefore, the statement’s volume is off by a factor of 10 and is thus incorrect if the reaction goes to completion in a typical way.

$ \boxed{\text{Statement II is incorrect.}} $

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Conclusion

Statement I is correct.

Statement II is incorrect.

Hence, the best choice is:

$ \boxed{\text{Option D: Statement I is correct but Statement II is incorrect.}} $

Explanation:

Formation of more stable free radical intermediate.

Explanation:

4-Ethyl-3,4-dimethyloctane

Explanation:

Explanation:

Explanation:

Explanation:

	LIST-i Name reaction		LIST-II Product obtainable
(A)	Swarts reaction	(I)	$\mathrm{Et}-\mathrm{I} \xrightarrow[\mathrm{DMF}]{\mathrm{KF}} \mathrm{Et}-\mathrm{F}$
(B)	Sandmeyer's reaction	(II)	$\mathrm{PhN}_2^{\oplus} \mathrm{Cl}^{-} \xrightarrow{\mathrm{CuCN} / \mathrm{KCN}} \mathrm{PhCN}+\mathrm{N}_2$
(C)	Wurtz Fittig reaction	(III)	$ \mathrm{Ph}-\mathrm{Cl}+\mathrm{EtCl} \xrightarrow[\text { ether }]{\mathrm{Na}} $ $ \mathrm{Ph}-\mathrm{Et}+\mathrm{Ph}-\mathrm{Ph}+\mathrm{Et}-\mathrm{Et} $
(D)	Finkelstein reaction	(IV)	$\mathrm{Et}-\mathrm{Cl} \xrightarrow[\text { acetone }]{\mathrm{NaI}} \mathrm{Et}-\mathrm{I}+\mathrm{NaCl}$

Explanation:

Ozonolysis product

Explanation:

Explanation:

Option (B) and (D) reaction are not able to form alkene as a product.

Explanation:

St-I : Correct statement

St-II : In correct statement because product has chiral centre.

Explanation:

Both Statement I and Statement II are correct.

Explanation:

	List - I (Reaction)		List - II (Name of reaction)
A.		III.	Fittig reaction
B.	$\mathrm{ArN}_2^{+} \mathrm{X}^{-} \xrightarrow[\mathrm{HCl}]{\mathrm{Cu}} \mathrm{ArCl}+\mathrm{N}_2 \uparrow+\mathrm{CuX}$	IV.	Gatterman reaction
C.	$\mathrm{C}_2 \mathrm{H}_5 \mathrm{Br}+\mathrm{NaI} \xrightarrow[\text { Acetone }]{\text { Dry }} \mathrm{C}_2 \mathrm{H}_5 \mathrm{I}+\mathrm{NaBr}$	II.	Finkelstein reaction
D.	$\begin{aligned} & \mathrm{CH}_3 \mathrm{C}(\mathrm{OH})\left(\mathrm{CH}_3\right) \mathrm{CH}_3 \xrightarrow[\mathrm{ZnCl}_2]{\mathrm{HCl}} \\ & \mathrm{CH}_3 \mathrm{C}(\mathrm{Cl})\left(\mathrm{CH}_3\right) \mathrm{CH}_3\end{aligned}$	IV.	Lucas reaction

Explanation:

Explanation:

Explanation:

Homoleptic complexes are complexes with identical ligands attached to the central metal atom.

(A) ${[Fe{O_4}]^{2 - }}$ - Homdeptic

Central metal : Fe

IN this complex, Fe is in +6 oxidation state.

{oxidation state : $x + 4( - 2) = - 2$

$x - 8 = - 2$

$x = - 2 + 6 = + 6$

Configuration of $Fe - {d^6}{s^2}$

Configuration of $Fe( + 6) - c{l^2}$

Iron has even number of d electrons. So, it is not correct.

(B) ${[Fe{(N)_6}]^{3 - }}$ - Homoleptic

Central metal : Fe

In this complex, the oxidation state of Fe is +3

Oxidation state :

$x + 6( - 1) - 3$

$x - 6 = - 3$

$x = - 3 + 6 = + 3$

Configuration of $Fe - {d^6}{s^2}$

Configuration of $Fe( + 3) - {d^5}$

Iron has odd number of d electrons.

So, it is correct.

(C) ${[Fe{(CN)_5}NO]^{2 - }}$ - Not homoleptic

This complex has different type of ligands.

${[CoC{l_4}]^{2 - }}$ - Homoleptic

Central metal - Co

In this complex, the oxidation state of Co is +2

Oxidation state:

$x + 4( - 1) = - 2$

$x - 4 = - 2$

$x = - 2 + 4 = + 2$

Configuration of $Co - {d^7}{s^2}$

Configuration of $Co( + 2) - {d^7}$

Cobalt has odd number of d electrons.

So, it is correct.

(E) $[Co{({H_2}O)_3}{F_3}]$ - Not homoleptic,

It has different type of ligands.

Complexes B and D are homoleptic complexes with odd number of d electrons in the central metal. So, correct answer is

(B) and (D) only.