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p-Block Elements and Their Uses

Description: p-block elements B, Al, Si, N, P, and S
Number of Questions: 20
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Tags: p-block elements B, Al, Si, N, P, and S
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Which of the following wrought alloys has maximum percentage of aluminium?

  1. Alloy 1050

  2. Alloy 1060

  3. Alloy 1100

  4. Alloy 1199

  5. Alloy 2014


Correct Option: D
Explanation:

Alloy 1050 contains 99.5% of aluminium by mass, alloy 1060 contains 99.6% aluminium by weight, alloy 1100 contains 99.0% aluminium by mass, alloy 1199 contains 99.9% of aluminium by weight and alloy 2014 contains 90.9% of aluminium by mass. Hence, alloy 1199 has the maximum percentage of aluminium by weight.

Standard electrode potential values for Al3+/Al is –1.66 V and that of Tl3+/Tlis + 1.26 V. About the formation of M+/M3+ ion in solution and the electropositive character of these metals, the true statement(s) is/are A. Al and Tl form stable Al3+ and Tl3+ ions in aqueous solution. B. Tl+ is more stable than Tl3+. C. Al is more electropositive element than that of thallium.

  1. A only

  2. B only

  3. C only

  4. A and B

  5. B and C


Correct Option: E
Explanation:

Standard electrode potential values for two half cell reactions suggest that aluminium has high tendency to make Al3+(aq) ions, whereas Tl3+ is not only unstable in solution but is a powerful oxidising agent also. Thus, Tl+ is more stable in solution than Tl3+. Aluminium being able to form Al3+ ions easily, is more electropositive than thallium.

Which of the following statement(s) is/are false?

P. Boron and Gallium form BF63- ions. Q. Alclad is the alloy of aluminium used in the formation of corrosion resistant sheet. R. Boron fibres are used in making bullet-proof vest and light composite material for aircraft.

  1. P only

  2. Q only

  3. R only

  4. P and Q

  5. Q and R


Correct Option: E
Explanation:

P. Due to non-availability of d orbitals, boron is unable to expand its octet. Therefore, the maximum covalence of boron cannot exceed 4. Al, Ga, In and Tl have vacant d orbitals and form such type of ions. Q. Alclad is a trademark of Alcoa used as a generic term to describe corrosion-resistant aluminium sheet formed from high-purity aluminium surface layers metallurgically bonded to high-strength aluminium alloy core material. R. Boron fiber exhibits a unique combination of high strength, high modulus and large diameter. Boron fibres are used in making bullet-proof vest and light composite material for aircraft.

Among the following the more stable complexes of thalium is/are

P. Tl(OH) Q. TlCl3 R. TlBr S. TlN3

  1. P only

  2. Q only

  3. R only

  4. P and Q

  5. Q and R


Correct Option: D
Explanation:

Thallium tends to oxidize to the +3 and +1 oxidation states as ionic salts. The +3 state resembles that of the other elements in thallium's group (boron, aluminum, gallium, indium). Thallium represents +1 oxidation state due to the inert pair effect. Hence, Tl(OH) and TlBr are more stable complexes than TlN3 and TlCl3.

Identify the incorrect statement(s) for boron.

P. Kernite is a mineral of boron. Q. 10B is more abundant isotope than11B. R. Boron has lower second ionization enthalpy than aluminium.

  1. P only

  2. Q only

  3. R only

  4. P and Q

  5. Q and R


Correct Option: E
Explanation:

P. Kernite (Na2B4O7.4H2O) is the mineral of boron. Q. There are two isotopic forms of boron 10B (19%) and 11B (81%). R. Boron (2427 KJ/mol) has higher second Ionization enthalpy than aluminium (1816 KJ/mol).

The general formula BnHn+6 for boranes, corresponds to

  1. closo-borane

  2. nido-boranes

  3. arachno-borane

  4. hypho-borane

  5. none of these


Correct Option: C
Explanation:

Arachno-borane has the structural formula- BnHn+6.

Pyrex glass is mainly a mixture of sodium borosilicate and aluminium borosilicate. Green coloured glass is made by the addition of ________.

  1. MnO2

  2. colbaltous oxide

  3. CuO

  4. CdS

  5. Cu2O


Correct Option: C
Explanation:

Green coloured glass is made by the addition of CuO.

Match the entries in Group - I (elements) with those in Group - II (frequent uses).

 
Group - I Group - II
P. B 1. construction materials
Q. Al 2. coatings
R. Ga 3. fibreglass production
S. In 4. LED lighting
  1. P - 3, Q - 1, R - 4, S - 2

  2. P - 2, Q - 1, R - 4, S - 3

  3. P - 3, Q - 4, R - 1, S - 2

  4. P - 4, Q - 2, R - 3, S - 1

  5. P - 2, Q - 4, R - 3, S - 1


Correct Option: A
Explanation:

The correct sequence of codes is P - 3, Q - 1, R - 4, S – 2.

Which of the following statement(s) is/are false?

P. Gallium has higher melting and boiling point than indium. Q. Tusionite is the mineral of boron. R. Gallium arsenide is used in the making of tunnel diodes for FM transmitter circuits.

  1. P only

  2. Q only

  3. R only

  4. P and Q

  5. Q and R


Correct Option: A
Explanation:

P. The melting and boiling points of gallium are 29.8oC and 2204oC whereas the melting and boiling points of indium are 156oC and 2072oC respectively.

Q. Tusionite is a rare colourless to transparent to translucent yellow brown trigonal borate mineral with chemical formula-MnSn(BO3)2. The mineral is composed of 18.86% manganese, 40.76% tin, 7.42% boron, and 32.96% oxygen.

R. Gallium and its derivatives have only found applications in recent decades. Gallium arsenide has been used in semiconductors, in amplifiers, in solar cells (for example in satellites) and in the manufacturing of tunnel diodes for FM transmitter circuits.

Which of the following wrought aluminium alloys contain gallium?

  1. Alloy 7178

  2. Alloy 7116

  3. Alloy 3003

  4. Alloy 1050

  5. None of these


Correct Option: B
Explanation:

Alloy 7116 is composed of 93.87-95.17% Al, 0.15% Si, 0.3% Fe, 0.5-1.1% Cu, 0.05% Mn, 0.8-1.4% Mg, 4.2-5.2% Zn, 0.05% V, 0.05% Ti and 0.03% Ga.

Regarding N, Al, P and Si, the second ionization enthalpy follow the order of

  1. N>P>Al>Si

  2. N>Al>P>Si

  3. Si>P>Al>N

  4. Al>P>N>Si

  5. N>Si>Al>P


Correct Option: A
Explanation:

Second ionization enthalpy of N = 2856 KJ/mol Second ionization enthalpy of P = 1903 KJ/mol Second ionization enthalpy of Al = 1816 KJ/mol Second ionization enthalpy of Si = 1577 KJ/mol Hence, regarding N, Al, P and Si, the second ionization enthalpy follow the order of N>P>Al>Si.

Which of the following pair of ‘compounds - their respective nature’ is not correctly matched?

  1. Carbon dioxide - acidic

  2. SiO2 - acidic

  3. Alumina - Amphoteric

  4. Caro’s acid - dibasic

  5. Sulphuric acid - dibasic


Correct Option: D
Explanation:

Caro’s acid (H2SO5) is monobasic on account of the presence of one replaceable hydrogen atom. The presence of one replaceable hydrogen atom is explained by the example of potassium peroxymonosulfate (KHSO5) which is formed by the reaction of H2SO5 with alkali. It is the potassium acid salt of peroxymonosulphuric acid and is widely used as an oxidising agent.

Which of the following group 13 elements has the highest melting point?

  1. B

  2. Al

  3. Ga

  4. In

  5. Tl


Correct Option: A
Explanation:

Boron (B) is metalloid and has melting point of 2076oC. Al has melting point 660oC, Ga has melting point of 29.8oC, In has the melting point of 156oC and Tl has the melting point of 303oC.  Hence, boron has the highest melting point among the given elements.

Match the Column - I (interhalogen compounds) with Column - II (corresponding geometry) and select the correct answer using the codes given below.

 
Column - I
Column - II
P. BrCl 1. square pyramidal
Q. IF7 2. linear
R. BrF3 3. T-shaped
S. IF5 4. pentagonal
  1. P - 2, Q - 1, R - 3, S - 4

  2. P - 3, Q - 4, R - 2, S - 1

  3. P - 2, Q - 4, R - 3, S - 1

  4. P - 3, Q - 4, R - 1, S - 2

  5. P - 4, Q - 3, R - 1, S - 2


Correct Option: C
Explanation:

BrCl has 3 lone pairs and linear geometry. IF7 has zero lone pair and pentagonal geometry. BrF3 has 2 lone pairs and T-shaped geometry. IF5 has one lone pair and square pyramidal geometry. Hence, the correct sequence of codes is P - 2, Q - 4, R - 3, S - 1.

Which of the following statements are not correct?

P. Decaborane-14 has regular icosahedron geometry. Q. Silicon do not form [SiCl6]2- and [SiF6]2-. R. n-Type semiconductors are made by the doping of Si atoms in Ga atoms.

  1. P and Q

  2. Q and R

  3. P and R

  4. All of these

  5. None of these


Correct Option: A
Explanation:

P. Decaborane-14 has ten boron atoms. This number is two short of being able to form a regular icosahedron geometry. Orthocarborane (B10C2H10R2) has 12 corners and 20 faces and has regular icosahedron geometry. Q. [SiF6]2– is known whereas [SiCl6]2– is not. The main reasons are: (i) six large chloride ions cannot be accommodated around Si4+ due to limitation of its size. (ii) interaction between lone pair of chloride ion and Si4+ is not very strong. R. p-Type semiconductors are made by the doping of Si atoms in group 15 elements and n-Type semiconductors are made by the doping of Si atoms in group 13 elements.

Match the entries in Column - I with those in Column - II.

 
Column - I
Column - II
P. turquoise 1. ZnS + BaSO4
Q. lithopone 2. AlPO4.Al(OH)3.H2O
R. talc 3. 3MgO.4SiO2.H2O
S. diaspore 4. Al2O3.H2O
  1. P - 2, Q - 1, R - 4, S - 3

  2. P - 2, Q - 1, R - 3, S - 4

  3. P - 3, Q - 2, R - 1, S - 4

  4. P - 2, Q - 4, R - 3, S - 1

  5. P - 4, Q - 3, R - 2, S - 1


Correct Option: B
Explanation:

Turquoise (AlPO4.Al(OH)3.H2O) is the mineral of aluminium. Lithopone is a mixture of ZnS and BaSO4. It is a brilliantly fine white powder and has a tendency to turn grey on exposure to the sunlight. Talc is a pure magnesium silicate in the form of 3MgO.4SiO2.H2O found naturally in soap stones. Diaspore (Al2O3.H2O) is the mineral of aluminium. Hence, the correct sequence of codes is P - 2, Q - 1, R - 3, S - 4.

The ionic radii of B3+, Al3+ Si4+ and N3- varies in the order of

  1. N3- > Al3+ > B3+ > Si4+

  2. N3- > B3+ > Si4+ > Al3+

  3. N3- > B3+ > Al3+ > Si4+

  4. N3- > Al3+ > Si4+ > B3+

  5. Al3+ > B3+ > Si4+ > N3-


Correct Option: D
Explanation:

The ionic radii of B3+ = 27 pm, Al3+ = 53.5 pm, Si4+ = 40 pm and N3- = 171 pm. Hence, the ionic radii of B3+, Al3+, Si4+ and N3- varies in the order of N3- > Al3+ > Si4+ > B3+.

The oxidation state of central metal atom in ‘Lunar caustic’ is

  1. +1

  2. +2

  3. +3

  4. -2

  5. +5


Correct Option: E
Explanation:

Silver nitrate (AgNO3) is known as Lunar caustic. The central metal in silver nitrate is nitrogen atom. Then,  +1 + x +3 (-2) = 0 x - 5 = 0 and x = +5 Hence, the oxidation state of N atom in silver nitrate is +5.

Identify the incorrectly matched pair of ‘compound - its specific name’.

  1. KMnO4 - Baeyer’s reagent

  2. C6H5NO2 - Oil of mirbane

  3. K2[HgI4] - Calgon

  4. Na3(PO3)3 - Graham’s salt

  5. SiO2 - Quartz


Correct Option: C
Explanation:

K2[HgI4] is Nessler’s reagent. Calgon is Na4(PO3)6.

Assertion (A): BF3 is weaker Lewis acid than that of BBr3. Reason (R): BF3 has planar geometry and is stabilized to a greater extent than BBr3 by pπ-back bonding. Select the correct opinion.

  1. A and R are both true and R is the correct explanation of A.

  2. A and R are both false and R is not the correct explanation of A.

  3. A is true but R is false.

  4. Both A and R are false.

  5. None of these


Correct Option: A
Explanation:

BF3 is a weaker Lewis acid than BCl3 because of the greater pπ-pπ back bonding effect in BBr3. BF3 and BBr3 both have sp2 hybridisation with trigonal planar geometry.

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