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CHEMISTRY THE P-BLOCK ELEMENTS-11th
QUIZ-13

Dear Readers,

IIT JEE exam which consists of JEE Main and JEE Advanced is one of the most important entrance exams for engineering aspirants. The exam is held for candidates who are aspiring to pursue a career in the field of engineering and technical studies.
Chemistry is important because everything you do is chemistry! Even your body is made of chemicals. Chemical reactions occur when you breathe, eat, or just sit there reading. All matter is made of chemicals, so the importance of chemistry is that it's the study of everything.
.

Q1.Aqueous solution of the following are matched according to their properties shown. Select the correct matching
  • Cl2O7 - acidic
  • Al2Cl6 - white fumes
  • Al2O3 - amphoteric
  • All of the above are correct matching
Solution
All of the above are correct matching
Q2.The wrong statement about N2O is
  •  It is nitrous oxide
  •  It is least reactive oxide of nitrogen
  •  It is not a linear molecule
  •  It is known as laughing gas
Solution
It is not a linear molecule
Q3.The dipole moment of NF3 is less than that of NH3 because
  •  F is more reactive than H
  •  NH3 forms associated molecules
  •  The resultant of the individual bond polarities is less
  •  The resultant of the individual bond polarities is opposed by the polarity of lone pair
Solution
The resultant of the individual bond polarities is opposed by the polarity of lone pair
As we know that electronegativity is increasing from left to right when we move in the period of periodic table. As F is right of N in the periodic table so the electronegativity of F is greater than the electronegativity of N.
So each of N – F bonds has its polarity towards F, which results in a net polarity which opposes the polarity of lone pair electrons. But in the case of NH3 the electronegativity of N is lesser than the electronegativity of H.
So each N – H bonds has its polarity towards N, which results in a net polarity which does not oppose the polarity of lone pair electrons. That’s why the dipole moment of NF3 less than that of NH3.

Q4.Consider the following boron halides
1. BF3 2. BCl3 3. BBr3 4. BI3
The Lewis acid characters of these halides are such that
  •  1 < 2 < 3 < 4
  •  1 < 3 < 2 < 4
  •  4 < 3 < 2 < 1
  •  4 < 2 < 3 < 1
Solution
1 < 2 < 3 < 4
The trend is due to π back bonding. All halides have lone pairs, they donate one lone pair to boron by overlapping with the p orbitals of Boron.
Effective overlap between orbitals decreases due to large size of p orbitals of halides. Hence electron deficiency of boron atom in its halides follow the trend BI3 > BBr3 > BCl3 > BF3 and so is the lewis acidity order.

Q5.Concentrations of the atmospheric CO2 have been rising because of
  •  Use of fossil fuels
  •  Acid rain
  •  Photochemical smog
  •  Ozone depletion
Solution
Use of fossil fuels
There are both natural and human sources of carbon dioxide emissions. Natural sources include decomposition, ocean release and respiration.
Human sources come from activities like cement production, deforestation as well as the burning of fossil fuels like coal, oil and natural gas.

Q6.Carbon and silicon belong to (IV) group. The maximum coordination of carbon is commonly occurring compounds is 4, whereas that of silicon is 6. This is due to
  •  Large size of silicon
  •  More electropositive nature of silicon
  • Availability of d-orbital in silicon
  •  Both (a) and (b)
Solution
Availability of d− orbital in silicon.
C − 2s22p2
Si − 3s23p2
Carbon has no d-orbital to expand but Si has vacant d-orbitals and it can expand its valency using these vacand d-orbitals and forms 6 coordinated compounds.

Q7.N2O is formed when
  •  Moist Fe reacts with NO
  •  Sn2+ reacts with conc. HNO3 in presence of HCl
  •  Cold dil. HNO3 reacts with Cu and Zn
  •  By all the reactions
Solution
By all the reactions

Q8.Hot conc. HNO3 converts graphite into
  •  Graphite oxide
  •  Benzene hexacarboxylic acid
  •  Both (a) and (b)
  •  None of the above
Solution
Benzene hexa-carboxylic acid.
Graphite reacts with hot conc. HNO3 and forms benzene hexa-carboxylic acid.



Q9.One of the acid listed below is formed only from P2O3; the rest area formed from P2O5. Acid formed from P2O3 is
  •  HPO3
  •  H2P2O7
  •  H2PO4
  •  H3PO3
Solution
H3PO3
Reaction is:
P2O3 + 3H2O → 2H3PO3
Q10. PbO2 → PbO ∆G298 < 0
SnO2 → SnO ∆G298 > 0 Most probable oxidation state of Pb and Sn will be

  •  Pb4+, Sn4+
  •  Pb4+, Sn2+
  •  Pb2+, Sn2+
  • Pb2+, Sn4+
Solution
Due to inert-pair effect
Pb2+ > Pb4+
Sn4+ > Sn2+

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