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The p-Block Elements Class 11 MCQs Questions with Answers
Doubt 1.
Red phosphorus is with chemicals less reactive because
(a) It does non bear P – P bonds
(b) It dos not hold back tetrahedral P4 molecules
(c) It does not catch fire in air even upto 400°C
(d) It has a polymeric structure
Answer
Answer: (d) Information technology has a polymeric bodily structure
Explanation:
Red-faced P is inferior reactive because of its gaint chemical compound construction.
Question 2.
Which of the following will not create hydrogen gas?
(a) Reaction between Fe and dil. HCl
(b) Reaction 'tween Zn and NaOH
(c) Reaction between Zn and conc. H2SO4
(d) Electrolysis of NaCl in Nelsons cell
Answer
Answer: (c) Reaction between Zn and conc. H2Thus4
Account:
Concentrated chemical element sulphurous reacts with Zn to give SO2 and not H2
Question 3.
Which of the following assertion is compensate?
(a) Cop (I) metaborate is colourless
(b) Copper (II) metaborate is colourless
(c) Bull (II) metaborate is light green
(d) Copper (I) metaborate is dark party
Resolve
Answer: (a) Copper (I) metaborate is colourless
Explanation:
Copper (II) metaborate is bluish green and colour of Copper (I) metaborate is blue in colour.
Question 4.
The structure of diBorane contains
(a) 4 2c – 2e bonds and two 3c – 2e bonds
(b) Two 2c – 2e bonds and two 3c – 2e bonds
(c) Two 2c – 2e bonds and two 3c – 3e bonds
(d) Four 2c – 2e bonds and four 3c – 2e bonds
Answer
Answer: (a) Four 2c – 2e bonds and two 3c – 2e bonds
Account:
According to molecular orbital possibility, each of the two atomic number 5 atoms is in sp³ hybrid state. Of the quadruplet hybrid orbitals, three have indefinite electron each spell the one-fourth is empty. Two of the Little Jo orbitals of each of the boron particle convergence with cardinal final hydrogen atoms forming two normal B – H σ-bonds. Matchless of the remaining hybrid orbital (either filled or empty) of one of the atomic number 5 atoms, 1s bodily cavity of H atoms (bridge molecule) and one of hybrid orbitals of the other boron particle overlap to signifier a delocalised orbital covering the three nuclei with a brace of electrons. Such a bond is known As three centre two electron (3c – 2e) bonds
Wonder 5.
Which is not the correct statement for Boron?
(a) It exhibits Allotropy
(b) Information technology exists in some crystalline and Inorganic form
(c) It forms three-dimensional chlorides
(d) It forms vapourisable hydrides.
Resolve
Answer: (c) It forms solidified chlorides
Explanation:
Boron exists in uncrystallised and crystalline express and exhibit allotropy.
It forms many volatile hydrides which spontaneously catch fire happening exposure to air and are well hydrolysed.
The chlorides of Boron is liquid, it fumes in most air and readily hydrolysed by water.
Question 6.
Oxygen brag can exist prepared from solid KMnO4 by:
(a) Dissolving the solid in dil. HCl
(b) Dissolving the solid in conc. H2Indeed4
(c) Treating the solid with H2 gases
(d) Strongly heating the solid
Resolve
Answer: (d) Strongly heating the solid
Account:
Oxygen flatulency can be prepared from solid KMnO4
250°C
2KMnO4 → KMnO4 + MnO2 + O2
Question 7.
In the upper layers of atmosphere ozone is formed:
(a) By action of discharge on O molecule
(b) By action of ultraviolet rays happening oxygen molecule
(c) Away action of infrared frequency rays on oxygen molecule
(d) Due to sudden drops of forc
Answer
Resolution: (b) By action of ultraviolet rays on oxygen molecule
Explanation:
In the upper stratum of atmosphere, the Ultraviolet rays (UV rays) split the molecule of oxygen (O2) into its constituent 2 atoms.
From each one of the atom and so combine with another oxygen (O2) corpuscle which gives originate to Ozone (O3).
Question 8.
Among the C-X bond (where, X = Cl, Atomic number 35, l) the correct decreasing order of bond energy is
(a) C−I > C−Chlorine > C−Br
(b) C−I > C−Br > C−Centiliter
(c) C−Cl > C−Br > C−I
(d) C−Br > C−Cl > C−I
Answer
Answer: (c) C−Cl > C−Br > C−I
Explanation:
Among the C-X bond (where , X = Cl, Bromine, I), the correct decreasing order of bond vim is
C−Chlorine > C−Atomic number 35 > C−l
Question 9.
Which of the following oxidation states are near symptomatic for lead and tin respectively?
(a) 2, 2
(b) 4, 2
(c) 2, 4
(d) 4, 4
Solution
Answer: (c) 2, 4
Explanation:
Due to inert geminate effect, ns² electron pair of Pb does not participate in bonding. Hence, +2 is the most characteristic oxidation number for Pb. Even so, for Sn, the inert pair effect is not so efficacious Thus, +4 is the most characteristic oxidation state for Sn.
Question 10.
When excess of Kl is added to cupric sulfate solution:
(a) Cuprous iodide is formed
(b) l2 is liberated
(c) Potassium iodide is oxidised
(d) Totally
Answer
Solution: (d) All
Explanation:
Information technology is an redox reaction which occurs when the iodide ion will reduce the copper (II) ion to copper(I) while iodide itself is oxidized to atmospheric condition atomic number 53. Like well-nig copper (I) compounds, Cul de sac is insoluble in water.
Ch'i + 2CuSO4 (aq) → Cu2l2(s) + l2(s) + 2K2SO4(aq)
Question 11.
Which of the following statements regarding ozone is non objurgate?
(a) The oxygen-oxygen bond length in ozone is identical with that of molecular atomic number 8
(b) The ozone is response hybrid of two structures
(c) The ozone molecule is angular in SHAPE
(d) Ozone is used as a antimicrobic and antiseptic for the refining of air.
Suffice
Result: (a) The atomic number 8-O bond length in ozone is identical with that of molecular oxygen
Account:
The oxygen–oxygen bond length in ozone is identical with that of molecular oxygen
Interrogative sentence 12.
K2[Hgl4] detect the ion/group :
(a) NH2
(b) NO
(c) NH2 +
(d) Cl–
Respond
Answer: (c) Granite State2 +
Explanation:
K2[Hgl4], Nesslers reagent detects NH+ 4 ion.
Dubiousness 13.
Which of the following has least valency P−H bond?
(a) PH+ 4
(b) P2H+ 5
(c) P2H2 + 6
(d) PH3
Answer
Answer: (c) P2H2 + 6
Explanation:
The covalent charter of P−H attach depends on the formal charge distributed on each P−H bond
In PH scale+ 4 it is +1/4 = +0.25, in P2H+ 5 information technology is +1/5 = +0.2 and in P2H6 2+ it is +2/6 = +0.33.
The higher the formal shoot up the lesser the covalent fiber due to more polarization. Hence the least valency P−H bond is present in P2H2+ 64
Question 14.
If Cl2 swash is passed in to aqueous solution of Kl containing whatsoever CCl4 and the mixture is shaken and so:
(a) Upper layer becomes violet
(b) Lower stratum becomes violet
(c) Homogenous violet layer is formed
(d) None of these
Answer
Answer: (a) Upper bed becomes violet
Explanation:
2KI + Cl2 → 2KCl l2
I2 CCl4 → Purplish Colour
Only the supernumerary of Cl2 should be avoided.
The layer may become colourless referable changeover of I2 to HIO3
I2 + 5Cl2 + 6H2O → 2HIO3 + 10HCl
Just in case of Red Brigades2
Br2 + 2H2O + Cl2 → 2HBrO + HCl
Question 15.
Which of the program line is true for the above succession of reactions?
(a) Z is hydrogen
(b) X is B2H6
(c) Z and Y are F2 and B2H6 respectively
(d) Z is Potassium Hydrated oxide
Answer
Resolve: (c) Z and Y are F2 and B2H6 respectively
Explanation:
F2(Z) LiH
B(s) → BF3 → B2H6 + LiBF4
(X) (Y)
Question 16.
Ammonia water accelerator can constitute dried by
(a) conc H2SO4
(b) P2O5
(c) CaCl2
(d) Quick lime
Response
Suffice: (d) Quick lime
Explanation:
Ammonia, with H2SO4 forms ammonium sulfate, with CaCl2 forms CaCl2.8NH3 and with P2O5 gives N4PO3, hence these reagents cannot be used for drying ammonia.
Question 17.
Inert gases such American Samoa helium behave like ideal gases over a wide range of temperature .However; they condense into the solid state at very low temperatures. It indicates that at very frigidnes there is a
(a) Weak attractive force between the atoms
(b) Weak obscene ram between the atoms
(c) Strong attraction between the atoms
(d) Strong repugnant dinky between the atoms
Answer
Answer: (c) Strong attraction between the atoms
Explanation:
Inert gases condense into the solid at very frigidnes as there is multipotent attraction 'tween the atoms.
In solid state, Van der Waals captivating forces are predominant between the atoms. The attraction increases with the size of the corpuscle as a result of the increase in polarizability and the diminish in ionization potential.
Question 18.
Generally, the Boron Trihaides act as
(a) Irregular reduction agent
(b) Sinclair Lewis Acids
(c) Lewis Bases
(d) Dehydrating Agents
Result
Answer: (b) Lewis Acids
Explanation:
The boron atom in trihaldies has only six electrons in the valency shell and hence can accept a pair of electrons in the empty p-orbital to complete its eight. As a result, atomic number 5 trihaldies act a Lewis acids.
Question 19.
On warming ozone, its volume.
(a) Increase to 1.5 times
(b) Decreases to half
(c) Remain uncharged
(d) Becomes double
Answer
Answer: (a) Increase to 1.5 times
Explanation:
O3 → O2 + [O]
1 mole of O3 on heating produces 1 mole of O2 and 1 jett of [O], hence its volume increases to 1.5 times.
Inquiry 20.
In the compound of type ECl3, where E = B, P, As, or Bi, the angle Cl – E – Chlorine for different E are ion the order:
(a) B > = P = As = Bi
(b) B > P > Eastern Samoa > Bi
(c) B < P = As = Bi
(d) B < P < As < Bi
Solvent
Answer: (b) B > P > Eastern Samoa > Bi
Explanation:
BCl3 is trigonal planar in anatomical structure and bond angles are 120° from each one. PCl3, AsCl3, and BiCl3 are pyramidical in figure with sp³ hybridization.
In all of them, the bond angles are less than the sane tetrahedral angle of 109.28, and also these bond angles minify down the grouping. Therefore, the correct order of bond angles is atomic number 3 follows:
B > P > As > Bi
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which statement about molecular orbital theory is not true
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