JEE Advanced · Chemistry · 21. p Block (G15-18)

The reaction of $P_2$ with $X$ leads selectively to $\mathrm{P}_4 \mathrm{O}_6$. The $\mathrm{X}$ is

A dry $\mathrm{O}_2$
B a mixture of $\mathrm{O}_2$ and $\mathrm{N}_2$
C moist $\mathrm{O}_2$
D $\mathrm{O}_2$ in the presence of aq. $\mathrm{NaOH}$

Verified Solution

Answer & Solution

Correct Answer

(B) a mixture of $\mathrm{O}_2$ and $\mathrm{N}_2$

Step-by-step Solution

Detailed explanation

White phosphorus on reaction with limited supply of oxygen gives lower oxide $P_4 \mathrm{O}_6$. Therefore, air $\left(\mathrm{O}_2+\mathrm{N}_2\right)$ is a good source for controlled supply of oxygen and the best choice for controlled oxidation of white phosphorus into lower oxide $\mathrm{P}_4 \mathrm{O}_6$.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Let

p, q, r

be non-zero real numbers that are, respectively, the

10^{th}, 100^{th}

and

1000^{th}

terms of a harmonic progression. Consider the system of linear equations

x + y + z = 1

10 x + 100 y + 1000 z = 0

q r x + p r y + p q z = 0

	List- $I$		List- $II$
$(I)$	If $\frac{q}{r} = 10$ , then the system of linear equations has	$(P)$	$x = 0, y = \frac{10}{9}, z = - \frac{1}{9}$ as a solution
$(II)$	If $\frac{p}{r} \neq 100$ , then the system of linear equations has	$(Q)$	$x = \frac{10}{9}, y = - \frac{1}{9}, z = 0$ as a solution
$(III)$	If $\frac{p}{q} \neq 10$ , then the system of linear equations has	$(R)$	infinitely many solutions
$(IV)$	If $\frac{p}{q} = 10$ , then the system of linear equations has	$(S)$	no solution
		$(T)$	at least one solution

The correct option is:

JEE Advanced 2022 Medium

Based on VSEPR model, match the xenon compounds given in List-I with the corresponding geometries and the number of lone pairs on xenon given in List-II and choose the correct option.

List - I	List - II
(P) $XeF _2$	(1) Trigonal bipyramidal and two lone pair of electrons
(Q) $XeF _4$	(2) Tetrahedral and one lone pair of electrons
(R) $XeO _3$	(3) Octahedral and two lone pair of electrons
(S) $XeO _3 F_2$	(4) Trigonal bipyramidal and no lone pair of electrons
	(5) Trigonal bipyramidal and three lone pair of electrons

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Reaction of iso-propylbenzene with $\mathrm{O}_2$ followed by the treatment with $\mathrm{H}_3 \mathrm{O}^{+}$forms phenol and a by-product $\mathbf{P}$. Reaction of $\mathbf{P}$ with 3 equivalents of $\mathrm{Cl}_2$ gives compound $\mathbf{Q}$. Treatment of $\mathbf{Q}$ with $\mathrm{Ca}(\mathrm{OH})_2$ produces compound $\mathbf{R}$ and calcium salt $\mathbf{S}$.
The correct statement(s) regarding $\mathbf{P}, \mathbf{Q}, \mathbf{R}$ and $\mathbf{S}$ is(are)

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The reaction of \(P_2\) with \(X\) leads selectively to \(\mathrm{P}_4 \mathrm{O}_6\). The \(\mathrm{X}\) is

Answer & Solution

Step-by-step Solution

See the Complete Solution

List - I	List - II
(P) \(XeF _2\)	(1) Trigonal bipyramidal and two lone pair of electrons
(Q) \(XeF _4\)	(2) Tetrahedral and one lone pair of electrons
(R) \(XeO _3\)	(3) Octahedral and two lone pair of electrons
(S) \(XeO _3 F_2\)	(4) Trigonal bipyramidal and no lone pair of electrons
	(5) Trigonal bipyramidal and three lone pair of electrons