Test: General Properties of 15th Group (Nitrogen & Phosphorus)


21 Questions MCQ Test Chemistry for JEE | Test: General Properties of 15th Group (Nitrogen & Phosphorus)


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QUESTION: 1

Hot conc. H2SO4 acts as moderately strong oxidising agent. It oxidises both metals and nonmetals. Which of the following element is oxidised by conc. H2SO4 into two gaseous products?

Solution:

Option A: Cu + 2H2​SO4​ → 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.

QUESTION: 2

Correct order of 2nd ionisation energy of C, N, O and F is

Solution:

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

QUESTION: 3

Which trend enthalpy is correct ? 

Solution:

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.

QUESTION: 4

Boiling Point of liquid nitrogen is 

Solution:

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.

QUESTION: 5

Oxidation of ammonia with CuO produce a gaseous chemical which can also be obtained by:

Solution:

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.

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

QUESTION: 6

Which of the following molecular species has unpaired electron(s) ?

Solution:

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

QUESTION: 7

Which is the correct order w.r.t the given property? 

Solution:

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

QUESTION: 8

The ratio of bond pairs and lone pairs in a P4 molecule is 

Solution:

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

*Multiple options can be correct
QUESTION: 9

One or More than One Options Correct Type
This section contains 5 multiple type questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Regarding black phosphorus correct statements are:

Solution:

Black phosphorus

  • Most stable allotrope of P.
  • Good conductor of heat and electricity.
*Multiple options can be correct
QUESTION: 10

Which are correct statements ? 

Solution:

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.

QUESTION: 11

 In the third period of the periodic table the element having smallest size is        

Solution:

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

QUESTION: 12

The correct statements among the given are :

Solution:

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.

*Multiple options can be correct
QUESTION: 13

Comprehension Type
This section contains a passage describing the theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)

Passage

Phosphorus was discovered by Brand (1669), Scheele isolated from bone ash and Lavoisier proved its elemental nature (1777). The principal minerals are phosphate rock, fluoroacetate, and chloroacetate. Phosphorus is prepared by the direct reduction of phosphorite by carbon in the presence of silica. It exists in different allotropic forms such as yellow or white, red, a-black,f3-black, etc. White P is most reactive, poisonous, glows in dark, and readily catches fire due to unstable discrete P4 molecules. Red P is inert, non-poisonous, does not glow, etc., due to its polymeric structure. a-black, f3 -black allotropes are also chemically inert, do not ignite at normal temperature. It has a layer structure like graphite and acts as a conductor. 

Q. Which of the following statements is/are correct?

Solution:

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.

*Multiple options can be correct
QUESTION: 14

Passage

Phosphorus was discovered by Brand (1669), Scheele isolated from bone ash and Lavoisier proved its elemental nature (1777). The principal minerals are phosphate rock, fluoroacetate, and chloroacetate. Phosphorus is prepared by the direct reduction of phosphorite by carbon in the presence of silica. It exists in different allotropic forms such as yellow or white, red, a-black,f3-black, etc. White P is most reactive, poisonous, glows in dark, and readily catches fire due to unstable discrete P4 molecules. Red P is inert, non-poisonous, does not glow, etc., due to its polymeric structure. a-black, f3 -black allotropes are also chemically inert, do not ignite at normal temperature. It has a layer structure like graphite and acts as a conductor. 

Q. The allotrope of phosphorus with low ignition temperature is:

Solution:

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

QUESTION: 15

Passage

Phosphorus was discovered by Brand (1669), Scheele isolated from bone ash and Lavoisier proved its elemental nature (1777). The principal minerals are phosphate rock, fluoroacetate, and chloroacetate. Phosphorus is prepared by the direct reduction of phosphorite by carbon in the presence of silica. It exists in different allotropic forms such as yellow or white, red, a-black,f3-black, etc. White P is most reactive, poisonous, glows in dark, and readily catches fire due to unstable discrete P4 molecules. Red P is inert, non-poisonous, does not glow, etc., due to its polymeric structure. a-black, f3 -black allotropes are also chemically inert, do not ignite at normal temperature. It has a layer structure like graphite and acts as a conductor. 

Q. The allotrope of phosphorus that has layer lattice-like graphite is:

Solution:

Black P has layer lattice like graphite.

QUESTION: 16

Matching List Type
Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c), and (d), out of which one is correct

Match column I with Column II and mark the correct option from the codes given below:

Solution:

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

Hence, option A is correct.

QUESTION: 17

Match the Column I with Column II and mark the correct option from the codes given below :

Solution:

(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.

QUESTION: 18

A brown ring is formed in the ring test for NO3 ion. It is due to the formation of

Solution:

When freshly prepared solution of FeSOis 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.

*Answer can only contain numeric values
QUESTION: 19

Number of chemical species having negative oxidation state for nitrogen among 

NF3, NCI3, NH2OH, NH3,CH3NH2, NH-2, L13N, N20, HCN, HNC, NO-2


Solution:

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 CH3NH= -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.

*Answer can only contain numeric values
QUESTION: 20

According to molecular orbital theory, number of electrons present in the antibonding orbitals of nitrogen.


Solution:

QUESTION: 21

Statement Type
This section is based on Statement I and Statement II. Select the correct answer from the codes given below.

Statement I: N2 is less reactive than P4.
Statement II: Nitrogen has more electron gain enthalpy than phosphorus.

Solution:

P — P single bond (213 kJ mol-1)  in Pmolecule 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.