An organic compound $'A'$ is oxidized with ${Na2O2}$ followed by boiling with ${HNO3}$. The resultant solution is then treated with ammonium molybdate to yield a yellow precipitate.
Based on above observation, the element present in the given compound is :
- Sulphur
- Nitrogen
- Fluorine
- Phosphorus
The Correct Option is D
Solution and Explanation
The phosphorus present in the compound in oxidised to phosphate.
The solution is boiled with nitric acid and then treated with ammonium molybdate to produced canary yellow precipitate.
${ Na3PO4 + 3HNO3 -> H3PO4 + 3NaNO3}$ ${H3PO4 + $\underset{\text{ (Ammonium molybdate)}}{{12 (NH4)2MoO4 }}$ + 21 HNO3 -> (NH4)3PO4.12MoO3 v + 21 NH4NO3 + 12 H2O }$
(Ammonium phosphomolybdate)
(canary yellow precipitate)
Top Questions on Group 15 Elements
- The maximum covalency of a non-metallic group 15 element 'E' with the weakest E-E bond is:
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[II.] Wastes generated by cotton mills
[III.] Wastes produced by food processing units
[IV.] Slag produced from blast furnace
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Identify the correctly matched sets:
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Questions Asked in JEE Main exam
In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
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- Let C be the circle of minimum area enclosing the ellipse E: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with eccentricity \( \frac{1}{2} \) and foci \( (\pm 2, 0) \). Let PQR be a variable triangle, whose vertex P is on the circle C and the side QR of length 29 is parallel to the major axis and contains the point of intersection of E with the negative y-axis. Then the maximum area of the triangle PQR is:
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- Coordinate Geometry
- Let circle $C$ be the image of
$$ x^2 + y^2 - 2x + 4y - 4 = 0 $$
in the line
$$ 2x - 3y + 5 = 0 $$
and $A$ be the point on $C$ such that $OA$ is parallel to the x-axis and $A$ lies on the right-hand side of the centre $O$ of $C$.
If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $\frac{1}{6}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to:
Concepts Used:
Group 15 Elements
Group 15 elements are also known as nitrogen group elements or nitrogen family.
It consists of:
- Nitrogen (N)
- Phosphorus (P)
- Arsenic (As)
- Antimony (Sb)
- Bismuth (Bi)
- Moscovium (Mc)
They are present on the right side of the periodic table. All the elements are arranged based on their atomic weight. The elements with similar properties reside in a column called the group.
All Group 15 elements have the electron configuration ns2np3 in their outer shell, where n is the principal quantum number.
Periodic Trends in Group 15 Elements
