Test: General Properties of Group-15 (Nitrogen & Phosphorus) - NEET MCQ

Option A: Cu + 2H_2​SO_4​ → CuSO₄​ + SO₂​ + 2H₂​O
Option B: S + 2H₂SO₄ → 2SO₂ +2H₂O
Option C: C + 2H₂SO₄ → CO₂ + 2SO₂ + 2H₂O
Option D: Zn + 2H₂SO₄ → ZnSO₄ + SO₂ + 2H₂O

Hence, option C is correct.

The second ionization energy refers to the energy required to remove the electron from the corresponding monovalent cation of the respective atom.

It is expected to increase from left to right in the periodic table with the decrease in atomic size.

Since the Oxygen atom gets a stable electronic configuration, 2s²2p³ after removing one electron, the O⁺ shows greater ionization energy than F⁺ as well as N⁺. 

Thus, correct order will be: O > F > N > C

Bonds → Bond enrgy
N-O → 201 kJ/mol
P-O → 340 kJ/mol
N-F → 272 kJ/mol
P-F → 490 kJ/mol
N≡N → 946 kJ/mol
P≡P → 490 kJ/mol
N-N → 160 kJ/mol
P-P → 209 kJ/mol

Hence option C is correct.

Liquid nitrogen is a cryogenic fluid that can cause rapid freezing on contact with living tissue. 
The temperature is held constant at 77 K by slow boiling of the liquid, resulting in the evolution of nitrogen gas.

Actually when ammonia is passed through a solution of calcium hypochlorite (bleaching powder), bromide water or passed over heated Cu oxide, it is oxidized to dinitrogen gas.

2NH₃ + 3CuO → 3Cu + 3H₂O + N₂
8NH₃ (excess) + 3Cl₂ → 6NH₄Cl + N₂

 contains two unpaired electrons and is paramagnetic in nature. On the other hand, ,  and  contains all paired electrons and are diamagnetic in nature. 

Only electronegativity of the given 4 decreases down the group whilst all the others increase or stay the same down the group.

P₄ has six P-P single bonds and four lone pair of electrons.
Ratio: 6/4 = 3/2 i.e. 3:2

Black phosphorus

• Most stable allotrope of P.
• Good conductor of heat and electricity.

Option A: Bismuth has 6 electron shells, whereas Antimony has 5 electron shells. Because of this, the attractive force between two Bismuth atoms is less due to electron shielding, resulting in bismuth possessing a lower boiling point than antimony.
Option B: In a period of moving from left to right, the ionization energy increases. Since N has a table half-filled 2p subshell which requires large energy.
Thus, the correct order of ionization energy is C < O < N.
Option C: Except N and Bi, all the group 15 elements exhibits allotropy. The allotropes of phosphorous are rather complex but essentially, there are three allotropic forms known as white, red, and black phosphorous.
Option D: Maximum covalency of N & P are 4 and 5.

Hence, option A,B,C,D is correct.

Atomic size decreases across the period.
Cl has a smaller size than Ar due to the inert nature atoms exist as single atoms.

Option A: Group 5A (or VA) of the periodic table are the pnictogens:  the nonmetals nitrogen (N), and phosphorus (P), the metalloids arsenic (As) and antimony (Sb), and the metal bismuth (Bi).
Option B: The electron gain enthalpy of P< N< S< O.
Option C: Minimum and maximum oxidation number of phosphorus are -3 and +5 respectively.
Option D: Fluorapatite is a phosphate mineral with the formula Ca₅(PO₄)₃F .

Hence, option A is correct.

The correct order of the element in abundance in the earth's crust:
P > N > As > Sb 

A molecule of N₂ has the pπ - pπ bonding.

White P is an allotrope of phosphorus with low ignition temperature.

Black P has layer lattice like graphite.

(i) Nitrogen is a non-metal.
(ii) Phosphorus is a non-metal.
(iii) Arsenic is a metalloid and shows Sublimation.
(iv) Bismuth is metal and shows the Inert pair effect.

Hence, option A is correct.

(i) Phosphine Oxide, R₃P=O, presents an important example of the participation of d-atomic orbitals of nonmetallic elements in π bonding. The presence of π bonding is detected with the help of evidence, such as reduction in the bond length, increase in the bond strength, and the stabilization of charge distribution. On these grounds, compared to ammine oxide phosphine oxide presents strong evidence of the presence of dπ - pπ bond, in very high stability of P = O.
(ii) Nitrogen has a unique ability to form Pπ−Pπ multiple bonds with itself and with other elements due to its small size and high electronegativity. 
(iii) N₂O₃ doesn't contain protons, so it is not a Brønsted acid. If N₂O₃ is dissolved in water, you get HNO₂ which is a weak acid, so it is an anhydride of an acid.
(iv) Ca₃N₂: The oxidation state is -3.

Hence, option B is correct.

When freshly prepared solution of FeSO₄ is added in a solution containing NO₃^– ion, it leads to formation of a brown coloured complex. This is known as brown ring test of nitrate.

Oxidation state of N in NF₃ = +3
Oxidation state of N in NCl₃ = +3
Oxidation state of N in NH₂OH = -1
Oxidation state of N in NH₃ = -3
Oxidation state of N in CH₃NH₂ = -3
Oxidation state of N in NH₂^- = -3
Oxidation state of N in Li₃N = -3
Oxidation state of N in N₂O = +1
Oxidation state of N in HCN = -3
Oxidation state of N in HNC = -3
Oxidation state of N in NO₂^- = +3

Hence, the number of chemical species having a negative oxidation state for nitrogen is 7.

P — P single bond (213 kJ mol^-1)  in P₄ molecule is much weaker than N ≡ N triple bond (941.4 kJ mol^-1) in N₂. 

Thus, Statement I is correct and Statement II is incorrect.